Isomorphisms of Pin groups

Question

My goal is to identify the $Pin$ group $$ 1 \to Spin(n) \to Pin^{\pm}(n) \to \mathbb{Z}_2 \to 1 $$ such that $Pin^{\pm}(n)$ are isomorphisms to other more familiar groups.

My trick is that to look at the center $Z$ of $Pin$ group, say $Z(Pin(n))$.

We see that, for $k \in \mathbb{Z}^+$,

when $n=4k+1$, $$ Z(Pin^+(4k+1))=\mathbb{Z}_2 \times \mathbb{Z}_2, $$ $$ Z(Pin^-(4k+1))=\mathbb{Z}_4. $$

when $n=4k+3$, $$ Z(Pin^+(4k+3))=\mathbb{Z}_4, $$ $$ Z(Pin^-(4k+3))=\mathbb{Z}_2 \times \mathbb{Z}_2. $$

when $n=4k$ or when $n=4k+2$, $$ Z(Pin^+(n))=Z(Pin^-(n))=\mathbb{Z}_2. $$

So my naive attempt is that we have the following $Pin$ group isomorphisms:

$$Pin^+(1)=\mathbb{Z}_2 \times \mathbb{Z}_2, \quad Pin^-(1)=\mathbb{Z}_4.$$

$$Pin^+(3)=SO(3) \times \mathbb{Z}_ 4 \text{ or } \frac{Spin(3) \times \mathbb{Z}_4}{\mathbb{Z}_2}, \quad Pin^-(3)=Spin(3) \times \mathbb{Z}_2.$$

$$Pin^+(5)=Spin(5) \times \mathbb{Z}_2,\quad Pin^-(5)=SO(5) \times \mathbb{Z}_4 \text{ or } \frac{Spin(5) \times \mathbb{Z}_4}{\mathbb{Z}_2}.$$

Question 1: Am I correct about the above statements? More generally, do we have*

$$Pin^+(4k+1)\cong Spin(4k+1) \times \mathbb{Z}_2,\quad Pin^-(4k+1)\cong SO(4k+1) \times \mathbb{Z}_4.$$

$$Pin^+(4k+3)\cong SO(4k+3) \times \mathbb{Z}_ 4, \quad Pin^-(4k+3)\cong Spin(4k+3) \times \mathbb{Z}_2.$$

I am not so sure, $Pin^+(3)=SO(3) \times \mathbb{Z}_ 4$, could it be that instead $Pin^+(3)=\frac{Spin(3) \times \mathbb{Z}_4}{\mathbb{Z}_2}?$ (It looks that mine agrees with Wikipedia, but I doubt Wikipedia may be wrong about $Pin^+(3)=SO(3) \times \mathbb{Z}_ 4$...) possibly only for the Lie algebra, not for the Lie group.

Could it be that:

$$Pin^+(4k+1)\cong Spin(4k+1) \times \mathbb{Z}_2,\quad Pin^-(4k+1)\cong \frac{Spin(4k+1) \times \mathbb{Z}_4}{\mathbb{Z}_2}.$$

$$Pin^+(4k+3)\cong \frac{Spin(4k+3) \times \mathbb{Z}_ 4}{\mathbb{Z}_ 2}, \quad Pin^-(4k+3)\cong Spin(4k+3) \times \mathbb{Z}_2.$$

Question 2: Are there other $Pin$ group isomorphisms for $n=4k$, or $4k+2$, say $$Pin^{+/-}(4k)\cong ?$$ and $$Pin^{+/-}(4k+2)\cong?$$

Thanks for your comments in advance.

Answer 1

The second version given in the question is correct: \begin{align} \mathrm{Pin}^+(4k+1) &\cong \mathrm{Spin}(4k+1) \times \mathbb{Z}_2 \\ \mathrm{Pin}^-(4k+1) &\cong (\mathrm{Spin}(4k+1) \times \mathbb{Z}_4)/\mathbb{Z}_2 \\ \mathrm{Pin}^+(4k+3) &\cong (\mathrm{Spin}(4k+3) \times \mathbb{Z}_ 4)/\mathbb{Z}_2 \\ \mathrm{Pin}^-(4k+3) &\cong \mathrm{Spin}(4k+3) \times \mathbb{Z}_2. \end{align}

When $n$ is odd, the element $e_1 e_2 \cdots e_n$ of the Clifford algebra commutes with all elements. Every element of the pin group is either an element of the spin group, or is $e_1 \cdots e_n$ times an element of the spin group. If $(e_1 \cdots e_n)^2 = 1$, then $\mathrm{Pin}$ is the direct product $\mathrm{Spin} \times \langle e_1 \cdots e_n \rangle \cong \mathrm{Spin} \times \mathbb{Z}_2$. If $(e_1 \cdots e_n)^2 = -1$, then instead of a direct product, we could say $\mathrm{Pin} \cong (\mathrm{Spin} \times \mathbb{Z}_4)/\mathbb{Z}_2$. This can potentially be described neatly in other ways. For $\mathrm{Pin}^+(3)$, Harvey states that $$\mathrm{Pin}^+(3) \cong \{ A \in U(2) : \det A = \pm 1 \}$$ which makes sense along the same lines as $\mathrm{Spin}(3) \cong SU(2)$.