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I am trying to understand connections in fibre bundles. I thought of the following problem:

Let $\Gamma$ be the discrete group generated by

\begin{pmatrix} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 6 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}

Then $\Gamma$ acts properly discontinuously on the Heisenberg group $H$ with compact quotient. The extension $0\to C(\Gamma) \to \Gamma \to \Gamma/C(\Gamma) \to 0$ shows that $H/\Gamma$ is a circle bundle over the torus.

The numbers $3,4$ seem to determine a Riemannian structure on the torus, as when one quotients $\mathbb R^2$ by $(3\mathbb Z,4\mathbb Z)$ & similarly $6$ determines a Riemannian structure on the circle.

The standard flat connection on $H \cong \mathbb R^2 \times \mathbb R$ should induce a connection on the quotient $H / \Gamma$. Is there a way to see the numbers $3,4,6$ explicitly in the connection form of the quotient?

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  • $\begingroup$ May I ask how do you get the "compact quotient"? $\endgroup$ – Bombyx mori Jul 22 '14 at 15:38
  • $\begingroup$ @Bombyxmori I think it is reasonable, considering that I "act in every direction". The formal argument I know invokes Poincaré duality to conclude: if a group $\Gamma$ acts (faithfully) on a contractible manifold $X$ and $\mathrm{cd} \, \Gamma = \dim X$, then $X / \Gamma$ is compact. ($\mathrm{cd}$ being the "cohomological dimension".) $\endgroup$ – Earthliŋ Jul 22 '14 at 15:48

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