Fundamental group of $Spin^c(2)$?

Question

Is the fundamental group of $Spin^c(2)$, the second complex spin group, also $\mathbb{Z}$? If so, how does one see this?

Just to avoid any confusion, my definition is: $$Spin^c(2) = (SO(2) \times \mathbb{S}^1)/\mathbb{Z}_2,$$ where $\mathbb{Z}_2$ is the group $\{(1, 1), (-1, -1)\}$.

I know that $Spin(2) \simeq SO(2) \simeq \mathbb{S}^1$, and I have the exact sequence

$$0 \rightarrow Spin(2) \rightarrow Spin^c(2) \rightarrow \mathbb{S}^1 \rightarrow 0,$$

where the first (non-trivial) map is given by inclusion and the second by the covering, so from this, going to homotopy, I get

$$0 \rightarrow \mathbb{Z} \rightarrow \pi_1(Spin^c(2)) \rightarrow \mathbb{Z} \rightarrow 0,$$

and this sequence splits, so it should be the case that $\pi_1(Spin^c(2)) = \mathbb{Z} \oplus \mathbb{Z}$. Is this correct? It doesn't seem to match other references that I have (Wikipedia, for instance).

Thanks!

Answer 1

In fact $\mathrm{Spin}^{\mathbb{C}}(2)\cong S^1\times S^1$ as Lie groups, so it does have $\pi_1\cong\mathbb{Z}\times\mathbb{Z}$.

The reason is that $\mathrm{Spin}(2)\cong S^1$ and the homomorphism $S^1\times S^1\to S^1\times S^1$ given by the formula $(z,w)\mapsto (zw,zw^{-1})$ has kernel $\pm(1,1)$ and is surjective. (So by the first isomorphism theorem the image $S^1\times S^1$ is isomorphic to the factor group $(S^1\times S^1)/S^0\cong\mathrm{Spin}^{\mathbb{C}}(2)$.)