bio  website  normalesup.org/~cornulier 

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3h

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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

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Torsionfree 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 (torsionfree + virtually F implies F). 
23h

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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

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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

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Finitely generated groups nonembeddable into $L_1(0,1)$
To answer your question, it would be a good PhD work to prove that any nonvirtuallyabelian nilpotent f.g. group (or, more naturally, any nonabelian simply connected nilpotent Lie group) has no bilipschitz embedding into $L^1$. 
2d

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Finitely generated groups nonembeddable 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 
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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 
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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 
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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 2cocycle, which remains nontrivial on the lattice. (If it were trivial on the lattice, it would be trivial on any finite index subgroup). One argument that the 2cocycle 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 2cocycle 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 
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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 $P(X)=X^pXc$ are separable. 
Aug 9 
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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 
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Finite extension of local fields
OK: so according to these definition, a local field is a nondiscrete 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 
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Finite extension of local fields
Pablo, would you define "higher local field"? even "local field" has several definitions. 
Aug 8 
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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 
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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 
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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 
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Compact subgroups of linear groups over nonarchimedean fields
No they construct countably many: as many as the cokernel of $P(x)=xx^p$, with $P:K\to K$ ($K=F_p((t))$. The image of $P$ contains the open 1ball, since for $x<1$ and $y=\sum_{n\ge 0} x^{p^n}$ we have $P(y)=x$. Hence $coker(P)$ is countable. 