Skip to main content

All Questions

Filter by
Sorted by
Tagged with
9 votes
0 answers
420 views

Hochschild-Serre spectral sequence via explicit filtration

Let $$1 \longrightarrow K \longrightarrow G \longrightarrow Q \longrightarrow 1$$ be a short exact sequence of groups and let $M$ be a $\mathbb{Z}[G]$-module. The Hochschild--Serre spectral ...
Laura's user avatar
  • 163
9 votes
0 answers
439 views

(Torsion in) homology of free nilpotent groups

It is known that for free $k$-step nilpotent group on $r$ generators $N(r, k)$ its integral homology is torsion-free in degrees $\leq 3$ (obvious for 1 and 2, Igusa&Orr computations for 3). ...
Denis T's user avatar
  • 4,600
9 votes
0 answers
376 views

Explicit description of a subgroup of the braid group $\mathsf{B}_2(C_2)$

This is related to my previous MathOverflow question Fundamental group of $\mathrm{Sym}^2(C_g)$ minus the diagonal. Let $C_2$ be a smooth curve of genus $2$ and $X:=\mathrm{Sym}^2(C_2)$ its second ...
Francesco Polizzi's user avatar
8 votes
1 answer
672 views

Can the Kan-Thurston theorem be turned into some kind of equivalence between groups and spaces?

I not really familiar with these subjects. I read this question and I was really surprised by the answer. My question is probably vague (so please do bear with me). The cited question/answer ...
google's user avatar
  • 277
8 votes
2 answers
632 views

Are torus knot groups linear?

The fundamental group $T(p,q)$ of the complement of a $(p,q)$-torus knot (in $S^3$) admits the presentation $\langle a, b \mid a^p=b^q \rangle $. Is $T(p,q)$ linear, i.e., is there a faithful ...
user41793's user avatar
8 votes
2 answers
507 views

Contractible Rips complex from non-hyperbolic group

I heard that the Rips complexes associated to the Cayley graphs of hyperbolic groups are contractible for a sufficiently large radius. Is the converse true? Namely, if a group is non-hyperbolic, then ...
Uzu Lim's user avatar
  • 903
8 votes
1 answer
1k views

Almost-direct product and 1-formality

Let $G$ be a finitely presented group. To $G$ is associated in a functorial way a Malcev Lie algebra which can be constructed in several equivalent ways. Roughly speaking, it is the quotient of the ...
Adrien's user avatar
  • 8,524
8 votes
2 answers
596 views

Infinite loop space maps into or out of BAut(F_n)

There is an inclusion $S_n \to Aut(F_n)$ from the symmetric group into the automorphism group of a free group. After applying the Quillen +-constriction, both $BS_{\infty}$ and $BF_{\infty}$ become ...
Romeo's user avatar
  • 2,734
8 votes
1 answer
1k views

Beyond an intro to topological graph theory...

I'm looking to find out what active areas of research there are in topological graph theory, particularly those that interface strongly with other areas of math (say, group theory, algebraic topology, ...
Dr Shello's user avatar
  • 1,180
8 votes
1 answer
382 views

Second homology of mapping class group of genus 3

In a survey paper of Korkmaz it is stated that $H_2(\mathrm{Mod}_3)$ is either $\Bbb Z$ or $\Bbb Z \oplus \Bbb Z_2$, but I was not able to find out a precise computation of this group (resolving the ...
Daniele Zuddas's user avatar
8 votes
1 answer
414 views

Augmentation ideal of a free group

If $F$ is a free group then it has cohomological dimension one, which implies that the augmentation ideal $IF=\operatorname{ker}(\epsilon:\mathbb{Z}G\to \mathbb{Z})$ of its group ring is a projective $...
Mark Grant's user avatar
  • 35.9k
8 votes
1 answer
420 views

Can we algorithmically contract loops in a simply connected space?

It is well-known that the question whether a given connected simplicial complex (or simplicial set) is simply connected, is algorithmically undecidable as it can model the word problem. Assuming ...
Peter Franek's user avatar
8 votes
1 answer
739 views

Transgressions commute with the Steenrod operations on the base and fiber in a central group extension?

The following sentence is quoted from the paper ON THE COHOMOLOGY OF SPLIT EXTENSIONS by D. J. BENSON AND M. FESHBACH: In general, the differentials in the Lyndon-Hochschild-Serre spectral sequence ...
Minghui's user avatar
  • 83
8 votes
1 answer
387 views

Outer automorphism group of Brieskorn homology sphere?

