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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
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
3 votes
1 answer
432 views

Relation between conjugacy class, quotient isomorphism class, and signature of Fuchsian groups

Let $\Gamma\le SL(2,\mathbb{Z})$ be a finite index subgroup, not necessarily "congruence". Let $c_4,c_6$ be the number of conjugacy classes of elements of order 4 and 6 respectively, let $c_{-1}$ be ...
stupid_question_bot's user avatar
13 votes
1 answer
218 views

The finiteness criterium $F$ under quasi-isometry

A group $G$ is defined to have $F$ if there exists a finite $K(G,1)$. This property is clearly not invariant under quasi-isometry as one can see from the trivial group and $\mathbb{Z}_2$. My question:...
Sven's user avatar
  • 131
13 votes
0 answers
586 views

Finite groups inside an infinite group with the same homology

Suppose we have a triple of groups $G,H,K$ satisfying the following conditions: $G$ and $H$ are finite groups and $K$ is an infinite group. there exist two monomorphisms $G \rightarrow K \leftarrow H$...
Ilias A.'s user avatar
  • 1,974
24 votes
1 answer
968 views

Groups whose finite index subgroups of fixed index are isomorphic

I am interested in finitely generated groups $G$ that are residually finite and have the following property: For each $d \geq 1$, $G$ has subgroups of finite index $d$, and all such subgroups are ...
Jens Reinhold's user avatar
15 votes
3 answers
926 views

Lower central series quotients in terms of (co)homology

Let $G$ be a group. It is well-known that $H_1(G,\mathbb{Z})=G/[G,G]$. Also (at least up to torsion) $[G,G]/[G,[G,G]]=\Lambda^2H^1(G,\mathbb{Z})/H_2(G,\mathbb{Z})$ as explained, for example, in this ...
SashaP's user avatar
  • 7,377
14 votes
0 answers
414 views

Does the category of G-spectra know G?

I was recently in the situation of having access to the category of $G$-modules (for some group $G$ which I had forgotten), as just a category, i.e. no monoidal structure, together with the forgetful ...
Vivek Shende's user avatar
  • 8,723
4 votes
0 answers
239 views

The homotopy type of the mapping space $Map_{B\rho}(BS^1,BG)$? for $G$ a compact Lie group

Given a homomorphism $\rho:S^1\rightarrow G$ with $G$ a compact Lie group there is an induced map of classifying spaces $B\rho:BS^1\rightarrow BG$. What is known about the homotopy type of the mapping ...
Tyrone's user avatar
  • 5,596
3 votes
0 answers
113 views

Have locally principal crossed homomorphisms been studied?

Take a (multiplicative finite) group $H$ acting on the left (by automorphisms) on an (additive finite) abelian group $A$, and recall that the abelian (additive) group of crossed homomorphisms from $H$ ...
Pablo's user avatar
  • 11.3k
13 votes
2 answers
810 views

Torsion-freeness of two groups with 2 generators and 3 relators and Kaplansky Zero Divisor Conjecture

Let $G_1$ and $G_2$ be the groups with the following presentations: $$G_1=\langle a,b \;|\; (ab)^2=a^{-1}ba^{-1}, (a^{-1}ba^{-1})^2=b^{-2}a, (ba^{-1})^2=a^{-2}b^2 \rangle,$$ $$G_2=\langle a,b \;|\; ...
Alireza Abdollahi's user avatar
14 votes
2 answers
416 views

Schur multiplier of $Sp(2g, \mathbb{Z}/2)$ for $g \geq 3$

This question is about the computation of $H_2(Sp(2g, \mathbb{Z}/2), \mathbb{Z})$, where $Sp(2g, \mathbb{Z}/2)$ is the group of symplectic $2g \times 2g$ matrices over $\mathbb{Z}/2$. With respect to ...
Carmen Rovi's user avatar
3 votes
0 answers
257 views

