All Questions
61 questions
62
votes
9
answers
9k
views
Fundamental groups of noncompact surfaces
I got fantastic answers to my previous question (about modern references for the fact that surfaces can be triangulated), so I thought I'd ask a related question. A basic fact about surface topology ...
35
votes
3
answers
1k
views
Second Betti number of lattices in $\mathrm{SL}_3(\mathbf{R})$
We fix $G=\mathrm{SL}_3(\mathbf{R})$.
Let $\Gamma$ be a torsion-free cocompact lattice in $G$. Is $b_2(\Gamma)=0$?
Here the second Betti number $b_2(\Gamma)$ is both the dimension of the ...
22
votes
2
answers
1k
views
The image of the point-pushing group in the hyperelliptic representation of the braid group
Let $B_{2g+1}$ be the Artin braid group on $2g+1$ strands. There is a symplectic representation
$\rho: B_{2g+1} \rightarrow Sp_{2g}(\mathbf{Z})$
called the "hyperelliptic representation," which ...
22
votes
1
answer
719
views
What is the cohomological dimension of the commutator subgroup of the pure braid group?
I'm interested in computing the cohomological dimension of the commutator subgroup $[P_n,P_n]$ of the pure braid group $P_n$. I wasn't able to find a reference in the literature.
Because $[P_n,P_n]$ ...
21
votes
2
answers
622
views
Morphism from a surface group to a symmetric group, lifted to the braid group
Let $\Sigma_g$ be the fundamental group of the closed orientable surface of genus $g\ge 2$; let $B_n$ be the braid group on $n\ge 3$ braids; let $S_n$ be the symmetric group on $n$ letters; let $p:B_n\...
18
votes
0
answers
1k
views
What is the strongest nerve lemma?
The most basic nerve lemma can be found as Corollary 4G.3 in Hatcher's Algebraic Topology:
If $\mathcal U$ is an open cover of a paracompact space $X$ such that every nonempty intersection of ...
17
votes
1
answer
683
views
Relationship between Smith's special homology groups and equivariant homology theory
EDIT: Tyler Lawson's answer was so nice that I was inspired to rewrite the notes discussed below to use Bredon homology in the definition of the Smith special homology groups. The original version is ...
17
votes
3
answers
1k
views
The second homotopy group of a simple CW-complex
Let $X$ be a CW-complex with
one 0-cell
two 1-cells
three 2-cells
no cells in dimensions 3 or higher.
Is it always true that $\pi_2(X)\ne 1$?
16
votes
1
answer
505
views
How many cells needed to build the classifying space $BG$?
Let $G$ be a finitely presented group of cohomological dimension $n$.
Apart from the unresolved ambiguity pertaining to the Eilenberg--Ganea conjecture, it is known that we can find an $n$-dimensional ...
15
votes
2
answers
968
views
Semidirect product decomposition of the Borromean rings group
Let $X=S^3\setminus B$ be the link complement of the Borromean rings.
(source)
Then $G=\pi_1(X)$ has a presentation of the form
$$
G = \langle \; a,b,c \mid [a,[b^{-1},c]],\; ...
14
votes
2
answers
906
views
Acyclic group and finite CW-complex
Is there a nontrivial example of an acyclic group $G$ such that its corresponding Eilenberg space $K(G,1)$ is homotopy equivalent to a finite CW-complex ?
14
votes
3
answers
683
views
Compact manifolds with big mapping class group
I was wondering if compact surfaces were the only compact manifolds with a "big" or "complicated" mapping class group.
Are there higher dimensional manifolds (which are not in some way reducible to ...
14
votes
2
answers
789
views
Restriction of a branched cover to its branch locus
Assume that we have a smooth, compact, complex surface $X$, and a smooth and irreducible divisor $B \subset X$. Let $G$ be a finite group. For every group epimorphism $$\varphi \colon \pi_1(X-B) \to G,...
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
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}\...
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
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
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 ...
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
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
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
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
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
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 ...
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, ...
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 ...
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 ...
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 ...
6
votes
2
answers
331
views
Epimorphisms from the genus $2$ surface braid group to finite groups
This question is somehow related to my previous MO question Explicit description of a subgroup of the braid group $\mathsf{B}_2(C_2)$; for the reader convenience, let me write down again the relevant ...
6
votes
1
answer
456
views
Restrictions on $\pi_1(X)$ of geometric origin (Kähler groups as example)
There's and old and extensively studied question about characterisation of fundamental groups of smooth compact Kähler manifolds. Restrictions imposed by Kählerness are somewhat fragile, and if we ...
6
votes
1
answer
658
views
Generalized Birman exact sequence for surfaces with boundaries
Let $S_g^n$ be a surface of genus g with n boundaries and let $Mod(S_g^n)$ be its mapping class group.
We will also denote by $S_{g,m}^n$ a surface of genus g with n boundaries and m punctures.
The ...
6
votes
1
answer
237
views
Example similar to the Griffiths twin cone but with fundamental group that allows surjection onto $\mathbb Z$
The Griffiths twin cone is an example of a wedge sum of two contractible spaces being non-contractible. Namely, it is the wedge sum $\mathbb G=C\mathbb H\vee_p C\mathbb H$ of two coni over the ...
6
votes
0
answers
484
views
Does a finitely generated aspherical group have an aspherical presentation with a finite generating set?
Let $G$ be a finitely generated group. Suppose $G$ has an aspherical presentation with a countably infinite generating set. Does $G$ have an aspherical presentation with a finite generating set?
Here ...
5
votes
2
answers
666
views
HNN extensions which are free products
which HNN-extensions are free products? this question is related with another still unsolved about Nielsen-Thruston-reducibility and connected-sum-irreducibility of 3d-torus- bundles...
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 ...
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 ...
5
votes
1
answer
336
views
"Simplicial complex" product of groups?
Let $X=(V,E)$ be a graph, and to each vertex $v \in V$, associate a group $G_v$. The graph product of the groups $G_v$ (as defined e.g. here) is $F/R$; the quotient of the free product of the $G_v$ by ...
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 ...
5
votes
2
answers
252
views
Monoid of continuous self-maps of (real) surfaces
Let $S$ be a closed surface of genus $g > 0$ and $[S,S] = Hom(\pi_{1}(S),\pi_{1}(S))$ be the monoid of (homotopy classes of) continuous maps from $S$ to itself. Consider the semi-group $A$ of ...
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 ...
5
votes
1
answer
232
views
Finite index subgroups of the mapping class group with geometric meaning
I have got a question that is perhaps not precise in a mathematical sense.
Is there a classification of all coverings of the moduli space of Riemann surfaces which are moduli spaces themselves, that ...
5
votes
0
answers
249
views
Aspherical space whose fundamental group is subgroup of the Euclidean isometry group
Let $M$ be a smooth, compact manifold without a boundary, with its universal covering $\tilde{M} = \mathbb{R}^n$. If there exists an injective homomorphism $h: \pi_1(M) \rightarrow O(k) \ltimes \...
3
votes
1
answer
267
views
In what sense is every element of $H_2(G)$ "represented by a free action on some surface"
(This is a cross-post of this unanswered math.stackexchange question)
In Edmond's 1982 paper Surface Symmetry II, at the bottom of page 145, he writes:
"Corollary - If $G$ is a split nonabelian ...
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 ...
3
votes
0
answers
115
views
Finite homology of a homogeneous space
Let $\Gamma$ be a cocompact lattice in $\operatorname{SL}(2,\mathbb R)$ and $X=\operatorname{SL}(2,\mathbb R)/\Gamma$ be the underlying homogeneous space. Can the homology group $H_1(X,\mathbb Z)$ be ...