Homotopy groups of projective Lie groups PO(N), PSO(N), and PSpin(N) Previously, we have learned from Homotopy groups O(N) and SO(N): $\pi_m(O(N))$ v.s. $\pi_m(SO(N))$ that:

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*$\pi_m(SO(N))$: a table consisting of the groups $\pi_m(SO(N))$ for $1 \leq m \leq 15$ and $2 \leq N \leq 17$ can be found on the nLab page for the orthogonal group in the section on homotopy groups.


*$\pi_m(O(N))$: $O(N)$ 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 $m \geq 1$, $\pi_m(O(N)) = \pi_m(SO(N))$.


*$\pi_m(\operatorname{Spin}(N))$: note that $\operatorname{Spin}(N)$ is 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 $m \geq 2$, $\pi_m(\operatorname{Spin}(N)) = \pi_m(SO(N))$.
This post means to ask the homotopy groups of projective Lie groups $PO(N)\equiv\frac{O(N)}{Z(O(N))}$,  $PSO(N)\equiv\frac{SO(N)}{Z(SO(N))}$, and $PSpin(N)\equiv\frac{Spin(N)}{Z(Spin(N))}$ where $Z(G)$ means the center subgroup of $G$.
Note that (please double check):
$$Z(SO(n))=\begin{cases}\mathbb{Z}_2,&n \text{ even but $n \neq 2$}\\
SO(2), & n=2.\\
0,&n \text{ odd}\end{cases}$$
$$Z(O(n))=\mathbb{Z}_2.$$
$$Z(Spin(n))=\begin{cases}\mathbb{Z}_2,&n\equiv1,3\,\mod  4\\ \mathbb{Z}_4&n\equiv 2\,\mod  4\\
\mathbb{Z}_2\oplus\mathbb{Z}_2&n\equiv0\;\mod 4\end{cases}$$

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*$\pi_m(PSO(N))$:

$$\pi_m(PSO(N))=?$$

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*$\pi_m(PO(N))$:
$$\pi_m(PO(N))=?$$


