# $\operatorname{Spin}^c(n)$ is a Lie group?

Assume that $n\geq 3$. (This implies $\pi_1(\operatorname{SO}(n))=\mathbb{Z}/2$.)

Let $\rho\colon \operatorname{Spin}(n)\to \operatorname{SO}(n)$ be the 2-fold universal cover. Identify $\ker(\rho)$ with $\{\pm 1\}$. Then, we can define $\operatorname{Spin}^c(n) =\operatorname{Spin}(n)\times S^1/ \{(1,1),(-1,-1)\}$.

Question 1. The subgroup {(1,1),(-1,-1)} is a normal subgroup of $\operatorname{Spin}(n)\times S^1$?

If question 1 is true, then I think the Lie group structures of $\operatorname{Spin}(n)$ and $S^1$ induces Lie group structure of $\operatorname{Spin}^c(n)$. But I couldn't prove Question 1.

Rather this, we have two smooth maps $\overline{\rho}\colon \operatorname{Spin}^c(n) \to \operatorname{SO}(n)$ and $\psi\colon \operatorname{Spin}^c(n)\to S^1$, defined by $\overline{\rho}([x,z])=\rho(x)$, $\psi([x,z])=z^2$. (Note that this is well-defined since we only have to check that $\overline{\rho}([x,z])=\overline{\rho}([-x,-z])$, $\psi([x,z])=\psi([-x,-z])$ which are trivial.)

Therefore, we have 2-fold covering space, $\overline{\rho}\times \psi\colon \operatorname{Spin}^c(n)\to \operatorname{SO}(n)\times S^1$.

Question 2. Since $\operatorname{SO}(n)\times S^1$ is a Lie group, is it true that $\overline{\rho}\times \psi$ induces a Lie group structure of $\operatorname{Spin}^c(n)$?

Question 3. If both Question 1 and Question 2 are true, then is it true that the induced Lie group structures of $\operatorname{Spin}^c(n)$ from both questions are same?

Finally, if you interpret this for low-dimensional case (i.e. $n=3,4$), then it would be really appreciated. Thank you in advance.

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It is not only normal but central. – Qiaochu Yuan Dec 28 '12 at 23:01