why the orthogonal group $O(k,l)$ is homotopy equivalent to $SO(K)\times SO(l)$

Question

I want to prove that the orthogonal group $O(k,l)$ (http://en.wikipedia.org/wiki/Indefinite_orthogonal_group)is homotopy equivalent to $SO(k)\times SO(l)$, so that $\pi_1(O(k,l))=\pi_1(SO(k))\times\pi_1(SO(l))$. I've searched in the referenced books of the wiki page, but no one seems to prove this statement. Do you have any idea of the proof or where i can find it? thank you very much

Answer 1

$O(p,q)$ obviously has the same fundamental group as $SO(p,q)$. The wikipedia commented that $SO(p,q)$ as a maximal compact subgroup $SO(p)\times SO(q)$. However is not a proof.

Now if $x$ is an element in $SO(p,q)$, then we can Gram-Schimdt $x$ into the standard form. Since Gram-Schimdt is a continuous process, in the end $x$ changes into an element with diagonal entries. Thus it is possible to change the top part of $x$ into $SO(p)$ and the bottom part into $SO(q)$. Therefore $SO(p,q)$ is homotopically equivalent to $SO(p)\times SO(q)$. I guess you do not really need Gram-Schimdt in the proof, but for now I do not know how.