Homotopy group of $O(p,q)$

Question

I am interested in knowing the homotopy group of $O(p,q)$ as the orthogonal group of indefinite quadratic form over the reals.

Here $O(p,q)$ is defined as $$ O(p,q) ={O}(p,q; \mathbb{R}) = \left\{Q \in \operatorname{GL}(p+q, \mathbb{R}) \mid Q^\mathsf{T}I_{p,q} Q = Q I_{p,q} Q^\mathsf{T} = I_{p,q}\right\} $$ with the field $F=\mathbb{R}$. Here $$ I_{p,q} \equiv \begin{bmatrix} \begin{pmatrix} 1 & & \\ & \ddots & \\ & & 1 \end{pmatrix}_{p \times p} & 0 \\ 0 & \begin{pmatrix} - 1 & & \\ & \ddots & \\ & & -1 \end{pmatrix}_{q \times q}\\ \end{bmatrix} $$

So

1. what is the homotopy group: $\pi_d(O(1,d-1))=?$

2. what is the homotopy group: $\pi_d(O(p,q))=?$

If you can answer $\pi_d(O(1,d-1))$, then it is enough to post your thoughts as an answer. I do appreciate it.

I am particularly interested in $\pi_0(O(1,d-1))$, $\pi_1(O(1,d-1))$, and $\pi_2(O(1,d-1))$, and $d=0,\dots,5$.

The $\pi_2(G)=0$ for Lie group $G$. I suppose it still holds for $G$ is a noncompact nonabelian Lie group.

See also useful info here Homotopy groups O(N) and SO(N): $\pi_m(O(N))$ v.s. $\pi_m(SO(N))$.

Answer 1

As hinted by the comments and links therein, we have $$ \pi_d((,))=\pi_d((p) \times O(q))=\pi_d((p)) \times \pi_d(O(q)). $$ Then $$\pi_0((p))=\mathbb{Z}/2,$$ $$\pi_{d>1}((p))=\pi_{d>1}(S(p)).$$

Thus to give some example, say $(3,1)$, $$ \pi_0((3,1))=\pi_0((3)) \times \pi_0(O(1))=(\mathbb{Z}/2)^2. $$ $$ \pi_1((3,1))=\pi_1((3)) \times \pi_1(O(1))=\mathbb{Z}/2. $$ $$ \pi_2((3,1))=\pi_2((3)) \times \pi_2(O(1))=0. $$ $$ \pi_3((3,1))=\pi_3((3)) \times \pi_3(O(1))=\mathbb{Z}. $$ $$ \pi_4((3,1))=\pi_4((3)) \times \pi_4(O(1))=\mathbb{Z}/2. $$ $$ \pi_5((3,1))=\pi_5((3)) \times \pi_5(O(1))=\mathbb{Z}/2. $$ $$ \pi_6((3,1))=\pi_6((3)) \times \pi_6(O(1))=\mathbb{Z}/12. $$