Braids with an infinite number of strings

Has anyone developed a theory for braids with an infinite number of strings?
Martin Peters's user avatar
5 votes
1 answer
207 views

homological 2 dimensional groups

In a Commentarii Mathematici Helvetici paper by Benno Eckman and Heinz Müller in 1980 (volume 50, pages 510-520) proved that poincaré Duality Groups of dimension 2 with positive first ...
Nicolas Boerger's user avatar
1 vote
0 answers
137 views

Acyclicity of covering space

Suppose we have some 2-dimensional non-aspherical finite CW-complex $K$ with $\pi_1(K)=G$. Is there any sufficient condition on $H\leq G$ (and maybe on the group $G$ itself) which allows to conclude ...
Samarkand's user avatar
  • 1,129
5 votes
1 answer
314 views

abelian and nonabelian parts of Aut($\widehat{F_2}$)

Let $F$ be the free profinite group on two generators. Let $\text{IA}(F) := \ker\left(\text{Aut}(F)\rightarrow GL_2(\widehat{\mathbb{Z}})\right)$, the group of "IA automorphisms" of $F$. (I'm also ...
Will Chen's user avatar
  • 10.7k
3 votes
1 answer
267 views

Homology of solvable Lie groups made discrete

In what follows "homology" will mean group homology, i.e. $H_*(BG^\delta;{\mathbf R})$ for the group $G$ with the discrete topology. It is well-known how to compute the homology of abelian groups, ...
ThiKu's user avatar
  • 10.4k
2 votes
0 answers
128 views

Divisible fundamental group [duplicate]

I apologize if this question seems trivial or elementary. Is there any concrete topological space with divisible fundamental group? For example, is there any such a space the fundamental group in ...
Sh.M1972's user avatar
  • 2,233
13 votes
1 answer
552 views

Realizing symmetric groups by diffeomorphisms

Let $M$ be a (closed, smooth) manifold of dimension $d$. For $n$ a positive integer, fix $n$ points $x_1, \dots, x_n \in M$. The group of diffeomorphisms of $M$ that permutes the points $x_i$ surjects ...
Jens Reinhold's user avatar
10 votes
2 answers
647 views

Groups with trivial rational homology and their finite index subgroups

For a short exact sequence $0 \to G \to H \to K \to 0$ of (discrete) groups with $K$ finite we have, as a consequence of the Hochschild-Serre spectral sequence, that $H^{\ast}(H;\mathbb Q) = H^{\ast}(...
Jens Reinhold's user avatar
4 votes
0 answers
135 views

Exotic 2-adic lifts of mod $2$ Steinberg idempotent

Denote $B_n$ the Borel subgroup of $Gl_n(Z/2)$, i.e., the subgroup of upper triangular matrices, $\Sigma _n$ the subgroup of permutation matrices. The (conjugate) Steinberg idempotent is defined to be ...
user43326's user avatar
  • 3,051
10 votes
1 answer
536 views

Inducing up the group homomorphism between mapping class groups

There are many ways to embed the braid group into the mapping class group of a surface. To describe one of them, let ${C}_{2g+2}(\mathbb{D}^2)$ be the configuration of unordered $2g+2$ points in the ...
Sam Nariman's user avatar
  • 1,003
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
6 votes
1 answer
406 views

Connection between Stalling's end theorem and Seifert-van Kampen Theorem

Stalling`s end Theorem (a group has more than one end iff it splits over a finite subgroup) and the Seifert-van Kampen Theorem (the fundamental group of a 'decomposable' space is a free amalgamated ...
M.U.'s user avatar
  • 721
3 votes
2 answers
699 views

What is the $\mathbb Z/2$-cohomology of $\mathrm B^n(\mathbb Z/2)$?