*$\pi_m(PSpin(N))$:
$$\pi_m(PSpin(N))=?$$
It seems that we have the group isomorphism $PSO(N)\cong PO(N) \cong PSpin(N)$, so we can answer all questions at once.
 A: Let's start with $PSO(N)$. First, the center of $SO(N)$ is not quite correct. Instead
$$Z(SO(N)) = \begin{cases} \{I\} & N\ \text{odd}\\ SO(2) & N = 2\\ \{\pm I\} & N > 2\ \text{even}.\end{cases}$$
As $PSO(1) = PSO(2) = 1$, from now on we will consider only $N \geq 3$. As $SO(N)$ is either equal to $PSO(N)$, or a double cover of it, we have $\pi_k(PSO(N)) \cong \pi_k(SO(N))$ for $k \geq 2$. For $N$ odd, we have $PSO(N) = SO(N)$ so $\pi_1(PSO(N)) = \pi_1(SO(N)) \cong \mathbb{Z}_2$. In order to also determine $\pi_1(PSO(N))$ for $N > 2$ even, we need a different argument.
Let $\pi : G \to H$ be a covering of topological groups, i.e. a covering map which is also a group homomorphism. If $K := \ker\pi$, then $G/K \cong H$, so the group of deck transformations of $\pi$ is isomorphic to $K$. Note that $K$ is a discrete normal subgroup, so if $G$ is connected, it follows that $K$ is central, i.e. $K < Z(G)$. Moreover, $Z(H) \cong Z(G/K) \cong Z(G)/K$, see this question. As $SO(N) \to PSO(N)$ is a covering (which is the identity if $N$ is odd) and $SO(N)$ is connected, we see that $Z(PSO(N)) = Z(SO(N)/Z(SO(N))) = Z(SO(N))/Z(SO(N)) = 1$, i.e. $PSO(N)$ is centreless. We also have the universal covering map $\pi : Spin(N) \to PSO(N)$ and $Spin(N)$ is connected, so $1 = Z(PSO(N)) \cong Z(Spin(N)/K) \cong Z(Spin(N))/K$, so $K \cong Z(Spin(N))$. As $K$ is also isomorphic to the group of deck transformations of $\pi : Spin(N) \to PSO(N)$, we see that
$$\pi_1(PSO(N)) \cong K \cong Z(Spin(N)) \cong \begin{cases}\mathbb{Z}_2\oplus\mathbb{Z}_2 & N = 4k\\
\mathbb{Z}_4 & N = 4k + 2, k > 0\\
Spin(2) & N = 2\\ 
\mathbb{Z}_2 & N = 2k + 1.\end{cases}$$
Now let's consider $PO(N)$. It is not the case that $PO(N) \cong PSO(N)$. First note that $Z(O(N)) = \{\pm I\}$. If $N$ is odd, then $-I \not\in SO(N)$ and $PO(N) \cong SO(N) = PSO(N)$. If $N$ is even, then $-I \in SO(N)$ and $PO(N)$ has two connected components, and the identity component is isomorphic to
$$SO(N)/\{\pm I\} \cong \begin{cases} PSO(N) & N > 2\ \text{even}\\
SO(2) & N = 2.\end{cases}$$
Finally, we have $PSpin(N) \cong PSO(N)$ for all $N$. To see this, note that
$$PSpin(N) \cong \frac{Spin(N)}{Z(Spin(N))} \cong \frac{Spin(N)/\{\pm 1\}}{Z(Spin(N))/\{\pm 1\}} \cong \frac{SO(N)}{Z(SO(N))} \cong PSO(N).$$
The penultimate isomorphism follows from the isomorphism $Z(Spin(n))/\{\pm 1\} \cong Z(SO(N))$. This is automatic for $N = 2$ as $Spin(2)$ and $SO(2)$ are abelian. For $N \neq 2$, note that the map $\pi : Spin(N) \to SO(N)$ restricts to a map $\pi|_{Z(Spin(N))} : Z(Spin(N)) \to Z(SO(N))$ with kernel $\{\pm 1\}$. As $|Z(Spin(N))/\{\pm 1\}| = |Z(SO(N))|$, we see that $\pi|_{Z(Spin(N))}$ is surjective* and hence the claim follows from the first isomorphism theorem.
These identifications can now be used to calculate the homotopy groups of $PO(N)$ and $PSpin(N)$ in terms of the homotopy groups of $PSO(N)$ established above.

*In general, if $\pi : G_1 \to G_2$ is a group homomorphism, then it restricts to a group homomorphism $\pi|_{Z(G_1)} : Z(G_1) \to Z(G_2)$ with kernel $\ker\pi\cap Z(G_1)$. It need not be surjective though, even if $\pi$ is surjective and $\ker\pi$ is a central subgroup (as is the case above). In particular, the observation $|Z(Spin(N))/\{\pm 1\}| = |Z(SO(N))|$ was necessary in order to deduce the surjectivity of $\pi|_{Z(Spin(N))}$.
For example, there is a surjective homomorphism $\pi : Q_8 \to \mathbb{Z}_2\oplus\mathbb{Z}_2$ given by $\pm 1 \mapsto (0, 0)$, $\pm i \mapsto (1, 0)$, $\pm j \mapsto (0, 1)$, and $\pm k \mapsto (1, 1)$. The kernel of $\pi$ is $\{\pm 1\}$ which is a central subgroup of $Q_8$; in fact $Z(Q_8) = \{\pm 1\}$. But the map $\pi|_{Z(Q_8)} : Z(Q_8) \to Z(\mathbb{Z}_2\oplus\mathbb{Z}_2)$ cannot be surjective as $|Z(\mathbb{Z}_2\oplus\mathbb{Z}_2)| = |\mathbb{Z}_2\oplus\mathbb{Z}_2| = 4 > 2 = |Z(Q_8)|$.
