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3h
comment Are infinite groups “locally topologizable”?
I don't know, but the class of groups admitting no such topology is closed under taking subgroups, because if $H\subset G$ we can extend the topology on $H$ so that $G\smallsetminus H$ is open and discrete. In particular, a group as in your question will be torsion, and the question boils down to finitely generated groups.
4h
comment Torsion-free group that is not of type F but is virtually of type F
Still, in Sarah's question finite presentability is not an issue, and thus from FP plus finite presentability we can deduce F, answering negatively the question (torsion-free + virtually F implies F).
23h
comment Semidirect products with braid groups and type $F_\infty$
Then the question does not seem to change when we replace the group by a finite index subgroup, and $B_k\ltimes F^k$ admits the direct product $P_k\times F^k$ as finite index subgroup.
1d
comment Semidirect products with braid groups and type $F_\infty$
How does $B_k$ act on $F^k$?
2d
revised Embeddings of finitely generated groups into uniformly convex Banach spaces
added 35 characters in body
2d
answered Embeddings of finitely generated groups into uniformly convex Banach spaces
2d
comment Finitely generated groups non-embeddable into $L_1(0,1)$
To answer your question, it would be a good PhD work to prove that any non-virtually-abelian nilpotent f.g. group (or, more naturally, any non-abelian simply connected nilpotent Lie group) has no bilipschitz embedding into $L^1$.
2d
comment Finitely generated groups non-embeddable into $L_1(0,1)$
There is not one Gromov random group, there is a recipe providing plenty of groups (and moreover it sometimes means those finitely presented groups containing these infinitely presented groups when they are arranged to be recursively presentable).
Aug
10
comment vanishing higher cohomology group for property T group?
No, the definition of $H^*(G,\pi)$ for a unitary representation (such as $\ell^2(\Gamma)$) does not involve the topology when $\Gamma$ is a discrete group. Then there is a notion of reduced topology $H^*(\Gamma,\pi)$ taking into account the topology, where $\bar{H^n}(\Gamma,\pi)$ is the quotient of $H^n(\Gamma,\pi)$ by the closure of $\{0\}$. A standard reference for this is Guichardet's book.
Aug
9
comment vanishing higher cohomology group for property T group?
PS: in this example, you both have $H^2(\Gamma,\mathbf{Z})$ and $H^2(\Gamma,\mathbf{R})$ nonzero.
Aug
9
comment vanishing higher cohomology group for property T group?
Why do you need a finite index subgroup? if you have any lattice in $Sp(n,1)$ then the universal covering yields a 2-cocycle, which remains non-trivial on the lattice. (If it were trivial on the lattice, it would be trivial on any finite index subgroup). One argument that the 2-cocycle is nontrivial is that the extension, being a lattice in the universal covering of $Sp(n,1)$ which also has T by S.P.Wang/Serre's theorem, has T, which implies that the 2-cocycle is nontrivial on the lattice in $Sp(n,1)$. Anyway this is very far from the original question.
Aug
9
answered Finite extension of local fields
Aug
9
comment Finite extension of local fields
I may miss something, but it seems to me that there is only one inseparable extension of degree $p$ for $F_q((t))$, namely $F_q(t^{1/p})$. The extensions defined by irreducible polynomials $P(X)=X^p-X-c$ are separable.
Aug
9
comment Finite extension of local fields
It would be nice to have a rigorous proof. If $P$ is a monic irreducible polynomial in $K[x]\smallsetminus K[x^p]$, of degree $n$ ($K=F_q((t))$, $q=p^k$), is it true that in $K[x]/P$, every polynomial close enough to $P$ has a root? If yes this would prove the countability result.
Aug
8
comment Finite extension of local fields
OK: so according to these definition, a local field is a non-discrete locally compact field. But their general definition of higher local field (HLF) makes little sense, even for dimension 2, let alone infinity. According to them, a HLF of dimension 2 is a complete discrete valuation field whose residual field is a local field... but the residual field is a discrete field and a local field is a topological notion. So it's not rigourous, and prone to ambiguity. If it means that the residual field admits a topology of local field, it becomes rigorous although it sounds pretty artificial.
Aug
8
comment Finite extension of local fields
Pablo, would you define "higher local field"? even "local field" has several definitions.
Aug
8
comment Compact subgroups of linear groups over nonarchimedean fields
Yes you miss something: it's an argument showing there are at least infinitely countably many extensions of degree $p$. It doesn't show there are at most countably many.
Aug
8
comment vanishing higher cohomology group for property T group?
the second $\ell^2$-Betti number of $SL_3(\mathbf{Z})$ is zero. This maybe implies that $H^2$ of $\ell^2(\Gamma)$ vanishes but I'm not 100% sure.
Aug
8
comment vanishing higher cohomology group for property T group?
If $\Gamma$ is a lattice in a symmetric space of dimension $2d$, then I think $H^d(\Gamma,\ell^2(\Gamma))$ is known nonzero, for the $d$-th $\ell^2$ Betti number is nonzero. For instance, if $\Gamma$ is a lattice in $SL_3(\mathbf{R})$ then $H^4(\Gamma,\ell^2(\Gamma))$ is nonzero whilst it has Property T. For $n=2$ and $\ell^2$, this does not work but maybe we should look hyperbolic groups with T and cohomological dimension 2.
Aug
8
comment Compact subgroups of linear groups over nonarchimedean fields
No they construct countably many: as many as the cokernel of $P(x)=x-x^p$, with $P:K\to K$ ($K=F_p((t))$. The image of $P$ contains the open 1-ball, since for $|x|<1$ and $y=\sum_{n\ge 0} x^{p^n}$ we have $P(y)=x$. Hence $coker(P)$ is countable.