question_id string | site string | title string | body string | link string | tags list | votes int64 | creation_date timestamp[s] | comments list | comment_count int64 | category string | diamond int64 |
|---|---|---|---|---|---|---|---|---|---|---|---|
1 | bioacoustics | How do animals adapt to a partially or fully deaf individual of their group? | In eusocial animals for which hearing is important to communicate between individuals, is there any species apart humans for which there are some evidence that the group adapts their acoustic communication to a fully or partially deaf individual of the group?
For instance, by systematically vocalizing louder/closer to a partially deaf individual, or switching the communication modality (e.g. acoustics to touching) for a fully deaf individual, or by showing more support to them when an alarm call is raised, etc.
I have not found any information after a quick search on the web and Google Scholar.
| https://bioacoustics.stackexchange.com/questions/1235/how-do-animals-adapt-to-a-partially-or-fully-deaf-individual-of-their-group | [
"hearing",
"adaptation"
] | 6 | 2022-11-14T07:01:24 | [
"Thanks @DanStowell, done!",
"Did you search for possible literature? If so it would be good to give some indication of that. Also, I think \"eurosocial\" should be \"eusocial\", right?"
] | 2 | Science | 0 |
2 | codegolf | Minimum cost of solving the Eni-Puzzle | You're tasked with writing an algorithm to efficiently estimate cost of solving an [Eni-Puzzle](http://www.enipuzzles.com/buy/eni-braille-puzzle-with-bold-colors) from a scrambled state as follows:
You're given m lists of containing n elements each(representing the rows of the puzzle). The elements are numbers between 0 and n-1 inclusive (representing the colors of tiles). There are exactly m occurrences of each integers across all m lists (one for each list).
For example:
m=3, n=4 :
[[3, 0, 3, 1], [[1, 3, 0, 1],
[1, 0, 2, 2], or [0, 2, 3, 1],
[3, 0, 1, 2]] [0, 3, 2, 2]]
You can manipulate these lists in two ways:
1: Swapping two elements between circularly adjacent indices in (non circularly) adjacent lists. Cost=1.
Ex:
m=3, n=4 :
Legal:
Swap((0,0)(1,1))
Swap((1,0)(2,3)) (circularly adjacent)
Illegal:
Swap((0,0)(0,1)) (same list)
Swap((0,0)(2,1)) (lists are not adjacent)
Swap((0,0)(1,0)) (indices are not circularly adjacent (they're the same)
Swap((0,0)(1,2)) (indices are not circularly adjacent)
2. Circularly shifting one of the lists (Cost=number of shifts)
Your algorithm must efficiently calculate minimum cost required to manipulate the lists such that the resulting lists are all rotations of each other (meaning the puzzle can be fully solved from this state using only rotation moves) i.e.:
[[0, 1, 2, 3] [[2, 1, 0, 3]
[3, 0, 1, 2] and [0, 3, 2, 1]
[1, 2, 3, 0]] [3, 2, 1, 0]]
...are both valid final states.
Instead of lists, you may use any data structure(s) of your choice to represent the puzzle, so long as the cost of simulating a valid move (sliding or rotating) on the puzzle with this representation is O(n*m). The setup cost of initializing this data structure can be disregarded.
A winning solution will compute the cost in the lowest asymptotic runtime in terms of m and n. Execution time will be assessed as a tie breaker.
| https://codegolf.stackexchange.com/questions/182647/minimum-cost-of-solving-the-eni-puzzle | [
"puzzle-solver",
"fastest-algorithm",
"sliding-puzzle"
] | 6 | 2019-04-03T19:53:28 | [
"@Jonah Because The costs I've outlined here are not the actual number of moves required to make equivalent moves on the puzzle. \"Swapping\" moves have a higher cost. To solve the actual puzzle, you first reach the state I described, or a state that is only one sliding move away. Because my specifications don't al... | 5 | Technology | 0 |
3 | crypto | Share Conversion between Different Finite Fields | Let us have any linear secret sharing scheme (LSSS) that works on some field $Z_{p}$, where p is some prime or a power of a prime e.g., Shamir Secret Sharing, Additive secret Sharing. The problem at hand is simple, for any secret shared value in $Z_{p}$, is it possible to convert it (and its shares) to elements on $Z_{q}$, where q could be either greater or smaller than p? Note that **I do not need** conversion from $Z_{2^n}$ to $Z_{2}$.
I am aware of the results on [PRSS](http://link.springer.com/chapter/10.1007%2F978-3-540-30576-7_19), and well as in this [2008](https://eprint.iacr.org/2008/221.pdf) work, however the former does not address field conversion, and the latter, have explanations that are not clear nor it offers perfect security.
Any help is more than welcome!
| https://crypto.stackexchange.com/questions/47554/share-conversion-between-different-finite-fields | [
"secret-sharing",
"multiparty-computation",
"finite-field",
"garbled-circuits",
"function-evaluation"
] | 11 | 2017-05-19T06:58:27 | [
"@Dragos... Btw.... Are you from Bristol's group? that Dragos?",
"I see..Even converting shares from $Z_p \\rightarrow Z_2^{r}$ is a hard problem - at least I have encountered it several times and don't know how to solve it efficiently other than bit decompose and simulate the binary circuit in $Z_p$.",
"@Drago... | 6 | Technology | 0 |
4 | crypto | RSA key such that pi deciphers to your name per RSA-OAEP | Can you efficiently construct an RSA public/private key pair with $8k$-bit public modulus such that $C=\left\lfloor\pi\,2^{8k-2}\right\rfloor$ deciphers per RSA-OAEP to your name as a bytestring in ASCII or UTF-8?
The decryption must be per RSAES-OAEP of [PKCS#1v2.2](http://mpqs.free.fr/h11300-pkcs-1v2-2-rsa-cryptography-standard-wp_EMC_Corporation_Public-Key_Cryptography_Standards_\(PKCS\).pdf#page=21), SHA-256 hash, no label, MGF1 with SHA-256, and sucessful. The key pair must conform¹ to [section 3](http://mpqs.free.fr/h11300-pkcs-1v2-2-rsa-cryptography-standard-wp_EMC_Corporation_Public-Key_Cryptography_Standards_\(PKCS\).pdf#page=6). It could come as $n$, $e$, $d$, or as a PEM text encoding of the private key per the format of [section A.1.2](http://mpqs.free.fr/h11300-pkcs-1v2-2-rsa-cryptography-standard-wp_EMC_Corporation_Public-Key_Cryptography_Standards_\(PKCS\).pdf#page=41) acceptable by an [ASN.1 decoder](https://lapo.it/asn1js/). Kudos if it passes² OpenSSL's [`RSA_check_key`](https://gitlab.com/gitlab-org/build/omnibus-mirror/openssl/blob/706457b7bda7fdbab426b8dce83b318908339da4/crypto/rsa/rsa_chk.c), more kudos if $8k\ge1024$, the higher the better.
$C$ is the first $k$ bytes of [these](http://mpqs.free.fr/C.txt) (in hex) for $8k$ up to $2^{16}$ bits.
* * *
Update: [this](https://crypto.stackexchange.com/a/80322/555) answers in the affirmative, for whatever usual $k$, with $p$ and $q$ equal-size primes, and $N$ hard to factor. The next frontier is a low $e$, say in $[3,2^{256})$, ideally $65537$.
Further stunts are possible, like having $C$ or $2$ the signature of anything per any usual RSA signature scheme, even deterministic and/or with message recovery.
As a consequence: verifying a signature of a file against Alice's RSA public key (including, certified and used for other purposes) is proof neither that the file is unchanged since signature, nor that Alice was the signer.
Example where that matters: the signature of a file is on a trusted site, the file is not. Alice manages to get her public key (which she uses normally) in the verifier's keystore (with the key identifier of the legitimate signer, if there is such thing in the signature). The verifier, who just uses crypto as a magic tool, sees that the signature verifies, and (rightly) trusts the secure site to hold only trusted signatures of trusted files. There is a chance that a file alteration by Alice goes uncaught, or/and that she can wrongly convince the verifier that she made the signature at the date notarized by the trusted site.
Update 2: Some consequences of the same possibility are discussed in Dennis Jackson, Cas Cremers, Katriel Cohn-Gordon and Ralf Sasse's [_Seems Legit: Automated Analysis of Subtle Attacks on Protocols that Use Signatures_](https://eprint.iacr.org/2019/779), originally in [proceedings of CCS 2019](https://doi.org/10.1145/3319535.3339813).
* * *
¹ That is: public modulus $n$ must be the product of $u\ge2$ distinct odd primes, odd public exponent $e$ with $3\le e<n$, private exponent $d$ with $0<d<n$ and $e\,d\equiv1\pmod{r-1}$ for each prime $r$ divisor of $n$.
² For keys otherwise per the above, it limits $u$ (including $u=2$ for $n$ less than 1024-bit, $u\le3$ for $n$ less than 4096-bit, see [`rsa_multip_cap`](https://gitlab.com/gitlab-org/build/omnibus-mirror/openssl/-/blob/706457b7bda7fdbab426b8dce83b318908339da4/crypto/rsa/rsa_mp.c#L100)). It also requires the full factorization of $n$ to be given, as well as reduced exponents and multiplicative inverses.
| https://crypto.stackexchange.com/questions/80211/rsa-key-such-that-pi-deciphers-to-your-name-per-rsa-oaep | [
"rsa",
"oaep"
] | 13 | 2020-04-23T04:22:27 | [] | 0 | Technology | 0 |
5 | crypto | Finding $x$ such that $g^x\bmod p<p/k$? | In a Schnorr group as used for DSA, of prime modulus $p$, prime order $q$, generator $g$ (with $p/g$ small), how can we efficiently exhibit an $x$ with $0<x<q$ such that $g^x\bmod p<p/k$, for sizable $k$ but $k\ll\sqrt q$ ?
I see that for small $k$, it is enough to try incremental values of $x$ until finding an $x$ that fits, with expected cost $O(k)$ modular multiplications in $\mathbb Z_p^*$. Is there better?
Assume $p$ is an $l$-bit prime; $q$ is an $n$-bit prime with $q$ dividing $p-1$; and $g=h^{(p-1)/q}$ for some (random?) $h$ in $\mathbb Z_p^*$. For concrete values, assume $l\approx1024$, $n\approx160$, $k\approx2^{64}$.
Update: I'm expecting the problem to be hard, and thus that the more difficult problem of exhibiting $x$ with $g^x\bmod p$ in a range of width $p/k$ must be hard.
| https://crypto.stackexchange.com/questions/48503/finding-x-such-that-gx-bmod-pp-k | [
"discrete-logarithm",
"group-theory"
] | 19 | 2017-06-20T08:57:11 | [
"Can you please provide me with some real values of variables in question instead of me attempting with random values.",
"@pintor: no, I have made no progress nor got more feedback than the above comments.",
"@fgrieu, did you get any more updates on the problem? I'm also curious",
"This seems to reduce to a d... | 8 | Technology | 0 |
6 | crypto | Time-memory tradeoffs in Shor's algorithm | Can a quantum computer with insufficient qubits to factor an integer of a given size make _any_ progress in factoring it? For example, what if a quantum computer is only one qubit short of what is necessary to attack a specific integer? Is it capable of making any progress in factoring it, or would it be just as useless as a 2 qubit machine? Assume qubits refer to logical qubits in a general-purpose cryptanalytic quantum computer with sufficient quantum error correction to solve the problem of decoherence.
Given that qubits can be thought of as the memory for quantum computers, I am essentially asking if time-memory tradeoffs are possible with Shor's algorithm when too few qubits are available.
| https://crypto.stackexchange.com/questions/67910/time-memory-tradeoffs-in-shors-algorithm | [
"factoring",
"quantum-cryptanalysis",
"shors-algorithm"
] | 8 | 2019-03-09T22:20:47 | [
"Interesting question. One problem certainly could occur when you have less qubits than the number of the period $r$ ($a^r \\bmod N$). This would probably mess up the inverse quantum fourier transform which amplifies the correct period (quantum wave interference), thus rendering the whole computation impractical. I... | 3 | Technology | 0 |
7 | crypto | Does the bias in RC4 drop asymptotically further in the keystream? | It's well-known that the RC4 keystream has significant biases that become less prominent later in the keystream. The most severe bias is in the second byte, which has a 128-1 bias towards zero. Other biases remain, and it's typically recommended to drop between 768 and 3072 bytes of the keystream.
Will dropping one more byte _always_ reduce the bias, or is there a point when the bias is as low as it is going to get and it won't get lower as the keystream goes on? It doesn't matter if the bias is negligible.
Note that I am only talking about short-term biases, not long-term ones which RC4-dropN can't solve.
**Does an _N_ exist that results in minimum bias in RC4-dropN?**
| https://crypto.stackexchange.com/questions/72353/does-the-bias-in-rc4-drop-asymptotically-further-in-the-keystream | [
"stream-cipher",
"rc4",
"distinguisher"
] | 7 | 2019-08-03T01:44:25 | [
"@MaartenBodewes I have not yet, but I also haven't done much research into it. There's a lot of academic text on the (in)security of RC4 which I should familiarize myself with.",
"Hey Forest. This question popped up again. Unfortunately I don't know the answer, but maybe you've found one in the mean time?",
"@... | 7 | Technology | 0 |
8 | cstheory | Partial circulant matrices: Is there a non-zero vector $v\in \{-1,0,1\}^n$ such that $Mv=0$? | The following question arose as a side product of some work I have been part of recently. An $m$ by $n$ $(0,1)$-matrix $M$ is partial circulant if it can be formed by taking the first $m$ rows of a [circulant matrix](http://en.wikipedia.org/wiki/Circulant_matrix) and all its entries are either $0$ or $1$. We assume that $m\leq n$.
Given such a partial circulant matrix $M$, what is the computational complexity of the following question?
> Is there a non-zero vector $v\in \\{-1,0,1\\}^n$ such that $Mv=0$?
The problem is clearly in NP but my guess is that it is not NP-hard (unlike [this related question](https://cstheory.stackexchange.com/questions/20277/decide-whether-a-matrixs-kernel-contains-any-non-zero-vector-all-of-whose-entri)). I have not, however, found a poly time solution.
| https://cstheory.stackexchange.com/questions/12060/partial-circulant-matrices-is-there-a-non-zero-vector-v-in-1-0-1-n-such | [
"cc.complexity-theory",
"ds.algorithms",
"matrices"
] | 20 | 2014-11-04T06:46:47 | [
"The circulant case was in fact considered at mathoverflow.net/questions/168474/… . However, it seems that there is fatal bug in one of the answers and the other answer is not a complete solution.",
"would it be more natural to study the circulant case 1st vs the partial circulant? also (idea) the other problem a... | 2 | Science | 1 |
9 | cstheory | Are monotone Boolean functions in P well-approximated by monotone polynomial-size circuits? | **Question 1:** Is it true that for every polynomial $p(n)$ and $\epsilon >0$ there is a polynomial $q(n)$ such that every monotone Boolean function on $n$ variables that can be expressed by a Boolean circuit of size $p(n)$ can be $\epsilon$-approximated by a _monotone_ Boolean circuit of size at most $q(n)$.
A function $f$ is $\epsilon$-approximated by a function $g$ if they agree on at least $(1-\epsilon)$ fraction of inputs.
We can ask a slightly more general question. Let $\mu_p$ denote the Bernoulli measure on $\\{0,1\\}^n$ where every variable is '1' with probability $p$. For a (non-constant) monotone function $f$ let $p_c(f)$ be the value such that $\mu_{p_c}(\\{x: f(x)=1\\})=1/2$.
**Question 2:** Is it true that for every polynomial $p(n)$ and $\epsilon >0$ there is a polynomial $q(n)$ such that every monotone Boolean function $f$ on $n$ variables, there is a Boolean function $g$ that can be expressed by a monotone Boolean circuit of size $q(n)$ such that for $p=p_c(f)$, $\mu_p(\\{x: f(x) \ne g(x) \\}) \le \epsilon$.
Of course, the instinctive thought is that the answers for both these question is **NO**. Is it known? Razborov famously proved that matching cannot be expressed by a monotone Boolean circuit of polynomial size. But matching can be approximated in its critical probability by the property that there are no isolated vertices and this property admits a monotone (low depth) circuit.
I asked the analog questions (Problems 1,2,3) for $AC^0$ and $TC^0$ in [this blog post](https://gilkalai.wordpress.com/2010/02/10/noise-stability-and-threshold-circuits/).
| https://cstheory.stackexchange.com/questions/31473/are-monotone-boolean-functions-in-p-well-approximated-by-monotone-polynomial-siz | [
"cc.complexity-theory",
"circuit-complexity",
"pr.probability"
] | 18 | 2015-05-12T23:58:06 | [
"Dear Gil, now I see the point: I just wrongly interpreted your question! We indeed have that every monotone circuit, which needs only to coincide with b-Clique on an extremely small, but special set of inputs, must be large. You, however, do not specify this subset of \"hard\" inputs, only its ratio $1-\\epsilon$.... | 8 | Science | 0 |
10 | cstheory | Weighted Hamming distance | Basically my question is, what kind of geometry do we get if we use a "weighted" Hamming distance. This is not necessarily Theoretical Computer Science but I think similar things come up sometimes, for instance in randomness extraction.
Define:
$d(x,y)=$ the Hamming distance between binary strings $x$ and $y$ of length $n$, $=$ the cardinality of $\\{k: x(k)\ne y(k)\\}$.
For a set of strings $A$,
$d(x,A)=\min \\{ d(x,y): y\in A\\}$.
The _$r$ -fold boundary of_ $A\subseteq \\{0,1\\}^n$ is
$$\\{x\in\\{0,1\\}^n: 0 < d(x,A)\le r\\}.$$
Balls centered at $0$ are given by
$$ B(p)=\\{x: d(x,0)\le p\\}, $$ where $0$ is the string of $n$ many zeroes.
A _Hamming-sphere_ is a set $H$ with $B(p)\subseteq H\subseteq B(p+1)$. (So it's more like a ball than a sphere, but this is the standard terminology...)
Now, Harper in 1966 showed that for each $k$, $n$, $r$, one can find a Hamming-sphere that has minimal $r$-fold boundary among sets of cardinality $k$ in $\\\\{0,1\\\\}^n$. So a ball is a set having minimal boundary -- just like in Euclidean space.
The cardinality of $B(p)$ is ${n\choose 0}+\cdots {n\choose p}$.
The $r$-fold boundary of $B(p)$ is just the set $B(p+r)\setminus B(p)$, which then has cardinality ${n\choose p+1}+\cdots+{n\choose p+r}$.
So far, so good. But now suppose we replace $d$ by a different metric $D$: first let $d_j(x,y)$ be the Hamming distance between the prefixes of $x$ and $y$ of length $j$, and then $$ D(x,y)=\max_{j\le n}\ \frac{d_j(x,y)}{f(j) }$$ where $0\le f(j)\le j$. (For example we could have $f(j)=\sqrt{j}$, or $f(j)=j/\log j$.)
This is supposed to make $D(x,y)$ small if the differences of $x$ and $y$ do not clump together at small values of $j$.
# Questions
> Is the minimum $r$-fold boundary (under $D$) realized by a $D$-ball?
>
> Is there a better definition of $D$?
>
> Under the metric $D$, what's the minimum size of the $r$-fold boundary of a subset of $\\\\{0,1\\\\}^n$ having cardinality $k$? (A reasonable lower bound would be nice.)
(Cross-[posted](https://mathoverflow.net/questions/38221/geometry-in-a-hamming-box) on MathOverflow).
| https://cstheory.stackexchange.com/questions/2637/weighted-hamming-distance | [
"co.combinatorics",
"randomness"
] | 20 | 2010-11-01T22:06:48 | [] | 0 | Science | 1 |
11 | cstheory | To what extent MSO = WS1S, when adding relations? | [This question has been asked on MathOverflow with no luck a month ago.]
Let me first clarify my definitions. For a word $w \in \Sigma^*$, with $\Sigma =\\{a_1, \ldots, a_n\\}$, I define two structures:
$\mathbb{N}(w) = \langle \mathbb{N}, <, Q_{a_1}, \ldots, Q_{a_n} \rangle$,
and the more usual _word model_ :
$\mathbb{N}^r(w) = \langle \\{0, \ldots, |w|-1\\}, <, Q_{a_1}, \ldots, Q_{a_n} \rangle$,
where $Q_{a_i} = \\{p \mid w_p = a_i\\}$.
Then WS1S is the set of second order formulas with models of the form $\mathbb{N}(w)$, with order, and for which second order quantification is limited to finite subsets of the domain. MSO is the set of monadic second order formulas with models of the form $\mathbb{N}^r(w)$, with order.
The usual proof that REG = WS1S proves at the same time that MSO = WS1S. My question is then, for which first or second order relations can we keep this to be true?
For instance, if we add a unary predicate $E(X)$ which says that a (monadic) second order variable contains an even number of objects, we add no power, as $E(X)$ is expressible as "there exists $X_1$ and $X_2$ that partition $X$, in such a way that if an element is in $X_i$ the next one in $X$ is in $X_j$, $i \neq j$, and the first element of $X$ is in $X_1$ and the last is in $X_2$."
Now, if we add a predicate $|X| < |Y|$, then WS1S becomes undecidable (see Klaedtke & Ruess, 10.1.1.7.3029), while MSO stays trivially decidable.
Thank you.
* * *
**Edit:** As a side question, ... is this question of interest? I mean, I'm no expert in the field, so I'm not sure this question is relevant.
| https://cstheory.stackexchange.com/questions/15/to-what-extent-mso-ws1s-when-adding-relations | [
"lo.logic",
"automata-theory",
"descriptive-complexity"
] | 19 | 2010-08-16T13:32:53 | [
"Your model $\\mathbb{N}(w)$ is very unusual : what is the need of considering a finite word on an infinite underlying structure ? If you forget the letter predicates (which do not change the problem a lot), you are in fact comparing MSO on finite linear orders, and WS1S on the order $\\omega$. This is strange, bec... | 6 | Science | 0 |
12 | cstheory | Descriptive complexity of communication complexity classes | It is well known that some major complexity classes, like P or NP, admit a full logical characterization (e.g NP = existential 2nd order logic by Fagin's theorem). On the other hand, one can also define complexity classes in communication complexity (where P = problems solvable with poly(logN) communication etc. - see [Complexity classes in communication complexity theory](http://dl.acm.org/citation.cfm?id=1382962) for more).
My question is - are any descriptive complexity characterizations known for communication complexity classes (or do any such results from standard complexity classes transfer to communication setting easily)?
| https://cstheory.stackexchange.com/questions/9041/descriptive-complexity-of-communication-complexity-classes | [
"cc.complexity-theory",
"lo.logic",
"communication-complexity"
] | 18 | 2011-11-19T14:15:36 | [
"You may want to limit the computational power of the parties, which is not done in the usual communication complexity models."
] | 1 | Science | 0 |
13 | cstheory | Does $EXP\neq ZPP$ imply sub-exponential simulation of BPP or NP? | By simulation I mean in the Impaglazzio-Widgerson [IW98] sense, i.e. sub-exponential deterministic simulation which appears correct i.o to every efficient adversary.
I think this is a proof: if $EXP\neq BPP$ then from [IW98] we get that BPP has such a simulation. Otherwise we have that $EXP=BPP$, which implies $RP=NP$ (because $NP \subseteq BPP$) and $EXP \in PH$. Now if $NP=RP=ZPP$ we have that $PH$ collapses to $ZPP$ and as a result $EXP$, but this cannot happen because of the assumption, so $RP\neq ZPP$ and this by the Kabanets paper "Easiness assumptions and hardness tests: trading time for zero error" implies that RP has such a simulation and as a result also NP.
This sounds like it could be a useful gap theorem. Does anyone know if it appears anywhere?
| https://cstheory.stackexchange.com/questions/430/does-exp-neq-zpp-imply-sub-exponential-simulation-of-bpp-or-np | [
"cc.complexity-theory",
"reference-request",
"complexity-classes",
"conditional-results"
] | 29 | 2010-08-23T13:21:04 | [
"ZPP is closed under complement, and contained in NP. So, $\\mathrm{NP\\subseteq ZPP}$ implies NP = ZPP = coNP, hence PH = NP = ZPP.",
"Why does NP in ZPP imply PH in ZPP?",
"(Just for reference): Identical question on MO: mathoverflow.net/questions/35945. Maybe someone finds the comments there inspiring.",
"... | 4 | Science | 1 |
14 | cstheory | Interesting PCP characterization of classes smaller than P? | The PCP theorem, $\mathsf{NP} = \mathsf{PCP}(\mathsf{log}\, n, 1)$, involves probabilistically checkable proofs with polynomial time verifiers, so the smallest class that can be characterized in this way (that is, $\mathsf{PCP}(0, 0)$) must be $\mathsf{P}$. There are also PCP characterizations of larger complexity classes (for example, $\mathsf{NEXP} = \mathsf{PCP}(\mathsf{poly}, \mathsf{poly})$), also using polynomial time verifiers.
Can we achieve an interesting (that is, not immediately following from the definitions) PCP characterization of smaller complexity classes by restricting the time or space used by the verifier? For example, by using a logarithmic space verifier, or an $\mathsf{NC}$ circuit verifier?
| https://cstheory.stackexchange.com/questions/12060/interesting-pcp-characterization-of-classes-smaller-than-p | [
"cc.complexity-theory",
"complexity-classes",
"pcp"
] | 20 | 2012-07-18T11:54:42 | [
"Your comment is contradictory to your question where you count “PCP(0, 0) = P” as a PCP characterization of P.",
"I suppose I meant are there any characterizations that don't follow immediately from the definition, in the same way that PCP(log n, 1) is non-obvious characterization of NP.",
"Why not? If you res... | 3 | Science | 1 |
15 | cstheory | Is Hankelability NP-hard? | I asked this question on [SO](https://stackoverflow.com/questions/29484864/an-algorithm-to-detect-permutations-of-hankel-matrices) on April 7 and added a bounty which has now expired but no poly time solution has been found yet.
I am trying to write code to detect if a matrix is a permutation of a Hankel matrix. Here is the spec.
