All Questions
Tagged with at.algebraic-topology gr.group-theory
288 questions
14
votes
4
answers
677
views
Computing homotopy groups of X such that pi_1(X) has solvable word problem
The paper
E. H. Brown, Jr., Finite computability of Postnikov complexes, Ann. of Math. (2) 65 (1957), 1-20.
proves that if $X$ is a finite simply-connected simplicial complex, then there is an ...
14
votes
0
answers
341
views
Is this class of groups already in the literature or specified by standard conditions?
In recent work
Lifting $N_\infty$ operads from conjugacy data on homotopical combinatorics / $N_\infty$ operads in equivariant homotopy theory, collaborators
Scott Balchin, Ethan MacBrough, and I ...
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 ...
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 \;|\; ...
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 ...
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$ ...
13
votes
2
answers
795
views
Which finite groups have low-degree essential cohomology?
Let $G$ be a finite group, $A$ some coefficients (e.g. $A = \mathbb{F}_2$ or $\mathbb{Z}$), and write $\mathrm{H}^\bullet_{\mathrm{gp}}(G; A)$ for the (ordinary) group cohomology of $G$ with ...
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:...
13
votes
2
answers
801
views
Does there exist smooth circle action on manifolds M^{4n} with exactly three fixed points such that n\neq 1
I have a question in mind for some time. That is, does there exist a smooth circle action on a closed manifold $M^{4n}$ ($n\geq 2$) with exactly three fixed points?
Remarks:(1) For n=1, the examples ...
13
votes
1
answer
289
views
Powers of the Euler class, torsion free subgroup of Homeo($S^1$)
For any subgroup $G$ of $\text{Homeo}(S^1)$, we have the Euler class $\chi$ in the group cohomology $H^2(G;\mathbb{Z})$. One can think of this class as the pullback of the generator of $H^2(\mathrm{B}\...
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$...
12
votes
1
answer
625
views
Any group is a quotient of an acyclic group?
As far as I know, for any group $G$ there exists an acyclic group $H$ such that $G$ is a subgroup of $H$.
I am wondering about the dual situation. Is any group $A$ a quotient of an acyclic group $B$ ...
12
votes
2
answers
583
views
Do there exist acyclic simple groups of arbitrarily large cardinality?
Recall that a group $G$ is acyclic if its group homology vanishes: $H_\ast(G; \mathbb Z) = 0$. Equivalently, $G$ is acyclic iff the space $BG$ is acyclic, i.e. $\tilde H_\ast(BG;\mathbb Z) = 0$.
In ...
12
votes
1
answer
522
views
Realizing inner automorphisms on Eilenberg-MacLane spaces
Let $G$ be a discrete group and let $(X,x_0)$ be a based Eilenberg-MacLane space for $G$, so there is a fixed isomorphism $\pi_1(X,x_0) = G$ and the universal cover $\widetilde{X}$ is contractible. ...
12
votes
2
answers
781
views
Where does the term "torsor" come from?
Is there a heuristic reason why principal homogeneous spaces of a group (object) $G$ (in some categories) are called $G$-torsors? Does it have anything to do with the idea of "torsion", ...
12
votes
2
answers
1k
views
Cohomological dimension of a homomorphism
Let $G$ and $\Gamma$ be discrete groups, and let $\phi\colon\thinspace G\to \Gamma$ be a homomorphism.
Define its cohomological dimension $\operatorname{cd}\phi$ to be the least integer $d$ such that $...
11
votes
9
answers
1k
views
Proving the impossibility of an embedding of categories
A number of topological invariants take the form of functors $\mathscr{T}\to\mathscr{G}$, where $\mathscr{T}$ is the category of all topological spaces and continuous functions, and $\mathscr{G}$ is ...
11
votes
1
answer
620
views
Is $SL(n,\mathbb{Z})$ a CAT(0) group?
Is it possible to find a CAT(0) space on which the matrix group $SL(n,\mathbb{Z})$ acts properly discontinuously and cocompactly? Note: when the cocompactness is dropped , it is possible.
11
votes
1
answer
455
views
Asking whether there is a compact Lie group containing affine symplectic group
The affine symplectic group is interesting and important in physics. However, the Lie group is noncompact. In order to have some good properties (Basically, we need some good behavior of Haar measure) ...
