## New answers tagged topological-groups

1

Taking for granted the
Lemma. Let $(X,\mathcal{A},\mu)$ and $(Y,\mathcal{B},\nu)$ be measure spaces, and let $(\mathbb{B},\|\;\|)$ be a Banach space. Let $F: X\times Y\to B$ be strongly measurable, with $\sigma$-finite support. Then there exists a sequence of simple functions $(F_n)$ converging a.e. to $F$ (w.r.to $\mu\otimes\nu$), and such that ...

1

(edit. I turned the preceding incorrect answer into a partial answer: yes with an ugly technical assumption).
Let $(X,\mathcal{A},\mu)$ and $(Y,\mathcal{B},\nu)$ be measure spaces, and let $(\mathbb{B},\|\;\|)$ be a Banach space.
Lemma. Let $F: X\times Y\to B$ be strongly measurable, with $\sigma$-finite support. Then there exists a sequence of simple ...

2

I think the answer is no, even in the case of $G:=\mathbb{R}$ with the Lebesgue measure and $B: =\mathbb{R}$ as a Banach space.
Let $F: \mathbb{R}\times \mathbb{R} \to \mathbb{R} $ be the characteristic function of the half-plane above the diagonal: $F(x,y):=\chi_\mathbb{{R}_+}(y-x)$. So $F\in L^\infty(\mathbb{R}\times \mathbb{R} )$; however the ...

3

If $G$ is finitely generated, then $G$ is isomorphic to its profinite completion by a result of Nikolov and Segal mentioned by Ian Agol. Thus, what are you asking is equivalent to the goodness introduced by Serre (see J. P. Serre, Galois cohomology, I.2.6).
The only good finitely generated profinite groups that I know are virually polycyclic (it can be ...

0

As I already told at the workshop, Taras Banakh and me solved this problem. But, surprisingly and converse to that I told at the workshop, the answer is affirmative. This is why is good to write a common paper: if one of the authors is trying to prove a conjecture and another is trying to refute it then one of them should be right. :-)

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Two non-discrete group topologies which together generate the discrete topology are sometimes called transversal. It is proved in "On transversal group topologies" by Dikranjan, Tkachenko and Yaschenko (see Theorem 3.13) that no totally bounded group topology on an infinite abelian group admits a transversal group topology. In particular, the answer to your ...

4

For such a general assertion, independent of classification, you'd need some one of the F. Bruhat and J. Tits papers, but (since I do not have copies nearby) I could not point you to any precise location within them.
F. Bruhat and J. Tits, BN-paires de type affine et donnees radicielles, C.R. Acad. Sci. Paris serie A, vol 263 (1966), pp. 598-601.
F. ...

1

It is not the answer but an important point, too long for a comment. There is no Hausdorff group topology on Z so that the sequence of prime numbers $2, 3, 5 ...$ converges to zero. If there were such topology, there would be finitely many pairs of primes of each given gap $k$. But that contradicts Zhang's Gap Theorem

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