**Input:** An n by n matrix M whose entries are 1 or 0.
**Output:** A permutation of the rows and columns of M so that M is a [Hankel matrix](http://en.m.wikipedia.org/wiki/Hankel_matrix) if that is possible. A Hankel matrix has constant skew-diagonals (positive sloping diagonals).
When I say a permutation, I mean we can apply one permutation to the order of the rows and a possibly different one to the columns.
A very nice $O(n^2)$ time algorithm [is known](https://stackoverflow.com/questions/20704900/determine-if-some-row-permutation-of-a-matrix-is-toeplitz) for this problem if we only allow permutation of the order of rows.
Peter de Rivaz pointed out this [paper](http://www.mat.ucsb.edu/~g.legrady/academic/courses/15w259/d/re_orderableMatrix.pdf) as a possible route to proving NP-hardness but I haven't managed to get that to work.
| https://cstheory.stackexchange.com/questions/31174/is-hankelability-np-hard | [
"cc.complexity-theory"
] | 28 | 2015-04-16T10:44:55 | [
"Cross-posted now to mathoverflow.net/questions/204294/is-hankelability-np-hard",
"@IgorShinkar I mean first some permutation on the order of the rows and then a possibly different permutation on the order of the columns, then stop.",
"Do you mean apply first some permutation on the rows and then some permutati... | 6 | Science | 1 |
16 | cstheory | Model-checking for three-variable logics and restricted structures | Denote the $k$-variable fragment of logic $L$ by $L^{(k)}$. The model-checking problem for a logic $L$ with respect to a class of structures $C$, denoted $MC(L,C)$, is the decision problem
> $MC(L,C)$
> _Input:_ formula $\phi$ of $L$, structure $S$ from $C$
> _Question:_ does $S$ satisfy $\phi$?
Is there a logic $L$, with associated class of structures $C$, and a subclass $D$ of $C$, such that
1\. $MC(L^{(3)},D)$ is "easy",
2\. $MC(L,D)$ is "hard", and
3\. $MC(L^{(3)},C)$ is "hard"?
With "easy" vs. "hard" I mean decidable vs. undecidable, PTIME vs. NP-hard, LogCFL vs. P-complete, or similar dichotomies.
# Motivation
For many logics the three-variable fragment is enough to express "hard" properties (this is true for MSO and FO over ordered structures). So results about undecidability often carry across from the logic to its three-variable fragment. There is also often a way to reduce the arity of predicates used in formulas efficiently (for instance, SAT can be reduced to 3SAT with a linear increase in formula size). Reducing arity facilitates expressing some formulas with fewer variables.
On the other hand, there exist logics which are "easy" for some classes of structures yet "hard" in general. For instance, the model-checking problem for MSO is "easy" (linear time) on unordered graphs satisfying the property "has treewidth less than $k$" for any fixed $k$, but "hard" (NP-complete) on unordered graphs in general.
I am asking whether there are any known examples where the restriction to 3 variables interacts with the property defining the subclass $D$ in an essential way to yield tractability, yet where neither of these restrictions on their own is enough.
Apologies for the vagueness of the question: I am trying to leave this general because I can't think of any examples. As soon as I restrict $D$ enough to ensure condition 1 holds, condition 2 seems to break. I suspect there are some unnatural classes which would work, but ideally $D$ should be "nice", at least hereditary and closed under isomorphism. As a further non-example, model-checking for FO is PSPACE-complete while model-checking for $FO^{(3)}$ is PTIME-complete; here the restriction to a finite number of variables is already enough to ensure an "easy" model-checking problem and condition 3 fails.
Definitions for completeness: FO is [first-order logic](http://en.wikipedia.org/wiki/First-order_logic), MSO is monadic [second-order logic](http://en.wikipedia.org/wiki/Second-order_logic), where quantification over sets of individuals is allowed, as well as quantification over individuals.
| https://cstheory.stackexchange.com/questions/2637/model-checking-for-three-variable-logics-and-restricted-structures | [
"cc.complexity-theory",
"co.combinatorics",
"lo.logic",
"descriptive-complexity"
] | 20 | 2010-11-03T11:47:19 | [
"@MichaëlCadilhac alas, no.",
"Have you made any progress on that question, András?"
] | 2 | Science | 1 |
17 | cstheory | In an $m$ by $n$ Boolean matrix, can you find a square block whose four corners are ones in $O(m \cdot n)$ time? | **Decision Problem**
Input: An $m$ by $n$ Boolean matrix $M$.
Decision Question: Does there exist a square block within $M$ such that upper-left corner entry == upper-right corner entry == lower-left corner entry == lower-right corner entry == 1? That is, all four corners of the square block are 1's.
**Cubic Time Solution**
We know that this problem can be solved in $O(m \cdot n \cdot min\\{m,n\\})$ time. The approach involves scanning through the matrix row by row. For each row, for each pair of 1's in that row, we check whether those 1's form an edge of a square block whose four corners are 1's.
**Our Question**
What is the time complexity of this problem? Is there a quadratic time solution? In particular, can we solve this decision problem in $O(m \cdot n)$ time?
**Extra Background**
1. If you're just looking for a rectangular block whose four corners are ones, this can be solved in $O(m \cdot n)$ time. Many variations to this problem can be solved in $O(m \cdot n)$ time as well. I co-authored a paper on this subject.
2. The paper ["Finding squares and rectangles in sets of points"](https://doi.org/10.1007/BF01931281) investigates a related problem from computational geometry where you're given a set of points in a 2D plane and you want to know if there are four points that form an axis-parallel square.
Written together with Eevvoor.
| https://cstheory.stackexchange.com/questions/47588/in-an-m-by-n-boolean-matrix-can-you-find-a-square-block-whose-four-corners | [
"ds.algorithms",
"matrices",
"computational-geometry",
"search-problem",
"boolean-matrix"
] | 18 | 2020-09-18T11:04:38 | [
"It feels somehow like an FFT would help, although I haven't been able to puzzle out details. You're looking for something translation-invariant, and working in frequency space often helps with that sort of thing. Tossing out this comment in case someone else can spot something.",
"@MichaelWehar: ok, let me know ... | 14 | Science | 0 |
18 | cstheory | Problem unsolvable in $2^{o(n)}$ on inputs with $n$ bits, assuming ETH? | If we assume the Exponential-Time Hypothesis, then there is no $2^{o(n)}$ algorithm for $n$-variable 3-SAT, and many other natural problems, such as 3-COLORING on graphs with $n$ vertices. Notice though that, in general, encoding the input for $n$-variable 3-SAT or $n$-vertex 3-COLORING takes something like $O(n\log n)$ bits. For example, to describe a sparse graph as input to 3-COLORING, for each edge we would have to list its endpoints. So the lower bound is not exponential in the length of the input. Therefore, my question is the following:
Is there a problem for which no $2^{o(n)}$ algorithm exists for inputs of length $n$ bits (assuming ETH)?
Ideally, the problem would be in NP (no cheating with succinct NEXP-hard problems!) and be reasonably natural, but I won't be picky.
Let me also note that after digging around I found that there are efficient ways to encode planar graphs with $O(n)$ bits. So, if one could find a problem that takes time exponential in the number of vertices even for planar graphs, the question would be settled. However, because planar graphs have treewidth $O(\sqrt{n})$, most natural problems have sub-exponential algorithms in this case.
| https://cstheory.stackexchange.com/questions/16148/problem-unsolvable-in-2on-on-inputs-with-n-bits-assuming-eth | [
"cc.complexity-theory",
"sat",
"planar-graphs",
"succinct"
] | 48 | 2013-01-19T08:24:59 | [
"I believe that it is an open question if we can have a lower bound $2^{\\Omega(n)}$ under ETH, where $n$ is the bit size. We know that proving lower bound $\\Omega(c^n)$ under SETH would disprove SETH.",
"How about the following problem, for some large constant $c>0$? Given the encoding of a non-deterministic T... | 8 | Science | 1 |
19 | cstheory | Sylver Coinage Game | A game in which the players alternately name positive integers that are not sums of previously named integers (with repetitions being allowed). The person who names 1 (so ending the game) is the loser.
The question is: If player 1 names ‘16’, and both players play optimally thereafter, then who wins?
It has been known that if player 1 name "5n," then player 1 wins. If player name "5n+2", the result is known(maybe palyer 2 wins, but I haven't found the source yet). 16 is the minimum number of "5n+1".
I guess that this problem is PSPACE-hard, but I haven't proved it yet.
| https://cstheory.stackexchange.com/questions/22112/sylver-coinage-game | [
"co.combinatorics",
"combinatorial-game-theory"
] | 18 | 2014-04-14T04:03:06 | [
"@NealYoung The algorithm in Winning Ways only gives whether an opening move is winning or not -- in accordance with Conway's comment, it relies on a theorem saying finitely moves of the form 2^a 3^b win, and I imagine (Conway having told me of that fact before) that this is what he meant when he said that an algor... | 7 | Science | 0 |
20 | cstheory | Complexity of the homomorphism problem parameterized by treewidth | The _homomorphism problem_ $\text{Hom}(\mathcal{G}, \mathcal{H})$ for two classes $\mathcal{G}$ and $\mathcal{H}$ of graphs is defined as follows:
> **Input:** a graph $G$ in $\mathcal{G}$, a graph $H$ in $\mathcal{H}$
>
> **Output:** decide if there is a homomorphism from $G$ to $H$, i.e., a mapping $h$ from the vertices of $G$ to those of $H$ such that, for any edge $\\{x, y\\}$ of $G$, $\\{h(x), h(y)\\}$ is an edge of $H$.
For each $k \in \mathbb{N}$, I will call $\mathcal{T}_k$ the class of the graphs of [treewidth](http://en.wikipedia.org/wiki/Treewidth) at most $k$. I'm interested in the problem $\text{Hom}(\mathcal{T}_k, \mathcal{T}_k)$, which I see as a _parameterized_ problem (by the treewidth bound $k$). My question is: **what is the complexity of this parameterized problem?** Is it known to be FPT? or is it W[1]-hard?
Here are some things that I found about the $\text{Hom}$ problem, but which do not help me answer the question. (I write $-$ for the class of all graphs.)
* <http://www.sciencedirect.com/science/article/pii/009589569090132J>: If $\mathcal{H}$ is bipartite then $\text{Hom}(-, \mathcal{H})$ is in PTIME, otherwise it is NP-complete, but of course the NP-hardness relies on allowing arbitrary $G$.
* [http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.86.9013&rep=rep1&type=pdf](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.86.9013&rep=rep1&type=pdf): If the treewidth of $\mathcal{G}$ (modulo homomorphic equivalence) is bounded by a constant then $\text{Hom}(\mathcal{G}, -)$ is in PTIME (and otherwise it isn't, assuming FPT != W[1]). Hence, in particular my problem $\text{Hom}(\mathcal{T}_k, \mathcal{T}_k)$ is in PTIME for fixed $k$, but this doesn't tell me what is the dependency on the parameter.
* From Flum and Grohe's book _Parameterized Complexity Theory_ , Corollary 13.17: The problem $\text{Hom}(\mathcal{T}_k, -)$ is FPT when parameterized by the _size_ of $G$ (but I am parameterizing by the treewidth)
* <http://users.uoa.gr/~sedthilk/papers/homo.pdf>, Corollary 3.2: When fixing a specific graph $H$, the problem $\text{Hom}(\mathcal{T}_k, \\{H\\})$, parameterized by k, is FPT (this even holds for more complicated counting variants), but I do not want to restrict to fixed $H$.
| https://cstheory.stackexchange.com/questions/34877/complexity-of-the-homomorphism-problem-parameterized-by-treewidth | [
"graph-algorithms",
"parameterized-complexity",
"treewidth",
"fixed-parameter-tractable",
"homomorphism"
] | 18 | 2016-06-02T03:37:34 | [
"This question is still open, but one remark: there is an FPT algorithm parameterized by treewidth for the graph isomorphism problem, here: epubs.siam.org/doi/abs/10.1137/… (Daniel Lokshtanov, Marcin Pilipczuk, Michał Pilipczuk, and Saket Saurabh, \"Fixed-Parameter Tractable Canonization and Isomorphism Test fo... | 1 | Science | 0 |
21 | cstheory | Is Node Multiway Cut NP-complete on planar graphs when all terminals lie on the outer face? | I am interested in the following problem.
**Node Multiway Cut on Planar Graphs with terminals on the outer face**
* Instance: A plane graph G, and integer k, and a set $S \subseteq V(G)$ of terminals which are all incident on the outer face of G.
* Question: Is there a set of vertices $X \subseteq V(G)$ of size at most $k$ such that all vertices of $S \setminus X$ belong to different connected components of $G - X$?
Informally, the problem asks whether all terminals can be separated with k vertex deletions, knowing that all terminals lie on the outer face. If we replace vertex deletions by edge deletions, then this is known to be possible in polynomial time by considering multiway cuts in the planar dual, and relating them to certain kind of Steiner trees ([Efficient Algorithms for k-Terminal Cuts on Planar Graphs](https://doi.org/10.1007/3-540-45678-3_29)). Since the edge-deletion version reduces to the vertex-deletion version, the edge-deletion version seems easier; no reduction in the reverse direction is known.
The question is: is Node Multiway Cut on Planar Graphs with terminals on the outer face NP-complete? The reason I am interested in this problem is that it would provide me with a starting point for another complexity lower bound if this is indeed NP-complete.
| https://cstheory.stackexchange.com/questions/8969/is-node-multiway-cut-np-complete-on-planar-graphs-when-all-terminals-lie-on-the | [
"cc.complexity-theory",
"np-hardness",
"planar-graphs"
] | 17 | 2011-11-14T09:06:13 | [
"I know for sure that the algorithm I mentioned works for edge weighted Steiner tree, but one should be able to adapt it to work for vertices.",
"I think that this is equivalent to solving vertex weighted Steiner tree where all terminals on the outer face. The very loose idea would be something like: consider the... | 2 | Science | 0 |
22 | cstheory | Can short-distance connectivity be harder than connectivity? | Has anybody seen the following (or similar) question being considered:
> Can it be **easier** to determine the presence/absence of $s$-$t$ paths than to determine the presence/absence of _short_ $s$-$t$ paths?
A bit more formally, the _distance_ -$k$ _connectivity_ problem STCON(n,k) is, given a subgraph of a complete undirected graph $K_n$ on $[n]=\\{1,\ldots,n\\}$, to decide whether there is a path from node $s=1$ to node $t=n$ of length $\leq k$. By the _length_ , I mean the number of inner nodes (just for notational convenience). The _connectivity_ problem STCON(n) corresponds to the case when $k=n-2$: is there any $s$-$t$ path at all? These problems are monotone boolean functions of $\binom{n}{2}$ variables $x_{i,j}$ corresponding to the edges of $K_n$.
It is well known, that $O(kn^2)$ fanin-$2$ AND and OR gates are enough to compute STCON(n,k). The desired monotone circuit is given by [Bellman-Ford](http://en.wikipedia.org/wiki/Bellman%E2%80%93Ford_algorithm): first compute $F_j^{(0)}=x_{s,j}$ for all nodes $j$, and then use the recursion $F_j^{(l+1)}=\bigvee_{i\in[n]} F_i^{(l)}\land x_{i,j}$. Then $F_j^{(l)}$ accepts a given graph iff it has an $s$-$j$ path of length $\leq l$. Hence, $F_n^{(k)}=$STCON(n,k).
> **Question 1:** Can STCON(n) have more than constant times **smaller** monotone circuits than STCON(n,k), for some $k\ll n$?
A boolean function $f(x_1,\ldots,x_n)$ is a _monotone projection_ of a boolean function $g(y_1,\ldots,y_m)$, if the function $f$ can be obtained from $g$ by replacing each $y$-variable by $0$ or $1$ or by some $x$-variable. Say, we trivially have that every THRESHOLD(n,k) is a monotone projection of MAJORITY(2n).
By taking $k$ copies of the input graph for STCON(n,k), one can easily show that STCON(n,k) is a monotone projection of STCON(kn). Thus, STCON(kn) cannot have smaller circuits than STCON(n,k). Can the factor "k" here be replaced by some constant?
> **Question 2:** Is STCON(n,k) a monotone projection of STCON(m) for some $m < kn$?
**ADDDED** (Apr.7 ): Among related results I know are the following:
1. _STCON and monotone**NC$^1$**_ : if a boolean function $f$ has a monotone formula of size $s$, then $f$ is a monotone projection of STCON(m) for $m=O(s)$; this is easy: just view formulas as parallel-sequential networks. In particular, this implies that _every_ monotone boolean function of $n$ variables _is_ a monotone projection of STCON(m) for some $m\leq n2^n$ (take a monotone DNF).
2. _STCON itself is not in monotone**NC$^1$**_ : every monotone circuit for STCON(n,k) requires depth about $(\log k)(\log n)$ (independent on circuit size!). This was shown by [Karchmer and Wigderson](http://www.math.ias.edu/~avi/PUBLICATIONS/MYPAPERS/KW90/KW90.pdf). As observed by [Grigni and Sipser](http://www.mathcs.emory.edu/~mic/papers/4.ps), this holds even when AND gates are allowed to have unbounded fanin! If, however, OR gates can have large fanin and AND gates have fanin-$2$, then depth $2\log k$ and size $O(n^3\log k)$ is already enough: repeated squaring of the adjacency matrix $\log k$ times.
3. _$s$-$t$ Connectivity vs. Connectivity_ : If $f=x_1x_2\lor x_3x_4\lor\cdots\lor x_{n-1}x_n$ is a monotone projection of CONN(m), then $m\geq 2^n+1$. This was proved by [Skyum and Valiant](http://dl.acm.org/citation.cfm?id=3158&CFID=497574256&CFTOKEN=86163804). Here CONN is a "sibling" of STCON: accepts a graph iff it is connected. Note that this particular function $f$ _is_ a monotone projection of STCON(m) even for $m=n$: take a bunch of $n/2$ parallel paths of length two.
4. _Formulas vs. Circuits_ : If the depth is restricted to $o(\log n)$ then STCON(n,k) requires even non-monotone(!) unbounded fanin _formulas_ of size $n^{\Omega(\log k)}$ for all $k\leq \log\log n$. This was proved by [Rossman](http://eccc.hpi-web.de/report/2013/169/download), and separated for the first time bounded-depth circuits from formulas, because for such values of $k$, repeated squaring gives the desired small (even monotone) _circuit_ of depth $2\log k\leq 2\log\log\log n=o(\log n)$.
5. _Directed vs. Undirected Connectivity_ : The directed reachability problem DSTCON(n) (where input graphs are directed) is a monotone projection of STCON(m) only if $m=n^{\Theta(\log n)}$. This was proved by [Potechin](http://eccc.hpi-web.de/report/2009/142/revision/1/download), and separated monotone versions of **L** and **NL**.
Does anybody know some results more tightly related to the STCON(n,k) vs. STCON(n) question itself?
* * *
**CORRECTION** (Apr 9): Unfortunately, just taking $k$ copies of the input graph $G$ for STCON(n,k), only shows that STCON(n,k) is a monotone projection of **D** STCON(kn), the **directed** (even acyclic) version of STCON. The difficult direction is to ensure that the modified graph $H=H_G$ has no s-t paths, when $G$ _has_ s-t paths, but all they are longer than $k$. If $H$ is allowed to be directed, this is easy to ensure by arranging the copies of $G$ into $k$ layers, and drawing _directed_ edges between layers according to $G$. But if the edges $H$ are required to be _undirected_ , this simple layering trick won't work. So, I must refine my Question 2 as (cf. with the 5-th result by Potechin above):
> **Question 2':** Is STCON(n,k) a monotone projection of **D** STCON(m) for some $m < kn$?
Both answers would be interesting. If YES, this would show that layering is not the best way to "count". If NO, this would show that every monotone switching network for STCON(n,k) requires $kn$ nodes. (By the Bellman-Ford algorithm - which itself is just the "layering trick" in disguise - $kn$ nodes are also enough.)
| https://cstheory.stackexchange.com/questions/31005/can-short-distance-connectivity-be-harder-than-connectivity | [
"cc.complexity-theory",
"graph-algorithms",
"circuit-complexity",
"lower-bounds",
"dynamic-programming"
] | 17 | 2015-04-03T10:00:23 | [] | 0 | Science | 0 |
23 | cstheory | Sequences with sublogarithmic concat and approximate split | Is there a data structure for representing sequences that supports the operations:
* **concat** takes two sequences of size $m$ and $n$ and produces a new sequence of size $m+n$ by joining them in time $o(\lg \min(n,m/n))$ (or $o(\lg \min(m,n/m))$ if $n>m$).
* For some constants $c$ and $N$, **approximate split** takes a sequence of size $n > N$ and divides it into two contiguous non-overlapping sequences of sizes $p$ and $q$ such that $p+q = n$ and $1/c < p/q < c$ in time $o(\lg n)$
* **singleton** takes a single element and produces a sequence containing only that element in $O(1)$ time.
* **linearize** takes a sequence of size $n$ and produces an array of size $n$ containing the elements of the sequence in time $O(n)$.
I'm really looking for worst-case or expected bounds on the time, but amortized bounds would also be interesting. Results in the word RAM would be interesting as well.
Here are some results I am aware of. In this table, **sloppy split** is like approximate split, but the bounds are $1/c < (\lg p)/(\lg q) < c$.
structure | concat | approx split | sloppy split
==================================================================================
doubly-linked lists | O(1) | O(n) | *
arrays | O(n+m) | O(1) | *
weight balanced trees | O(lg (n+m)) | O(1) | *
treaps and skip lists | O(lg min(n,m)) ** | O(1) ** | *,**
functional finger trees | O(min(lg lg m/n, lg lg n) | O(lg n) | O(1)
*: same as approx split
**: expected
By functional finger trees I mean "[Purely functional representations of catenable sorted lists](http://www.math.tau.ac.il/~haimk/papers/loglog23.ps)" by Kaplan and Tarjan.
| https://cstheory.stackexchange.com/questions/5964/sequences-with-sublogarithmic-concat-and-approximate-split | [
"ds.data-structures"
] | 17 | 2011-04-08T22:01:48 | [] | 0 | Science | 0 |
24 | cstheory | Linear-time algorithm to test if clique number equals degeneracy bound? | Given a connected simple graph $G=(V,E)$, let $d$ denote its [degeneracy](https://en.wikipedia.org/wiki/Degeneracy_\(graph_theory\)) and let $\omega$ denote the size of a [maximum clique](https://en.wikipedia.org/wiki/Clique_problem).
A well-known bound on the clique number is $\omega\le d+1$, which is helpful when solving the maximum clique problem or when [enumerating all maximal cliques](https://dl.acm.org/doi/abs/10.1145/2543629?casa_token=fxzBWpiA1ZsAAAAA:CrMNWZ-0GU9DU3L0-UE43AniI2H-2gvXWFzx341kOOA8MvUU2QnCVAQUxcpoQQGtfriWot8YZGi9TA).
How fast can one test whether this bound is tight, $\omega=d+1$? An algorithm that runs in time $O(dm)=O(m^{1.5})$ is [known](https://pubsonline.informs.org/doi/10.1287/opre.2019.1970).
Is there a faster algorithm, say, running in linear time?
Or, is $O(m^{1.5})$ best possible under something like [SETH](https://en.wikipedia.org/wiki/Exponential_time_hypothesis)?
| https://cstheory.stackexchange.com/questions/47947/linear-time-algorithm-to-test-if-clique-number-equals-degeneracy-bound | [
"graph-algorithms",
"parameterized-complexity",
"clique",
"fixed-parameter-tractable"
] | 17 | 2020-11-30T14:49:56 | [
"Maybe you are aware, but this problem recently appeared in codeforces codeforces.com/contest/1439/problem/B. It seems nobody in the discussion mentions anything better than $O(m^{1.5})$.",
"Um...it looks like their article was just published and they list your question as Open Problem 1 (minus the possible conne... | 5 | Science | 0 |
25 | cstheory | Is it possible to boost the error probability of a Consensus protocol over dynamic network? | Consider the binary consensus problem in a synchronous setting over dynamic network (thus, there are $n$ nodes, and some of them are connected by edges that may change round to round). Given a randomized $\delta$-error Monte Carlo protocol for Consensus in this setting, is it possible to use this protocol as a black box to obtain another Consensus protocol with smaller error?
Note that for problems with deterministic outcome (such as Sum), this can be done by repeatedly executing the protocol and taking the majority as the answer. However I am unable to figure out how for Consensus, so I wonder whether this is actually impossible to do.
\--- Details ---
**Dynamic Network.** There are $n$ nodes on a dynamic network over synchronous setting (thus the protocol will proceed in rounds). Each node has a unique ID of size $\log(n)$. There are edges connecting pairs of nodes, and this set of edges may change from round to round (but remains fixed throughout any particular round). There may be additional restrictions to the dynamic network that allows the problem to be solved by some protocol (such as having a fixed diameter), but it is irrelevant to our discussion.
The only information known to each node is their unique ID. They do not know $n$ or the topology of the network. In particular, they do not know who their neighbors are in the current round (neighbors are nodes that are adjacent to them with an edge). The only way to figure out the neighbors is after receiving messages from them.
**How nodes communicate.** The only method of communication between nodes is by broadcasting messages under the $CONGEST$ model (D. Peleg, "Distributed Computing, A Locality-Sensitive Approach"). More specifically, in each round, every node may broadcast an $O(\log n)$ sized message. Then, every node will receive ALL messages that were broadcasted by all of its neighbors in that particular round in an arbitrary order. In particular, nodes may attach their IDs onto their messages, since their IDs are of size $O(\log n)$.
**Binary Consensus.** The binary consensus problem over this network is as follows. Each node has a binary input ($0$ or $1$), and each must decide either $0$ or $1$ under the the validity, termination, and agreement for consensus with $\delta$ error:
* validity means that for $z \in \\{0, 1\\}$, if all nodes have $z$ as its initial input, then the only possible decision is $z$.
* termination means that each node must eventually decide.
* agreement means that all nodes must have the same decision.
* The error rate means that under worst case input and worst case network, the probability that the protocol does not satisfy any of the three conditions is at most $\delta$ over average coin flips of the protocol's coin.