I would like to know the cohomology groups $\mathrm H^\bullet(\mathrm B^n(\mathbb Z/2);\mathbb Z/2)$. I assume that this is a standard computation, but I'm not sure where to look up the answer (and, ...
Theo Johnson-Freyd's user avatar
85 votes
23 answers
11k views

Solving algebraic problems with topology

Often, topologists reduce a problem which is - in some sense - of geometric nature, into an algebraic question that is then (partiallly) solved to give back some understanding of the original problem. ...
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
2 votes
1 answer
387 views

Generators of the colored braid group (two colors), reference

I consider the group $B_{n,n}$, the braids, colored in two colors, say all odd strings are black and all even strings are white. It is easy to find a set of generators for $B_{n,n}$: $$ \begin{cases} \...
Nikita Kalinin's user avatar
3 votes
0 answers
100 views

project limit on $n$- simplical complex which is principal homogeneous with respect to an action

The setting: Let G be compact locally $\Bbb{Q}_p$ analytic group. We fix a countable basis of open normal subgroups $G\supset G_1\supset ...G_r\supset...$ We suppose that we are given a system of ...
MathStudent's user avatar
20 votes
3 answers
1k views

Center of a simply-connected simple compact Lie group and McKay correspondence

Let $G$ be a simply-connected simple compact Lie group. Its center $Z(G)$ is a finite abelian group, say $Z(G) = \mathbb Z/k\mathbb Z$ for $G=SU(k)$. I find the following interpretation of $Z(G)$ in ...
Hiraku Nakajima's user avatar
5 votes
1 answer
428 views

Centralizers in the universal central extensions of the alternating groups?

For $n \ge 8$ the Schur multiplier $H_2(BA_n, \mathbb{Z})$ (where $A_n$ denotes the alternating group) stabilizes to $\mathbb{Z}_2$, and hence there is a universal central extension $\widetilde{A}_n$ ...
Qiaochu Yuan's user avatar
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
3 votes
1 answer
445 views

Co-rank of a group with $a^2b^2c^2=1$ (fundamental group of non-orientable surface)

What is the co-rank of a group $$G=\langle a_1,a_2,\dots,a_h\mid a_1^2a_2^2\dots a_h^2=1\rangle,$$ that is, finitely generated group with $h$ generators and one relation? By co-rank, I mean the ...
Irina's user avatar
  • 504
9 votes
2 answers
650 views

Action of the homotopy braid groups on reduced free groups

Firstly some definitions: $B_n$ is the braid group with $n$ strands. $\widetilde{B_n}$ is "homotopy braid group", which is a factor group of $B_n$ by adding the relation that $A_{j,k}$ ...
Zuriel's user avatar
  • 1,108
5 votes
2 answers
573 views

Are homotopy braid groups residually nilpotent?

A group is called residually nilpotent if given any non-identity element, there is a normal subgroup not containing that element, such that the quotient group is nilpotent. It is known that pure braid ...
Zuriel's user avatar
  • 1,108
1 vote
0 answers
275 views

Explicitly showing that a free group is LERF [closed]

Let $F$ be a free group on a finite set $X$, and let $M$ be a finitely generated subgroup. Marshall Hall's theorem states that $M$ is closed in the profinite topology on $F$. That is, $M$ is the ...
Pablo's user avatar
  • 11.3k
13 votes
3 answers
2k views

Which groups are LERF?

A finitely generated group $G$ is called LERF if every finitely generated $H \leq G$ is closed in the profinite topology on $G$ (equivalently, there is a family of finite index subgroups of $G$ ...
Pablo's user avatar
  • 11.3k
3 votes
0 answers
421 views

Marshall Hall's theorem for surface groups [closed]

Let $\Gamma_g$ be a surface group of genus $g \geq 2$, that is we have a presentation: $$\Gamma_g = \langle x_1,y_1 \dots, x_g,y_g \vert \prod_{i = 1}^g [x_i,y_i] = 1\rangle$$ Let $H \leq \Gamma_g$ ...
Pablo's user avatar
  • 11.3k
4 votes
2 answers
337 views

