## New answers tagged topological-groups

1

The answer is Yes, provided that for every $\alpha$ and every $x\in G$, $f_\alpha(x^{-1})=f_\alpha(x)$.
Since we already completed the construction of the Haar measure we can actually use it in our proof. I suppose that one can show this also elementarily, but why bother?
Choose a left Haar measure on $G$.
Let me assume, as the expression in consideration ...

3

Since it is a metric space, your question is equivalent to asking whether it is separable (i.e. has a dense countable subset). Now, under some assumption on the manifold (compact is ok for sure, but second-countable is probably enough too), $\mathrm{Ham}_c(M,\omega)$ is second-countable (hence separable) for the $C^1$-topology, as a subspace of the ...

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