**Question.** For a particular set of dynamic networks, suppose we have a lower bound on the average number of rounds incurred by any randomized $\delta$-error protocol $P$ for solving binary consensus under this settings such that $\delta < \frac{1}{3}$, under worst case dynamic network from this set, worst case nodes' inputs, and on average coin flips of the protocol. Does this bound automatically hold for any $\delta'$-error protocol for $\delta < \delta' < \frac{1}{3}$?.
| https://cstheory.stackexchange.com/questions/32442/is-it-possible-to-boost-the-error-probability-of-a-consensus-protocol-over-dynam | [
"ds.algorithms",
"reductions",
"dc.distributed-comp"
] | 16 | 2015-08-23T23:24:43 | [
"@Peter Also this question is more of a whether it is possible to obtain a better protocol using a black box protocol. For example, in Sum, each node can run the protocol multiple times (either in parallel or in a row) and takes a majority. Using Chernoff's bound, for any $0 < \\delta' < \\delta < \\frac{1}{3}$, on... | 7 | Science | 0 |
26 | cstheory | complexity of checking if a subspace is a Euclidean section of L1 | If $X$ is a linear subspace of ${\mathbb R}^n$, $X$ is high-dimensional, and for every $x\in X$ we have
$(1-\epsilon) \sqrt n ||x||_2 \leq ||x||_1 \leq \sqrt n ||x||_2$
for some small $\epsilon >0$, then we say that $X$ is an almost-Euclidean section of $\ell_1^n$, and (the matrix whose image is) X is useful in compressed sensing. A random subspace works excellently, and there is a huge research program devoted to the **explicit construction** of such spaces.
Is it known what is the complexity of approximating the "Euclidan-sectionness" of $X$? That is, given a subspace $X$, say presented via a basis, consider the problem of finding the unit (in $\ell_2$ norm) vector in $X$ of smallest $\ell_1$ norm.
> What is the complexity of this problem? Are hardness of approximation results known?
Apart from specific applications, these seem to be interesting problems. Is it known what is the complexity of finding the vector of **maximum** $\ell_1$ norm among the unit vectors of a given subspace?
| https://cstheory.stackexchange.com/questions/4992/complexity-of-checking-if-a-subspace-is-a-euclidean-section-of-l1 | [
"cc.complexity-theory",
"cg.comp-geom",
"norms",
"compressed-sensing"
] | 17 | 2011-02-17T18:27:44 | [
"This is not an answer to your question. But in the vein of computational problems arising from compressed sensing applications, Koiran and Zouzias have a recent paper on checking whether a matrix satisfies the restricted isoperimetry property (RIP) and related problems.",
"Thanks Sariel, I had inverted the plac... | 3 | Science | 0 |
27 | cstheory | Looking for an operator on polynomials | I have a small, self-contained, math question, whose motivation is from theoretical computer science (specifically, list decoding of algebraic codes, derivative/multiplicity codes, etc). I wonder whether someone might have an idea.
I'm looking for an operator T that can be applied to m-variate polynomials over a finite field. When applied to a polynomial, the operator should yield a polynomial of not much higher degree. The operator should satisfy the following property: for every m-variate polynomial $F$, and m univariate non-constant polynomials $g_1,...,g_m$, if you know $g_1,...,g_m$ and $F(g_1(t),...,g_m(t))$ for a parameter t, then you can also compute $TF(g_1(t),...,g_m(t))$. The operator should be non-trivial, in the sense that for a fixing $x_1,..,x_m$, the value $F(x_1,...,x_m)$ does not determine the value of $TF(x_1,...,x_m)$.
For $m=1$, the derivative operator gives exactly that: By the chain rule $(F(g(t)))' = F'(g(t))\cdot g'(t)$, which implies that $F'(g(t)) = (F(g(t)))'/g'(t)$. My question is whether there is an operator that works for general m.
| https://cstheory.stackexchange.com/questions/10881/looking-for-an-operator-on-polynomials | [
"it.information-theory",
"coding-theory",
"algebra",
"polynomials"
] | 16 | 2012-03-28T08:17:46 | [
"Honestly, I asked this question so long ago, I only barely remember what I needed it for :-)... I believe that the issue is that multiplying by a fixed polynomial is \"trivial\" in the sense that to compute the multiplication at a point, you only need to know the value of F at the point (and nothing about F's inne... | 8 | Science | 0 |
28 | cstheory | When does adding edges decrease the cover time of a graph? | When first learning about random walks on a graph $G$, one may have an intuitive feeling that adding edges to $G$ will decrease its cover time $C(G)$. However, this is not the case. The [path graph](https://en.wikipedia.org/wiki/Path_graph) $P_n$ has cover time $C(P_n) = \Theta(n^2)$ while the [($\frac{n}{2}$, $\frac{n}{2}$)-lollipop graph](http://mathworld.wolfram.com/LollipopGraph.html) $L_{\frac{n}{2}, \frac{n}{2}}$ has cover time $C(L_{\frac{n}{2}, \frac{n}{2}}) = \Theta(n^3)$. Eventually adding edges does decrease the cover time because the complete graph $K_n$ has cover time $C(K_n) = \Theta(n \log n)$ by analogy to the [coupon collector problem](https://en.wikipedia.org/wiki/Coupon_collector%27s_problem).
Under what assumptions though would the supposed intuition hold true? Specifically, what about [vertex transitivity](https://en.wikipedia.org/wiki/Vertex-transitive_graph)?
> Let $G$ be a vertex-transitive graph with $n$ vertices. Consider any vertex-transitive graph $G^+$ obtained from $G$ by adding edges (i.e. both $G$ and $G^+$ are vertex transitive and $G$ is a spanning subgraph of $G^+$).
>
> Is it the case that $C(G^+) \le C(G)$?
| https://cstheory.stackexchange.com/questions/33071/when-does-adding-edges-decrease-the-cover-time-of-a-graph | [
"graph-theory",
"pr.probability"
] | 16 | 2015-11-12T18:35:47 | [
"My bad, I confused spanning with induced.",
"@chazisop Those subgraphs are not spanning.",
"Wouldn't the complete graph $K_{n}$ and any of its subgraphs $K_{k}$, $k < n$ be a counterexample for general vertex transitivity?"
] | 3 | Science | 0 |
29 | cstheory | Complexity of approximating the range of a matrix | Given an $m$ by $n$ matrix $M$ with $m \leq n$ and elements from $\\{-1,1\\}$, let us define:
$$S_M = |\\{Mx : x \in \\{-1,1\\}^n\\}|.$$
I believe that it is NP-hard to compute $S_M$ exactly, by applying the reductions from [Decide whether a matrix's kernel contains any non-zero vector all of whose entries are -1, 0, or 1](https://cstheory.stackexchange.com/questions/20277/decide-whether-a-matrixs-kernel-contains-any-non-zero-vector-all-of-whose-entri) to the following decision problem: does $S_M = 2^n$?
> Is it possible to approximate $S_M$ to within a constant factor in polynomial time? If not, what is the best one can do in polynomial time?
[Cross-posted to <https://mathoverflow.net/questions/229852/complexity-of-approximating-the-size-of-the-range-of-a-matrix> ]
| https://cstheory.stackexchange.com/questions/33676/complexity-of-approximating-the-range-of-a-matrix | [
"cc.complexity-theory",
"approximation-algorithms",
"linear-algebra"
] | 15 | 2016-01-28T02:00:10 | [
"(1) If there is a poly-time $2^{n^{1-\\epsilon}}$-approximation for any $\\epsilon>0$, then there is a poly-time $(1+1/q(n))$-approximation for any polynomial $q$. To see this, given $M$, consider the block-diagonal matrix $M(k)$ formed by $k$ independent copies of $M$ along the diagonal, so $S_M=(S_{M(k)})^{1/k}... | 1 | Science | 0 |
30 | cstheory | Intersecting Complexity Classes with Advice | In [on hiding information from an oracle](http://linkinghub.elsevier.com/retrieve/pii/0022000089900184), the authors (Abadi, Feigenbaum, and Kilian) wrote:
> $(\mathsf{NP/poly} \cap \mathsf{co\text-NP}{/poly})$ ... is **not known** to be equal to $(\mathsf{NP} ∩ \mathsf{co\text-NP}){/poly}$.
They highlighted that in the conference paper, they mistook the two classes. Apparently, the latter is a subset of the former, but we don't know if the containment is strict.
Assuming $X$ and $Y$ are complexity classes, and $F$ is a set of functions specifying the length of the advice strings, are there recent results comparing $(X_{/F} \cap Y_{/F})$ and $(X \cap Y)_{/F}$, resolving issues like the one pointed above?
| https://cstheory.stackexchange.com/questions/5839/intersecting-complexity-classes-with-advice | [
"cc.complexity-theory",
"reference-request",
"complexity-classes",
"advice-and-nonuniformity"
] | 15 | 2011-04-03T03:17:43 | [
"@HenryYuen: I think that a Language $L$ is in $(NP \\cap coNP)/poly$ iff there is a language $K$ in $NP \\cap coNP$ and $a_i$ advice of polynomial length s.t. $x \\in L$ iff $(x, a_{|x|}) \\in K$.",
"@HenryYuen: Oh, I got your point. Strangely, the definition of (NP ∩ coNP)/poly in Complexity Zoo (qwiki.stanford... | 7 | Science | 0 |
31 | cstheory | Intermediate problems between PSPACE and EXPTIME | Intermediate problems between P and NP are quite famous, and are sometimes considered as complexity classes by themselves.
Do you know of any problem that is known to be PSPACE-hard and in EXPTIME, and resisting all efforts to be proved complete for one of these classes ?
Lifted (succinct) versions of problems between LOGSPACE and PTIME are accepted, but even more interesting would be problems that are not of this form.
Here are some I found, this area of complexity theory seems to be the realm of games:
* Phutball: <https://arxiv.org/abs/0804.1777>
* Go (with Chinese rules): D. Lichtenstein and M. Sipser, Go is polynomial-space hard
| https://cstheory.stackexchange.com/questions/38393/intermediate-problems-between-pspace-and-exptime | [
"cc.complexity-theory",
"complexity-classes",
"exp-time-algorithms",
"pspace"
] | 17 | 2017-06-09T06:50:47 | [
"I think the more correct way of stating the question is asking for candidate problems in ExpTime - PSpace - ExpTime-hard. Note that e.g. in the case of P vs. NP we don't know if Factoring or GI is P-hard."
] | 1 | Science | 0 |
32 | cstheory | Mutual information vs. Product sets | Suppose we have two _dependent_ random variables $X$ and $Y$, each of which is uniform over $\\{0,1\\}^n$, such that their mutual information $I(X;Y)$ is small, say, at most $\sqrt{n}$. Does this imply that there exist large sets $\mathcal{X}, \mathcal{Y} \subset \\{0,1\\}^n$ such that the product set $\mathcal{X} \times \mathcal{Y}$ is contained in the support of the distribution $(X,Y)$?
A variant of this question can be phrased in graph-theoretic terms: Suppose we have a sufficiently dense bipartite graph G. Does this imply that $G$ contains a large complete sub-graph $G'$? (i.e., $G'$ is required to be a complete bipartite graph with bi-partition $(A,B)$ such that $A$ and $B$ are relatively dense in the corresponding parts of $G$)
| https://cstheory.stackexchange.com/questions/10225/mutual-information-vs-product-sets | [
"co.combinatorics",
"it.information-theory"
] | 15 | 2012-02-15T17:49:06 | [
"If $(X,Y)$ is a product of iid trials $(X_i,Y_i)$ then I am inclined to say yes.",
"If my calculation is correct, a standard use of the probabilistic method shows that for sufficiently large N, there exists a bipartite graph on N+N vertices with at least (1/2)N^2 edges that does not contain K_{k,k} as a subgraph... | 3 | Science | 0 |
33 | cstheory | Computability of a "weird" set | The starting point of this question is the observation that the smallest positive integers $a,b,c$ satisfying
$$\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = 4$$
are [absurdly high](https://plus.google.com/+johncbaez999/posts/Pr8LgYYxvbM). This leads to the following general question: Is the set $C\subseteq {\mathbb N}$ defined by $$ C = \left\\{n\in\mathbb{N}\setminus\\{0\\}: \big(\exists a,b,c \in\mathbb{N}\setminus\\{0\\}\big):\frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} = n\right\\}$$ computable?
| https://cstheory.stackexchange.com/questions/39383/computability-of-a-weird-set | [
"computability"
] | 14 | 2017-10-26T07:49:40 | [
"Cross-posted from mathoverflow.net/questions/278747/is-this-set-computable",
"I'm not an expert, but if you simplify the equation you get a cubic diophantine equation in three variables, and I think it's an open problem if universality can be achieved in such a setting. In every case, quadratic diophantine equat... | 2 | Science | 0 |
34 | cstheory | Can a Penrose tile cellular automaton be Turing-complete? | This question was based on an incorrect premise ... see Colin's comment below. Forget it.
This was inspired by the discussion on [this Math Overflow question](https://mathoverflow.net/questions/45378/undecidability-in-conways-game-of-life/45382). First, I need to define our terms.
In a Penrose tiling, there are generally several different legal patterns of tiles which will cover a given region. A Penrose tile cellular automaton should be some rule that deterministically takes a Penrose tiling $T$ to a Penrose tiling $T'$, where each tile in $T'$ is deterministically given by looking at the tiles in a radius $r$ around the corresponding spot in $T$. You should be able to implement these by having a set of rules of the form: when a region of shape $S$ is covered by a pattern $X$ of tiles, replace it with a pattern $Y$ of tiles. In order to do this deterministically, you either have to make sure no two of your replaceable regions can overlap or have some priority ranking on these replacement rules. But certainly you can ensure this, and define lots of different Penrose tiling cellular automata.
What does it mean for a Penrose tiling celluar automaton to be Turing-complete. Here's what I think it should mean (although I won't insist on it in the answers): the halting problem should be reducible to the behavior of such an automaton. Let's take some canonical Penrose tiling $T_0$ that provides constraints on our starting position. We start with an input $G$ to a universal Turing machine, and we'd like to decide whether the machine halts on input $G$. I would like some computable map from $G$ to a starting Penrose tiling $S$ such that $S$ differs from $T_0$ on only a finite set of tiles. We then run the Penrose tiling cellular automaton, starting with $S$, and wait for some finite canonical configuration $C$ at position $0$ in the Penrose tiling. I want configuration $C$ to appear at position $0$ if and only if the universal Turing machine halts on input $G$.
The strange thing about Penrose tiling cellular automata is that every finite legal configuration of tiles appears infinitely often in every Penrose tiling of the plane. Thus, our universal Penrose tiling is not only computing the behavior of our universal Turing machine on input $G$, it's also simultaneously carrying out the first steps of every other possible computation. This makes me doubt whether Penrose tiling cellular automata could be Turing complete. On the other hand, if they're not Turing complete, what is the class of computations that Penrose tiling cellular automata can perform?
| https://cstheory.stackexchange.com/questions/2883/can-a-penrose-tile-cellular-automaton-be-turing-complete | [
"computability",
"machine-models",
"cellular-automata"
] | 15 | 2010-11-11T06:14:56 | [
"Yes, that is true for tilings formed via the deflation process. But an arbitrary planar tiling might have a \"rift\" along a line in the plane, along which the higher level tiles do not match up (like a tiling of the plane by squares in which the upper half plane is shifted right by a fraction of a unit). This is ... | 14 | Science | 0 |
35 | cstheory | NP-Hardness of 4-cycle packing problem in complete bipartite digraph? | A directed complete bipartite graph is a bipartite graph where there is exactly one directed edge between any two vertices from its two different parts. In other words, it's an orientation of a complete bipartite graph.
Given a directed complete bipartite graph, we are asked to find the largest set of edge-disjoint 4-cycles. Note that these 4-cycles can share node(s).
Is this problem NP-Hard? I tried to reduce from 2P2N-3SAT to this problem, but failed. On the other hand, I found no polynomial algorithm could solve it optimally.
| https://cstheory.stackexchange.com/questions/45998/np-hardness-of-4-cycle-packing-problem-in-complete-bipartite-digraph | [
"graph-theory",
"np-hardness"
] | 14 | 2019-12-10T00:40:50 | [] | 0 | Science | 0 |
36 | cstheory | Is there a P-complete language X such that succinct-X is in P? | I came across a paper called "A Note on Succinct Representation of Graphs". It seems that in the discussion section they claim that for any problem $X$ that is $\mathrm{P}$-hard under projections, $\mbox{succinct-X}$ is $\mathrm{EXP}$-hard.
This made me wonder if the theorem fails for more complicated kinds of reductions. This led me to the following.
> **My Question**
>
> Are there any problems $X$ such that $X$ is $\mathrm{P}$-complete under logspace reductions and $\mbox{succinct-X}$ is in $\mathrm{P}$?
**Additional Questions:**
(1) If this is an open problem, then what are the implications if such an $X$ existed?
(2) If this is an open problem, is it still open is we weaken the requirement to $\mbox{succinct-X}$ in $PH$?
(3) It seems that such an $X$ would also satisfy that $X$ is $\mathrm{P}$-hard under logspace reductions, but not $\mathrm{P}$-hard under projections. Are such problems known to exist?
(4) I might be mistaken, but I think that I have a construction for such an $X$ assuming that $NP = L$. Would this be interesting?
**Note:** I'm still learning about projections so please let me know if I made any mistakes. Thank you!
| https://cstheory.stackexchange.com/questions/42185/is-there-a-p-complete-language-x-such-that-succinct-x-is-in-p | [
"cc.complexity-theory",
"complexity-classes",
"reductions",
"succinct"
] | 14 | 2019-01-09T22:34:19 | [
"Werent p-projections used in Valiants' original completeness and algebra paper?",
"@EmilJeřábek Yes, I noticed this too. :)",
"Seeing that projections are a restricted class of polylogtime reductions, let me add that what I wrote above about logtime reductions also applies to polylogtime reductions.",
"@Emi... | 13 | Science | 0 |
37 | cstheory | Fourier spectrum of the parity of two monotone Boolean functions | This is a question that I've been pondering, on and off, for a while, and unsuccessfully. I'd be very interested in any insight regarding this conjecture. (Or rather, these conjectures.)
Recall that, given a Boolean function $f\colon \\{-1,1\\}^n \to \\{-1,1\\}$, the Kahn—Kalai—Linial theorem states that $$\max_{i\in[n]} \operatorname{Inf}_i f \geq \operatorname{Var}[f]\cdot \Omega\\!\left(\frac{\log n}{n}\right) \tag{1} $$ from which we get that, for any _monotone_ Boolean function $f\colon \\{-1,1\\}^n \to \\{-1,1\\}$, $$\max_{i\in[n]} \widehat{f}(i) \geq \operatorname{Var}[f]\cdot \Omega\\!\left(\frac{\log n}{n}\right) = (1-\widehat{f}(0)^2)\cdot \Omega\\!\left(\frac{\log n}{n}\right) \tag{2} $$ (writing $\widehat{f}(i)$ for $\widehat{f}(\\{i\\})$, $i\in\\{0,\dots,n\\}$). Moreover, (2) is tight, as shown by considering the $\textsf{Tribes}_n$ function. In particular, this implies that $$ W^{(0)}[f]+W^{(1)}[f] = \sum_{i=0}^n \widehat{f}(i)^2 = \Omega\\!\left(\frac{\log^2 n}{n^2}\right) \tag{3} $$ for any monotone Boolean function $f\colon \\{-1,1\\}^n \to \\{-1,1\\}$, where $W^{(k)}[f] \stackrel{\rm def}{=} \sum_{S: \lvert S\rvert =k} \widehat{f}(S)^2$.
Now, consider _two_ monotone Boolean functions $f,g\colon \\{-1,1\\}^n \to \\{-1,1\\}$, and let $h\stackrel{\rm def}{=}fg$ be their parity. It is easy to see that we could have $W^{(0)}[h]+W^{(1)}[h]=0$, e.g. by considering $f,g$ to be two different dictator functions (but in that very specific case, $W^{(2)}[h]=1$). _But must there be some non-negligible Fourier mass on the first 3 Fourier levels, then?_
> What can we say about $W^{(0)}[h]+W^{(1)}[h]+W^{(2)}[h] = \sum_{S: \lvert S\rvert \leq 2} \widehat{h}(S)^2$?
**Conjecture 1.** For any two monotone Boolean functions $f,g\colon \\{-1,1\\}^n \to \\{-1,1\\}$, one must have $W^{(0)}[fg]+W^{(1)}[fg]+W^{(2)}[fg] > 0$.
Actually, I'd be actually inclined to believe the following stronger statement:
**Conjecture 2.** For any two monotone Boolean functions $f,g\colon \\{-1,1\\}^n \to \\{-1,1\\}$, one must have $W^{(0)}[fg]+W^{(1)}[fg]+W^{(2)}[fg] \geq \frac{1}{\operatorname{poly}(n)}$.
_Side note:_ this would have implications about _weak learning_ of the class of "2-monotone" functions, which are exactly those functions obtained as parity or anti-parity of $2$ monotone functions. (See e.g. [BCOST15].) But more importantly, this is a very simple-looking question, that has been nagging at me for way too long.
* * *
[BCOST15] Eric Blais, Clément L. Canonne, Igor Carboni Oliveira, Rocco A. Servedio, Li-Yang Tan. _Learning Circuits with few Negations._ APPROX-RANDOM 2015: 512-527
| https://cstheory.stackexchange.com/questions/37722/fourier-spectrum-of-the-parity-of-two-monotone-boolean-functions | [
"boolean-functions",
"fourier-analysis",
"monotone"
] | 14 | 2017-03-07T06:55:43 | [
"@daniello As far as I can tell, not much, at least in terms of blackbox interpretation of the results. The question above would have applications to weak learning (learning to advantage $1/2+1/\\mathrm{poly}(n)$), while the paper you link shows lower bound on testing (and a relaxation, parameterized testing). As f... | 2 | Science | 0 |
38 | cstheory | Is it possible to find the median with a linear size sorting network? | Is there a sorting network that makes only $O(n)$ comparisons and finds the median?
The AKS sorting network sorts with $O(\log n)$ parallel steps, but here I am only interested in the number of comparisons. The median of medians algorithm finds the median with $O(n)$ comparisons, but it cannot be implemented as a sorting network.
Remark. In fact, in a recent work, we needed a version that is less powerful than pairwise comparison and resembles sorting networks. In our model one could input two elements, $a$ and $b$, and the output was either "a" or "b", such that the output is at least as big as the other number. (In case of equality, either one of them, and we are interested in worst case complexity.) In this variant we could prove that there is a solution with $O(n)$ comparisons. Of course I am also interested if anyone knows anything about this model ever being studied.
| https://cstheory.stackexchange.com/questions/27765/is-it-possible-to-find-the-median-with-a-linear-size-sorting-network | [
"ds.algorithms",
"sorting",
"sorting-network",
"selection"
] | 14 | 2014-12-09T06:40:04 | [
"Actually the comparator circuit model is a bit different, since you are allowed to repeat inputs and their negations. What you are looking for is a comparator network for majority.",
"@Yuval Your result seems to be the only hit on google for \"comparator circuit complexity\" and even there I cannot see the defin... | 3 | Science | 0 |
39 | cstheory | Exponential-time factorization of polynomials | Let an _explicit_ field be a field for which equality is decidable (in some standard model of computation). I am interested in the factorization of univariate polynomials over an explicit field.
It is known that over some explicit fields, testing irreducibility of polynomials is undecidable [1]. Though, over many fields, the irreducible factorization of polynomials is known to be computable in polynomial time, either deterministically (over $\mathbb Q$ [2] or any number field) or probabilistically (over finite fields [3]). Here the size of the input is the size of the _dense representation_ of the polynomial, that is the list of all its coefficients.
> Is there an explicit field for which
>
> * either the best known algorithm has an exponential running time,
> * or (better) there exists an exponential lower bound for irreducibility testing?
>
**Quick clarification:** There may be very different _explicit fields_ , depending on the complexity (for instance) on the equality test or of the standard operations (addition, multiplication). Actually, I am not aware of any result about polynomial factorization which is not a polynomial time upper bound, or undecidability. So I am interested in any kind of different result (exponential time is just an example). Conditional lower bounds are of interest too.
**References.**
[1] A. Fröhlich et J. C. Shepherdson (1955): [On the factorisation of polynomials in a finite number of steps](http://dx.doi.org/10.1007/BF01180640).
[2] A. K. Lenstra, H. W. Lenstra Jr., L. Lovász (1982): [Factoring polynomials with rational coefficients](http://dx.doi.org/10.1007/BF01457454).
[3] E.R. Berlekamp (1967): [Factoring Polynomials Over Finite Fields](http://www3.alcatel-lucent.com/bstj/vol46-1967/articles/bstj46-8-1853.pdf).
| https://cstheory.stackexchange.com/questions/18502/exponential-time-factorization-of-polynomials | [
"algebraic-complexity",
"polynomials",
"exp-time-algorithms"
] | 14 | 2013-07-30T01:12:45 | [
"Thanks @JoshuaGrochow! I found the paper incredibly easy to read, even though my German lessons are quite old now...",
"Depending on your definition of \"explicit field\", the result you cite as [1] was also proven by van der Waerden much earlier (1930). Amazingly, he proved this before the notion of computabili... | 4 | Science | 0 |
40 | cstheory | Pseudorandom functions in ACC^0? | In the lower bound result by Ryan Williams (Non-uniform $\mathsf{ACC}$ circuit lower bounds), there is a mention of "little evidence that Pseudorandom function generators exist in $\mathsf{ACC}^0$. Is there any development in this regard that might be of interest? (Even old results implying something along these lines will be great.)
| https://cstheory.stackexchange.com/questions/10220/pseudorandom-functions-in-acc0 | [
"cc.complexity-theory",
"reference-request",
"cr.crypto-security",
"circuit-complexity",
"natural-proofs"
] | 14 | 2012-02-15T01:26:37 | [
"The following paper doesn't answer your question but does show somewhat simpler candidate PRFs ccs.neu.edu/home/viola/papers/spn.pdf"
] | 1 | Science | 0 |
41 | cstheory | DPLL and Lovász Local Lemma | Let $\varphi$ be a CNF formula. Suppose that each of $\varphi$'s clauses consist of exactly $t$ literals (and, moreover, all literals within one particular clause correspond to different variables). It is well known that if every clause has less than $2^t / e$ clauses that share variables with it, then $\varphi$ is satisfiable (let us call such formulae _easy_). Satisfiability can be proved easily using [Lovász local lemma](http://en.wikipedia.org/wiki/Lov%C3%A1sz_local_lemma). Moreover, using [a recent result](http://arxiv.org/abs/0903.0544) by Moser and Tardos one can show that one of the satisfying assignments can be found in polynomial expected time using the following very simple procedure:
* Pick a random assignment
* While there exists an unsatisfied clause, resample all its variables.
