General Linear Group is(not) compactly generated

Question

We know that any connected Lie Group is compactly generated.

I have a feeling that $\mathbb{GL}_n\mathbb{R}$ is not compactly generated. Is it true?

If it is how can I prove this? If not, what would be a simple example?

Answer 1

You can show that $GL_{n}(\mathbb{R})$ has two connected components (the set of matrices with positive determinant and the set with negative determinant). Showing that these are exactly the connected components can be done with an exact sequence argument, which I'll show if you really want me to but is actually a really fun exercise to do (hint: try considering the sequence $GL_{1} \subseteq GL_{2} \subseteq \cdots \subseteq GL_{n}.$)

Now, if you take the set $S$ of positive determinant matrices whose elements have absolute value $\le 1$, then this set is compact and generates all of the positive determinant matrices. Then, to get all of $GL_{n}(R)$, we can take as compact generating set as the disjoint union of $S$ with any single matrix of negative determinant. I hope that is a nice enough example.