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I have known the data of $\pi_m(SO(N))$ from this Table: $$\overset{\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\quad\textbf{Homotopy groups of orthogonal groups}}{\begin{array}{lccccccccc} \hline & \pi_1 & \pi_2 & \pi_3 & \pi_4 & \pi_5 & \pi_6 & \pi_7 & \pi_8 & \pi_9 \\ \hline SO(2) & Z & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ \hline SO(3) & \boxed{Z_2} & 0 & Z & Z_2 & Z_2 & Z_{12} & Z_2 & Z_2 & Z_3 \\ \hline SO(4) & Z_2 & \boxed{0} & (Z)^{\times2} & (Z_2)^{\times2} & (Z_2)^{\times2} & (Z_{12})^{\times2} & (Z_2)^{\times2} & (Z_2)^{\times2} & (Z_3)^{\times2} \\ \hline SO(5) & Z_2 & 0 & \boxed{Z} & Z_2 & Z_2 & 0 & Z & 0 & 0 \\ \hline SO(6) & Z_2 & 0 & Z & \boxed{0} & Z & 0 & Z & Z_{24} & Z_2 \\ \hline SO(n),\;\ n>6 & Z_2 & 0 & Z & 0 & 0 & 0 & & & \\ \hline \end{array}}$$

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: enter image description here

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

Suicidal Person
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wonderich
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    $SO(N)$ is the connected component of the identity of $O(N)$, so all of their $\pi_k$ agree, $k \ge 1$. – Qiaochu Yuan Oct 03 '17 at 04:52
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    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.

     – Tyrone Oct 03 '17 at 13:17
  • @Tyrone, thanks, +1, this is a great summary. – wonderich Oct 03 '17 at 15:22

1 Answers1

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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.

Michael Albanese
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  • Can you happen to comment also $_k(pin())$? if we already know $_k(())=_k(pin()/\mathbb{Z}_2)$? Thank you! (There is a long exact sequence of homotopy groups, but I wonder can you say the explicit answer directly.) – wonderich Feb 14 '21 at 16:20
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    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 $Spin(2) \cong S^1$. – Michael Albanese Feb 14 '21 at 16:29
  • Many thanks, let me make sure $(\mathbb{Z}_n)$. Is that $_0(\mathbb{Z}_n)=\mathbb{Z}_n$, and $(\mathbb{Z}_n)=0$ for $k\geq 1$? – wonderich Feb 14 '21 at 16:56
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    Yes, that's correct. – Michael Albanese Feb 14 '21 at 17:29