On the other hand, most of modern SAT solvers are [DPLL](http://en.wikipedia.org/wiki/DPLL_algorithm)-based. It means that they try to find a satisfying assignment using brute force with two simple prunings:
* If a formula contains clause with one literal, then we can fix it.
* If one variable occurs in a formula only with (or without) negation, then we can fix it.
**The question: is it true that a version of DPLL that splits on random variables finds a satisfying assignment of any easy formula in polynomial expected time?**
| https://cstheory.stackexchange.com/questions/7720/dpll-and-lov%c3%a1sz-local-lemma | [
"cc.complexity-theory",
"ds.algorithms",
"sat"
] | 14 | 2011-08-13T02:32:29 | [
"@ilyaraz Still hoping you'll post an official answer to this question.",
"@Stasys Actually, Dmitriy Itsykson (who you probably know) convinced me that the answer is negative. I'll elaborate on it soon. Argument is more or less the following: we pick a hard unsatisfiable formula and pad it in order to make it eas... | 3 | Science | 0 |
42 | cstheory | Which monotone DNFs are evasive? | A Boolean function $\phi$ on variables $X$ is _[evasive](https://en.wikipedia.org/wiki/Evasive_Boolean_function)_ if every decision tree for $\phi$ has height $|X|$. In other words, for any strategy that picks variables of $X$ and asks for their value, an adversary can answer the queries such that the strategy needs to ask about the value of all variables of $X$ to know the value to which $\phi$ evaluates.
A _monotone DNF_ is a disjunction of conjunction of variables, representing a monotone Boolean function on the set $X$ of variables that it uses.
**Which monotone DNFs are evasive?** Is there a characterization of them? What is the complexity, given a monotone DNF, of determining whether it is evasive?
Examples of evasive monotone DNFs include single-clause DNFs, e.g., $x \land y \land z$, where we must query the value of every variable (and the adversary can answer $1$) until the last one which determines the value of the formula. For another example, take $(x \land y) \lor (y \land z)$: if I ask about $x$ or $z$ then the opponent answers $0$ and I'm left with evaluating the other clause, if I ask about $y$ the opponent answers $1$ and I must evaluate $x \lor z$ which requires me to query both variables.
However, not all monotone DNFs are evasive. For instance, for $(s \land t) \lor (t \land u) \lor (u \land v)$, I can query variable $t$. If it is false, then I know that clauses $s\land t$ and $t \land u$ are false, so I can evaluate the function just knowing $u$ and $v$ (I don't need to query $s$). If it is true, then I have to evaluate $s \lor u \lor (u \land v)$, and as the second clause subsumes the third, I just need to know $s$ and $u$ (I don't need to query $v$).
Playing a bit with the problem in the case of graph DNFs (with two variables per clause), it looks like for a path $(x_0 \land x_1) \lor \cdots \lor (x_{n-1} \land x_{n})$ with $n>0$ the function is not evasive iff $n$ is a multiple of 3, and this seems to generalize to trees if there is some kind of tree structure on nodes that are 3 steps apart. For general graphs, I have no idea. I have looked for related work about evasiveness and decision tree complexity for monotone CNFs and DNFs but I didn't find anything relevant.
On graph DNFs, the problem can be seen as a kind of game on the graph of the formula. You pick a vertex $v$ (= variable) on the graph, and the opponent can choose between:
* removing $v$ and all its incident edges (= $v$ evaluates to $0$, all clauses involving it are removed), or
* removing the open ball of radius 2 centered on $v$, i.e., $v$, its adjacent edges, the neighbors of $v$, and their adjacent edges (= $v$ evaluates to $1$, all clauses involving $v$ become singleton clauses involving $v$'s neighbors, you will need to ask about each of these neighbors because each has a singleton clause involving it, but these singleton clauses subsume all other clauses involving the neighbors so you get to remove them).
The question is about characterizing and recognizing the graphs $G$ for which you have a strategy that guarantees that you win, where "winning" means "reaching a graph with an isolated vertex" (= a variable that you do not need to ask about).
| https://cstheory.stackexchange.com/questions/44278/which-monotone-dnfs-are-evasive | [
"np-hardness",
"boolean-functions"
] | 13 | 2019-07-16T13:28:04 | [
"Since the Alexander dual of the corresponding simplicial complex consists of all the independent sets of $G$, a necessary condition for evasiveness is that the independence polynomial of $G$ vanishes at $-1$.",
"Possibly relevant references for the lazy: A survey of evasiveness: Lower bounds on the decision-tree... | 2 | Science | 0 |
43 | cstheory | Question on Products of Graphs | Let $S_{n}(G)$, $C_{n}(G)$, $T_{n}(G)$ be the $n$-fold Strong Product, Cartesian Product and Tensor Product of a graph $G$ on $|V|$ vertices.
Let the chromatic number ($\chi(G)$) and the independence number ($\alpha(G)$) of $G$ be known through a (possibly exponential time) algorithm. Is it known that the calculation of $\chi(S_{n}(G))$, $\chi(C_{n}(G))$, $\chi(T_{n}(G))$,$\alpha(S_{n}(G))$, $\alpha(C_{n}(G))$ and/or $\alpha(T_{n}(G))$ require exponential in $n$ complexity for any $n$ say even when restricted to some special graphs?
| https://cstheory.stackexchange.com/questions/7522/question-on-products-of-graphs | [
"cc.complexity-theory",
"graph-theory",
"graph-colouring"
] | 14 | 2011-07-26T18:45:56 | [
"@Andrew: The first one",
"By $n$-fold strong product do you mean $G\\boxtimes G\\boxtimes \\ldots \\boxtimes G$, or $G\\boxtimes K_n$?",
"Here is Wikipedia's article on Hedetniemi's conjecture.",
"Hi Hsien: Actually you answered part of the question. If you assume the conjecture, then we already have an $o(0... | 6 | Science | 0 |
44 | cstheory | Deterministic context-free languages that can be represented as the word problem of a group | Consider a group $G$. We call $G$ virtually free is it contains a free subgroup of finite index.
If $G$ is finitely generated by some set $X \subseteq G$ one can consider the _word problem_ $W\\!P(G)$ that is the formal language consisting of all words over the alphabet $X \cup X^{-1}$ that evaluate to the unit in $G$.
It is a famous result by Mueller and Schupp that the $W\\!P(G)$ is context-free if and only if $G$ is virtually free.
It is also known that if $W\\!P(G)$ is context-free, it is deterministic context-free.
My question is: Do we know more about how the class of deterministic context-free languages that can be represented as the word problem of some group, is contained in the class of all (deterministic) context-free languages?
| https://cstheory.stackexchange.com/questions/39954/deterministic-context-free-languages-that-can-be-represented-as-the-word-problem | [
"fl.formal-languages",
"gr.group-theory"
] | 13 | 2018-01-11T04:49:31 | [
"I'm not sure if you're still interested in this, but it is known that such languages are closed under deletion and have universal prefix. This is true for all word problems, not just DCFL. There's a proof here: D. W. Parkes and R. M. Thomas. Groups with context-free reduced word problem. Communications in Algebra,... | 6 | Science | 0 |
45 | cstheory | Complexity to compute the eigenvalue signs of the adjacency matrix | Let $A$ be the $n\times n$ adjacency matrix of a (non-bipartite) graph. Assume that we are given the amplitudes of its eigenvalues, i.e., $|\lambda_1|=a_1,\ldots, |\lambda_n|=a_n$, and we would like to calculate their signs. Is there a faster way of computing the signs of these eigenvalues, other than recomputing the eigenvalues themselves?
| https://cstheory.stackexchange.com/questions/16789/complexity-to-compute-the-eigenvalue-signs-of-the-adjacency-matrix | [
"cc.complexity-theory",
"ds.algorithms",
"linear-algebra",
"spectral-graph-theory"
] | 13 | 2013-03-07T12:38:48 | [
"Thanks, that's exactly the point. I was wondering if the knowledge of the amplitudes made the problem any simpler. I think that this could have interesting implications in problems where eigenvalues are well approximated by the squared degrees of the graph, etc.",
"Nice question! For comparison, the best known r... | 2 | Science | 0 |
46 | cstheory | Name and references for balanced variant of the long code? | The [long code](https://en.wikipedia.org/wiki/Long_code_%28mathematics%29) (arising in PCP theory etc) is an encoding of a set of $k$ values, using binary strings of length $2^k$ (double exponential in the number of bits needed to specify a value), with one coordinate position in the encoding for each of the $2^k$ different binary functions on the input values. Instead, I want to consider a shorter but still doubly exponential code, with one coordinate position for each of the $\tbinom{k}{k/2}/2$ partitions of the $k$ values into two equal-sized subsets (assuming $k$ is even). Each partition can be thought of as corresponding to two different binary functions (the functions whose inverse images form the given partition) and we choose arbitrarily which one of these two functions to use.
The minimum distance of the code (and in fact the distance between all pairs of codewords) is $\tbinom{k-2}{(k-2)/2}$, smaller than the length of the code by only a factor of $\tfrac{k}{2k-1}>\tfrac{1}{2}$. For instance for $k=8$ we get a code of length 35 with minimum distance 20, which <http://www.codetables.de/> tells me matches the [Griesmer bound](https://en.wikipedia.org/wiki/Griesmer_bound) for binary linear codes of the same size (rank 3 or size 8).
I am interested in this for reasons only vaguely related to coding theory, but it seems likely to me that the coding theorists have already studied this code. Unfortunately, I don't know of a good way of looking up coding schemas by the formulas for their parameters. If it has been previously studied, what is it called? And what are some references to it?
(BTW, the property I actually care about, which is easy enough to prove using Sperner's theorem, is that this is the longest possible binary code in which each two positions are independent of each other, in the sense that all four possible combinations of symbols in the two positions are realized by codewords. There appears to be something in the literature called a Sperner code but it's transposed from this — each codeword has equally many zeros and ones, rather than the property of the code I'm looking at, where each bit position has equally many codewords for which that position is a 0 or 1.)
| https://cstheory.stackexchange.com/questions/19602/name-and-references-for-balanced-variant-of-the-long-code | [
"reference-request",
"coding-theory"
] | 13 | 2013-10-31T16:11:44 | [
"Ok, if it wasn't in the literature before, it is now. See section 6 of arxiv.org/abs/1303.1136 — this code gives the smallest set of vertices to delete from a hypercube graph in order to make all remaining hypercube subgraphs have at most 1/8 of the number of vertices of the original graph."
] | 1 | Science | 0 |
47 | cstheory | Generalizing limit-colimit coincidence to Scott-continuous adjunctions: any uses? | In Abramsky and Jung's 1994 handbook chapter on denotational semantics, after proving that the limit and colimit of expanding sequences exist and coincide, they have the following to say about generalizing this theorem:
> We can generalize Theorem 3.3.7 [the limit-colimit coincidence of expanding sequences in DCPO] [...] we can use general Scott-continuous adjunctions instead of e-p-pairs. [...] By the passage from embeddings to, no longer injective, lower adjoints, we allow domains not only to grow but also to shrink as we move on in the index set. Thus points, which at some stage looked different, may at a later stage be recognized to be the same. [...] For the main text, we must remind ourselves that this generalization has so far not found any application in semantics.
It's been over fifteen years since this chapter was written. Have there been any applications of this generalization in semantics since then?
| https://cstheory.stackexchange.com/questions/2528/generalizing-limit-colimit-coincidence-to-scott-continuous-adjunctions-any-uses | [
"pl.programming-languages",
"semantics",
"denotational-semantics",
"domain-theory"
] | 13 | 2010-10-28T10:07:39 | [] | 0 | Science | 0 |
48 | cstheory | The best known upper bound for two-way probabilistic finite automata with one-counter | It is known that the class of languages recognized by two-way deterministic finite automata with one-counter (2D1CAs) is a proper subset of $ \mathsf{L} $ (deterministic log-space): A 2D1CA can run at most $ O(n^2) $ steps if it halts on a given input, and there is a language in $ \mathsf{L} $, i.e. \begin{equation} \mathtt{DG} = \lbrace w_0 \$ w_1 \$ \cdots \$ w_k \mid k \geq 1, w_{i \in \lbrace 0, \ldots, k \rbrace} \in \lbrace a,b\rbrace^* , \mbox{ and } w_0=w_i \mbox{ for some } 1 \leq i \leq k \rbrace, \end{equation} which cannot be recognized by any 2D1CA [(Duris and Galil, 1982)](http://www.sciencedirect.com/science/article/pii/0304397582900871).
What happens if such an automaton can use a coin?
More precisely, what is the best known upper bound for the class of languages recognized by two-way **probabilistic** finite automata with one-counter (2P1CAs) with bounded error?
| https://cstheory.stackexchange.com/questions/11080/the-best-known-upper-bound-for-two-way-probabilistic-finite-automata-with-one-co | [
"cc.complexity-theory",
"reference-request",
"automata-theory",
"upper-bounds"
] | 14 | 2012-04-13T13:36:31 | [] | 0 | Science | 0 |
49 | history | Are there any references to entombed animals in ancient India? | The 13th century Hindu philosopher Arulnandi Shivacharya wrote a work called the Shiva Jnana Siddhiyar, which among other things contains a refutation of Buddhist philosophy. In [this excerpt](https://i.sstatic.net/KxNh7.jpg), various Buddhist theories of what the ultimate cause of the body is are refuted:
> If you say bodies are formed from the mixture of four elements, then these cannot unite as their natures are opposed to each other. **If you say they are formed by the union of blood and semen, then account for toads being found in the heart of rocks, and worms in the heart of trees.** If you say the real cause is good and bad Karma, then these, being opposed, cannot join and form bodies. If food is the cause, then the food which in youth develops the body is not capable of preventing decay in old age. If intelligence is the cause, then that which is formless Chaitanya cannot assume Achaitanya (non-intelligent) form. If you assert that bodies are formed from nothing, then we could cull flowers from the sky.
My question is about the part in bold, which is arguing that intercourse cannot be the ultimate cause of the body, because there are toads which are found in the center of solid rock, and they couldn't have been produced by intercourse since there's no way that toads could have gone inside a solid rock and reproduced.
This statement struck me as rather odd, but then I found out from the Wikipedia article [Entombed animal](https://en.wikipedia.org/wiki/Entombed_animal) that claims of toads being found in solid rock are surprisingly common across many cultures:
> Entombed animals are animals reportedly found alive after being encased in solid rock, or coal or wood, for an indeterminate amount of time. The accounts usually involve frogs or toads. The reputed phenomenon, sometimes called "toad in the hole", has been dismissed by mainstream science, but has remained a topic of interest to Fortean researchers.... References to entombed animals have appeared in the writings of William of Newburgh, J. G. Wood, Ambroise Paré, Robert Plot, André Marie Constant Duméril, John Wesley, and others. Even Charles Dickens mentioned the phenomenon in his journal All the Year Round. **According to the Fortean Times, about 210 entombed animal cases have been described in Europe, North America, Africa, Australia, and New Zealand since the fifteenth century.**
Conspicuously absent from that list of places is Asia. So my question is, are there any references in ancient India to entombed animals? The work I quoted above was written in the 13th century. But I'm wondering whether there are earlier references to this phenomenon. Arulnandi Shivachrya refers to this phenomenon as if it's common knowledge, so I'm guessing older sources would mention it as well.
| https://history.stackexchange.com/questions/34768/are-there-any-references-to-entombed-animals-in-ancient-india | [
"india",
"science",
"animals",
"ancient-india",
"folklore"
] | 16 | 2017-01-05T23:06:24 | [
"THe proposed edit seems to replace a question with an answer. I'm confused.",
"Unless someone can say what's available in the article I linked to above, there's not much else that can be said—at least the last time I researched this, I spent a fair bit of time and didn't uncover anything useful, but it's so very... | 4 | Culture & Recreation | 0 |
50 | history | What was the first overland road from Sweden to Finland? | The [Swedish post road](https://en.wikipedia.org/wiki/King%27s_Road_\(Finland\)) from Norway, through Sweden, used the Åland archipelago to pass into Sweden, and this is easily found (evidence of) in the south of Finland to the present day. **When (and where) was the first overland route constructed overland from Sweden into (Swedish) Finland?**
The only (poor) evidence I have for roads existing in the north is by the [War of 1808–9](https://en.wikipedia.org/wiki/Finnish_War) where Russian forces were planning to advance overland into Sweden (along with an army group advancing across the Gulf of Bothnia). One of the WP article's references does say "In addition, several new good roads had been built into Finland greatly reducing the earlier dependency on naval support for any large operation in Finland." but it doesn't specify where these roads were.
I looked through all articles on [Swedish](https://en.wikipedia.org/wiki/Category:Roads_in_Sweden) and Finnish road networks on the English Wikipedia, and the most I found was a reference to a ['Finnmark path'](https://en.wikipedia.org/wiki/Arctic_Ocean_Highway) which was meant to have gone from Finnish Lapland to Finnmark in the 16th century. The [Finnish WP article](https://fi.wikipedia.org/wiki/J%C3%A4%C3%A4merentie) for the same page _does not_ mention the Finnmark path at all, and I couldn't find anything else on a road of that name.
I understand—from the comments—that the term "road" can be meaningless without further definition for a period much longer than a few centuries ago. For clarity, I'm defining road as purpose-built (or purpose-developed) and used regionally for that purpose, such as the post road mentioned above. This would mean hunting tracks that slowly developed don't count, while a merchant-led endeavour to expand (and maintain) the tracks between two townships would.
| https://history.stackexchange.com/questions/62286/what-was-the-first-overland-road-from-sweden-to-finland | [
"transportation",
"sweden",
"finland"
] | 6 | 2020-12-21T03:13:26 | [
"Carl Linnaeus's 1732 trip to Lapland was recorded in a fascinating book. Of course he was not interested in the speediest or easiest route, so much as the most interesting, but it does give first-hand detailed account of his travels & travel conditions. Wikipedia summary: \"Linnaeus travelled clockwise around ... | 7 | Culture & Recreation | 0 |
51 | history | Which misdeeds of F. Reineke was J.B.B. de Lesseps "obliged to suppress"? | F. Reineke was commandant of Kamchatka from about 1780 to 1784. According to Витер's _История символов как история административного деления государства_ he was responsible for moving the territorial administration from Bolsheretesk to Nizhnekamchatsk. According to J.B.B. de Lesseps, Reineke resigned in 1784 ["for reasons which I am obliged to suppress"](https://archive.org/details/travelsinkamtsch01less/page/140).
It was common for Siberian administrators to be venal. According to Forsyth's _A history of the peoples of Siberia_ , the very first governor of Siberia, Matvei Gagarin, was hanged for fraud, nepotism, and selling offices; another governor in Irkutsk was relieved from office for gross corruption and replaced with his second, who was "vicious" and "ruthless".
**Was Reineke guilty of the same old corruption and exploitation, or something worse?** I'm skeptical that de Lesseps, a Frenchman, had any reason to suppress a story of Russian corruption.
| https://history.stackexchange.com/questions/55995/which-misdeeds-of-f-reineke-was-j-b-b-de-lesseps-obliged-to-suppress | [
"russia",
"18th-century",
"government",
"crime",
"kamchatka"
] | 6 | 2019-12-31T20:47:10 | [
"One more thing: According to this source, in 1786 F. Reineke moves from Kamchatka to Irkutsk and in 1798 retires from the civil service. All this suggests that his record in Kamchatka was just fine. There was some controversy related to his subordinates, as well as some misdeeds in the Russian postal service in Ka... | 7 | Culture & Recreation | 0 |
52 | history | Who owned the ship Ensayo, and what were they doing near Baja California in 1842? | According to Plummer's _The Shogun's Reluctant Ambassadors_ , in 1842, sea drifters from the _Eijū-maru_ were picked up and by the _Ensayo_ , a "Spanish pirate ship" with a Philippine crew. It was "carrying illegal goods between Spain or Mexico and Manila," and "[b]ecause it was on a 'wanted list', it could not enter any harbors freely." The Japanese men were treated badly and forced ashore on a beach in Baja California.
I found the description confusing. Of Pacific ports, only the Philippines were still Spanish. The Manila galleon system had been over for decades. During Mexico's first twenty years of independence, as far as I know, it didn't trade with East Asia. The Spanish were a minor presence in the Pacific even when they controlled a lot of its coastline.
Who owned _Ensayo_ and what was it up to?
| https://history.stackexchange.com/questions/47476/who-owned-the-ship-ensayo-and-what-were-they-doing-near-baja-california-in-1842 | [
"19th-century",
"spain",
"age-of-sail",
"philippines",
"piracy"
] | 6 | 2018-08-05T23:12:12 | [
"@AlbertoYagos: I concur with T.E.D.'s view.",
"I had \"Vida en México de trece náufragos japoneses\" in my hands, but forgot to look for this point!!",
"@AlbertoYagos - I can't always speak for all our voters, but I for one would consider what you have there acceptable, with perhaps a GoodReads link to the boo... | 6 | Culture & Recreation | 0 |
53 | history | What would a criminal defense attorney charge in late 19th century America? | What would a criminal defense attorney charge on the [American frontier](https://en.wikipedia.org/wiki/American_frontier#The_Postbellum_West) in a murder trial in 1885? I am specifcally interested in what it would have cost three cowpokes to mount a murder defence in Rapid City, Dakota Territory 1885. I know they were defended by Colonel William Parker who was a successful lawyer and politician in the Territory at the time, but so far I have found no clue what his fees were.
| https://history.stackexchange.com/questions/68240/what-would-a-criminal-defense-attorney-charge-in-late-19th-century-america | [
"united-states",
"19th-century",
"law",
"old-west"
] | 5 | 2022-02-04T11:16:22 | [
"Would also significantly help if you documented your preliminary research. I imagine that the answer is highly variable - urban setting vs rural, proximity of a a railroad, the presence of skilled attorney vs having to recruit one from elsewhere, Of course, the cost will range from zero (self representation) t... | 2 | Culture & Recreation | 0 |
54 | linguistics | In which non-Sinitic languages do negative clauses retain older constituent order in SVC-derived complex predicates? | Many complex predicates are historically derived from serial verb constructions. This is not only true of the Sinitic family. For example, in Saramaccan (Byrne 1987, as cited in Givón 2009):
(1) a bi-fefi di-wosu kaba.
he TNS-paint the-house finish
He finished painting the house.
However, the Chinese constructions are more innovative in that the object has been moved to the end of the sentence, for the most part. Yet in most dialects of modern Chinese except Mandarin, complex predicates also sometimes preserve the older V + O + C order similar to the Saramaccan example in (1). This is most common in negative clauses. (The C is what is traditionally termed a _complement_ in Chinese linguistics, but it's more like a _satellite_ in Talmy's sense, and is developed from the 'lighter' verb of an asymmetric serial verb construction). An example from Cantonese is (2):
(2) nei wan ngo m dou
you find me NEG arrive
You cannot find me.
My question: I know negative clauses are more conservative in general, but are there specific cases outside the Sinitic branch where object placement in complex predicates is more conservative in negative clauses? Thanks!
Givón, T. (2009). _The genesis of syntactic complexity: Diachrony, ontogeny, neuro-cognition, evolution_. John Benjamins Publishing.
| https://linguistics.stackexchange.com/questions/21555/in-which-non-sinitic-languages-do-negative-clauses-retain-older-constituent-orde | [
"syntax",
"historical-linguistics",
"word-order",
"negation"
] | 7 | 2017-04-06T01:26:05 | [] | 0 | Science | 0 |
55 | linguistics | Are there any languages where the first person cannot be an object? | In some languages, nouns low on the animacy hierarchy, particularly inanimates cannot surface as A, and if a situation arises where they are underlyingly A, some reparative strategy such as a passive must be used. Some theorise that this was the case in early PIE, and a friend of mine mentioned Algonquian and Japhung Qiangic in the conversation that lead me to consider this:
Are there any languages where a similar rule is found at the other end of the animacy hierarchy, such that the 1st person is defective and cannot be O, and an underlying 1st person O requires a syntactic derivation to be used, moving the first person either to S (a passive) or oblique status (an antipassive)?
[Chukchi](https://en.wikipedia.org/wiki/Chukchi_language) seems to come close, as it has a verbal prefix _(i)ne- ~ (e)na-_ , which Comrie(1979) analyses as a "de-ergativising prefix" which in some cases acts like an antipassive is required with 1st person singular Os when the A is not 3rd person plural (in which case a 1sg.S/O suffix _is_ used), however it's not quite a full antipassive as in these cases, as while there is a change to verbal person agreement (absolutive forms are used to mark underlying A) there is not change to nominal case marking, and furthermore, in the "present II"-tense it's used with a much larger set of A/P combinations. Further examples of borderline cases like Chukchi would be fine, but what I'm really interested in is if there are any more clear-cut examples where the construction involved is clearly a syntactic derivation.
* * *
Comrie, B. 1979: _Degrees of Ergativity: Some Chukchee evidence_ , pp. 219-40 in Plank, F. 1979 (ed.): _[Ergativity - Towards a Theory of Grammatical Relations](https://drive.google.com/file/d/1f3eXlLDI2qC3yeMG1YStsmtH5RSXzHLE/view)_ , Academic Press
| https://linguistics.stackexchange.com/questions/27161/are-there-any-languages-where-the-first-person-cannot-be-an-object | [
"syntax",
"grammar",
"list-of-languages",
"grammatical-object",
"person"
] | 7 | 2018-02-16T13:02:04 | [
"Inanimate objects not being able to be agents makes some sense (if your language doesn't use a lot of metaphor), but why would any language prohibit 1st person objects? I can imagine this as a kind of taboo, but not as an actual ungrammaticality."