11
votes
2
answers
843
views
covers of $Z^\infty$
Is it possible to cover $Z^\infty$ (the infinite direct sum of $Z$'s with the $l_1$-metric) by a finite set of collections of subsets $U^0,...,U^n$ such that each collection $U^i$ consists of ...
11
votes
2
answers
475
views
What is a finite Haken cover of the Seifert–Weber space?
It's known that the Seifert–Weber space (obtained from a dodecahedron by gluing opposite faces with a 3/10 turn) is an example of a non-Haken 3-manifold. Since every closed 3-manifold is virtually ...
11
votes
2
answers
656
views
$G$ cocycle split and trivialized to a coboundary in $J$, given a group homomorphism $J \overset{r}{\rightarrow} G$
Consider a generic nontrivial 3-cocycle $\omega_3^G(g_1,g_2,g_3) \in H^3(G,U(1))$ in the cohomology group of $G$ with $U(1)=\mathbb{R}/\mathbb{Z}$ coefficient. In otherwords, here the 3-cocycle $\...
11
votes
1
answer
811
views
What is an interpretation of the relation in the cohomology of the pure braid groups?
In 1968, Arnol'd proved that the integral cohomology of the pure braid group $P_n$ is isomorphic to the exterior algebra generated by the collection of degree-one classes $\omega_{i,j}\ (1 \le i < ...
11
votes
1
answer
167
views
A group of type F that is an extension of type F-by-type F
Let us first recall that a group of type $F$ is a group admitting a compact classifying space.
Let $K$ and $Q$ be groups of type $F$. Consider the family $\mathcal{G}(K, Q)$ consisting of groups $G$ ...
11
votes
0
answers
221
views
On an Artin (?) subgroup of braid groups
While working on something apparently unrelated I encountered a "braid-like" group, which is a relatively geometric subgroup of a braid group and seems to be itself an Artin group. It seems ...
11
votes
0
answers
331
views
If an additive group of $\Bbb R^2$ contains a smoothly deformed circle, is it necessarily all of $\Bbb R^2$?
It can be shown that if an additive subgroup of $\Bbb R^2$ contains the unit circle, then it is necessarily all of $\Bbb R^2$. Does this also hold for a suitably smoothly deformed unit circle?
...
11
votes
0
answers
203
views
Fundamental groups of reduced subgroup lattices
Let $G$ be a group. Its subgroup lattice, denoted $\Sigma G$, consists of all subgroups of $G$ partially ordered by inclusion. The topology of this poset is quite trivial, since it always has a ...
10
votes
4
answers
6k
views
Commutativity of the fundamental group of any Lie Group [closed]
How do we formally prove that the fundamental group of any Lie group is always commutative?
10
votes
4
answers
2k
views
Proving that a countable group is not finitely generated
I would like to learn about techniques for proving that a countable group is not finitely generated. I am also interested in learning about examples. Finally, I am particularly, but not exclusively, ...
10
votes
1
answer
1k
views
Acyclic Finite Groups
A group is called acyclic if its classifying space has the same homology of a point. Examples of acyclic groups include Higman's group with four generators and relations, also ...
10
votes
2
answers
890
views
Are virtual cubulated groups cubulated?
Suppose $G$ has a finite index subgroup $N$ such that $N$ acts properly and cocompactly on a CAT(0)-cube complex. Does $G$ also act properly and cocompactly on a CAT(0)-cube complex?
Edit: After ...
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}(...
10
votes
1
answer
855
views
finite complex with non-finitely generated homology with local coefficients
I am looking for an explicit example, if one exists, of a (pointed) finite connected CW-complex $X$ such that some homology group with local coefficients $H_n(X,{\mathbb Z}[\pi_1 X])$ is not a ...
10
votes
2
answers
497
views
Equivariant cohomology of the complement to the arrangement $\bigcup_{i\neq j}\vec x_i = \vec x_j$?
$\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}\DeclareMathOperator\Conf{Conf}$Let $V=\mathbb{R}^d$ be a $d$-dimensional (Euclidean) vector space over real numbers.
Let $G=\SO(V)$ be the ...
10
votes
1
answer
635
views
Self-homomorphisms of surface groups
Let $X$ be a closed, orientable surface of genus at least 2, and let $\phi: \pi_1(X) \to \pi_1(X)$ be a surjective homomorphism. Is $\phi$ necessarily injective?
