On conditions to prove that a Lie group is compact.

Question

Can anyone help me with this problem: If G is a Lie group and H a compact subgoup of G, such that G/H is also compact, how to prove that G is compact? I have tried to push forward an open covering of G to G/H be the quotient map, apply compactness and then pull it back, but there are subtle accommodations that should be made and I can not figure out how. Thanks in advance.

Answer 1

Root idea: take an open cover $$ G = \bigcup_{\alpha} U_\alpha. $$ Since $H$ is compact, so is every coset $gH$. So, $G$ can be partitioned into compact components, so to speak. And, note that the quotient space $G/H$ is also compact, by assumption.

So, each coset $gH$ is also covered by the same covering, but by compactness you only need a finite number of them. Denote this cover by $$ \mathcal{U}_g = \bigcup_{i=1}^{n_g} U_{\alpha_i}^g $$ where I hope it is somehow clear that every $U_{\alpha_i}^g$ is one of the original $U_\alpha$'s. (There may very well be better notation and indexing, but I'm in a hurry.)

As $gH \subseteq \mathcal{U}_g$ and $q$ is a closed map there is an open set $W_g$ in $G/H$ such that $q^{-1}(W_g) \subseteq \mathcal{U}_g$. Now $\bigcup_{g \in G} W_g$ covers $G/H$ and by compactness this can be done finitely: $$ G/H = \bigcup_{i=1}^r W_{g_i}. $$ Now, if you take the preimage under $q$ of both sides you get $$ G = \bigcup_{i=1}^r q^{-1}(W_{g_i}). $$ But now we have $$ G = \bigcup_{i=1}^r q^{-1}(W_{g_i}) \subseteq \bigcup_{i=1}^r \mathcal{U}_{g_i} = \bigcup_{i=1}^r \bigcup_{k=1}^{n_{g_i}} U_{\alpha_k}^{g_i}. $$ This last union is a finite one derived from the original cover, so $G$ is compact.