] | 1 | Science | 0 |
56 | mathoverflow | 2, 3, and 4 (a possible fixed point result ?) | The question below is related to the classical [Browder-Goehde-Kirk fixed point theorem](https://www.researchgate.net/publication/38323269_An_elementary_proof_of_the_fixed_point_theorem_of_Browder_and_Kirk).
Let $K$ be the closed unit ball of $\ell^{2}$, and let $T:K\rightarrow K$ be a mapping such that $$\Vert Tx-Ty\Vert _{\ell^{4}}\leq\Vert x-y\Vert _{\ell^{3}}$$ for all $x,y\in K$.
Is it true that $T$ has fixed points?
| https://mathoverflow.net/questions/18264/2-3-and-4-a-possible-fixed-point-result | [
"fa.functional-analysis",
"open-problems",
"banach-spaces"
] | 79 | 2010-03-15T06:26:45 | [
"$FP(p,p,\\infty)$ is false for any $p$. We can let $(Tx)_i = x_{i-1}$ for $i<1$ and $(Tx)_0 =(1/2)( 1 - \\sum_i x_i^2)$. Then $||Tx||_{\\ell_2}^2 = (1/4)(1-||x||_{\\ell_2}^2)^2+ x_{\\ell_2}^2 \\leq 1$ since $x_{\\ell_2}\\leq 1$ and, because each coordinate is a $1$-Lipschitz function in the $\\ell_2$ norm, the who... | 13 | Science | 1 |
57 | mathoverflow | Grothendieck's Period Conjecture and the missing p-adic Hodge Theories | Singular cohomology and algebraic de Rham cohomology are both functors from the category of smooth projective algebraic varieties over $\mathbb Q$ to $\mathbb Q$-vectors spaces. They come with the extra structure of Weil cohomology theories - grading, cup product, orientation map, and cycle class map. We can consider the space of all invertible natural transformations from singular cohomology to algebraic de Rham cohomology that respect this structure. For purely formal reasons, this is an affine scheme. The coordinates are the matrix coefficients of the natural transformation for each algebraic variety, and the relations are given by the various commutative diagrams that must hold. Conjecturally, this scheme is a torsor for the motivic Galois group of $\mathbb Q$.
de Rham's theorem, together with GAGA, gives us an isomorphism between these two functors when tensored with $\mathbb C$. In other words, it gives us a $\mathbb C$-point of this affine scheme.
Grothendieck's period conjecture says that this point is a generic point - so the ring of periods in $\mathbb C$ is equal to the whole ring of functions on this scheme.
* * *
On the other hand, consider the $p$-adic analogue. $p$-adic etale cohomology and algebraic De Rham cohomology are both functors from the category of smooth projective algebraic varieties over $\mathbb Q_p$ to $\mathbb Q_p$-vector spaces. They are Weil cohomology theories.
We can consider the space of all isomorphisms between these two functors as an affine scheme over $\mathbb Q_p$.
$p$-adic Hodge theory gives a $B_{dR}$-valued point of this scheme.
However, it is not the generic point. The reason is that the isomorphism described by $p$-adic Hodge theory naturally factors through the category of $\operatorname{Gal}(\overline{\mathbb Q}_p|\mathbb Q_p)$-representations. Hence it lies in a closed subscheme that is a torsor under, not the full motivic Galois group, but the Zariski closure of $\operatorname{Gal}(\overline{\mathbb Q}_p|\mathbb Q_p)$ inside it. This group is much smaller because the Tate conjecture fails over $\mathbb Q_p$, and a map of Tannakian categories that is not a full functor corresponds to a map of Tannakian groups that is not surjective.
* * *
So the analogue of the period conjecture fails badly over $\mathbb Q_p$. This leads me to my question: Grothendieck's period conjecture essentially gives us a description of the scheme of isomorphisms betweeen singular and de Rham cohomology - it's the spectrum of the period ring. However, we do not have a similar picture of the scheme of isomorphisms between $p$-adic etale cohomology and de Rham cohomology. The period ring only gives us one point.
> Can we explicitly describe any other points of this scheme of isomorphisms?
| https://mathoverflow.net/questions/200083/grothendiecks-period-conjecture-and-the-missing-p-adic-hodge-theories | [
"nt.number-theory",
"tannakian-category",
"p-adic-hodge-theory"
] | 61 | 2015-03-15T14:18:19 | [
"I should correct myself: to obtain that \"over a finite field mixed motives are pure\" we must assume that such mixed motives exist (which follows from the Tate conjecture) and that they have a weight filtration respected by Frobenius. Then for each mixed motive $M$ appropriate polynomials in Frobenius are idempot... | 7 | Science | 1 |
58 | mathoverflow | Does every triangle-free graph with maximum degree at most 6 have a 5-colouring? | ## A very specific case of Reed's Conjecture
Reed's $\omega$,$\Delta$, $\chi$ conjecture proposes that every graph has $\chi \leq \lceil \tfrac 12(\Delta+1+\omega)\rceil$. Here $\chi$ is the chromatic number, $\Delta$ is the maximum degree, and $\omega$ is the clique number.
When restricted to triangle-free graphs, the equivalent question is, Does every triangle-free graph have chromatic number $\leq \frac \Delta 2 +2$?
This is known for $\Delta\leq 4$. In general for triangle-free graphs, $\chi \leq O(\Delta/\log \Delta)$, so the conjecture is also true for very large $\Delta$.
How about $\Delta=5$? $\Delta=6$? Because of parity, $\Delta=6$ is the easier of these two cases (and actually easily implies the $\Delta=5$ case. Can anyone prove it?
Kostochka proved that every triangle-free graph has $\chi \leq \frac 2 3 \Delta +2$. He also proved that $\chi\leq \frac \Delta 2 +2$ for graphs of sufficiently large girth depending on $\Delta$. Can anyone prove it for girth $\geq 5$? $4$?
This would at least provide some hope for proving Reed's Conjecture for triangle-free graphs.
> Does every triangle-free graph with $\Delta\leq 6$ have $\chi \leq 5$? What about every graph with girth at least five?
| https://mathoverflow.net/questions/37923/does-every-triangle-free-graph-with-maximum-degree-at-most-6-have-a-5-colouring | [
"graph-theory",
"graph-colorings",
"co.combinatorics"
] | 55 | 2010-09-06T12:58:11 | [
"Another possible tangential attack is to establish the conjecture for the \"warmth\" $w$ of the graph, because Brightwell and Winkler proved that $w \\le \\chi$.",
"Good question; it is known to hold, without the round-up. Reed never published the result in a paper but it's in Graph Colouring and the Probabilist... | 3 | Science | 1 |
59 | mathoverflow | Class function counting solutions of equation in finite group: when is it a virtual character? | Let $w=w(x_1,\dots,x_n)$ be a word in a free group of rank $n$. Let $G$ be a finite group. Then we may define a class function $f=f_w$ of $G$ by
$$ f_w(g) = |\\{ (x_1,\dots, x_n)\in G^n\mid w(x_1,\dots,x_n)=g \\}|. $$ The question is:
> Can we characterize the words $w$ for which $f_w$ is a virtual character for any group $G$?
(A virtual character is a difference of two characters, sometimes also called generalized character.) That $f_w$ is a virtual character of $G$ is equivalent to
$$ (f_w,\chi) = \frac{1}{|G|} \sum_{x_1, \dots,x_n\in G} \chi(w(x_1, \dots,x_n)) \in \mathbb{Z} $$ for all $\chi \in \operatorname{Irr}(G)$.
Some examples and non-examples:
1. $n=1$, $w= x^k$. Then $(f_w,\chi)\in \mathbb{Z}$ is the $k$-th Frobenius-Schur-indicator.
2. $w=[x,y]=x^{-1}y^{-1}xy$. Then $(f_w,\chi)= \frac{|G|}{\chi(1)}$, so $f_w$ is in fact a character.
3. For $w=x^{-k}y^{-1}x^ly$, it can be shown that $(f_w,\chi)= \frac{|G|}{\chi(1)} a$ for some integer $a$.
4. $f_w$ is a virtual character for words of the form $w= x_1^{e_1}x_2^{e_2} \dots x_n^{e_n}$.
5. For $w=[x^2,y^2]$ and $G= SL(2,7)$, the class function $f_w$ is not a generalized character.
It is always the case that $(f_w, \chi)$ is an algebraic integer in $\mathbb{Q}(\chi)$. This follows from an old result of Solomon saying $f_w(g)$ is divisible by $|C_G(g)|$ when the number of variables is $>1$. (See also [this paper](http://de.arxiv.org/abs/1105.6066v1) by Rodriguez Villegas and Gordon, where my question is discussed very briefly.) So for rational-valued groups, $f_w$ is a virtual character for any word. (In fact, it suffices that $G$ is a normal subgroup of a rational-valued group.) In view of 4., the same is true for abelian groups, and I don't know examples of solvable groups where $f_w$ is not a virtual character.
Another remark is that $f_w$ does not change if we replace $w$ by its image under an automorphism of the free group in question. Using this, one can of course find rather complicated looking words for which $f_w$ is trivial, that is, $f_w(g)\equiv |G|^{n-1}$.
Added later:
The proofs that $(f_w,\chi)\in \mathbb{Z}$ in 2.-4. above basically use the same "trick". I'm also interested in other examples where perhaps another idea of proof is used than in the examples 1.-4., or "more conceptual" explanations why $f_w$ sometimes is a (virtual) character. (I admit that this is somewhat vague.)
A related question is when $f_w$ is a character, not only a virtual character. For $G$ the symmetric group, this is discussed in Exercise 7.69k of "Enumerative Combinatorics 2" by Richard Stanley. (Note that in this case, $f_w$ is always a virtual character.)
| https://mathoverflow.net/questions/105854/class-function-counting-solutions-of-equation-in-finite-group-when-is-it-a-virt | [
"gr.group-theory",
"rt.representation-theory",
"finite-groups",
"characters"
] | 54 | 2012-08-29T08:38:56 | [
"To add to HJRW's comment a function is a $\\mathbb Q$-linear combination of characters iff it is constant on elements generating conjugate subgroups. In your set up because it is an algebraic integer combination of characters you just need it to be a $\\mathbb Q$-linear combination to be a virtual character. So a... | 2 | Science | 1 |
60 | mathoverflow | How many algebraic closures can a field have? | Assuming the axiom of choice given a field $F$, there is an algebraic extension $\overline F$ of $F$ which is algebraically closed. Moreover, if $K$ is a different algebraic extension of $F$ which is algebraically closed, then $K\cong\overline F$ via an isomorphism which fixes $F$. We can therefore say that $\overline F$ is _the_ algebraic closure of $F$.
(In fact, much less than the axiom of choice is necessary.)
Without the axiom of choice, it is consistent that some fields do not have an algebraic closure. It is consistent that $\Bbb Q$ has two non-isomorphic algebraically closed algebraic extensions.
It therefore makes sense to ask: Suppose there are two non-isomorphic algebraically closed algebraic extensions. Is there a third? Are there infinitely many? Are there Dedekind-infinitely many?
> What is provable from $\sf ZF$ about the spectrum of algebraically closed algebraic extensions of an arbitrary field? What about the rational numbers?
| https://mathoverflow.net/questions/325756/how-many-algebraic-closures-can-a-field-have | [
"set-theory",
"ac.commutative-algebra",
"lo.logic",
"axiom-of-choice",
"field-extensions"
] | 54 | 2019-03-19T02:41:22 | [
"@DavidRoberts: Should, yes. One day...",
"Shouldn't you add an answer about this now, pointing to your own paper?",
"@AsafKaragila thanks!",
"@Holo: arxiv.org/abs/1911.09285",
"@Holo: The one I am putting in arxiv later today. Check Monday's Replacements.",
"@AsafKaragila hi, may I ask which paper of you... | 22 | Science | 1 |
61 | mathoverflow | Atiyah's paper on complex structures on $S^6$ | M. Atiyah has posted a preprint on arXiv on the non-existence of complex structure on the sphere $S^6$.
<https://arxiv.org/abs/1610.09366>
It relies on the topological $K$-theory $KR$ and in particular on the forgetful map from topological complex $K$-theory to $KR$.
Question: what are the nice references to learn about $KR$?
**Edit :** Thank you very much for the comments and suggestions, M. Atiyah's paper "K-theory and reality." Quart. J. Math., Oxford (2), 17 (1966),367-86 is a fantastic paper. However I have more questions.
Question 1: how to build the morphism $$KSp(\mathbb{R}^6)\rightarrow K^{7,1}(pt)?$$
Question 2: why do we use $\mathbb{R}^{7,1}$?
In fact I do not understand the sentence "The 6-sphere then appears naturally as the base of the light-cone". And why it suffices to look at this particular model of $S^6$.
| https://mathoverflow.net/questions/253577/atiyahs-paper-on-complex-structures-on-s6 | [
"dg.differential-geometry",
"at.algebraic-topology",
"complex-geometry",
"kt.k-theory-and-homology",
"index-theory"
] | 52 | 2016-10-31T13:01:46 | [
"I think question should be edited to remove any implication that it is asking about possible correctness of Atiyah's paper. If the title were to say something about your actual question on KR, that would be good.",
"Regarding the map you wish to understand, I think that the KSp should be seen as coming from one ... | 16 | Science | 0 |
62 | mathoverflow | Concerning proofs from the axiom of choice that ℝ³ admits surprising geometrical decompositions: Can we prove there is no Borel decomposition? | This question follows up on a [comment](https://mathoverflow.net/questions/92919/filling-mathbbr3-with-skew-lines#comment238532_92919) I made on Joseph O'Rourke's recent [question](https://mathoverflow.net/questions/92919/filling-mathbbr3-with-skew-lines), one of several questions here on mathoverflow concerning surprising geometric partitions of space using the axiom of choice.
1. Joseph O'Rourke: [Filling $\mathbb{R}^3$ with skew lines](https://mathoverflow.net/questions/92919/filling-mathbbr3-with-skew-lines)
2. Zarathustra: [Is it possible to partition $\mathbb{R}^3$ into unit circles?](https://mathoverflow.net/questions/28647/is-it-possible-to-partition-mathbb-r3-into-unit-circles)
3. Benoît Kloeckner: [Is it still impossible to partition the plane into Jordan curves without choice?](https://mathoverflow.net/questions/21327/is-it-still-impossible-to-partition-the-plane-into-jordan-curves-without-choice)
The answers to these questions show that the axiom of choice implies, by rather easy and flexible arguments, that space admits several surprising geometrical decompositions. To give a few examples:
* $\mathbb{R}^3$ is the union of disjoint circles, all with the same radius. Proof: Well-order the points in type $\frak{c}$. At stage $\alpha$, consider the $\alpha^\text{th}$ point. If it is not yet covered, we may find a unit circle through this point that avoids the fewer-than-$\mathfrak{c}$-many previously chosen circles. QED
* $\mathbb{R}^3$ is the union of disjoint circles, with all different radii. Proof: at stage $\alpha$, find a circle of a previously unused radius that avoids the fewer-than-$\mathfrak{c}$-many previously chosen circles. QED
* $\mathbb{R}^3$ is the union of disjoint skew lines, each with a different direction. Proof: at stage $\alpha$, put a line through the $\alpha^\text{th}$ point with a different direction than any line used previously. QED (Edit: it appears that one might achieve this one constructively by using the $z$-axis and a nested collection of [hyperboloids](http://en.wikipedia.org/wiki/Ruled_surface), which unless I am mistaken would give lines of different directions.)
* $\mathbb{R}^3$ is the union of disjoint lines, each of which pierces the unit ball.
* And so on.
The excellent article _Jonsson, M.; Wästlund, J._ , [**Partitions of $\mathbb{R}^3$ into curves**](http://dx.doi.org/10.7146/math.scand.a-13850), Math. Scand. 83, No. 2, 192–204 (1998). [ZBL0951.52018](https://zbmath.org/?q=an:0951.52018). [JStor](https://www.jstor.org/stable/24493132), proves among other things that $\mathbb{R}^3$ can be partitioned into unlinked unit circles, either all with the same radius or all with different radii, and they have a very general theorem about partitioning $\mathbb{R}^3$ into isometric copies of any family of continuum many real algebraic curves.
So the general situation is that the axiom of choice constructions are both easy and flexible and not entirely ungeometrical.
Meanwhile, there are also a few concrete constructions, which do not use the axiom of choice. For example, Szulkin (see theorem 1.1 in [the Jonsson, Wästlund article](http://dx.doi.org/10.7146/math.scand.a-13850)) shows that $\mathbb{R}^3$ is the union of disjoint circles by a completely explicit method. And there are others. But my question is not about these cases where there is a concrete construction. Rather, my question is:
**Question 1.** Can we prove, in any of the cases where there is no concrete construction, that there is none?
For example, taking Borel as the measure of "explicit", the question would be: can we prove, for any of the kinds of decompositions I mention above or similar ones, that there is no Borel such decomposition? This would mean a decomposition of $\mathbb{R}^3$ such that the relation of "being on the same circle" would be a Borel subset of $\mathbb{R}^6$, and in this case, the function that maps a point to the nature of the circle on which it was to be found (the center, radius, and so on) would also be a Borel function.
A closely related question would be:
**Question 2.** Can we prove for any of the decomposition types that it requires the axiom of choice?
For example, perhaps there is an argument that in one of the standard models of $\text{ZF}+\neg\text{AC}$ that there is no such kind of decomposition.
My only idea for an attack on question 1 is this: if there were a Borel decomposition of $\mathbb{R}^3$ into circles or lines of a certain kind, then the assertion that this particular Borel definition was such a partition would be $\Pi^1_2$ and hence absolute to all forcing extensions. So the same Borel definition would work to define such a partition even in forcing extensions of the universe having additional reals. Furthermore, the ground model partitions would be a part of that new partition. And this would seem to be a very delicate geometric matter to fit the new reals into the partition in between the ground model objects. But I don't know how to push an argument through.
| https://mathoverflow.net/questions/93601/concerning-proofs-from-the-axiom-of-choice-that-%e2%84%9d%c2%b3-admits-surprising-geometrical | [
"lo.logic",
"set-theory",
"axiom-of-choice",
"mg.metric-geometry",
"descriptive-set-theory"
] | 51 | 2012-04-09T13:32:24 | [
"@LSpice Thanks!",
"Sounds like a good idea. Perhaps one could make such a kind of argument work with determinacy or $\\text{AD}_{\\mathbb{R}}$?",
"I really like these questions and every so often I go back and think about them. For me the obvious strategy to attempt is analytic - under some convenient assumpti... | 17 | Science | 1 |
63 | mathoverflow | Enriched Categories: Ideals/Submodules and algebraic geometry | While working through Atiyah/MacDonald for my final exams I realized the following:
The category(poset) of ideals $I(A)$ of a commutative ring A is a closed symmetric monoidal category if endowed with the product of ideals
$$I\cdot J:=\langle\\{ab\mid a\in I, b\in J\\}\rangle$$
The internal Hom is given by the ideal quotient; let's denote it $[I, J]:=(J:I)=\lbrace a\in A\mid aI\subset J\rbrace$. We have
$$IJ\subset K \Leftrightarrow I\subset [J,K].$$
Furthermore the category(poset) of submodules $U(M)$ of an A-Module $M$ is enriched over the ideal category via
$$[U,V]:=\lbrace a\in A\mid aU\subset V\rbrace.$$
Where does this enrichment come from? More specifically: As in algebraic geometry we are dealing with the set of prime ideals ( wich is more or less the subcategory
$$\mathrm{Int^{op}}/A\subset\mathrm{CRing^{op}}/A$$
of ring-maps $A\to B$ with $B$ an integral domain), I'd like to see the category of Ideals as some "flattened" version of $\mathrm{CRing^{op}}/A$ by taking the kernel. Or more generally i'd like to see like to see it as a "flattened" version of the category of modules (by taking the annihilator?). So:
**Question:** Is it possible to build the enriched categories $I(A)$ and $U(M)$ out of $\mathrm{CRing^{op}}/A$ and $\mathrm{Mod}_A/M$ in a nice way? So: Can we for example write $I\cdot J$ down as a kernel of some $A\to B$? Or $\mathrm{Ann}(M)\cdot \mathrm{Ann}(N)$ just by using $N,M$ and constructions in/on $\mathrm{Mod}_A$?
| https://mathoverflow.net/questions/76443/enriched-categories-ideals-submodules-and-algebraic-geometry | [
"ac.commutative-algebra",
"ag.algebraic-geometry",
"ct.category-theory",
"monoidal-categories",
"enriched-category-theory"
] | 48 | 2011-09-26T11:24:59 | [
"\"The category(poset) of ideals $I(A)$ of a commutative ring A is a closed symmetric monoidal category if endowed with the product of ideals\" That's how I'll explain commutative algebra to my son!",
"Very interesting question. $I(A)$ is equivalent to the full subcategory of the comma-category which consists of ... | 2 | Science | 0 |
64 | mathoverflow | Extending a line-arrangement so that the bounded components of its complement are triangles | Given a finite collection of lines $L_1,\dots,L_m$ in ${\bf{R}}^2$, let $R_1,\dots,R_n$ be the connected components of ${\bf{R}}^2 \setminus (L_1 \cup \dots \cup L_m)$, and say that $\\{L_1,\dots,L_m\\}$ is _triangulating away from infinity_ iff every $R_j$ that is bounded is triangular.
Can every finite line-arrangement be augmented (by adding more lines) to yield one that is triangulating away from infinity?
(The old and new arrangements need not be generic; they may have parallel lines and/or multiple intersections.)
Some background: A slightly different (and, in my view, less natural) question was raised by me about twenty years ago; it has never been answered either. That problem first appeared in Richard K. Guy and Richard J. Nowakowski’s American Mathematical Monthly column on open problems (see “Bite-Sized Combinatorial Geometry Problems”, AMM, Vol. 103, No. 4, Apr. 1996, p. 342 and “Monthly Unsolved Problems”, AMM, Vol. 104, No. 10, Dec. 1997, pp. 969-970). Stan Wagon invited the readers of his on-line column to completely solve a one-parameter class of problems of this kind in which three lines symmetrically divide an equilateral triangle; see [Link](https://web.archive.org/web/20171230114418/http://mathforum.org/wagon/spring96/p812.html) .
A related problem that might be simpler to solve in the negative is a version in which we not only require that all bounded components have 3 sides but also require that all unbounded components have 2 or 3 sides. If nobody can solve my original problem affirmatively or negatively, a negative solution to this problem would deserve the bounty.
(Historical aside: Many years ago I proposed the original version of this problem to John Conway. He said “Let me think about that,” and as the curiosity-bug bit into his brain, he began to draw pictures, make observations, and formulate conjectures. But he was no fool; he could see that I was deliberately trying to entice him into working on the problem, and he was too proud to want to be seen as one who is so easily seduced. He shuddered as if shaking off an unpleasant memory and said “You know, I don’t have to work on just ANY damned problem!”)
| https://mathoverflow.net/questions/155196/extending-a-line-arrangement-so-that-the-bounded-components-of-its-complement-ar | [
"discrete-geometry",
"plane-geometry",
"hyperplane-arrangements"
] | 47 | 2014-01-20T11:45:57 | [
"the dual version of Silvester-Caley thm in a stronger version shows that a simplicial argangment has to satisfy an optimal incidency condition that cannot hold if one move some lines in a certain neigbourhood while some other will remain fixed. In another hand a set of lines construced with the condition of my fi... | 30 | Science | 0 |
65 | mathoverflow | Homotopy type of TOP(4)/PL(4) | It is known (e.g. the Kirby-Siebenmann book) that $\mathrm{TOP}(n)/\mathrm{PL}(n)\simeq K({\mathbb Z}/2,3)$ for $n>4$. I believe it is also known (Freedman-Quinn) that $\mathrm{TOP}(4)/\mathrm{PL}(4)\to K({\mathbb Z}/2,3)$ is 5-connected. Is it known whether $\mathrm{TOP}(4)/\mathrm{PL}(4)$ is not equivalent to $K({\mathbb Z}/2,3)$?
**Edit:** The following lists the relevant definitions.
* The topological group $\mathrm{TOP}(n)$ is the group of self-homeomorphisms of ${\mathbb R}^n$ with the compact-open topology.
* The topological group $\mathrm{PL}(n)=|\mathrm{PL}_S(n)|$ is defined as the geometric realization of the simplicial group $\mathrm{PL}_S(n)$. The $k$-simplices of $\mathrm{PL}_S(n)$ are the piecewise linear homeomorphisms $\Delta^k\times{\mathbb R}^n\to\Delta^k\times{\mathbb R}^n$ which commute with the projection onto $\Delta^k$.
* With the above definitions, there exists a canonical map of topological groups $\mathrm{PL}(n)\to\mathrm{TOP}(n)$. Then the space $\mathrm{TOP}(n)/\mathrm{PL}(n)$ is defined as the homotopy fibre of the induced map $B\mathrm{PL}(n)\to B\mathrm{TOP}(n)$. It is _not_ actually a quotient of a group by a subgroup.
* Here is a way of recovering the homotopy type of $\mathrm{TOP}(n)/\mathrm{PL}(n)$ as an actual quotient. Let $\mathrm{TOP}_S(n)$ be the singular complex of $\mathrm{TOP}(n)$: $\mathrm{TOP}_S(n)$ is the simplicial set whose $k$-simplices are continuous maps $\Delta^k\to\mathrm{TOP}(n)$; these are in canonical bijection with the homeomorphisms $\Delta^k\times{\mathbb R}^n\to\Delta^k\times{\mathbb R}^n$ commuting with the projection onto $\Delta^k$. Hence we obtain an inclusion of simplicial groups $\mathrm{PL}_S(n)\hookrightarrow\mathrm{TOP}_S(n)$, which induces by adjunction the previous map of topological groups $\mathrm{PL}(n)\to\mathrm{TOP}(n)$. The space $\mathrm{TOP}(n)/\mathrm{PL}(n)$ is weakly homotopy equivalent to the geometric realization of the simplicial set $\mathrm{TOP}_S(n)/\mathrm{PL}_S(n)$ (which is levelwise given by taking cosets).
| https://mathoverflow.net/questions/106971/homotopy-type-of-top4-pl4 | [
"at.algebraic-topology",
"gt.geometric-topology",
"homotopy-theory"
] | 45 | 2012-09-11T19:38:57 | [
"@Ricardo: yes, that's what I meant.",
"@Peter: Thank you for remarking that $\\mathrm{PL}(n)$ is not defined as a subgroup of $\\mathrm{TOP}(n)$. In fact, by $\\mathrm{TOP}(n)/\\mathrm{PL}(n)$ I meant the homotopy fibre of the map $B \\mathrm{PL}(n)\\to B \\mathrm{TOP}(n)$, which I believe is the usual meaning. ... | 9 | Science | 0 |
66 | mathoverflow | A kaleidoscopic coloring of the plane | > **Problem.** Is there a partition $\mathbb R^2=A\sqcup B$ of the Euclidean plane into two Lebesgue measurable sets such that for any disk $D$ of the unit radius we get $\lambda(A\cap D)=\lambda(B\cap D)=\frac12\lambda(D)$?
