Can $\operatorname{Spin}(n)$ be defined through the short exact sequence?

Question

Define spin group $\operatorname{Spin}(n)$ to be the double cover group of $SO(n)$. Then it fits into the short exact sequence $0 \to \mathbb{Z}_2 \to \operatorname{Spin}(n) \to SO(n) \to 0$. My question is whether the converse is correct:

If for some Lie group $G$, sequence $0 \to \mathbb{Z}_2 \to G \to SO(n) \to 0$ is exact, then is $G$ necessarily isomorphic to $\operatorname{Spin}(n)$? If not, what condition need to be added?

Answer 1

No: consider the short exact sequence $0 \to \mathbb{Z}/2 \to \mathbb{Z}/2 \times SO(n) \to SO(n) \to 0$. However, these are the only two possibilities.

If $0 \to \mathbb{Z}/2 \to G \to SO(n) \to 0$ is a short exact sequence of Lie groups for $n > 2$, then $G = Spin(n)$ or $G = SO(n) \times \mathbb{Z}/2$.

For consider $\pi_1(SO(n)) = \mathbb{Z}/2$. The short exact sequence of groups gives a long exact sequence in homotopy: $$\cdots \to \pi_1(\mathbb{Z}/2) \to \pi_1(G) \to \pi_1(SO(n)) \to \pi_0(\mathbb{Z}/2) \to \pi_0(G) \to \pi_0(SO(n)) \to 0.$$ Now $\pi_1(\mathbb{Z}/2) = 0$ and $\pi_1(SO(n)) = \mathbb{Z}/2$.¹ We thus have two cases:

1: $\pi_1(G) = 0$. In this case, the exact sequence above gives $\pi_1(SO(n)) \to \pi_0(\mathbb{Z}/2)$ is injective and thus surjective, so $\pi_0(G) = 0$. This shows $G$ is the double cover of $SO(n)$.

2: $\pi_1(G) = \mathbb Z/2$. In this case, there is a splitting of the map $G \to SO(n)$ given by lifting $$\begin{array}{ccc} & & G \\ & & \downarrow \\ SO(n) & \rightarrow^{id} & SO(n) \end{array}$$ and sending $1$ to $1$. This lift $\mu$ is a group homomorphism by uniqueness of lifting: both $x\mapsto \mu(a)\mu(x)$ and $x \mapsto \mu(ax)$ are lifts of left-multiplication by $a$. Thus, the short exact sequence for $G$ splits.

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1. See this question for computation of $\pi_1(SO(n))$ for $n >2$.