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Let $H$ be a closed subgroup of a Lie group $G$. Then $H$ acts on $G$ by left translation. The action is:

(i): free

(ii): proper

Proof: (i) is clear. For (ii), we need to show that the preimage of compact sets under $p: H \times G \rightarrow G \times G, (h,g) \mapsto (hg,g)$ are compact. The image $S$ of $p$ is the set of $(g_1,g_2) \in G \times G$ such that $Hg_1 = Hg_2$. Then $S$ is closed, as it the preimage of the diagonal map of

$$G \times G \rightarrow H \backslash G \times H \backslash G$$

Now $p: H \times G \rightarrow S$ is a bijection with continuous inverse $p^{-1}: S \rightarrow H \times G$ given by $(g_1,g_2) \mapsto (g_1g_2^{-1},g_2)$. Thus if $K \subseteq G \times G$ is compact, so is $S \cap K$, then so is $p^{-1}(S \cap K) = p^{-1}(K)$.

General theory of Lie groups then tells us that the quotient $H \backslash G$ has a unique manifold structure such that the quotient map

$$\pi: G \rightarrow H \backslash G$$

is a submersion, and $\pi$ becomes a principal fibre bundle with $H$ as a fibre. This should imply that local trivial sections exist everywhere: $H \backslash G$ is covered by open sets $U$ such that $\pi^{-1}(U)$ looks like $H \times U$ with the action $h'.(h,u) = (h'h,u)$.

I have never thought about the quotient map of a group onto a coset space as looking anything like a principal fibre bundle (unless we have something like $G = H \times H'$). Suppose I did not know anything about $H \backslash G$ having a manifold structure. Is it possible to see directly that $\pi: G \rightarrow H \backslash G$ has local trivial sections?

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  • $\begingroup$ Just a comment about notation, $H/G$ is more common notation for the quotient space, whereas $H \backslash G$ usually represents set difference (also commonly notated as $H - G$). $\endgroup$ – Osama Ghani Jul 23 '18 at 8:49
  • $\begingroup$ For me, $H/G$ would mean the set of left cosets of $G$ in $H$, which doesn't make sense here because $H$ is a subgroup of $G$. I'm used to $G/H$ being the left cosets of $H$ in $G$, and $H \backslash G$ being the right cosets of $H$ in $G$. Unfortunately confusing with set difference, I know. $\endgroup$ – D_S Jul 24 '18 at 23:33
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The construction of such a section is actually part of the standard proof that $H\backslash G$ is a smooth manifold. Consider a linear subspace $V$ in the Lie algebra $\mathfrak g$ of $G$ that is complementary to the Lie subalgebra $\mathfrak h$ corresponding to $H$. Then one shows that there is a neighborhood $W$ of $0$ in $V$ such that $\exp(W)$ intersects $H$ only in the neutral element $e$. Possibily shrinking $W$, $\psi(X):=H\exp(X)$ defines a diffeomorphism from $W$ onto an open neighborhood of $o:=He\in H\backslash G$ (which is the inverse of a standard chart for the manifold structure on that space). The trivializing section around $o$ is then defined by $\sigma(\psi(X)):=\exp(X)$. Then you just transport this around using right translations on $G$ and $H\backslash G$ by elements of $G$.

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  • $\begingroup$ How do we know that for small enough $W$, $X \mapsto H \operatorname{exp}(X)$ is a homeomorphism onto an open set of $H \backslash G$? $\endgroup$ – D_S Oct 23 '18 at 5:08
  • $\begingroup$ In the actual construction of the set $W$, what show is that $W$ can be chosen small enough so that $(X,h)\mapsto \exp(X)h$ defines a diffeomorphism from $W\times H$ onto an open neighborhood of $H$ in $G$. (The map is easily seen to be bijective and a local diffeomorphism around $(0,e)$ and with a bit of work you show that it is a diffeomorphism onto its image and that the image is open). This readily implies that $\{H\exp(X)\}$ is open in $H\backslash G$, and the inverse to $X\mapsto H\exp(X)$ can be easily expressed in terms of the inverse of the above diffeomorphism. $\endgroup$ – Andreas Cap Oct 23 '18 at 10:00

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