Definition of $\operatorname{Spin}^c(n)$ group

Question

I have trouble with the definition of the $\operatorname{Spin}^c(n)$-group.

First off (for example in this paper), the group $\operatorname{Spin}^c(n)$ is defined as the double cover of $\operatorname{SO}(n)\times\operatorname{U}(1)$.

I.e., one wants (?) the exact sequence $$1\rightarrow \mathbb{Z}_2\rightarrow \operatorname{Spin}^c(n) \rightarrow \operatorname{SO}(n)\times\operatorname{U}(1)\rightarrow 1.$$

However, this is then followed by the formula $$\operatorname{Spin}^c(n) = \operatorname{Spin}(n)\times_{\mathbb{Z}_2} \operatorname{U}(1)= (\operatorname{Spin}(n)\times \operatorname{U}(1))/\{\pm (1,1)\},$$

which I don't understand in this relation: By dividing out the $\pm 1$ from $\operatorname{Spin}(n)$, shouldn't we just get $\operatorname{SO}(n)$? Or in other words, this formula leads to the exact sequence $$1\rightarrow \mathbb{Z}_2\rightarrow \operatorname{SO}(n)\times\operatorname{U}(1)\rightarrow\operatorname{Spin}^c(n)\rightarrow 1.$$

So where is my mistake in thinking?

Answer 1

There are three different quotients of $\operatorname{Spin}(n)\times U(1)$ by central subgroups of order two:

1. $\dfrac{\operatorname{Spin}(n)\times U(1)}{\{(1, 1), (-1, 1)\}} \cong \dfrac{\operatorname{Spin}(n)}{\{\pm 1\}}\times U(1) \cong SO(n)\times U(1)$,
2. $\dfrac{\operatorname{Spin}(n)\times U(1)}{\{(1, 1), (1, -1)\}} \cong \operatorname{Spin}(n)\times \dfrac{U(1)}{\{\pm 1\}} \cong \operatorname{Spin}(n)\times U(1)$, and
3. $\dfrac{\operatorname{Spin}(n)\times U(1)}{\{(1, 1), (-1, -1)\}} =: \operatorname{Spin}^c(n)$.

In the first two cases, we are just taking a $\mathbb{Z}_2$ quotient of one of the factors, while in the last case, we're taking the so-called 'diagonal' $\mathbb{Z}_2$ quotient (i.e. the $\mathbb{Z}_2$ subgroup is not coming from a subgroup of either factor). All three of these quotients are distinct (at least for $n > 2$).

There is also the quotient of $\operatorname{Spin}(n)\times U(1)$ by a central subgroup of order four (isomorphic to $\mathbb{Z}_2\times\mathbb{Z}_2$):

$$\dfrac{\operatorname{Spin}(n)\times U(1)}{\{(1, 1), (-1, 1), (1, -1), (-1, -1)\}} \cong \dfrac{\operatorname{Spin}(n)}{\{\pm 1\}}\times\dfrac{\operatorname{U(1)}}{\{\pm 1\}}\cong SO(n)\times U(1).$$

The group $SO(n)\times U(1)$ arises as a $\mathbb{Z}_2$ quotient of all three groups above (in particular, they are all double covers of $SO(n)\times U(1)$). In the first two cases, the groups are obtained by quotienting $\operatorname{Spin}(n)\times U(1)$ by a $\mathbb{Z}_2$ subgroup in one of the factors, then the next quotient is by the $\mathbb{Z}_2$ subgroup in the other factor. For the final case, let $[(a, z)]$ denote the element of $\operatorname{Spin}^c(n)$ corresponding to $(a, z) \in \operatorname{Spin}(n)\times U(1)$ under the quotient map $\operatorname{Spin}(n)\times U(1) \to \operatorname{Spin}^c(n)$. Note that $[(-a, -z)] = [(a, z)]$. Now, by the third isomorphism theorem, we have

\begin{align*} SO(n)\times U(1) &\cong\dfrac{\operatorname{Spin}(n)\times U(1)}{\{(1, 1), (-1, 1), (1, -1), (-1, -1)\}}\\ &\cong \frac{\frac{\operatorname{Spin}(n)\times U(1)}{\{(1, 1), (-1,-1)\}}}{\frac{\{(1, 1), (-1, 1), (1, -1), (-1, -1)\}}{\{(1, 1), (-1, -1)\}}}\\ &\cong \dfrac{\operatorname{Spin}^c(n)}{\{[(1, 1), (1, -1)]\}}. \end{align*}

So we do indeed have the short exact sequence

$$1 \to \mathbb{Z}_2 \to \operatorname{Spin}^c(n) \to SO(n)\times U(1) \to 1$$

where the map $\mathbb{Z}_2 \to \operatorname{Spin}^c(n)$ is given by $1 \mapsto [(1, 1)]$ and $-1 \mapsto [(1, -1)]$.