Some context:
I'm trying to compute a fundamental group of $O(p,q)$ (the indefinite orthogonal group of signature $(p,q)$).
I'm aware that it is not connected so I've proven explicitly that it has four connected components. So the problem reduces to $SO^+(p,q).$
By the method that was shown to us, we should find a homogeneous space of this group.
Then the stabilizer should be a group of a similar type (like $SO^+(p-1,q)$ or something). And then we can apply an exact sequence of fibration to reduce the problem to a smaller dimension. (It should work since this homogeneous space is supposedly contractible).
It works for example for $SO^+(3,1)$ and for Lobachevsky space $\mathbb{H}^3$ that is contractible homogeneous space of the said group.
Yet in more general setting I was unable to determine which space to use.
Perhaps I should take $\{(x_0)^2+\ldots+(x_{p-1})^2-x_{p}^2-\ldots-x_{p+q-1}^2=1\}?$
However I don't know how to construct a contracting homotopy (if $p,q>0$) or how to correctly determine the stablizer of the point.
Any help would be appreciated!
P.S. I know there is a solution to this problem using Iwasawa decomposition and yet I don't want to go there. (However, it might be useful since it gives an answer immediately). This method suggests that $SO^+(p,q)$ contracts to the compact subgroup $SO(p)\times SO(q).$
UPD: I found out that I was actually quite wrong about contractability of the set $\{(x_0)^2+\ldots+(x_{p-1})^2-x_{p}^2-\ldots-x_{p+q-1}^2=1\}.$ It is in fact homotopy equivalent to $S^{p-1}.$ So using an exact sequence of fibration I managed to compute almost all fundamental groups except for the case of
$SO^+(3,3).$
Here is a part of long exact sequence I'm struggling with:
$$
\dots\to\pi_2(SO^+(3,3))\to \pi_2(S^2) \to \pi_1(SO^+(2,3))\to\pi_1(SO^+(3,3))\to \pi_1(S^2)=0 \to \dots
$$
I know that $\pi_2(SO^+(3,3))$ is zero, but I don't want to involve it.
Also, I computed $\pi_1(SO(2,3))\cong\mathbb{Z}\oplus\mathbb{Z}/2.$
So I want to show that the map $\pi_2(S^2)\cong\mathbb{Z}$ is injective. Then I want to understand why it maps $\pi_2(S^2)$ exactly to $2\mathbb{Z}\subset\mathbb{Z}\oplus\mathbb{Z}/2.$
Best Answer
I'm going to ignore your approach to suggest one which gives you the maximal compact quickly. It won't use much different than your ideas of playing with fiber sequences.
Consider the space $\mathcal S_{p,q}$ of splittings $\Bbb R^{p,q}$ as $V \oplus W$, where $V$ is a $p$-dimensional subspace where the metric is positive definite and $W$ is a $q$-dimensional subspace on which the metric is negative definite. Topologize this as a subspace of the quotient $GL(p+q)/GL(p) \times GL(q)$.
Observe that $SO^+(p,q)$ acts properly and transitively on $\mathcal S_{p,q}$. (I will leave this part of the argument to you, it is straightforward.)
Less obvious is that $\mathcal S_{p,q}$ is contractible. This will use a fiber sequence argument.
Write $\mathcal V_{p,q}$ for the space whose elements are bases $\{x_1, \cdots, x_p, y_1, \cdots, y_q\}$ for $\Bbb R^{p,q}$, where the metric is positive definite on the span $\langle x_1, \cdots, x_p\rangle$ and negative definite on the span $\langle y_1, \cdots, y_q\rangle$, topologized as a subspace of $GL(p+q)$. Then we have a fiber sequence $$GL(p) \times GL(q) \to \mathcal V_{p,q} \to \mathcal S_{p,q}.\require{AMScd}$$
We also have a fiber sequence $\mathcal V_{p-1,q} \to \mathcal V_{p,q} \to (\Bbb R^p \setminus 0)$, compatible with the actions of $GL(p-1) \times GL(q)$, which descends to a map $\mathcal S_{p-1, q} \to \mathcal S_{p,q}$; send $(V,W) \mapsto (V \oplus \langle e_p\rangle, W)$, where $e_p$ is the last basis vector in $\Bbb R^p$. So we have obtained the diagram
$$\begin{CD} GL(p-1) \times GL(q) @>>> GL(p) \times GL(q) @>>> (\Bbb R^p \setminus 0) \\ @VVV @VVV @|\\ \mathcal V_{p-1,q} @>>> \mathcal V_{p,q} @>>> (\Bbb R^p \setminus 0)\\ @VVV @VVV\\ \mathcal S_{p-1,q} @>>> \mathcal S_{p,q} \end{CD}$$
Now induct. If $\mathcal S_{p-1,q}$ is contractible, then the top-left vertical arrow is an equivalence, so by the 5-lemma the middle vertical arrow is an equivalence, and hence $\mathcal S_{p,q}$ is contractible. A nearly identical argument allows you to reduce the value of $q$.
For the base case, $\mathcal S_{0,0}$ is the one-point space. Therefore $\mathcal S_{p,q}$ is contractible for all $(p,q)$.
Because the stabilizer of $(\Bbb R^p, \Bbb R^q)$ is $SO(p) \times SO(q)$, we find we have a fiber sequence $$SO(p) \times SO(q) \to SO^+(p,q) \to \mathcal S_{p,q}$$ with contractible base, so that the inclusion $SO(p) \times SO(q) \to SO^+(p,q)$ is an equivalence.
I suspect there is a non-inductive, geometric argument that $\mathcal S_{p,q}$ is contractible using that the subspaces must intersect the null-cone trivially.