10
votes
1
answer
580
views
Nonhyperbolic groups that contain no free abelian groups or Baumslag-Solitar groups
I've heard it conjectured that a finitely presentable group $G$ is hyperbolic if it satisfies the following two conditions.
$G$ contains no subgroup isomorphic to a Baumslag-Solitar group $BS(n,m)$ (...
10
votes
1
answer
274
views
A flatness result of Fiedorwicz for amalgamated free products of monoids in connection with classifying spaces of monoids
In Lemma 5.2(a) of Z. Fiedorowicz, Classifying Spaces of Topological Monoids and Categories American Journal of Mathematics Vol. 106, No. 2 (Apr., 1984), pp. 301-350 the author proves the following.
...
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 ...
10
votes
2
answers
337
views
Finitely dominated universal spaces for the family of solvable subgroups
$\DeclareMathOperator\PSL{PSL}\DeclareMathOperator\Sz{Sz}$In short, I am interested in the question which finite groups $G$ admit a finitely dominated universal space with respect to the family of ...
10
votes
0
answers
458
views
is a group $G$, that admits finite $k(G, 1)$ and has no Baumslag-Solitar subgroups, necessarily hyperbolic?
This is the first question asked in Bestvina's article "Questions in Geometric Group Theory". Does anyone know if there has been any progress made on this problem? Is the question answered if $G$ is ...
9
votes
3
answers
735
views
Judging whether a finitely presented group is a 3-manifold group?
Given a finitely presented group $G$, how many necessary conditions do people know for $G$ to be isomorphic to the fundamental group of some closed connected 3-manifold? (e.g. residually finite)
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}$ ...
9
votes
2
answers
939
views
A question on the fundamental group of a compact orientable surface of genus >1
Let $G=\pi(X,x)$ be the fundamental group of a compact orientable
surface of genus $g\ge 2$. It is well known that a presentation of
$G$ is
$$G=\langle x_1,y_1,\dots,x_g,y_g \ | \ [x_1,y_1]\cdots
[x_g,...
9
votes
1
answer
308
views
How small can the support of a nontrivial $\mathbb F_p$-cocycle on $C_p$ be?
Let $p$ be a prime, and let $\phi : C_p^n \to \mathbb F_p$ be an $\mathbb F_p$-valued $n$-cocycle on $C_p$ (the cyclic group of order $p$) which is not an $n$-coboundary, i.e. $\phi$ represents a ...
9
votes
1
answer
202
views
Are compact simple groups homotopically non-abelian?
Take a compact connected simple centreless Lie group $G$.
Can the commutator map $G\times G\to G$ sending $(x,y)$ to $[x,y]$ be homotopic to a constant map?
I am interested mostly in the case, ...
9
votes
1
answer
386
views
Different definitions of formality for groups
Let $X$ be a space with fundamental group $G$. Recall that the de Rham fundamental group of $X$ is the inverse limit of the Malcev completions of the nilpotent truncations of $G$. This has a Lie ...
9
votes
1
answer
309
views
Comparing cohomology of a total complex with the cohomology of semidirect product
$\DeclareMathOperator{\Tot}{Tot}$I have the following problem. Let $H$ and $G$ be groups such that $H$ acts on $G$, i.e., there exists a group homomorphism $H\to \mathrm{Aut}(G)$ and let $M$ be an ...
9
votes
1
answer
308
views
Projective resolutions of finite-dimensional representations of infinite groups
Let $G$ be a group and let $V$ be a finite-dimensional complex representation of $G$. Question: Under what circumstances can I find a projective resolution
$$ \cdots \longrightarrow P_3 \...
9
votes
0
answers
269
views
Colimits of symmetric groups
The infinite symmetric group $S_{\infty}$ of finitely supported permutations of $\mathbb{N}$ can be written as a colimit over the $S_n$'s with respect to the embedding $S_{n} \to S_{n+1}$ that maps $\...
9
votes
0
answers
372
views
Groups with trivial outer automorphism group and prescribed center?
Given an arbitrary abelian group $A$, can we find a group $G$ such that
$\mathrm{Out}(G)=\mathrm{Aut}(G)/\mathrm{Inn}(G)=1$, and
$Z(G)\simeq A$?
Why is this interesting? Given a group $G$, we have ...