Diffeomorphism theorem for Lie Groups?

Question

The integral lattice $\Bbb Z^n$ is a discrete subgroup of the Lie group $\Bbb R^n$. Therefore, it acts freely and properly discontinuously on $\Bbb R^n$ and the orbit space $\Bbb R^n/\Bbb Z^n$ has a smooth manifold structure. I wanted to show that $\Bbb R^n/\Bbb Z^n$ is diffeomorphic to the toral group $T^n = S^1\times\dots\times S^1$. I realized that I can show something more general:

Let $G$ be a Lie group and let $F:G \to H$ be a surjective Lie group homomorphim. If $\Gamma=\ker F$ is a discrete subgroup, then the orbit space $G/\Gamma$ is diffeomorphic to $H$.

Partial Proof: Let $\pi:G \to G/\Gamma$ be the quotient map. Define $\tilde{F}: G/\Gamma \to H$ by $\tilde{F}(\Gamma x) = F(x)$. This is a well defined bijection which is also a homeomorphism. Now since $\pi$ is a covering map, for each $p \in G/\Gamma$ there exists a connected neighbourhood $U$ of $p$ and a connected neighbourhood $\tilde{U}$ in $G$ such that $\pi|\tilde{U}: \tilde{U} \to U$ is a diffeomorphism. Therefore $\tilde{F}|U = F\circ\pi^{-1}$ and hence is a smooth map. Therefore $\tilde{F}$ is smooth because it's locally smooth.

This proof isn't complete since I haven't shown that $\tilde{F}^{-1}$ is smooth. This is where I'm stuck. It'll be helpful if someone gives a hint as to how to show that $\tilde{F}$ is a diffeomorphism.

I see why $\tilde{F}$ is continuous but why is it a homeomorphism? — Henry T., May 05 '24 at 00:47

score 2 · Accepted Answer · answered Jul 17 '20 at 12:18

2

One way to see this is as follows : you proved that $G/\Gamma \to H$ was a homeomorphism. $\Gamma$ is discrete so this implies that $G$ and $H$ have the same dimension.

Also, $G\to H$ is a submersion (you have to prove that), therefore on tangent spaces it is surjective, so by a dimension argument it is an isomorphism on tangent spaces, therefore $G\to H$ is a local diffeomorphism (by the local inversion theorem).

This should be enough to conclude

(By the way, you probably know this but of course the $T^n \cong \mathbb{R^n/Z^n}$ case is completely elementary and easier than the general case)

answered Jul 17 '20 at 12:18

Maxime Ramzi

45,086

Okay so despite several tries I'm not able to show that $F$ is a submersion. Can show $F$ is a submersion? – General Grievous Jul 18 '20 at 10:44
Do you know Sard's theorem ? – Maxime Ramzi Jul 18 '20 at 10:51
Thanks I think I can prove it now – General Grievous Jul 18 '20 at 11:09
So if the rank is not maximum at a point, it cannot be maximum at any other point(lie group homomorphisms have constant rank). This contradicts sards theorem. – General Grievous Jul 18 '20 at 11:45

score 1 · Answer 2 · answered Jul 17 '20 at 12:27

This is true in a more general setting: Suppose that $F:G\to H$ is a homomorphism of Lie groups whose derivative $F':\mathfrak g\to\mathfrak h$ is surjective and that $H$ is connected. Then $ker(F)$ is a closed normal subgroup of $G$ and thus a Lie subgroup. Moreover, the space $G/H$ of left cosets (or equivalently of $H$-orbits in $G$) canonically is a smooth manifold and hence a Lie group and the canonical map $p:G\to G/ker(F)$ is surjective homomorphism of Lie groups and a smooth submersion. Next, surjectivity of $F'$ and connectedness of $H$ imply that $F$ is surjective. Thus $F$ induces a bijective homomorphism $\underline{F}:G/ker(F)\to H$ such that $\underline{F}\circ p=F$. Since $p$ is a surjective submersion, this implies that $\underline{F}$ is a smooth bijective homomorphism of Lie groups. Finally, it is well known that the Lie algebra of $ker(F)$ is $\ker(F')$, which immediatey implies that $\underline{F}$ has bijective derivative. This easily implies that all tangent maps of $\underline{F}$ are linear isomorphisms. Hence $\underline{F}$ is a local diffeomorphism and the inverse function theorem shows that its inverse is smooth. Thus $\underline{F}:G/ker(F)\to H$ is an isomorphism of Lie groups.

Diffeomorphism theorem for Lie Groups?

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