A Karrass-Solitar theorem for surface groups

Let $\Gamma_g$ be a surface group of genus $g \geq 2$. That is, there is a presentation $$\Gamma_g = \langle x_1, y_1, \dots, x_g, y_g \vert \prod_{i = 1}^{g}[x_i,y_i] = 1\rangle$$ Is there a ...
Pablo's user avatar
  • 11.3k
5 votes
1 answer
264 views

Bases of surface groups

Let $\Gamma_g$ be a surface group of genus $g \geq 2$. A $2g$-tuple $(x_1,y_1, \dots,x_g,y_g) \in \Gamma_g^{2g}$ will be called a Surface Basis if we have the presentation $$\Gamma_g = \langle x_1, ...
Pablo's user avatar
  • 11.3k
33 votes
3 answers
6k views

(co)homology of symmetric groups

Let $S_n=\{\text{bijections }[n]\to[n]\}$ be the n-th symmetric group. Its (co)homology will be understood with trivial action. What are the $\mathbb{Z}$-modules $H_k(S_n;\mathbb{Z})$? Using GAP, we ...
Leo's user avatar
  • 1,589
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
15 votes
1 answer
640 views

Torsion-free group that is not of type F but is virtually of type F

Recall that a group $G$ is of type F if there exists a compact $K(G,1)$. There are many examples of groups which are not of type F but which are virtually of type F, that is, they have finite-index ...
Sarah's user avatar
  • 153
5 votes
1 answer
964 views

1D TQFT in Freed-Hopkins-Lurie-Teleman

In the first section of Freed-Hopkins-Lurie-Teleman they construct a one-dimensional Topological Quantum Field Theory. $F(\circ_+)$ is a vector space and $F(\circ_-)$ is the dual. $F(\circ-\circ)$ is ...
john mangual's user avatar
  • 22.8k
4 votes
0 answers
144 views

When does finite presentability of the associated graded Lie algebra of a group imply the group is finitely presented?

Let $G$ be a finitely generated group; let $L(G)$ denote the graded Lie algebra (over $\mathbb{Q}$) associated to the lower central series of $G$. I would like to know conditions for when the finite ...
Peter Goetz's user avatar
1 vote
1 answer
151 views

A formula for isotropy group $\pi_1(G_a)$

Let $G$ be a compact Lie group and $T$ be its maximal tours, and $a\in \mathfrak{g}^*$. and $G_a$ be the isotropy group of $G$ then $T\subset G_a$ and we know that $\pi_1(T)=\mathbb{Z}^n$. My ...
user avatar
4 votes
1 answer
365 views

Topological interpretation for groups of type $FP_2$

A group $G$ is of type $FP_2$ if it admits a partial projective resolution of $\mathbb{Z}G$-modules $$ P_2 \rightarrow P_1 \rightarrow P_0 \rightarrow \mathbb{Z} \rightarrow 0$$ with each $P_i$ being ...
Gaelan Hanlon's user avatar
17 votes
1 answer
575 views

Group cochains invariant under the action of the symmetric group

Let $G$ be a finite group and $A$ an abelian group. Recall the cochain groups $$ C^k = \{f: G^k \to A\} $$ and the coboundary map $$ \delta : C^k \to C^{k+1} $$ $$ (\delta f)(g_1, \ldots, ...
Kevin Walker's user avatar
  • 12.8k
5 votes
0 answers
222 views

Nilpotent Localization in Group Theory

Algebraic topologists have invented a very pretty technique of localizing nilpotent groups. (Garth Warner covers the topic in his book manuscript Topics in Topology and Homotopy Theory). For ...
arsmath's user avatar
  • 6,870
5 votes
1 answer
629 views

What is known about maximal free subgroups of surface groups?

Let $\Gamma_g=< a_1,...,a_g,b_1,...,b_g | \prod_{i=1}^g [a_i,b_i]>$ (a surface group). What is known about maximal free subgroups of $\Gamma_g$ for $g>1.$ (I.e. free subgroups which are not ...
Adam's user avatar
  • 2,390