In this post, it is discussed how a Brieskorn homology sphere $\Sigma(a_1,a_2,a_3)$ with $\displaystyle \frac{1}{a_1}+ \frac{1}{a_2}+ \frac{1}{a_3} < 1$ is an aspherical manifold with a ...
Jeffrey Rolland's user avatar
8 votes
3 answers
628 views

A question on Hawaiian earring group

I have asked this question in MSE but have not got any satisfactory answer, so I am asking it here. Any idea on how to approach this problem will be highly appreciated. Consider the Hawaiian earring. ...
Pritam Majumder's user avatar
8 votes
2 answers
594 views

Finiteness of non-abelian cohomology

Let $H$ be a finite group acting on a group $G$. Are there general conditions one can place on $G$ so that the set $H^1(H, G)$ is guaranteed to be finite? For instance, does finiteness hold if $G$ ...
Winnie's user avatar
  • 183
8 votes
1 answer
532 views

Localizations of non-nilpotent spaces

For simplicity let's talk about $p$-localizations of spaces for a fixed prime $p$. Every space $X$ has a well-defined $p$-localization which can be constructed by the small object argument and which ...
Karol Szumiło's user avatar
8 votes
1 answer
513 views

Equivariant (co)homology of flag manifolds, convolution algebra and nil hecke algebra?

For a complex reductive group $G$ and its Borel subgroup $B$, it seems to be well-known that the equivariant homology group $H^G_*(G/B\times G/B)$ forms a nil-Heck algebra $$NH=\Bbbk[y_i,\partial_{j}]...
Cubic Bear's user avatar
8 votes
1 answer
519 views

The extension class of a finite Heisenberg group

Let $\mathbb{K}$ be a field of characteristic $\neq 2$ and let $(V, \omega)$ be a symplectic vector space. Then the Heisenberg group $\mathsf{Heis}(V, \, \omega)$ is the central extension of the ...
Francesco Polizzi's user avatar
8 votes
0 answers
128 views

What are the stable cohomology classes of the "orthogonal groups" of finite abelian groups?

Let $A$ be a finite abelian group, and equip it with a nondegenerate symmetric bilinear form $\langle,\rangle : A \times A \to \mathrm{U}(1)$. Then you can reasonably talk about the "orthogonal ...
Theo Johnson-Freyd's user avatar
8 votes
0 answers
238 views

Is there a finite group with nontrivial $H^2$ but vanishing $H^4$, $H^5$, and $H^6$?

Is there a finite group $G$ such that the group cohomology $\mathrm{H}^2_{\mathrm{gp}}(G; \mathbb{Z}/2)$ is nontrivial but $\mathrm{H}^4_{\mathrm{gp}}(G; \mathbb{Z}/2)$, $\mathrm{H}^5_{\mathrm{gp}}(G;...
Theo Johnson-Freyd's user avatar
8 votes
0 answers
285 views

Are the braid groups good in the sense of Toën?

In this question it is asked whether the mapping class groups are 'good' in relation to pro-finite completion. Helpfully, one of the answers gives a link to a proof that the braid groups are good ...
Patrick Elliott's user avatar
8 votes
0 answers
204 views

Relationship between the p-radical subgroups and the parabolics in a BN-pair generality

A theorem of Quillen says that if $G$ is a finite Chevalley group over characteristic $p$, then the poset $\mathcal{A}_p(G)$ of nontrivial elementary abelian subgroups of $G$ is homotopy equivalent (I ...
Cihan's user avatar
  • 1,726
7 votes
2 answers
957 views

Computations in group cohomology

Hello, Given a finitely presentable group $G$, I'm interested in the cup-product from $H^1$ to $H^2$ with real coefficients. I want to know if this is explicitly computable (with a computer) with a ...
mister_jones's user avatar
7 votes
3 answers
586 views

Dimension of classifying space of a group

If $N$ is a normal subgroup of a group $G$ such that $G/N= \mathbb{Z}$. Suppose that the classifying space of $G$ is a finite CW-complex of dimension $n$. Does it follow that the classifying space of $...
lab's user avatar
  • 451
7 votes
3 answers
1k views

Is it possible to construct K(G, 1) out of a subgroup and its quotient?

