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Questions tagged [topological-groups]

A topological group is a group $G$ together with a topology on the elements of $G$ such that the group operation and group inverse function are both continuous (with respect to the topology).

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3 votes
1 answer
2k views

Infinite products of topological groups

While studying for a topological groups course, I wondered if we could define the product of uncountably many topological groups such that the product is still a topological group. That is: let $G_i$ ...
commonname's user avatar
19 votes
9 answers
6k views

Haar measure on a quotient, References for

I remember reading Weil's "Basic Number Theory" and giving up after a while. Now I find myself thinking of it (thanks to some comments by Ben Linowitz). Right from the very beginning, Weil ...
Anweshi's user avatar
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20 votes
4 answers
3k views

Find a "natural" group that contains the quotient of the infinite symmetric group by the alternating subgroup

Let $S_\infty$ the group of permutations of $\mathbb{N}$. It can be shown that there is no homomorphism $S_\infty \to \mathbf{Z}/2$ extending the sign on the finite symmetric groups. Is it possible to ...
Martin Brandenburg's user avatar
6 votes
3 answers
324 views

Inverses in convolution algebras

Let $G$ be a locally compact totally disconnected group, and to make life easy let's suppose its Haar measure is bi-invariant. Let $C_c(G)$ be the space of locally constant complex functions on $G$ ...
D. Savitt's user avatar
  • 2,713
14 votes
3 answers
1k views

Countable subgroups of compact groups

What is known about countable subgroups of compact groups? More precisely, what countable groups can be embedded into compact groups (I mean just an injective homomorphism, I don't consider any ...
Konstantin Slutsky's user avatar
9 votes
2 answers
699 views

Potential connected non-Lie subgroup

This painful question is inspired by the question "non-Lie subgroups" . Let $f$ be a discontinuous additive map from $\mathbb{R} \to \mathbb{R}$. Is it possible that the graph of $f$, inside ...
David E Speyer's user avatar

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