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I know how to prove the following result. However, my proof is a little bit long and complicated and only uses fairly low tech results in group cohomology. It would be nice if I could find a citation in the literature that directly implies this claim but I am not experienced in group theory or homological algebra and to be honest I'm not sure where to start looking for such a result.

Lemma: Suppose $\mathbb{T}^n = \mathbb{R}^n/ \mathbb{Z}^n $ is the standard $n$-dimensional torus and $\Gamma$ is a finite group of order $k$. Consider a short exact sequence: $$ 0 \to \mathbb{T}^n \to E \to \Gamma \to 1. $$ Suppose $ R := \{ t \in \mathbb{T}^n \mathrel| kt = 0 \} $. Then the following short exact sequence splits: $$ 0 \to \mathbb{T}^n/R \to E/R \to \Gamma \to 1. $$

The proof basically boils down to the fact that $ H^2( \Gamma, \mathbb{T}^n) $ is annihilated by $k$ and using that to show that the induced map $ H^2(\Gamma, \mathbb{T}^n) \to H^2(\Gamma, \mathbb{T}^n/R) $ is the zero map and therefore the curvature class of the top short exact sequence projects to zero.

I guess the things I would like to know are:

  • Is this 'obvious' to those experienced with finite group extensions or Lie theory?
  • Is there a textbook/paper where I can find a result or exercise that directly implies this lemma?
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  • $\begingroup$ Using \text inside math mode for formatting results in funny-looking text. Use stars **stars**` rather than $\textbf{textbf}$ $\textbf{textbf}$. I have edited accordingly. (Also, I think that this is an interesting question!) $\endgroup$
    – LSpice
    Commented Jun 25, 2020 at 17:29
  • $\begingroup$ Yes, it is almost obvious. $\endgroup$ Commented Jun 25, 2020 at 17:58
  • $\begingroup$ @MikhailBorovoi, as a sorta-Lie theorist for whom it is not obvious, would you be willing to explain why it should be obvious? $\endgroup$
    – LSpice
    Commented Jun 25, 2020 at 18:50
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    $\begingroup$ It is almost obvious, see my one-page answer. $\endgroup$ Commented Jun 25, 2020 at 19:11

1 Answer 1

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(I write an answer rather than a comment in order to accommodate exact sequences.)

Let $$0\to T\to E\to\Gamma\to 1\tag{$E_1$}$$ be your first group extension, where $T$ is the torus ${\Bbb R}^n/{\Bbb Z}^n$. Write $R_k\subset T$ for the kernel of multiplication by $k$ in $T$ and consider your second exact sequence $$0\to T/R_k\to E/R_k\to\Gamma\to 1.\tag{$E_2$}$$ To the extension $(E_1)$ we associate its cohomology class $\eta_1\in H^2(\Gamma,T)$, and to extension $(E_2)$ we associate its class $\eta_2\in H^2(\Gamma,T/R_k)$. Then it follows from the constructions of $\eta_1$ and $\eta_2$ that $\eta_2$ is the image of $\eta_1$ under the homomorphism $$\phi_*\colon H^2(\Gamma,T) \to H^2(\Gamma,T/R_k)$$ induced by the canonical homomorphism $$\phi\colon T\to T/R_k.$$ We have not yet used the assumption that $T$ is a torus and that $\#\Gamma=k$.

Now consider the surjective homomorphism $$\alpha\colon T\to T,\quad x\mapsto kx.$$ Its kernel is $R_k$, and so it induces an isomorphism $$\alpha_*\colon T/R_k\to T.$$ Identifying $T/R_k$ with $T$ using $\alpha_*$, we obtain that our $$\phi\colon T\to T$$ is multiplication by $k$. It follows that $$\phi_*\colon H^2(\Gamma,T) \to H^2(\Gamma,T)$$ is multiplication by $k$ as well. Since $\Gamma$ is a group of order $k$, multiplication by $k$ annihilates $H^2(\Gamma,T)$. See Corollary 1 of Proposition 8 in Section 6, page 105, of: Atiyah and Wall, Cohomology of groups, in: Cassels and Fröhlich (eds.), Algebraic Number Theory, Acad. Press 1967, pp. 94-115. It follows that $\eta_2=0$ and the sequence $(E_2)$ splits.

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  • $\begingroup$ Without details from @JoelVillatoro it's hard to know for sure, but this seems similar to the proof that they very briefly sketched in the post. $\endgroup$
    – LSpice
    Commented Jun 25, 2020 at 19:26
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    $\begingroup$ The argument uses only the following property of $T$: the map $$T\to T, \quad x\mapsto kx$$ is surjective. $\endgroup$ Commented Jun 25, 2020 at 19:26
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    $\begingroup$ @LSpice: OP complained that his proof is low tech. Mine is not. $\endgroup$ Commented Jun 25, 2020 at 19:34
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    $\begingroup$ @MikhailBorovoi, thank you very much for your reply. Your proof here is certainly much better than the one I wrote originally. Certainly more efficient. $\endgroup$ Commented Jun 25, 2020 at 19:57

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