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I have a question about short exact sequence.

Notation: $\mathbb{T}= S^1$.

Let $F$ be a finite abelian group, and let $G$ be a compact abelian group, and assume we have a short exact sequence $$1\rightarrow F\rightarrow G\rightarrow \mathbb{T}\rightarrow 1$$ Is this sequence necessarily splits? If not is there anything we can say about $G$, is it a Lie group?

Thanks!

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2 Answers 2

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I believe we can say $G$ is a Lie group. The translation action of $F$ on $G$ is free, and since $F$ is finite, it is properly discontinuous. Hence the quotient map $G \to \mathbb{T}$ is a covering map, and so you can lift the smooth structure up to $G$.

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  • $\begingroup$ Yes that's the answer I was looking for, thanks. $\endgroup$
    – Yanko
    Commented Jun 25, 2017 at 16:28
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The sequence is clearly not split in general. Identify $S^1$ with the complex numbers of norm $1$, and consider $$1 \longrightarrow \mu_n \longrightarrow S^1 \stackrel{f_n}{\longrightarrow} S^1 \longrightarrow 1,$$ where $f_n(z)=z^n$ and $\mu_n$ is the cyclic group of $n^{\mathrm{th}}$ roots of unity.

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  • $\begingroup$ Yes you are right! But is it true that $G$ is a Lie group? $\endgroup$
    – Yanko
    Commented Jun 25, 2017 at 16:05

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