Homotopy groups O(N) and SO(N): $\pi_m(O(N))$ v.s. $\pi_m(SO(N))$

I have known the data of $\pi_m(SO(N))$ from this Table:

I wonder whether there are some useful information that I can relate $\pi_m(SO(N))$ and $\pi_m(O(N))$?

Here is the difficulty somehow posted by MO to obtain $\pi_m(O(N))$, https://mathoverflow.net/questions/99663/homotopy-groups-of-on

But generally there seem to be some relations, like this:

Can I get the full Table of $\pi_m(O(N))$ from $m=1$~$10$ and $N=2$~$11$ precisely? Any literatures?

Solution 1:

As pointed out in the comments, $O(N)$ consists of two connected components which are both diffeomorphic to $SO(N)$. So $\pi_0(O(N)) = \mathbb{Z}_2$, $\pi_0(SO(N)) = 0$, and for $m \geq 1$, $\pi_m(O(N)) = \pi_m(SO(N))$.

As for $\operatorname{Spin}(N)$, note that is it a double cover of $SO(N)$. When $N = 1$, we see that $\operatorname{Spin}(1) = \mathbb{Z}_2$ so $\pi_0(\operatorname{Spin}(1)) = \mathbb{Z}_2$ and all its other homotopy groups are trivial, while for $N = 2$ we have $\operatorname{Spin}(2) = S^1$ which has first homotopy group $\mathbb{Z}$ and all higher homotopy groups trivial. When $N \geq 3$, $\operatorname{Spin}(N)$ is the universal cover of $SO(N)$ so $\pi_1(\operatorname{Spin}(N)) = 0$ and for $m \geq 2$, $\pi_m(\operatorname{Spin}(N)) = \pi_m(SO(N))$.

The groups $\pi_m(SO(N))$ for $1 \leq m \leq 15$ and $1 \leq N \leq 17$ are given in appendix A, section 6, part VII of the Encyclopedic Dictionary of Mathematics. The table can now be found on nLab.