Suppose we have a short exact sequence $1 \to K \to G \to Q \to 1$ and let $X_K$ and $X_Q$ be $K(K, 1)$ and $K(Q, 1)$, respectively. Is it possible to construct (in a "natural" way) a model of $K(G, 1)...
Janusz Przewocki's user avatar
7 votes
2 answers
494 views

How does the Steenrod algebra act on $\mathrm{H}^\bullet(p^{1+2}_+, \mathbb{F}_p)$?

Let $p$ be an odd prime. The $\mathbb F_p$ cohomology of the cyclic group of order $p$ is well-known: $\mathrm{H}^\bullet(C_p, \mathbb F_p) = \mathbb F_p[\xi,x]$ where $\xi$ has degree 1, $x$ has ...
Theo Johnson-Freyd's user avatar
7 votes
1 answer
2k views

Automorphism group of the special unitary group $SU(N)$

Let us consider the automorphism group of the special unitary group $G=SU(N)$. We know there is an exact sequence: $$ 0 \to \text{Inn}(G) \to \text{Aut}(G) \to \text{Out}(G) \to 0. $$ For $G=SU(2)...
annie marie cœur's user avatar
7 votes
2 answers
860 views

mapping space between classifying spaces

I wanted to ask a summary of known results and references about the homotopy type of the mapping space $\mathrm{Map}(BG,BK)$ (and specially the connected components) between the classifying spaces ...
Ilias A.'s user avatar
  • 1,974
7 votes
2 answers
537 views

Residually finite + torsion free + finite index = finite complex?

Suppose $G$ is a residually-finite group and $H < G$ a torsion-free subgroup of finite index. What characterizes such $G$ such that $BH$ is homotopic to a finite complex? I believe Serre showed ...
Romeo's user avatar
  • 2,734
7 votes
2 answers
376 views

are there finite nonabelian characteristic quotients $G$ of $F_2$ inducing a surjection $Aut(F_2)\twoheadrightarrow Aut(G)$?

Let $F_2$ be the free group of rank 2. Let $K\le F_2$ be a characteristic subgroup, such that $G := F_2/K$ is finite. Do there exist examples of such nonabelian $G$ such that the induced map $$Aut(...
stupid_question_bot's user avatar
7 votes
1 answer
463 views

Cohomology of the mapping class group of a non-orientable surface?

What is the low degree cohomology of the mapping class group of a non-orientable surface? More specifically, what is the universal central extension of the mapping class group of a non-orientable ...
Kevin Walker's user avatar
  • 12.8k
7 votes
2 answers
676 views

Is the Euler characteristic of aspherical connected 2-complexes at most 1? (No!) What can be said about subcomplexes of 2-complexes deformation retractible onto graphs.

I have several related questions, i do not know which one is more important to me, i think it would depend on their answers. Is it true that the Euler characteristic of a finite connected aspherical ...
Alexey Muranov's user avatar
7 votes
1 answer
672 views

How does this geometric description of the structure of PSL(2, Z) actually work?

There is a beautiful way to see that the congruence subgroup $\Gamma(2)$ is free on two generators: the action of $\Gamma(2)$ on $\mathbb{H}$ is free and properly discontinuous, and there is a modular ...
Qiaochu Yuan's user avatar
7 votes
2 answers
539 views

Injectivity of the cohomology map induced by some projection map

Given a (compact) Lie group $G$, persumably disconnected, there exists a short exact sequence $$1\rightarrow G_c\rightarrow G\rightarrow G/G_c\rightarrow 1$$ where $G_c$ is the normal subgroup which ...
Ye Weicheng's user avatar
7 votes
1 answer
1k views

Classifying space of semidirect product of groups

Assume that $G$ and $H$ are two groups and $G\rtimes _\phi H$ is their semidirect product. My question is, how does the classifying space $B(G\rtimes_\phi H)$ of $G\rtimes _\phi H$ relate to $BG$ and $...
Totoro's user avatar
  • 2,535
7 votes
2 answers
672 views

cohomological dimension of a group acting on a product

I recently came across an interesting result of Kobayashi [Corollary 5.5], a special case of which is the following: Suppose $\Gamma$ is a discrete torsion free subgroup of $SL_n(\mathbb{R})$ which ...
Andrew Zimmer's user avatar
7 votes
1 answer
288 views

A finitely presented group whose rational cohomology is not nilpotent

Does there exist a finitely presented (preferably $\text{FP}_{\infty}$) group $\Gamma$ and an element $\alpha \in \text{H}^{\ast>0}(B\Gamma;\mathbf{Q})$ that is not nilpotent? If non-discrete ...
Jens Reinhold's user avatar
7 votes
1 answer
519 views

When do covering spaces correspond to characteristic subgroups?

