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I want to prove that the orthogonal group $O(k,l)$ ( 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

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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 Jan 23 '13 at 1:20
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
up vote 2 down vote accepted

$O(p,q)$ 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.

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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 5 '13 at 20:41

I'm writing here in response to user32240 because my answer is too long.

I'm sorry, but i still don't understand what you mean. Let's suppose to have $A\in SO(p,q)$, let's set $e_1,\cdots, e_{p+q}$ an orthogonal basis for the vector space $V$ such that $\langle e_1,\cdots ,e_p\rangle$ and $\langle A(e_1),\cdots, A(e_p)\rangle$ are positive subspaces and $\langle e_{p+1},\cdots ,e_{p+q}\rangle$ and $\langle A(e_{p+1}),\cdots ,A(e_{p+q})\rangle$ are negative subspaces. If we write the matrix for $A$ with respect to the basis $e_1,\cdots ,e_{p+q}$ we get that the columns are still orthogonal and so i don't get why it would be useful to use gram schmidt at this point.

So i thought this other way: we call $p_1$ the projection onto $\langle e_1,\cdots e_p\rangle$ and $p_2$ the projection onto $\langle e_{p+1},\cdots,e_{p+q}\rangle$, they are both continuous. So we consider $p_1(\langle A(e_1),\cdots,A(e_p)\rangle)$ and we apply gram-schmidt on $p_1(A(e_1)),\cdots,p_1(A(e_p))$ obtaining let's say the vectors $f_1,\cdots,f_p$. Then we do the same thing on $A(e_{p+1}),\cdots,A(e_{p+q})$, i.e. we apply gram-schmidt on $p_2(A(e_{p+1})),\cdots,p_2(A(e_{p+q}))$ obtaining the vectors $f_{p+1},\cdots f_{p+q}$. Then the matrix which has for columns the components of the vectors $f_1,\cdots,f_{p+q}$ should be the required matrix.

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