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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

ciccio
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  • minor nitpick: $O(k,l)$ isn't connected, while $SO(k)\times SO(l)$ is so they can't be homotopy equivalent. That said, I believe each component of $O(k,l)$ is homotopy equivalent to $SO(k)\tims SO(l)$ and user32240 gives the reason why. – Jason DeVito - on hiatus Jan 23 '13 at 01:20
  • i'm sorry, i made a mistake! of course you're right. what i ment is that the connected component $O(k,l)^++$ is homotopy equivalent to $SO(k)\times SO(l)$. $O(k,l)^++$ is the connected component of the transformations which mantein the orientation on each maximal positive and negative space – ciccio Jan 23 '13 at 15:04

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$O(p,q)$ 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.

Bombyx mori
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  • i'm sorry, in the question i ment $SO^+(l,k)$ (using the wiki notation), the subgroup of $O(l,k)$ which manteins the orientation of both maximal positive and negative subspaces. i want to prove that $SO^+(l,k)$ is homotopy equivalent to $SO(l)\times SO(K)$. – ciccio Jan 23 '13 at 16:18
  • I don't understand how the gram-schmidt process should give me the homotopy..what do you mean by "standard form"? can you explain? thank you – ciccio Jan 23 '13 at 16:27
  • $SO(p,q)$ is connected, if I am not mistaken. The Gram-Schimdt process gives you a deformation retract from $SO(p,q)$ to $SO(p)\times SO(q)$. What it does is to change a given matrix into diagonal form - think about $SO(p,q)$ as $p+q$ orthogonal vectors. Suppose we pick up the first vector and change the rest according to Gram-Schmidt, in the end with the new basis we get a diagonal matrix much as standard diagonalization of symmetric non-degenerate bilinear form. This process is continuous in $SO(p,q)$ and hence give you a deformation. – Bombyx mori Jan 23 '13 at 21:11
  • i wrote an aswer in response to you – ciccio Feb 05 '13 at 20:41