(I.V.Protasov called such partitions _kaleidoscopic_).
Observe that for the $\ell_1$\- or $\ell_\infty$-norms on the plane such partitions exist: just take a suitable chessboard coloring.
The problem can be reformulated in terms of convolutions: _Is there a measurable function $f:\mathbb R^2\to\\{1,-1\\}$ such that its convolution with the characteristic function $\chi_D$ of the unit disk $D$ is identically zero?_
(The problem was posed 08.11.2015 by T.Banakh and I.Protasov on [page 19](http://www.math.lviv.ua/szkocka/viewpage.php?vol=0&page=19) of [Volume 0](http://www.math.lviv.ua/szkocka/viewbook.php?vol=0) of the [Lviv Scottish Book](http://www.math.lviv.ua/szkocka)).
| https://mathoverflow.net/questions/219860/a-kaleidoscopic-coloring-of-the-plane | [
"real-analysis",
"mg.metric-geometry",
"measure-theory",
"harmonic-analysis",
"geometric-measure-theory"
] | 44 | 2015-10-02T11:55:32 | [
"@AntonPetrunin No, I do not know such a figure. It seems that for any triangle a kaleidoscopic coloring does exist.",
"Do you know any plane figure (maybe nonconvex) without kaleidoscopic colorings? (wrt parallel translations)",
"I seem to see a 3-color kaleidoscope for the perfect hexagonal ball.",
"I'll re... | 5 | Science | 0 |
67 | mathoverflow | What does the theta divisor of a number field know about its arithmetic? | This question is about a remark made by van der Geer and Schoof in their beautiful article "Effectivity of Arakelov divisors and the theta divisor of a number field" (from '98) ([link](http://arxiv.org/pdf/math/9802121v3)).
Let me first of all recall that a curve $C$ over a function field can be recovered from its Jacobian $\operatorname{Pic}(C)$ and its theta divisor $\Theta \subset \operatorname{Pic}^{g-1}(C)$ ($g$ the genus of $C$).
In their article van der Geer and Schoof construct analogues of the Jacobian and the theta divisor in the case of number fields.
Namely they consider the Arakelov-Picard group $\operatorname{Pic}(F)$ of a number field $F$ and define its theta divisor as restriction of the function $h^0$ on $\operatorname{Pic}(F)$ given by $$h^0(D) = \log\left(\sum_{f \in I} e^{-\pi \|f\|^2_D}\right),$$ where $I$ is a lattice associated with the Arakelov divisor $D$, to the subspace $\operatorname{Pic}^{(d)}(F)$ of Arakelov divisors of degree $d$. Here $d$ is a suitable analogue of the genus of a curve.
Now they say that "it should be possible to reconstruct the arithmetic of the number field $F$ from $\operatorname{Pic}^{(d)}(F)$ together with $h^0$".
My question is now of course which parts of the arithmetic of $F$ are known to be recovered from their analogy? For example I would be very interested in the question if the units of $F$ can be recovered somehow.
| https://mathoverflow.net/questions/47920/what-does-the-theta-divisor-of-a-number-field-know-about-its-arithmetic | [
"nt.number-theory",
"arakelov-theory",
"theta-functions",
"reference-request"
] | 41 | 2010-12-01T08:17:15 | [
"Since I think it is useful information to have around, I summarize from memory from the deleted answer and subsequent discussion: the above mentioned paper is the thesis of R. Groenewegen \"Vector bundles and geometry of numbers\", more specifically its second part \"Torelli for number fields\", which explores the... | 2 | Science | 0 |
68 | mathoverflow | Homology of $\mathrm{PGL}_2(F)$ | **Update:** As mentioned below, the answer to the original question is a strong No. However, the case of $\pi_4$ remains, and actually I think that this one would follow from Suslin's conjecture on injectivity of $H_i(\mathrm{GL}_n(F))\to H_i(\mathrm{GL}_{n+1}(F))$ up to torsion (in the case $i=4$, $n=3$) together with some results of Dupont (specifically exact sequence (15.10) in Dupont's book "Scissors congruences, group homology and characteristic classes") and Goncharov's conjecture on motivic cohomology in terms of Bloch groups in weights $2$ and $3$. Concretely:
> Question. Let $F$ be any field. Consider the inclusion $F^\times\to \mathrm{PGL}_2(F)$ of the diagonal matrices. Is the induced map $$ \bigwedge^4 F^\times_{\mathbb Q} = H_4(F^\times,\mathbb Q)\to H_4(\mathrm{PGL}_2(F),\mathbb Q)$$ surjective?
(Explanation: In Dupont's (15.10), the kernel of the last map is precisely $H^2(F,\mathbb Q(3))$ under Goncharov's conjecture. Under Suslin's conjecture, $H_4(\mathrm{PGL}_3(F),\mathbb Q)\to H_4(\mathrm{PGL}_4(F),\mathbb Q)$ is injective, and the target is stable. Computing in terms of the homology of the $K$-theory space, one sees that the contributions to $H_4(\mathrm{PGL}_4(F),\mathbb Q)$ are $\mathrm{Sym}^2 K_2^M(F)_{\mathbb Q}$ and $H^2(F,\mathbb Q(3))$ and $K_4^M(F)_{\mathbb Q}$. The last term $K_4^M(F)_{\mathbb Q}$ is precisely the cokernel of the map $H_4(\mathrm{PGL}_3(F),\mathbb Q)\to H_4(\mathrm{PGL}_4(F),\mathbb Q)$ by Suslin's theorem, so $H_4(\mathrm{PGL}_3(F),\mathbb Q)$ should have contributions from $H^2(F,\mathbb Q(3))$ and $\mathrm{Sym}^2 K_2^M(F)_\mathbb{Q}$. The latter admits a surjection from $\mathrm{Sym}^2(\bigwedge^2 F^\times)_{\mathbb{Q}} = H_4((F^\times)^2\rtimes \Sigma_3)$, which is precisely the decomposable part of $H_4(\mathrm{PGL}_3(F),\mathbb Q)$ in Dupont's work. Looking back at (15.10), this shows that the first term should be zero, which leads to the question. There is some confusion between $\mathrm{PSL}$ and $\mathrm{PGL}$ in what I said, but at least assuming that all elements of $F$ have $6$-th roots, the argument should work.)
* * *
Here is a hopelessly naive question. Please point me to the relevant literature!
Let $F$ be any field. The Cartan subgroup of $\mathrm{PGL}_2(F)$ is $F^\times\rtimes \Sigma_2$. Let $X_2(F)$ be the homotopy cofiber of $B(F^\times\rtimes\Sigma_2)\to B\mathrm{PGL}_2(F)$, so there is a long exact sequence $$ \ldots\to H_i(F^\times\rtimes\Sigma_2)\to H_i(\mathrm{PGL}_2(F))\to H_i(X_2(F))\to \ldots $$ (I'm not sure if I should write $H_i(G)$ or $H_i(BG)$...). If I am not making a stupid mistake, then the known computations of the homology of $\mathrm{GL}_2(F)$ going up to the first unstable group $H_3(\mathrm{GL}_2(F))$ can be summarized by saying that $H_i(X_2(F))=0$ for $i=1,2$, and $H_3(X_2(F))=\mathfrak p(F)$ is the pre-Bloch group of $F$. (The pre-Bloch group is the quotient of $\mathbb Z[F^\times]$ by the ``$5$-term relations''.) (I might be off by some $2$-torsion.) By Hurewicz (and the check that $\pi_1 X_2(F)=0$), the same holds true for the homotopy groups of $X_2(F)$.
As far as I could find, little is known about the homology of $\mathrm{GL}_2(F)$ or $\mathrm{PGL}_2(F)$ in degrees $>3$. The rational structure could be determined by the following very naive question.
> Question. Are the homotopy groups $\pi_i X_2(F)$ bounded torsion for $i\neq 3$?
I might even expect the implicit bound to be independent of $F$ (but of course depend on $i$; the order of magnitude should be $i!$).
The only evidence I have is that a back-of-the-envelope calculation seems to suggest that this holds true for finite fields. In that case the order of $\mathrm{PGL}_2(\mathbb F_q)$ is the product of $q-1$, $q$ and $q+1$. Localized at $q-1$, the homology agrees with the homology of the Cartan; localized at $q$, it is only in degrees larger than $q$ (which is OK as the bound on torsion may depend on $i$), and localized at $q+1$, the homology seems to agree with the homology of a $K(\mathbb Z/(q+1),3)$ (up to bounded torsion in each degree), which is roughly $\mathbb Z/(q+1)$ in all degrees $\equiv 3\mod 4$ (and is bounded torsion in other degrees). The pre-Bloch group $\mathfrak p(\mathbb F_q)$ is also $\mathbb Z/(q+1)$ up to $2$-torsion, so things add up.
The case of $F=\mathbb Q$ might be amenable to computations, but I was unable to do those. Are there any relevant results about $H_4(\mathrm{GL}_2(F))$ that might shed light on $\pi_4 X_2(F)$?
A final remark: I expect that it is critical that $F$ is a field. The similar statement should be false already for discrete valuation rings, or $\mathbb Z[\tfrac 1n]$ (which is why the case of $F=\mathbb Q$ is not so easy to compute; I presume one would try to first compute for all $\mathbb Z[\tfrac 1n]$ and then pass to a colimit, and the desired structure should only appear in the colimit).
**Update:** As mentioned by Matthias Wendt in the comments, a special case of the work of Borel-Yang says that $H_i(\mathrm{PGL}_2(\mathbb Q),\mathbb Q)=0$ for $i>0$. Inserting this into the above long exact sequence, one immediately sees that $X_2(\mathbb Q)$ is in fact rather complicated, and for example has nontorsion $\pi_5$. Thus, the answer to the question is a strong No. Thanks for the help in sorting this out!
| https://mathoverflow.net/questions/298336/homology-of-mathrmpgl-2f | [
"gr.group-theory",
"algebraic-groups",
"kt.k-theory-and-homology",
"group-cohomology",
"algebraic-k-theory"
] | 39 | 2018-04-20T02:58:29 | [
"@Matthias Wendt: Thanks a lot for the Borel-Yang reference! Let's not worry right now about the exact shape of the torsion; I'd be happy enough with the statement rationally (but then my \"evidence\" would be vacuous). About homotopy vs. homology: One reason is that the homotopy of the K-theory space is usually s... | 10 | Science | 0 |
69 | mathoverflow | Correspondence between eigenvalue distributions of random unitary and random orthogonal matrices | In the course of a physics problem ([arXiv:1206.6687](http://arxiv.org/abs/1206.6687)), I stumbled on a curious correspondence between the eigenvalue distributions of the matrix product $U\bar{U}$, with $U$ a random unitary matrix and $\bar{U}$ its complex conjugate, on the one hand, and the random orthogonal matrix $O$ with determinant $-1$, on the other hand. The random matrices $U$ and $O$ are uniformly distributed with respect to the Haar measure on the unitary group $U(N)$ and the orthogonal group $O(N+1)$.
One eigenvalue of $O$ is fixed at $-1$ (to ensure that det $O=-1$). The other $N$ eigenvalues have a distribution $p_O$ which was [known](http://en.wikipedia.org/wiki/Circular_ensemble) since Girko (1985). We calculated the distribution $p_{U\bar{U}}$ of the eigenvalues of $U\bar{U}$ (which we did not find in the literature _\--- has anyone seen it before?_). We discovered to our surprise that $p_{U\bar{U}}=p_O$. This holds for both $N$ even and odd (in the latter case both $U\bar{U}$ and $O$ have an eigenvalue fixed at $+1$).
**Question:** Is there a more direct route to arrive at this identity between the two eigenvalue distributions, without going through a separate calculation of each one? (You can find two such separate calculations in the Appendix of arXiv:1206.6687, but this seems a rather unsatisfactory way of understanding the correspondence.)
* * *
Some intuition for what is going on: for both $U\bar{U}$ and $O$ the eigenvalues different from $\pm 1$ come in complex conjugate pairs $e^{\pm i\theta}$. The matrix $O$ has an unpaired eigenvalue at $-1$, which repels $\theta$ from $\pi$. The matrix $U\bar{U}$ cannot have an unpaired eigenvalue at $-1$ by construction and somehow this leads to a repulsion of $\theta$ from $\pi$ which is mathematically equivalent to what happens for the matrix $O$.
For example, when $N=2$ the eigenvalue $e^{i\theta}$ has the same distribution $P(\theta)=(2\pi)^{-1}(1+\cos\theta)$ for both $U\bar{U}$ and $O$. For $N=3$ the distribution is $P(\theta)=\pi^{-1}(1-\cos^2\theta)$, in addition to an eigenvalue fixed at $+1$, again the same for $U\bar{U}$ and $O$. The correspondence continues for larger $N$, when factors $(\cos\theta_k-\cos\theta_l)^2$ appear in both distributions.
| https://mathoverflow.net/questions/105512/correspondence-between-eigenvalue-distributions-of-random-unitary-and-random-ort | [
"pr.probability",
"random-matrices",
"haar-measure",
"mp.mathematical-physics",
"linear-algebra"
] | 37 | 2012-08-26T03:42:59 | [
"@student -- the first statement [$U^TU\\in{\\rm COE}(n)$] is correct, but the second statement is not, I think. The eigenvalue distribution of $U\\bar{U}$, with $U\\in{\\rm CUE}(n)$ is not the eigenvalue distribution of ${\\rm CRE}(n)$, it's the distribution of the eigenvalues $\\neq -1$ in ${\\rm CRE}(n+1)$. So t... | 13 | Science | 0 |
70 | mathoverflow | Is there any positive integer sequence $c_{n+1}=\frac{c_n(c_n+n+d)}n$? | In a [recent answer](https://mathoverflow.net/a/282154/17581) Max Alekseyev provided two recurrences of the form mentioned in the title which stay integer for a long time. However, they eventually fail.
> **QUESTION** Is there any (**added:** strictly increasing) sequence of positive integers $c_1,c_2,c_3,\dots$ satisfying the relation $$ c_{n+1}=\frac{c_n(c_n+n+d)}n $$ for for all $n\geq 1$ (and some integer constant $d$)?
(As Sam Hopkins notes, it would be also very interesting if the indices of the sequence start from some $k>1$ rather than from 1.)
| https://mathoverflow.net/questions/282159/is-there-any-positive-integer-sequence-c-n1-fracc-nc-nndn | [
"nt.number-theory",
"integer-sequences",
"recurrences"
] | 37 | 2017-09-27T08:35:39 | [
"@WillSawin: phrased this way, the transformation sort of looks like those found in the theory of cluster algebras, so maybe if there is a positive answer to the question it could be found via the \"Laurent phenomenon.\"",
"If we ignore the division by n, a necessary condition for the integrality of the original ... | 4 | Science | 0 |
71 | mathoverflow | What is the three-dimensional hyperbolic volume of a four-manifold? | Every smooth closed orientable 4-manifold may be constructed via a handle decomposition. Before asking a couple of questions, I recall some well-known facts about handle-decompositions of 4-manifolds.
We can of course order the handles according to their index. Handles of index 0 and 1 form a connected 4-dimensional handlebody, whose boundary is a closed 3-manifold, diffeomorphic to a connected sum $\\#_g(S^2\times S^1)$ of some $g$ copies of $S^2\times S^1$ (if $g=0$ we get $S^3$). Handles of index 2 are attached to some framed link $L\subset \\#_g(S^2\times S^1)$. Since 3- and 4-handles also form a 1-dimensional handlebody, after the attaching of the 2-handles we must necessarily obtain a 4-manifold whose boundary is again diffeomorphic to $\\#_h(S^2\times S^1)$, for some $h$ which is not necessarily equal to $g$. The new $\\#_h(S^2\times S^1)$ is obtained by _surgery_ along the framed link $L$.
Therefore, in some sense, constructing closed orientable 4-manifolds reduces to constructing framed links $L\subset \\#_g(S^2\times S^1)$ that produce some $\\#_h(S^2\times S^1)$ via surgery. Let us define the _3-dimensional hyperbolic volume_ of a handle decomposition as the Gromov norm of $(S^2\times S^1) \setminus L$ (which is in turn the sum of the hyperbolic volumes of its pieces according to geometrization, whence the name). Let us then define the _3-dimensional hyperbolic volume_ of a closed orientable 4-manifold as the infimum of all the hyperbolic 3-dimensional volumes among all its handle decompositions. The infimum is actually a minimum because the set of 3-dimensional hyperbolic volumes is well-ordered.
The general question is:
> What can we say about the 3-dimensional hyperbolic volume of a closed 4-manifold?
A more specific one:
> Which closed 4-manifolds have zero 3-dimensional hyperbolic volume?
which is equivalent to the following:
> Which closed 4-manifolds admit a handle decomposition such that $\\#_g(S^2\times S^1)\setminus L$ is a graph manifold?
Complex projetive plane belongs to this class, and also many doubles of 2-handlebodies: if you take any link $L\subset S^2\times S^1$ whose complement is a graph manifold, you can attach 2-handles to it, and then make the double of the resulting bounded 4-manifold. The resulting double has volume zero.
Finally, we have the following very specific question:
> Is there a 4-manifold with positive 3-dimensional hyperbolic volume?
I would expect that most (all?) aspherical 4-manifolds have positive volume, and maybe also many simply connected ones, but I don't know the answer.
| https://mathoverflow.net/questions/69819/what-is-the-three-dimensional-hyperbolic-volume-of-a-four-manifold | [
"gt.geometric-topology",
"4-manifolds",
"3-manifolds",
"knot-theory"
] | 37 | 2011-07-08T12:14:14 | [
"I think this might be related to this question: mathoverflow.net/q/19974/1345 In particular, if the 3-dimensional volume is zero, I suspect that the manifold admits an F-structure, and the simplicial volume is zero (and likely minimal volume =0). The point is that the circle action on a Seifert manifold M extends ... | 2 | Science | 0 |
72 | mathoverflow | Two-convexity ⇒ Lefschetz? | Assume that
* $\Omega$ is an open simply connected set in $\mathbb R^n$
* _(two-convexity)_ if 3 faces of a 3-simplex belong to $\Omega$ then whole simplex in $\Omega$.
Is it true that any component of intersection of $\Omega$ with any 2-plane is simply connected?
**Comments:**
* Note that an open connected set $\Omega\subset\mathbb R^n$ is convex (in the usual sense) if together with two sides of triangle $\Omega$ contains whole triangle.
* If the boundary of $\Omega$ is a smooth hypersurface then the answer is YES. [The above property implies that at most one principle curvature of the boundary is negative. Then the statement follows easily from a Morse-type argument; see Gromov's "Sign and geometric meaning of curvature", Lefschetz theorem in Section $\tfrac12$. In fact this argument, shows that in this case any component of intersection of $\Omega$ with any affine subspace (not necessary 2-dimensional) is simply connected.]
* The answer is YES if $n=3$. [In this case I can mimic the Morse-type argument.]
* The statement would follow if one could approximate any $\Omega$ by two-convex domains with smooth boundary. BUT the example below shows that such approximation does not exist for $n\ge 4$.
* I was surprised that my question has a lot in common with [this question](https://mathoverflow.net/questions/29373/minimal-length-embeddings-of-braids-into-r3-with-fixed-endpoints/)
**Example:** We will construct two-convex simply connected open set $\Omega$ in $\mathbb R^4$ such that intersection $L_{t_0}$ of $\Omega$ with some hyperplane is not simply connected. [This NOT a counterexample, but it shows that it is impossible to prove it using smoothing, as indicated in the comments.]
Set $$\Pi=\\{\,(x,y,z,t)\in\mathbb R^4\mid\,y< x^2\\}.$$ Let $\Pi'$ be the image of $\Pi$ under a generic rotation of $\mathbb R^4$, say $(x,y,z,t)\mapsto(z,t,x,y)$.
Note that $\Pi$ is open and two-convex. Therefore $\Omega=\Pi\cap \Pi'$ is also open two-convex set. One can choose coordinates so that $\Omega$ is an epigraph for a function $f\colon\mathbb R^3\to\mathbb R$ like $$f=\max\\{\alpha_1-\beta_1^2,\alpha_2-\beta_2^2 \\},$$ where $\alpha_i$ and $\beta_i$ are linear functions. In particular $\Omega$ is contactable.
Let $L_{t_0}$ be the intersection of $\Omega$ with hyperplane $t=t_0$; it is a complement of two convex parabolic cylinders in general position. If these cylidners have a point of intersection then $\pi_1 L_{t_0}=\mathbb Z$.
[In particular, the function $f$ can not be approximated by smooth functions which Hessian has at most one negative eigenvalue value at all points.]
| https://mathoverflow.net/questions/55788/two-convexity-%e2%87%92-lefschetz | [
"convexity",
"convex-geometry",
"morse-theory",
"tag-removed"
] | 36 | 2011-02-17T14:12:07 | [
"@WillSawin, I guess your question is about last par, starting with \"Let $L_{t_0}$...\". We are in hypereplane, i.e., in $\\mathbb R^3$. Imagine that your parabolic cylinders have exactly one point of intersection. Then the complement is homotoipically equivalent to $\\mathbb{S}^1$.",
"I don't understand the exa... | 4 | Science | 0 |
73 | mathoverflow | Orthogonal vectors with entries from $\{-1,0,1\}$ | Let $\mathbf{1}$ be the all-ones vector, and suppose $\mathbf{1}, \mathbf{v_1}, \mathbf{v_2}, \ldots, \mathbf{v_{n-1}} \in \\{-1,0,1\\}^n$ are mutually orthogonal non-zero vectors. Does it follow that $n \in \\{1,2\\} \cup \\{4k : k \in \mathbb{N}\\}$?
A few notes are in order:
1. If we do not allow "$0$" entries in the vectors then this is a well-known necessary condition for the existence of an $n \times n$ [Hadamard matrix](https://en.wikipedia.org/wiki/Hadamard_matrix), so the answer to my question is "yes" if we replace $\\{-1,0,1\\}$ with$\\{-1,1\\}$.
2. With some computer help, I have shown that there do not exist collections of vectors with these properties when $n = 3$, $5$, or $6$, so the answer to my question is "yes" when $n \leq 6$.
**Edit (May 19, 2023):** In the comments, Max Alekseyev has shown that the answer is "yes" when $n \leq 12$, and Ilya Bogdanov has shown that the answer is "yes" when $n$ is an odd prime.
| https://mathoverflow.net/questions/446995/orthogonal-vectors-with-entries-from-1-0-1 | [
"co.combinatorics",
"linear-algebra",
"matrices"
] | 35 | 2023-05-17T12:10:00 | [
"I've computationally verified the conjecture $n=15$ and together with Ilya's and Mikhail's arguments that makes it proved for all $n\\leq 20$. For $n=21$, I can tell for now that if a counterexample exists, it should have at least 6 vectors with exactly 14 nonzero components.",
"@MaxAlekseyev Same argument that ... | 28 | Science | 0 |
74 | mathoverflow | Is there a Mathieu groupoid M_31? | I have read something which said that the large amount of common structure between the simple groups $SL(3,3)$ and $M_{11}$ indicated to Conway the possibility that the Mathieu groupoid $M_{13}$ might exist.
Indeed, it exists, and the group of permutations of the other points generated by paths in which the hole starts and ends at the same point is the Mathieu group $M_{12}$.
It is natural to ask whether or not something analogous exists for the Mathieu group $M_{24}$. The corresponding linear group seems to be $GL(5,2)$.
$2^{4}:A_{8}$ is a vector (or hyperplane) stabilizer in $GL(5,2)$ and an octad stabilizer in $M_{24}$. $2^{6}:(GL(2,2)\times GL(3,2))$ is likewise the stabilizer of a 2-dimensional (or 3-dimensional) space in $GL(5,2)$, and the stabilizer of a 'trio' (to use the term from SPLaG) in $M_{24}$.
The natural setting, therefore, in which to look for something analogous to $M_{13}$ is $(\mathbb{Z}/(2))^{5} \backslash 0$. (This can also be regarded as $4$-dimensional projective space over $\mathbb{Z}/(2)$, but I will here think of it as $5$-dimensional space over $\mathbb{Z}/(2)$ with the origin removed.)
Then, to have $24$ points left over when the 'holes' are put in place, it would be natural to have a 'missing $3$-dimensional space' for a hole.
A move should consist of moving the $3$-dimensional 'hole' to some other $3$-dimensional subspace (minus the origin) of $(\mathbb{Z}/(2))^{5}$. The intersection of these spaces can have dimension $1$ or $2$, and there should be different rules for moving the hole to another location depending on the dimension of its intersection with the new location.
Who has tried this before? Are there any papers on it?
| https://mathoverflow.net/questions/70680/is-there-a-mathieu-groupoid-m-31 | [
"gr.group-theory",
"finite-groups",
"reference-request"
] | 34 | 2011-07-18T14:48:41 | [
"It's good to know. 31 vs. 13 for M12."
] | 1 | Science | 0 |
75 | mathoverflow | Subalgebras of von Neumann algebras | In the late 70s, Cuntz and Behncke had a paper
H. Behncke and J. Cuntz, _Local Completeness of Operator Algebras_ , Proceedings of the American Mathematical Society, Vol. 62, No. 1 (Jan., 1977), pp. 95- 100
the about the following question. Let $A$ be a $C^\star$-algebra and let $B \subset A$ be some $*$-subalgebra. Let $B$ be dense and such that every maximal abelian subalgebra of $B$ is norm-closed. Is it true that $B=A$? They proved this in various cases.
