# 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?

• $SO(N)$ is the connected component of the identity of $O(N)$, so all of their $\pi_k$ agree, $k \ge 1$. Commented Oct 3, 2017 at 4:52
• In the answer of the MO post you give there are also fibrations $SO(n-1)\rightarrow SO(n)\rightarrow S^{n-1}$ and $SO(n)\rightarrow O(n)\xrightarrow{det} \mathbb{Z}_2$ and $\mathbb{Z}_2\rightarrow Spin(n)\rightarrow SO(n)$. The first uses the action of $SO(n)$ on $S^{n-1}\subseteq \mathbb{R}^{n}$, the second is the determinant and the third is the universal cover. Hence all of $O(n)$, $SO(n)$, $Spin(n)$ have the same higher homotopy. Commented Oct 3, 2017 at 13:17
• @Tyrone, thanks, +1, this is a great summary. Commented Oct 3, 2017 at 15:22

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.
• Can you happen to comment also $𝜋_k(𝑆pin(𝑁))$? if we already know $𝜋_k(𝑆𝑂(𝑁))=𝜋_k(𝑆pin(𝑁)/\mathbb{Z}_2)$? Thank you! (There is a long exact sequence of homotopy groups, but I wonder can you say the explicit answer directly.) Commented Feb 14, 2021 at 16:20
• For $N \geq 3$, we have $\pi_1(Spin(N)) = 0$ and $\pi_k(Spin(N)) \cong \pi_k(SO(N))$ for $k \geq 2$. The remaining cases follow from the isomorphisms $Spin(1) \cong \mathbb{Z}_2$ and $Spin(2) \cong S^1$. Commented Feb 14, 2021 at 16:29
• Many thanks, let me make sure $𝜋_𝑘(\mathbb{Z}_n)$. Is that $𝜋_0(\mathbb{Z}_n)=\mathbb{Z}_n$, and $𝜋_𝑘(\mathbb{Z}_n)=0$ for $k\geq 1$? Commented Feb 14, 2021 at 16:56