Given a covering space $p \colon X \to Y$, we get an injection $p^* \colon \pi_1(X) \to \pi_1(Y)$, and we know that the image $p^*(\pi_1(X))$ is normal in $\pi_1(Y)$ if an only if $p$ is regular, that ...
Anschel Schaffer-Cohen's user avatar
7 votes
1 answer
197 views

Homology of a limit of semidirect products

Suppose I have two families of groups $A_k$ and $B_k$ indexed by the natural numbers and suppose $B_k$ acts on $A_k$. Suppose there are groups homomorphisms $A_{k+1} \rtimes B_{k+1} \to A_k \rtimes ...
2223's user avatar
  • 111
7 votes
1 answer
506 views

$G$ cocycle split to a coboundary in $J$, via a group extension

Consider a generic nontrivial $d$-cocycle $\omega_d^G \in H^d(G,U(1))$ in the cohomology group of a group $G$ with $U(1)=\mathbb{R}/\mathbb{Z}$ coefficient. In otherwords, here the $d$-cocycle $\...
wonderich's user avatar
  • 10.5k
7 votes
2 answers
590 views

Pairs of Permutations up to Simultaneous Conjugation

The conjugacy classes of $S_n$ are the cycle types since if $\tau = (\dots)(\dots)\dots(\dots)$, the conjugation $\tau \mapsto \sigma \tau \sigma^{-1}$ permutes the labels in the cycles of $\tau$. ...
john mangual's user avatar
  • 22.8k
7 votes
1 answer
415 views

Classifying abelian (but non-central) group extensions using homotopy theory

Let $G$ be a group and let $A$ be an abelian group equipped with an action of $G$. Group extensions $$1 \longrightarrow A \longrightarrow \Gamma \longrightarrow G \longrightarrow 1$$ inducing the ...
Andy Putman's user avatar
  • 44.8k
7 votes
1 answer
859 views

How commutative is Quillen's Plus-Construction?

This question is inspired by this question about the dependence of K-theory on the order of multiplication in the ring. I did not think long about it, so maybe the answer really lies on the surface; ...
Andreas Thom's user avatar
  • 25.5k
7 votes
1 answer
310 views

Homotopy between posets

This is entirely a new area for me and I apologise in advance if the questions are silly. In Quillen's paper "Homotopy properties of the posets of non-trivial $p$-subgroup of a group" (see ...
GURI920826's user avatar
7 votes
1 answer
433 views

Associativity with infinite nesting

I was trying to understand the Eilenberg-Mazur swindle (which I learned about here) especially as it could be used to show that if $A, B$ are compact (topological) $n$-manifolds whose connect sum is $...
Rbega's user avatar
  • 2,299
7 votes
1 answer
663 views

Explicit 2-cocycle from a 2nd cohomology group $H^2[Q_8 \times \mathbb{Z}/2\mathbb{Z}, U(1)]$

I would like to know the explicit expression of 2-cocycle from a 2nd cohomology group $H^2[Q_8 \times \mathbb{Z}/2\mathbb{Z}, U(1)]$ with $U(1)\equiv \mathbb{R}/\mathbb{Z}$ coefficient, or namely $H^2[...
wonderich's user avatar
  • 10.5k
7 votes
0 answers
333 views

Positive instances of the Eilenberg-Ganea conjecture with families

The original Eilenberg-Ganea conjecture, which remains unsettled, can be formulated as: any (discrete) group $G$ of cohomological dimension $\operatorname{cd}(G)=2$ has geometric dimension $\...
Mark Grant's user avatar
  • 35.9k
7 votes
0 answers
422 views

Very frustrated reading a proof of the faithfulness of Artin's representation of braid groups

I am reading BRAID GROUPS, FREE GROUPS, AND THE LOOP SPACE OF THE 2-SPHERE by F.R. Cohen and J. Wu and here is an extract of the paper: (The proof is not finished yet but I am very confused by now.) ...
Zuriel's user avatar
  • 1,108
6 votes
2 answers
765 views

Cup products and the transfer map

Let $G_1$ be a finite-index subgroup of $G_2$. Let $i : H^{\ast}(G_2) \rightarrow H^{\ast}(G_1)$ be the induced map of rings. There is then a transfer homomorphism $\tau : H^{\ast}(G_1) \rightarrow ...
Troy A's user avatar
  • 63