I want to ask a related question in the von-Neumann-setting. More precisely: Let $A$ be a von Neumann algebra and $B \subset A$ be some ultra-weakly dense $\star$-subalgebra such every MASA of $B$ is ultra-weakly closed in $A$. Is it true that $B=A$? A result of Gert Pedersen implies that once $B$ is a $C^\star$-algebra, then $B=A$. Hence, the two questions are closely related. Taking the work of Behncke-Cuntz and Pedersen together, it is known that $B=A$ if $A$ has no $II_1$-part.
> **Question:** Is it true for every von Neumann algebra?
or a little more modest
> **Question:** Is it true for the hyperfinite $II_1$-factor?
Another strenghtening of the the assumption (which could help) would be the following:
> **Question:** Let $A$ be a $II_1$-factor and $B \subset A$ be a ultra-weakly dense $\star$-subalgebra such that for every hyperfinite subalgebra $R \subset A$ one has that $R \cap B$ is ultra-weakly closed in $A$. Is it true that $B=A$?
| https://mathoverflow.net/questions/34692/subalgebras-of-von-neumann-algebras | [
"oa.operator-algebras",
"fa.functional-analysis"
] | 33 | 2010-08-05T14:40:37 | [
"If I understand correctly, you need to add a density hypothesis in your question?"
] | 1 | Science | 0 |
76 | mathoverflow | Isometric embeddings of finite subsets of $\ell_2$ into infinite-dimensional Banach spaces | Question: Does there exist a finite subset $F$ of $\ell_2$ and an infinite-dimensional Banach space $X$ such that $F$ does not admit an isometric embedding into $X$?
There are some results of the type: each finite ultrametrics admits an isomeric embedding into any infinite dimensional Banach space. See Shkarin, S. A. Isometric embedding of finite ultrametric spaces in Banach spaces. Topology Appl. 142 (2004), no. 1-3, 13-17 for this, and also the papers: Dekster, B. V. Simplexes with prescribed edge lengths in Minkowski and Banach spaces. Acta Math. Hungar. 86 (2000), no. 4, 343-358; Faver, Timothy; Kochalski, Katelynn; Murugan, Mathav Kishore; Verheggen, Heidi; Wesson, Elizabeth; Weston, Anthony Roundness properties of ultrametric spaces. Glasg. Math. J. 56 (2014), no. 3, 519-535.
There is a related question on MathOverflow: [Almost isometric embeddability implies isometric embeddability](https://mathoverflow.net/questions/209435/almost-isometric-embeddability-implies-isometric-embeddability)
Added on 9/29/2016: Related results were obtained by James Kilbane in <https://arxiv.org/abs/1609.08971>
Added on 4/3/2017: In another [recent paper](https://arxiv.org/pdf/1704.00319.pdf) James Kilbane proved that the set of possible counterexamples (if they exist) is small in a certain sense.
| https://mathoverflow.net/questions/221181/isometric-embeddings-of-finite-subsets-of-ell-2-into-infinite-dimensional-ban | [
"mg.metric-geometry",
"banach-spaces",
"metric-embeddings"
] | 33 | 2015-10-17T17:44:32 | [
"The last result I know in this direction is in Petrov, F. V.; Stolyarov, D. M.; Zatitskiy, P. B. On embeddings of finite metric spaces in ln∞. Mathematika 56 (2010), no. 1, 135–139",
"Is much known when $X$ is $\\ell^n_{\\infty}$ for $n<\\#F$?"
] | 2 | Science | 0 |
77 | mathoverflow | Peano Arithmetic and the Field of Rationals | In 1949 Julia Robinson showed the undecidability of the first order theory of the field of rationals by demonstrating that the set of natural numbers $\Bbb{N}$ is _first order definable_ in $(\Bbb{Q}, +, \cdot$).
It is not hard to see that Robinson's result can be reformulated in the following symmetric form.
**Theorem A.** _The structures_ ($\Bbb{N}, +, \cdot$) _and_ $(\Bbb{Q}, +, \cdot$) _are bi-interpretable._
The following generalization of Theorem A is considered folkore (I am not aware of a published reference).
**Theorem B.** _If $(M, +, \cdot)$ is a model of $PA$ (Peano arithmetic), then the field of rationals $\Bbb{Q}^M$ of $(M, +, \cdot)$ is bi-interpretable with $(M, +, \cdot )$._
Let $EFA$ denote the _exponential function arithmetic_ fragment of $PA$, a fragment also known as $I\Delta_{0}+exp$.
Based on _a posteriori_ evidence _classical_ theorems of Number Theory do not require the full power of $PA$ since they can be already verified in $EFA$ (indeed Harvey Friedman has conjectured that even FLT can be verified in $EFA$, with a proof that would be very different from Wiles').
This suggests that in Theorem B one should be able to weaken $PA$ to $EFA$, hence my question:
**Question**. _Is there a published reference for the strengthening of Theorem B, where $PA$ is weakened to $EFA$?_
P.S. The following paper provides an excellent expository account of Robinson's theorem (and related results).
D. Flath and S. Wagon, _How to Pick Out the Integers in the Rationals: An Application of Logic to Number Theory_ , American Mathematical Monthly, Nov. 1991.
| https://mathoverflow.net/questions/65664/peano-arithmetic-and-the-field-of-rationals | [
"lo.logic",
"nt.number-theory"
] | 32 | 2011-05-21T11:01:13 | [
"@Samuel Ried: No I have not (sorry for the tardy reply, I have been \"away\" for a long while).",
"@AliEnayat: Have you found an answer to your question yet? I would be very interested to know.",
"SJR: The answer to your first question is positive, i.e., nothing fancier is going on. Regarding the second one: ... | 4 | Science | 0 |
78 | mathoverflow | $f\circ f=g$ revisited | This may be related to [solving $f(f(x))=g(x)$](https://mathoverflow.net/questions/17614). Let $C(\mathbb{R})$ be the linear space of all continuous functions from $\mathbb{R}$ to $\mathbb{R}$, and let $\mathcal{S}:=\\{g\in C(\mathbb{R}) ; \exists\; f\in C(\mathbb{R})$ s.t. $f\circ f=g \\}$ . Is there some infinite dimensional (or, at least, bidimensional) linear subspace of $C(\mathbb{R})$ contained in $\mathcal{S}$ ?
P.S. As a remark, there is a [maybe] interesting connection between [How to solve $f(f(x)) = \cos(x)$?](https://mathoverflow.net/questions/17605) and [Borsuk pairs of Banach spaces ](https://mathoverflow.net/questions/11675). Namely, let $E$ be the closed subspace of $C[-1,1]$ consisting of all even functions, and let $K$ be the closed unit ball of $E$. Then the continuous mapping $\Psi:$ $K$ $\rightarrow$ $E$ expressed by $\Psi(f)$ $:=$ $f\circ f$ $+$ $\left(\left\Vert f\right\Vert _{\infty}-1\right)\cdot\cos$
is odd on $\partial K$, and has no zeros in $K$.
| https://mathoverflow.net/questions/18882/f-circ-f-g-revisited | [
"ca.classical-analysis-and-odes",
"fractional-iteration"
] | 32 | 2010-03-20T14:35:10 | [
"In the previous comment, the full stop was taken as a part of the url - however, the intention was clearly to link to this post: solving $f(f(x))=g(x)$.",
"Unfortunately, not even two vectors in that cone generate a subspace in ${\\mathcal S}$ since ${\\mathcal S}$ contains no decreasing functions (see mathoverf... | 4 | Science | 0 |
79 | mathoverflow | The central insight in the proof of the existence of a class of Kervaire invariant one in dimension 126 | I understand from [a helpful earlier MO question](https://mathoverflow.net/questions/101031/kervaire-invariant-why-dimension-126-especially-difficult) that the techniques leading to the celebrated resolution of the Kervaire invariant one problem in the other candidate dimensions yield no insight on dimension 126, to the extent that until recently there was not consensus as to which way it was likely to go.
For a few weeks now, I've been aware of [a manuscript of Minami](https://arxiv.org/pdf/2106.14604) which, if I'm reading correctly, asserts the existence of such an element in dimension 126 as a consequence of Wang and Xu's 2017 paper on the stable 61-stem. Presumably they would not have failed to remark on the resolution of the Kervaire invariant one problem if such was readily visible from what they had done, so it would seem that there is something that is not _obviously_ a consequence of it, but which can nevertheless be extracted in eight pages if one does somehow know what to look for. It's common that when an old conjecture like this is resolved, it is as an application of some newly discovered techique or heretofore-ungleaned insight into the inner workings of the encompassing theory.
**What is the new insight here?**
The work on the stable stems is forbiddingly technical work I am unlikely to ever grasp the full details of, but it's still possible sometimes to get the flavor of the ideas involved, and that's what I would hope for here.
| https://mathoverflow.net/questions/397905/the-central-insight-in-the-proof-of-the-existence-of-a-class-of-kervaire-invaria | [
"at.algebraic-topology",
"homotopy-theory",
"stable-homotopy"
] | 31 | 2021-07-19T16:48:31 | [
"Probably you aware by now but just in case, there is also this preprint \"On the Last Kervaire Invariant Problem\" by Weinan Lin, Guozhen Wang, and Zhouli Xu (arxiv.org/abs/2412.10879) which also claims to resolve the final case of dimension 126 in the affirmative.",
"You're most welcome. \"Suitable for MathOve... | 3 | Science | 0 |
80 | mathoverflow | On the definition of regular (non-noetherian, commutative) rings | All rings are commutative with unit. A ring $R$ is called _regular_ if it satisfies
**(Reg)** Every finitely generated ideal of $R$ has finite projective dimension.
Clearly this gives the usual definition if $R$ is noetherian. In the non-noetherian world, valuation rings are regular (this is the reason why I got interested), as are polynomial rings in arbitrary sets of variables over a field.
I am wondering about the reason for this definition, as opposed in particular to:
**(Reg')** Every finitely presented $R$-module has finite projective dimension.
Clearly (Reg') implies (Reg), and they are equivalent if $R$ is _coherent_ (every finitely generated ideal is finitely presented).
To me, (Reg') looks much more natural than (Reg), and from my limited knowledge of the literature it seems that the association "coherent+regular" is quite popular. So, my question is
> What are examples of regular rings not satisfying (Reg')?
Since my post is really about motivating the definition, I am of course more interested in "natural" examples (not just constructed for the purpose).
On the other hand, all examples are welcome since I don't know any.
| https://mathoverflow.net/questions/327031/on-the-definition-of-regular-non-noetherian-commutative-rings | [
"ag.algebraic-geometry",
"ac.commutative-algebra",
"homological-algebra",
"regular-rings",
"valuation-rings"
] | 31 | 2019-04-03T01:38:58 | [
"If I understand correctly, to show that regular Noetherian is regular in this sense, you have to use the fact that for Noetherian rings being proj dimension <n at a point implies it in a neighborhood?",
"@KestutisCesnavicius: Thanks for the reference. Both properties are stable under any localization, and are Za... | 3 | Science | 0 |
81 | mathoverflow | Todd class as an Euler class | Let $X$ be a relatively nice scheme or topological space.
In various physics papers I've come accross, the Todd class $\text{Td}(T_X)$ is viewed as the Euler class of the normal bundle to $X\to LX$. Here $LX$ is the loop space of $X$. They usually call it $e_{reg}$, the regularised Euler class.
**Question:** Is there a way of making this notion of regularised Euler class rigorous?
Whatever the correct definition is, it should work for
* finite dimension vector bundles, where it reduces to usual notion of Euler class,
* loop spaces, where it reduces to the Todd class.
| https://mathoverflow.net/questions/379590/todd-class-as-an-euler-class | [
"ag.algebraic-geometry",
"at.algebraic-topology",
"loop-spaces"
] | 31 | 2020-12-23T03:29:19 | [
"You can define $i_*$ for any closed embedding between smooths actually, and $i^*i_*=e$ is still true.",
"@Pulcinella The two questions are indeed related: in finite dimensions in the equivariant setting, after localization both $e(N_i)$ and $i^*$ become invertible, so that one can define $i_*$ from the formula $... | 14 | Science | 0 |
82 | mathoverflow | When are two C*-algebras isomorphic as Banach spaces? | We may consider each $C^*$-algebra as a Banach space (by forgetting the multiplication and adjoint). I wonder how drastic this step is, i.e., which properties of the $C^*$-algebra are reflected by its Banach space structure.
Clearly, finite-dimensionality and other cardinality properties can be detected. Thus, to avoid these cases, let us restrict to separable, infinite-dimensional $C^*$-algebras.
For commutative $C^*$-algebras the problem has been solved by Milutin. In particular, $C(X)$ and $C(Y)$ are isomorphic as Banach spaces for any uncountable, compact, metric spaces $X$ and $Y$.
In the paper
Hamana. **On linear topological properties of some $C^*$-algebras** , Tohoku Math. J., II. Ser. 29, 157-163 (1977).
it is shown that the Banach space structure also reflects if all irreducible representations of the $C^*$-algebra are finite-dimensional.
Let us ask what happens at the 'opposite' end of the scale:
> Question: Are all simple, separable, infinite-dimensional $C^*$-algebras isomorphic as Banach spaces?
| https://mathoverflow.net/questions/156621/when-are-two-c-algebras-isomorphic-as-banach-spaces | [
"fa.functional-analysis",
"oa.operator-algebras"
] | 31 | 2014-02-03T15:01:12 | [
"This has been answered many times in the comments but is the pair $c_0$ (the null sequences) and the algebra of continuous functions on the closed unit interval not a simple counter example to your initial question?",
"It is an open problem whether $C^∗_r(F_2)$ is Banach isomorphic to a nuclear $C^*$-algebra (sa... | 17 | Science | 0 |
83 | mathoverflow | "Three great cocycles" in Complex Analysis as cohomology generators | In his [lecture notes](http://www.math.harvard.edu/~ctm/home/text/class/harvard/213a/00/html/course/course.pdf), C. McMullen discusses "the three great cocycles" in Complex Analysis: the derivative $$f\mapsto\log f',$$ the non-linearity $$f\mapsto (\log f')'dz$$ and the Schwarzian derivative $$f\mapsto \left(\frac{f'''}{f'}-\frac32\left(\frac{f''}{f'}\right)^2\right)(dz)^2$$ I would like to understand what "cocycle" means here precisely, and in which sense these maps are "the only" cocycles.
For diffeomorphisms of the circle, this is very clearly explained, e. g., in the [book](http://www.math.psu.edu/tabachni/Books/BookPro.pdf) by Ovsienko and Tabachnikov. Namely, $\mathrm{Diff}(S^1)$ is a group acting naturally on each of the spaces $F_\lambda(S^1)$ of tensor densities of degree $\lambda$. The corresponding (group) 1-cohomology is trivial unless $\lambda$ equals $0$, $1$, or $2$, where it is one-dimensional and generated by the above cocycles.
Question: is there an equally simply formulated statement in the context of complex variable, e. g. treating the above cocycles as generators of a group (or perhaps groupoid?) cohomology? This [MO answer](https://mathoverflow.net/a/38133/56624) suggests that the group is $\mathrm{Aut}$(meromorphic functions), but, at least naively, this seems to coincide with a subgroup of the Mobius group.
**Update:** Let me restate the question in simpler terms.The maps above satisfy the cocycle equation $$ D(g\circ f)(z)=(f'(z))^\lambda D(g)(f(z))+D(f(z)), $$ for $\lambda=0,1,2$, respectively. For a tensor density $\psi$, one can define the coboundary $$ D_\psi:f\mapsto (f')^\lambda(\psi\circ f)-\psi, $$ satisfying the same equation. So, the question is: what are the explicit conditions guaranteeing that the three maps above are the only, up to coboundaries, solutions to the cocycle equation?
| https://mathoverflow.net/questions/231823/three-great-cocycles-in-complex-analysis-as-cohomology-generators | [
"reference-request",
"cv.complex-variables",
"group-cohomology"
] | 31 | 2016-02-22T05:57:23 | [
"I'm confused by the notation, especially $D$ and $\\circle$.",
"See perhaps \"Explicit formulae for cocycles of holomorphic vector fields with values in $ \\lambda$ densities\" by Wagemann (arxiv.org/pdf/math-ph/0003035.pdf).",
"@WillSawin, well, in the case of the circle the space of tensor densities of degre... | 4 | Science | 0 |
84 | mathoverflow | When is a compact topological 4-manifold a CW complex? | Freedman's $E_8$-manifold is nontriangulable, as proved on page (xvi) of the Akbulut-McCarthy 1990 Princeton Mathematical Notes "Casson's invariant for oriented homology 3-spheres". Kirby showed that a compact 4-manifold has a handlebody structure if and only if it is smoothable: [1](http://www.manifoldatlas.him.uni-bonn.de/index.php/Questions_about_surgery_theory) and [2](https://mathoverflow.net/questions/7921/failure-of-smoothing-theory-for-topological-4-manifolds/38739#38739). When is a compact topological 4-manifold a CW complex?
| https://mathoverflow.net/questions/73428/when-is-a-compact-topological-4-manifold-a-cw-complex | [
"at.algebraic-topology",
"open-problems",
"4-manifolds",
"cw-complexes",
"topological-manifolds"
] | 31 | 2011-08-22T11:32:18 | [
"I have good reason to believe that it is an open question! Apologies - I hadn't seen the earlier posting mathoverflow.net/questions/36838/… ",
"I take it this is an open problem? Small note, but your question was effectively asked in another form, here: mathoverflow.net/questions/36838/… I usually think of CW... | 2 | Science | 0 |
85 | mathoverflow | Cohomology of symmetric groups and the integers mod 12 | When $n \ge 4$, the third homology group $H_3(S_n,\mathbb{Z})$ of the symmetric group $S_n$ contains $\mathbb{Z}_{12}$ as a summand. Using the universal coefficient theorem we get $\mathbb{Z}_{12}$ as a summand of the cohomology group $H^3(S_n,\mathbb{Z}_{12})$.
**What's a nice 'formula' for a $\mathbb{Z}_{12}$-valued 3-cocycle on $S_n$ that generates this summand of the third cohomology?**
In more concrete terms, I'm looking for a nontrivial recipe to get an element $c(g,g',g'') \in \mathbb{Z}_{12}$ from three elements of $S_n$, such that
$$ c(g',g'',g''') - c(gg',g'',g''') + c(g,g'g'',g''') - c(g,g',g''g''') + c(g,g',g'') = 0 $$
for all $g,g',g'',g''' \in S_n$. Here "nontrivial" means that for all nonzero $\alpha \in \mathbb{Z}_{12}$, we do not have
$$ \alpha c(g,g',g'') = f(g',g'') - f(gg',g'') + f(g,g'g'') - f(g,g') $$
for some $f \colon S_n \times S_n \to \mathbb{Z}_{12}$.
| https://mathoverflow.net/questions/314626/cohomology-of-symmetric-groups-and-the-integers-mod-12 | [
"gr.group-theory",
"group-cohomology"
] | 30 | 2018-11-05T12:17:19 | [
"@DevSinha - can you explain that class and how it works in more detail, or point me toward an explanation? I wanted a \"formula\", but an explicit geometrical description like this would also be nice.",
"@PhillippLampe - another typo! I'll fix it, thanks.",
"@JohnBaez Why do the second and the third summand... | 27 | Science | 0 |
86 | mathoverflow | Is there an Ehrhart polynomial for Gaussian integers | Let $N$ be a positive integer and let $P \subset \mathbb{C}$ be a polygon whose vertices are of the form $(a_1+b_1 i)/N$, $(a_2+b_2 i)/N$, ..., $(a_r+b_r i)/N$, with $a_j + b_j i$ being various Gaussian integers. For any Gaussian integer $u+vi$, let $h_P(u+vi)$ be the number of lattice point in $(u+vi) \cdot P$. I would like to say that $h_P(u+vi)$ is in some sense "quasi-polynomial modulo $N$". To emphasize: **The reason that this does not follow from the ordinary Ehrhart polynomial theorem is that I am multiplying by a Gaussian integer, not just an ordinary integer.**
The case $N=1$ can be studied using [Pick's theorem](http://en.wikipedia.org/wiki/Pick%27s_theorem). The area of $P$ transforms simply under multiplication by Gaussian integers, the number of lattice points on the boundary is only a little messier. In particular, it is easy to show that there is some non-zero Gaussian integer $D$ such that, as long as $GCD(D, u+vi)=1$ and $GCD(u,v)=1$, then $h_P(u+vi) = (u^2+v^2) \mathrm{Area}(P) + \mbox{constant}$.
A result that would satisfy me is that there is some non-zero non-zero Gaussian integer $D$ such that, as long as $GCD(D, u+vi)=1$ and $GCD(u,v)=1$, then $$h_p(u,v) = (u^2+v^2) \mathrm{Area}(P) + a(u,v) u + b(u,v) v + c(u,v)$$ where $a$, $b$ and $c$ are periodic modulo $N$ in each input.
Motivation: This is enough to give an elementary deduction that the [quartic residue](http://en.wikipedia.org/wiki/Quartic_reciprocity#Quartic_residue_character) symbol $\left[ \frac{z}{u+vi} \right]$ is periodic as a function of $u+vi$.
Motivation behind the motivation: There are a series of papers: [Habicht](http://www.ams.org/mathscinet-getitem?mr=113867), [Kubota](http://www.ams.org/mathscinet-getitem?mr=921585), [Hill](http://www.ams.org/mathscinet-getitem?mr=1324545) which promise to prove $m$-th power reciprocity laws roughly by counting lattice points. I can't follow any of them, so I thought I would true to reverse engineer what might be true in order to make the $\mathbb{Z}[i]$ proof come out.
For this reason, I would also be interested in statements for other number fields, but I don't have a precise question worked out in that setting.
| https://mathoverflow.net/questions/165884/is-there-an-ehrhart-polynomial-for-gaussian-integers | [
"nt.number-theory",
"mg.metric-geometry",
"lattices"
] | 30 | 2014-05-11T18:46:47 | [
"Dear David. Sorry for the late question, but, did you eventually found some place (in addition to the articles you mention) where this is explained/worked, hopefully for other number fields?",
"Yeah, right, I didn't think straight. Sorry for noise.",
"I do not see this. A rotation can change the lattice in a q... | 6 | Science | 0 |
87 | mathoverflow | Nontrivial tangent bundle that is diffeomorphic to the trivial bundle | Is there an example of a smooth $n$-manifold $M$ whose tangent bundle is nontrivial as a bundle but is nonetheless (abstractly) diffeomorphic to the trivial bundle $M \times \mathbb{R}^n$?
(This question was inspired by [Trivial fiber bundle](https://mathoverflow.net/questions/58685/trivial-fiber-bundle).)
| https://mathoverflow.net/questions/58793/nontrivial-tangent-bundle-that-is-diffeomorphic-to-the-trivial-bundle | [
"dg.differential-geometry",
"differential-topology"
] | 30 | 2011-03-17T16:25:39 | [
"$T^*G\\cong G\\times \\mathfrak g^*\\cong G\\times \\mathbb R^{n}$",
"@Dylan and @Somnath: Tangent bundles over spheres never give such examples. Moreover, in general, if the homotopy-equivalence $M\\to M$ induced by a diffeomorphism $TM\\to M\\times {\\mathbb R}^n$ is homotopic to a diffeomorphism $M\\to M$, th... | 6 | Science | 0 |
88 | mathoverflow | Is there a field $F$ which is isomorphic to $F(X,Y)$ but not to $F(X)$? | > Is there a field $F$ such that $F \cong F(X,Y)$ as fields, but $F \not \cong F(X)$ as fields?
I know only an example of a field $F$ such that $F$ isomorphic to $F(x,y)$ : this is something like $F=k(x_0,x_1,\dots)$. But in this case we have $F \cong F(x)$.
This is related to [this](https://mathoverflow.net/questions/225896) question. This was previously [asked](https://math.stackexchange.com/questions/2000333) on MSE but after 2 months and a bounty, no answer was provided.
Thank you very much!
| https://mathoverflow.net/questions/258647/is-there-a-field-f-which-is-isomorphic-to-fx-y-but-not-to-fx | [
"ra.rings-and-algebras",
"examples",
"fields"
] | 30 | 2017-01-03T03:25:32 | [
"Oops, I forgot to mention that I've posted a related question here: mathoverflow.net/questions/259117/…",
"I have an example where I know $F\\cong F(X,Y)$ but I don't know whether $F\\cong F(X)$. There are torsion-free abelian groups $G$ such that $G\\cong G\\oplus\\mathbb{Z}\\oplus\\mathbb{Z}$ but $G\\not\\cong... | 3 | Science | 0 |
89 | mathoverflow | The field of fractions of the rational group algebra of a torsion free abelian group | Let $G$ be a torsion free abelian group (infinitely generated to get anything interesting). The group algebra $\mathbb{Q}[G]$ is an integral domain. Let $\mathbb{Q}(G)$ be its field of fractions.
> Are there non-isomorphic torsion free abelian groups $G$ and $H$ such that $\mathbb{Q}(G)\cong\mathbb{Q}(H)$? If so, are there easy examples?
This is motivated by [this question](https://mathoverflow.net/questions/258647/is-there-a-field-f-which-is-isomorphic-to-fx-y-but-not-to-fx). There are [examples](https://mathoverflow.net/questions/218113/a-is-isomorphic-to-a-oplus-mathbbz2-but-not-to-a-oplus-mathbbz) of torsion free abelian groups $G$ where $G\cong G\oplus\mathbb{Z}\oplus\mathbb{Z}$, but $G\not\cong G\oplus\mathbb{Z}$. Since $\mathbb{Q}(G\oplus\mathbb{Z}\oplus\mathbb{Z})\cong\mathbb{Q}(G)(X,Y)$ and $\mathbb{Q}(G\oplus\mathbb{Z})\cong\mathbb{Q}(G)(X)$, a negative answer to my question would give a positive answer to the motivating question.
