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Jan
6
awarded  Nice Answer
Dec
18
answered Residually Finite Braid Group
Dec
15
awarded  metric-spaces
Dec
11
comment Weak topology on an infinite-dimensional normed vector space is not metrizable
I wrote this sketch a long time ago, but if I am right, you may suppose without loss of generality that your countable basis has the form $\{ \mathrm{ker}(\zeta_n) \mid n \geq 1 \}$, just by definition of the weak topology. Then, use the fact that any $\zeta \in X^*$ is continuous with respect to the weak topology, and look at the link between $\mathrm{ker}(\zeta)$ and the $\mathrm{ker}(\zeta_n)$'s to deduce that $\zeta$ is a finite linear combination of the $\zeta_n$'s.
Dec
5
comment Proving the completeness theorem of metric spaces.
This is essentially true by definition: if $A \subset B$, then we say that $A$ is dense in $B$ precisely when $\overline{A}=B$. In particular, any subspace is dense in its closure.
Nov
15
asked Weakly relatively hyperbolic groups
Nov
13
accepted Centralizers in mapping class groups
Nov
12
awarded  Nice Question
Nov
8
asked Centralizers in mapping class groups
Oct
25
awarded  Guru
Sep
29
accepted Relative hyperbolicity and splittings
Sep
27
comment Infinite torsion CAT(0) groups
There exist infinite torsion groups acting properly discontinuously on a CAT(0) space. See for instance Example II.7.11 in Haefliger and Bridson's book, where the authors construct, for every increasing sequence $G_0 \subset G_1 \subset \cdots$ of finite groups, a proper action of $\lim G_n$ on a metric simplicial tree.
Sep
26
comment Commuting elements in ${\rm PSL}(2,\mathbb{Z})$ are powers of some element
Probably, the centralizer of $g$ may be virtually cyclic and not necessarily cyclic. I think an even more general justification comes from Bass-Serre theory, by looking at the action on the associated Bass-Serre tree.
Sep
26
revised On powers in free groups
added 1 character in body; edited tags
Sep
26
answered On powers in free groups
Sep
25
answered Writing specific word as product of commutators in free groups
Sep
25
comment No finitely generated abelian group is divisible
Related: math.stackexchange.com/questions/309120/…
Sep
23
comment Can $\mathbb{Z}_8$ be the commutator subgroup of a group?
Related: math.stackexchange.com/questions/361582/…
Sep
19
comment Hyperbolic groups and cohomology
A silly observation: for any group $G$, the free product $G \ast \mathbb{F}_n$ is a relatively hyperbolic group (and a fortiori an acylindrically hyperbolic group). Because the cohomological groups of $G \ast \mathbb{F}_n$ are essentially the sames as those of $G$, it seems that there are no restrictions on the possible cohomological groups of a relatively (or acylindrically) hyperbolic group.
Sep
19
comment Relative hyperbolicity and splittings
@DerekHolt: You are completely right, $A \underset{C}{\ast} B$ clearly cannot be hyperbolic relatively to $\{A,B\}$ if $C$ is infinite. I have just posted an answer, giving a potential example. (In fact, I wonder if $C$ malnormal in $A$ and $B$ implies $C$ malnormal in $A \underset{C}{\ast} B$...)