12

Suppose that $G$ is a topological group and that $H$ is a subgroup of $G$ so that $H$ and $G/H$ are compact. I am trying to show that $G$ must be compact.

The first idea is to use the natural map $f: G \rightarrow G/H$, this is a closed map since $H$ is compact. Thus if $\mathscr{C}$ is a collection of closed subsets of $G$ with finite intersection property (FIP), then $\{f(C): C \in \mathscr{C}\}$ a collection of closed subsets of $G/H$. This collection also has FIP because

$$f(C_1 \cap \ldots \cap C_n) \subseteq f(C_1) \cap \ldots \cap f(C_n)$$

so $\{f(C): C \in \mathscr{C}\}$ has nonempty intersection because $G/H$ is compact. So there exists some coset $gH$ that is contained in every $f(C)$, that is $C \cap gH$ is nonempty. My next idea would be to show that $\{C \cap gH: C \in \mathscr{C}\}$ is a collection of closed sets in $gH$ with FIP. Then this collection, and thus $\mathscr{C}$ would have nonempty intersection.

We know that $C \cap gH$ is closed in $gH$, but I don't know how to show that it has FIP. Right now I think that might not even be true in general, since our choice of $gH$ was arbitrary. How can I choose $gH$ in the intersection $\cap_{C \in \mathscr{C}} f(C)$ so that $\cap_{C \in \mathscr{C}} C \cap gH$ is nonempty? Does this proof idea work?

Mikko Korhonen
  • 26,156

2 Answers2

12

The quotient map $f: G \rightarrow G/H$ is perfect: it is continuous, closed (as $H$ is compact, as you say) and has compact fibres (inverse images of points are compact); the latter is clear as all fibres are homeomorphic to $H$.

And the inverse image of a compact set under a perfect map is compact. And $G$ is the inverse image of $G/H$, which is compact by assumption.

The proof of the latter is not too hard (some details left for you):

Suppose $f: X \rightarrow Y$ is perfect, and let $K \subset Y$ be compact. Let $U_i, i \in I$ be an open cover of $f^{-1}[K]$. For each $k \in K$, we can cover $f^{-1}[\{k\}]$ by finitely many of the $U_i$, call the union of these $U(k)$. Then $O(k) = Y\setminus f[X \setminus U(k)]$ is an open neighbourhood of $k$ in $Y$ (this uses $f$ is a closed map), so finitely many of them cover $K$. The corresponding $U(k)$, and thus the finitely many $U_i$ that compose them, are the required finite subcover of $f^{-1}[K]$.

Another idea is to use that $G/H \times H$ is also compact, and the product (the group product) map is probably (haven't checked the details) a map onto $G$ from this space. This is maybe more appropriate for a course in topological groups...

Henno Brandsma
  • 250,824
  • Thanks, this works. I wonder if there is a way to do it as I originally tried. I'm not sure about that last part.. how would you define the map $G/H \times H \rightarrow G$? It would have to be continuous and surjective, I have doubts whether this is actually possible. – Mikko Korhonen Mar 06 '13 at 21:17
  • I withdraw the last idea, I don't think it'll work. I don't know about making the FIP idea work. There should be a way, as FIP's are just dual to covers anyway. – Henno Brandsma Mar 06 '13 at 21:26
  • Hi, why is the quotient map closed if $H$ is compact? BTW there's an argument using FIP in Chevalley's Theory of Lie Groups. Basically just take neighborhoods of the identity, say $U$'s, and use $U^{-1}C\cap gH$'s as the family of sets that satisfy the FIP on $gH$. – ah-- Jan 30 '19 at 06:19
2

WLOG we can assume $\mathscr{C}$ to be closed under finite intersection (if not, then just consider $\overline{\mathscr{C}}$ the extension of $\mathscr{C}$ by adding all finite intersections of $\mathscr{C}$). So, in that case if $\mathscr{F}$ is a finite collection of sets in $\mathscr{C}$, we have that $C'=\bigcap_{C\in \mathscr{F} } C$ is in $\mathscr{C}$ and therefore $\bigcap_{C\in \mathscr{F}} (C\cap gH)=gH\cap C'\neq\emptyset$.

Proving that $\{C\cap gH: C\in\mathscr{C}\}$ has the finite intersection property.

Julian Mejia
  • 4,458