I've found a little about this construction in the literature, but not much. For example, in
_Gervasio G. Bastos and T. M. Viswanathan_ , MR 942063 [**Torsion-free abelian groups, valuations and twisted group rings**](http://dx.doi.org/10.4153/CMB-1988-021-x), _Canad. Math. Bull._ **31** (1988), no. 2, 139--146.
it is proved that $\mathbb{Q}(G)$ is a purely transcendental extension of $\mathbb{Q}$ if and only if $G$ is free abelian.
Any other relevant references would be welcome. Also, there are obvious variants: e.g., using a different field, such as $\mathbb{F}_2$ instead of $\mathbb{Q}$.
| https://mathoverflow.net/questions/259117/the-field-of-fractions-of-the-rational-group-algebra-of-a-torsion-free-abelian-g | [
"gr.group-theory",
"ac.commutative-algebra",
"fields",
"abelian-groups"
] | 29 | 2017-01-08T13:30:07 | [
"This question could be related.",
"Lovely question Jeremy. I only wish I could say something useful about it!",
"Just a remark: if $G$ has no primitive element (in the sense that for every $g\\in G$ there exists $n\\ge 2$ such that $n^{-1}g\\in G$) then I guess that $\\pm G$ is characterized within $\\mathbb{Q... | 3 | Science | 0 |
90 | mathoverflow | Linking formulas by Euler, Pólya, Nekrasov-Okounkov | Consider the formal product $$F(t,x,z):=\prod_{j=0}^{\infty}(1-tx^j)^{z-1}.$$ (a) If $z=2$ then on the one hand we get Euler's $$F(t,x,2)=\sum_{n\geq0}\frac{(-1)^nx^{\binom{n}2}}{(x;x)_n}t^n,$$ on the other we get Pólya's formula (the "cycle index decomposition", see Stanley's EC2, p.19) $$F(t,x,2)=\sum_{n\geq0}Z(S_n,(1-x)^{-1},\dots,(1-x^n)^{-1})t^n.$$ (b) If $t=x$ then we get Nekrasov-Okounkov's $$F(x,x,z)=\sum_{n\geq0}x^n\sum_{\lambda\vdash n}\prod_{\square\in\lambda}\left(1-\frac{z}{h_{\square}^2}\right);$$ where $h_{\square}$ is the hook-length (for notations and references you may see _[Theorems, problems and conjectures](http://dauns.math.tulane.edu/%7Etamdeberhan/conjectures.html)_).
> **My Question still waiting for an answer.** Is there a unifying combinatorial right-hand side in $$\prod_{j\geq0}(1-tx^j)^{z-1}=?$$
**Remark 1.** One possibility is perhaps a refinement in line with (b) involving hook-lengths.
**Remark 2.** Even special cases, such as evaluation for specific numerical values of $t$ and/or $z$, would be interesting.
**Remark 3.** Yet another direction is to find a hook-length expansion for $F(t,x,2)$. This could be fascinating because the result will connect the cycle index polynomial with hooks.
| https://mathoverflow.net/questions/251068/linking-formulas-by-euler-p%c3%b3lya-nekrasov-okounkov | [
"nt.number-theory",
"co.combinatorics",
"rt.representation-theory",
"power-series",
"gauge-theory"
] | 29 | 2016-09-29T21:18:19 | [
"Neither one follows from the other.",
"never mind -- NO formula generalizes the cycle index",
"does Nekrasov-Okounkov formula really fall out of cycle index for permutation groups??"
] | 3 | Science | 0 |
91 | mathoverflow | Non-linear expanders? | Recall that a family of graphs (indexed by an infinite set, such as the primes, say) is called an _expander family_ if there is a $\delta>0$ such that, on every graph in the family, the discrete Laplacian (or the adjacency matrix) has spectral gap $|\lambda_0-\lambda_1|\geq \delta$. (Assume all graphs in the family have the same degree (= valency) $d$.)
If, for every $p$, we specify some maps from $\mathbb{Z}/p\mathbb{Z}$ to itself, then we are defining a graph with vertex set $\mathbb{Z}/p\mathbb{Z}$ for each $p$: a vertex is adjacent to the vertices it is mapped to.
If we consider the maps $x\to x+1$, $x\to 3x$ (say), then we do not get expanders. On the other hand, if we take $x\to x+1$, $x\to x^{-1}$, then, as is widely known, we do get an expander family.
Consider the maps
$x\mapsto x+1$, $x\mapsto x^3$.
Do they (and their inverses, if you wish) give an expander family? (Let $p$ range only over primes $p\equiv 2 \mod 3$, so as to keep the map $x\mapsto x^3$ injective.)
What about the maps $x\mapsto x+1$, $x\mapsto 3 x$, $x\mapsto x^3$? Do they, taken together, give an expander family?
(Is there a way to relate these maps to the action of a linear group? Are there examples of sets of not neessarily linear maps giving rise to expander families?)
UPDATE: What if these were shown _not_ to be expanders? What would be some interesting consequences?
| https://mathoverflow.net/questions/168426/non-linear-expanders | [
"nt.number-theory",
"gr.group-theory",
"graph-theory"
] | 29 | 2014-05-28T07:17:01 | [
"Can you say anything on the structure of those $A\\subset{\\mathbb F}_p$ with $|(3A)\\setminus(A+1)|<c|A|$ (or smaller than $|A|^\\delta$ at least)? If so, it might be possible to use this information to show that such $A$ cannot be stable under the map $x\\mapsto x^3$.",
"Yes, as I said it should do something l... | 10 | Science | 0 |
92 | mathoverflow | Can every 3-dimensional convex body be trapped in a tetrahedral cage? | Can every 3-dimensional convex body be trapped in a tetrahedral cage?
Although the question is fairly unambiguous, I give all relevant definitions:
$\bullet$ A subset $C$ of $\mathbb{R}^n$ is an $n$-dimensional _**convex body**_ if $C$ is convex, compact, and has non-empty interior.
$\bullet$ A _**polyhedral cage**_ $P^{(1)}$ in $\mathbb{R}^3$ is the union of all edges (_i.e._ , the 1-skeleton) of a convex $3$-dimensional polyhedron $P$. In particular, a _**tetrahedral cage**_ is the union of the six edges of some tetrahedron.
$\bullet$ A convex $3$-dimensional body $C$ is _**trapped**_ by the tetrahedral cage $T^{(1)}$, that is, by the $1$-skeleton of the tetrahedron $T$, if the cage is fixed (motionless), and if for every continuous rigid motion (rotations allowed) $f_t(C);\ 0\le t\le 1$, either $f_t(C)$ intersects $T$ for every $t\in [0,1]$ or $T^{(1)}$ contains an interior point of $f_t(C)$ for some $t_0\in [0,1]$.
In other words, $C$ cannot be moved arbitrarily far from $T$ while, during the entire motion, avoiding the cage's bars penetrating $C$'s interior.
$\bullet$ Remark. In dimension $n>3$, one can consider analogous questions, with a variety of types of a simplicial cage, by taking the $i$-skeleton of a convex $n$-dimensional simplex, with $1\le i\le n-2$.
A trivial example: A ball is trapped in the cage consisting of the $1$-skeleton of the regular tetrahedron edge-tangent to the ball. Also, obviously, every convex body can be trapped in some polyhedral cage.
For an interested reader, I suggest a few of somewhat less trivial exercises: each of the following convex bodies can be trapped in a tetrahedral cage: the cube, the circular cylinder of any (finite) height, the circular cone, the regular tetrahedron.
| https://mathoverflow.net/questions/379560/can-every-3-dimensional-convex-body-be-trapped-in-a-tetrahedral-cage | [
"mg.metric-geometry",
"convex-geometry"
] | 28 | 2020-12-22T20:17:05 | [
"@Anton The half-circle, for example, and many other, related.",
"In 2-dimensional case, one can ask which convex figures can be trapped by three nails. Rectangle cannot be trapped, but I do not see other examples.",
"@DanRomik Romik : Yes, it is quite a bit different.",
"The paper How to cage an egg by Oded... | 6 | Science | 0 |
93 | mathoverflow | Mathieu group $M_{23}$ as an algebraic group via additive polynomials | An elegant description of the Mathieu group $M_{23}$ is the following: Let $C$ be the multiplicative subgroup of order $23$ in the field $F=\mathbb F_{2^{11}}$ with $2^{11}$ elements. Then $M_{23}$ is the group of additive maps of $F$ to itself which permute the set $C$. The restriction to $C$ then is the natural action of $M_{23}$ of degree $23$.
Any additive map of $F$ to itself has the form $x\mapsto\sum_{i=0}^{10}a_ix^{2^i}$ for $a_i\in F$. The condition that such maps preserve $C$ can be expressed easily as an equivalent set of polynomial conditions on the coefficients $a_0,\dots,a_{10}$. For instance, for a variable $t$, compare coefficients of $$\prod_{c\in C}(t-\sum_{i=0}^{10}a_ic^{2^i})=t^{23}-1.$$ In addition, we need the equations $a_i^{2^{11}}=a_i$ to ensure that the $a_i$ are in $F$.
However, the polynomials we obtain this way (or some other way, there are somewhat smarter possibilities) are extremely complicated.
Of course, any finite subgroup of a linear group is an algebraic set, that is a solution of a system of polynomial equations. And in general such a system will be messy.
Here, I somehow expect that there should be a simple system of polynomials discribing $M_{23}$ as an algebraic set. The obvious attempt would be to first take a system as above, and then hope that a Groebner basis looks better. After some naive attempts it seems that the systems are too complex for Magma or Singular to compute the Groebner bases.
Before trying to refine the approach, here is my question: Has anyone seen this version to describe $M_{23}$?
| https://mathoverflow.net/questions/264513/mathieu-group-m-23-as-an-algebraic-group-via-additive-polynomials | [
"ag.algebraic-geometry",
"gr.group-theory",
"polynomials",
"finite-groups"
] | 28 | 2017-03-13T06:48:14 | [
"@DaveBenson Indeed, I had also tried something along this idea. The upper bound of the Galois group of $Y^{23}+XY^3+1$ is achieved by showing that this polynomial divides an additive polynomial of degree $2^{11}$. So actually this additive polynomial has Galois group $M_{23}$, and its groups acts on an $11$-dimens... | 14 | Science | 0 |
94 | mathoverflow | Derivative of Class number of real quadratic fields | Let $\Delta$ be a fundamental quadratic discriminant, set $N = |\Delta|$, and define the Fekete polynomials $$ F_N(X) = \sum_{a=1}^N \Big(\frac{\Delta}a\Big) X^a. $$ Define $$ f_N(X) = \frac{F_N(X)}{X^N-1}. $$ Canceling common factors it can be shown that $$ f_N(X) = \frac{H_N(X)}{\Phi_N(X)}, $$ where $\Phi_N$ is the $N$-th cyclotomic polynomial and $H$ has integral coefficients. (These definitions seem rather ad hoc, but in fact the objects above are all quite "natural".)
If $\Delta < 0$, then $f_N(1) = \frac{2h}w$, where $h$ is the class number of $\Delta$ and $w$ the number of roots of unity in the corresponding quadratic number fields. In fact we get $$ f_N(1) = - \frac{\sum_{a=1}^{N-1} a \Big( \frac{\Delta}a\Big)}N . $$
If $\Delta > 0$, then $f_N(1) = 0$, which suggests looking at the first derivative. Here we find that $f_N'(1)$ is always even, so we set $f_N'(1) = - 2 h$ and get $$ h = \frac{\sum_{a=1}^{N-1} \frac{a(a-1)}2 \Big( \frac{\Delta}a\Big)}N . $$ Here is a small table:
$ $ $ \begin{array} {c|ccccccccccc} N & 12 & 13 & 17 & 21 & 24 & 28 & 29 & 33 & 37 & 41 & 44 \\\ \hline h & 1 & 1 & 2 & 2 & 3 & 4 & 3 & 6 & 5 & 8 & 7 \end{array} $
It is quite remarkable that $h$ seems to possess, for positive discriminants, properties analogous to class numbers. In particular, $h$ is odd if and only if $\Delta = p > 5$ is a prime $\equiv 5 \bmod 8$, and if $\Delta$ is divisible by $n$ distinct primes, then $h$ is divisible by $2^{n-2}$ etc.
In the case $\Delta < 0$, Hasse and Bergstrom have shown how to extract the "genus factor" from the analytic class number formula, but the necessary calculations were horrible. Thus if we want to avoid going through similar technicalities in the case $\Delta > 0$ then we would need a natural interpretation of $h$ as the order of an analogue of a "class group".
So I guess my question is whether there is such a mysterious group (connected to real quadratic number fields) with order $h$. Even a proof that $h \ne 0$ would be welcome, since in the case of negative discriminants this is equivalent to the nonvanishing of Dirichlet's L-function at $s = 1$.
**Edit.** As Rene Schoof has pointed out, the numbers in question are essentially generalized Bernoulli numbers $B_{2,\chi}$ for (quadratic) Dirichlet characters with conductor $N$. These numbers are defined by $$ B_{2,\chi} = \sum_{a=1}^N \chi(a) B_2(a/N), $$ where $B_2(X) = X^2 - X + \frac16$ is the second Bernoulli polynomial. The constant term $\frac 16$, in which this expression differs from the one I have used above, is irrelevant since $\sum \chi(a) = 0$.
These numbers $B_{2,\chi}$ show up as factors of eigenspaces of cuspidal divisor class groups as studied in the book by Kubert-Lang (more exactly, the $p$-part of the group is reflected in the divisibility by $p$ of the Bernoulli numbers). The Birch-Tate-Lichtenbaum conjecture also predicts that the $B_{2,\chi}$ are factors of $K_2({\mathcal O})$, where ${\mathcal O}$ is the ring of integers of the maximal real subfield of the $N$-th roots of unity (Kubert-Lang, p. 151).
Although these numbers have such a prominent place in number theory, their elementary properties apparently have not been investigated (I've looked into Washington's and Lang's books on cyclotomic number fields). Numerical experiments suggest e.g. the following:
Let $\chi$ be an even Dirichlet character with conductor $N$. Then $h_2(N) = \frac{N}4 B_{2,\chi}$ is an integer for all $N > 8$.
Moreover, $h_2(N)$ is odd if and only if
1. $N = p$ for primes $p \not \equiv 1 \bmod 8$;
2. $N = 4p$ for primes $p \equiv 3 \bmod 8$;
3. $N = 8p$ for odd primes $p \not \equiv \pm 1 \bmod 8$.
In absence of any kind of genus theory for cuspidal divisor class groups it seems that the only option for proving such results is working directly with the definition of these numbers. As mentioned above, this becomes technical for integers with many prime factors.
There are similar results for $h_3(N) = \frac{N^2}6 B_{3,\chi}$, and there can be no doubt that most of this generalizes in some form to all generalized Bernoulli numbers $B_{n,\chi}$.
This whole set of question came out of my approach to quadratic reciprocity in Chapter 7 of my notes on [Pell conics](http://www.rzuser.uni-heidelberg.de/~hb3/pell.html).
| https://mathoverflow.net/questions/114434/derivative-of-class-number-of-real-quadratic-fields | [
"nt.number-theory",
"algebraic-number-theory"
] | 28 | 2012-11-25T10:14:45 | [
"Hi! I think these are generalized Bernoulli numbers $B_{2,\\chi}$ associated the quadratic character $\\chi$ of conductor $\\Delta$. Apart from a power of $2$ perhaps, they are equal to the order of the $\\chi$-part of the cuspidal divisor class group of the modular curve $X_1(\\Delta)$. See the book by Kubert-La... | 1 | Science | 0 |
95 | mathoverflow | derived category of equivariant coherent sheaves and fixed points | The K-group $K^T(X)$ of $T$(torus)-equivariant coherent sheaves on a variety $X$ is isomorphic to $K^T(X^T)$, that of the fixed point locus via the inclusion homomorphism, when we tensor the quotient field of the representation ring $R(T)$ of $T$.
Is there a similar result for $D_T(Coh X)$ and $D_T(Coh X^T)$, derived categories of $T$-equivariant coherent sheaves ? I do not know even how to formulate `the quotient field of $R(T)$' in the derived category case.
| https://mathoverflow.net/questions/48407/derived-category-of-equivariant-coherent-sheaves-and-fixed-points | [
"kt.k-theory-and-homology",
"ag.algebraic-geometry"
] | 28 | 2010-12-05T16:09:13 | [
"@Ben, you're absolutely right, somehow I missed the salient word \"coherent\" in the question. But even for constructible sheaves, is there a localization statement at the level of derived categories?",
"It doesn't seem like that to me at all. Is there anything about coherent sheaves in GKM?",
"This seems li... | 3 | Science | 0 |
96 | mathoverflow | Computational complexity of topological K-theory | I am a novice with K-theory trying to understand what is and what is not possible.
Given a finite simplicial complex $X$, there of course elementary ways to quickly compute the cohomology of $X$ with field coefficients. However, it is known that computing the rational homotopy groups of $X$ is at least $NP$-hard (in fact, at least $\\# P$-hard), simply because it is possible to reduce NP-hard problems to the computations of some rational homotopy groups.
For any of: (1) Real K-theory, (2) Complex K-theory, (3) p-adically completed K-theory, is there an algorithm to compute $K^0$ of a finite simplicial complex? Is this algorithm polynomial-time?
If there is no known algorithm, is there at least evidence that any theoretical algorithm, should it exist, must be at least $NP$-hard? In other words, is there any way to reduce an $NP$-hard problem to the calculation of some $K$ group?
| https://mathoverflow.net/questions/125745/computational-complexity-of-topological-k-theory | [
"kt.k-theory-and-homology",
"at.algebraic-topology",
"computational-complexity"
] | 28 | 2013-03-27T10:15:43 | [
"Just thought I'd add a citation for the fact that rational homotopy groups of simply-connected spaces are NP hard here. In particular, this is not just some $\\pi_1$ issue.",
"Link: jointmathematicsmeetings.org/meetings/national/jmm2012/…",
"Have you looked at Gunnar Carlsson's stuff? They have this software f... | 8 | Science | 0 |
97 | mathoverflow | Finite-dimensional subalgebras of $C^\star$-algebras | Let $A$ be a unital $C^\star$-algebra and let $a_1,\dots,a_n$ be a finite list of normal elements in $A$ which (together with their adjoints) generate a norm-dense $\star$-subalgebra $B \subset A$. Clearly, if $A$ is finite-dimensional, then every element in $A$ (and hence $B$) has finite spectrum. I am asking for the converse.
> **Question:** Assume that each element of $B$ has finite spectrum. Is it true that $A$ has to be finite-dimensional?
The existence of finitely-generated infinite torsion groups shows that this might be a highly non-trivial problem. In this case one would consider the reduced group $C^\star$-algebra and note that all monomials in the generators of the group and their inverses (which are equal to the adjoints of the generators) would have finite spectrum. However, the generated algebra would still be infinite-dimensional. I do not know of any simpler way to come up with such an example. In this case it is conceivable that the random-walk operator associated with the generating set (which is an element in the real group ring) has infinite spectrum, even though I did not prove this.
Maybe there is also need to consider the spectra of elements in matrices over $B$ (which of course follow to be finite if $A$ is finite-dimensional.) In view of this, I am not only asking for an answer to the question but also for the right question (or a better one) if the answer to the original question is negative.
A stronger assumption would be to assume that $A$ itself consists only of elements with finite spectrum. This case seems much easier to approach and the answer seems to be positive. In fact every infinite-dimensional $C^\star$-algebra should contain an element with infinite spectrum.
Just to get started, a more concrete instance of the question above is:
> **Question:** Let $p_1, p_2$ and $p_3$ be three projections in a $C^*$-algebra with the property that every (non-commutative) polynomial in $p_1, p_2$ and $p_3$ has finite spectrum. Is it true that the projections generate a finite-dimensional algebra?
| https://mathoverflow.net/questions/35207/finite-dimensional-subalgebras-of-c-star-algebras | [
"oa.operator-algebras",
"c-star-algebras",
"von-neumann-algebras",
"open-problems"
] | 28 | 2010-08-11T02:49:03 | [
"Every finitely generated *-algebra is generated by a finite list of normal elements (the real and imaginary parts of its generators) so the assumption that the $a_i$ are normal is superfluous.",
"@AliTaghavi: Yes, in this case, the spectrum of $A$ is equal to the joint spectrum of $a_1,\\dots,a_n$ (which is cont... | 10 | Science | 0 |
98 | mathoverflow | Can one divide by the cardinal of an amorphous set? | This question arose in a discussion with Peter Doyle.
It is provable in ZF that one can divide by any positive finite cardinal $k$: if $X \times \\{1,\ldots,k\\} \simeq Y \times \\{1,\ldots,k\\}$ then $X \simeq Y,$ where I use $\simeq$ to denote the existence of a bijection between two sets.
Assuming the Axiom of Choice, the positive finite cardinals are precisely the cardinals that one can divide by since $1\cdot\mathfrak{m} = 2\cdot\mathfrak{m}$ for any infinite cardinal $\mathfrak{m}$. The ultimate question is whether this is actually provable in ZF:
> If $\mathfrak{m}$ is an infinite cardinal, must there be two cardinals $\mathfrak{p} \neq \mathfrak{q}$ such that $\mathfrak{p}\cdot\mathfrak{m} = \mathfrak{q}\cdot\mathfrak{m}$?
As a candidate for a counterexample, I thought it might be possible to divide by the cardinal of a (strongly) amorphous set. Specifically, in the First Fraenkel model:
> Is it true that if $X \times A \simeq Y \times A$ then $X \simeq Y$, where $A$ is the set of atoms?
| https://mathoverflow.net/questions/235798/can-one-divide-by-the-cardinal-of-an-amorphous-set | [
"lo.logic",
"set-theory",
"axiom-of-choice"
] | 28 | 2016-04-09T15:05:50 | [
"@Lorenzo: A hint would imply that this is easy... it's not! Peter Doyle has written up division by three and division by four - math.dartmouth.edu/~doyle",
"Can someone give me a hint on how to prove that in $\\text{ZF}$ we can divide by finite cardinals?",
"I like this question!",
"There is Lovasz's unique ... | 4 | Science | 0 |
99 | mathoverflow | Unital $C^{*}$ algebras whose all elements have path connected spectrum | A unital $C^{*}$ algebra is called a "Path connected algebra" if the spectrum of all its elements is a path connected subset of $\mathbb{C}$.
> **What is an example of a non commutative path connected $C^*$ algebra? What is an example of a simple path connected $C^{*}$ algebra? In particular does $C^{*}_{red} F_{2}$ satisfiy this property?**
>
> As another question: Is the tensor product of two path connected algebra, a path connected algebra?(For spatial norm).
| https://mathoverflow.net/questions/195900/unital-c-algebras-whose-all-elements-have-path-connected-spectrum | [
"fa.functional-analysis",
"oa.operator-algebras",
"operator-theory",
"sp.spectral-theory",
"c-star-algebras"
] | 28 | 2015-02-06T16:24:12 | [
"@JohannesHahn Thanks for your comment. As you pointed out a minimal dynamical system gives a simple C* algebra. But what about the spectrum of its elements(Path connected spectrum)?",
"@HannesThiel Thank you very much for your comment and your attention. The motivation for the second part of my que... | 13 | Science | 0 |
100 | mathoverflow | A question on simultaneous conjugation of permutations | Given $a,b\in S_n$ such that their commutator has at least $n-4$ fixed points, is there an element $z\in S_n$ such that $a^z=a^{-1}$, and $b^z=b^{-1}$? Here $a^z:=z^{-1}az$.
Magma says that the answer is yes up to $n\leq 11$ (computed by M. Zieve). The answer is also yes if $a$ and $b$ commute via the argument below. It would be great to have an answer in case a,b do not commute, even under the assumption that $\langle a,b\rangle$ contains the alternating group $A_n$. If $\langle a,b\rangle$ contains $A_n$ with $n\geq 4$, then any $z$ as above is unique.
The commutative case: If $a^b=a$, the orbits (and cycles) of $a$ are permuted by $b$. Say $b$ permutes under conjugation cyclicly the orbits $A_1,A_2,\ldots,A_r$ of $a$. Let $a_i\in S_n$ be the cycle induced by $a$ on $A_i$, for $i=1,\ldots,r$. Choose $c\in S_n$ to be element of order $r$ such that $b=c$ on $A_i$ for $i=1,...,r-1$. (This forces $a_r^c=a_1$). Since $b=c$ outside $A_1$ and $a_1^{bc^{-1}}=a_1$, we have $b=a_1^e c$ for some integer $e$.
Choose $v$ to be an involution on $A_1$ such that $a_1^v=a_1^{-1}$ on $A_1$, and trivial outside $A_1$. Choose $w$ to be the involution in the symmetric group on $1,\ldots,r$ such that $(1,\ldots,r)^w=(r,\ldots,1)$ and $1^w=2$. Define $u$ to be a permutation such that if $(i,j)$, $i\leq j$, is an orbit of $w$, then $\alpha^u = \alpha^{c^{j-i}}$ for every $\alpha\in A_i$, $i=1,\ldots,r$. This way $c^u =c^{-1}$. Let $z:=vv^c\cdots v^{c^{r-1}}u$, so that $c^z=c^{-1}$, and $a^z=a^{-1}$. (Moreover, $a_i^z=a_{i^w}^{-1}$ for $i=1,\ldots,r$). It follows that $b^z=a_{1^w}^{-e}c^{-1} = a_2^{-e}c^{-1}=c^{-1}a_1^{-e} = b^{-1}$.
Some might consider this proof (in the commutative case) to be simpler in the language of wreath products.
| https://mathoverflow.net/questions/267191/a-question-on-simultaneous-conjugation-of-permutations | [
"co.combinatorics",
"gr.group-theory",
"permutations",
"algebraic-combinatorics"
] | 27 | 2017-04-14T04:11:08 | [
"For the record: This is proved in Theorem 4.9 of arXiv:1810.08971v4 by Junyao Pan. (I have not checked the proof.)",
"Consider the polygonal mesh with vertices $\\{1,\\dots,n\\}$, directed edges $i\\to a(i)$ and $i\\to b(i)$ for each $i$, and for each (possibly trivial) cycle $(k\\;\\cdots\\;k+m-1)$ of $[a,b]$ a... | 7 | Science | 0 |
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