Why does the pin group satisfy $\mathrm{Pin}_+(2)=U(1)\oplus U(1)$?

Question

I am looking for resources on the explicit constructions of the low dimension $\text{Pin}_{\pm}(n)$ groups, especially $\text{Pin}_{\pm}(2)$. I am looking through "Clifford Algebras, Clifford Groups, and a Generalization of the Quaternions: The Pin and Spin Groups" by Jean Gallier and I would especially like to see why $\text{Pin}_+(2) = U(1) \oplus U(1)$ and if there is a classical group isomorphism for $\text{Pin}_-(2)$.

Answer 1

I found a method of solution from Moore's Quantum Symmetry Book. I adopted notation from Gallier, which is opposite of many standard notations, so $\text{Pin}^+(2)$ here might be $\text{Pin}^-(2)$ in other texts:

$\text{Pin}^+(2)$: This group is generated by elements of $S^1$, i.e. $\text{Pin}^+(2) = \langle ae_1 + be_2 \rangle$ where $a^2 + b^2 = 1$. Thus every element can be written as products of such vectors. By the grading it is clear we have two different types of elements of $\text{Pin}^+(2)$, $\langle \alpha e_1 + \beta e_2 \rangle, \langle \gamma + \delta e_1 e_2 \rangle$, with $\alpha^2 + \beta^2 = \gamma^2 + \delta^2 = 1$ We cannot get other combinations due to the grading, i.e. we can only get even or odd elements, nothing mixed. Thus we can parametrize $\text{Pin}^+(2)$ with $$E(\theta) = \cos(\theta)e_1 + \sin(\theta)e_2$$ $$F(\theta) = \cos(\theta) + \sin(\theta) e_1 e_2$$ By strategic use of trig identities we get a very nice group structure, we will demonstrate the first one fully, the others follow similarly: $$E(a)E(b) = (\cos a e_1 + \sin b e_2)(\cos b e_1 + \sin b e_2) = -(\cos a \cos b + \sin a \sin b) + (\sin b \cos a - \cos b \sin a) e_1 e_2 $$ $$= - \cos(a - b) + \sin (b-a) e_1 e_1 = - \cos(a - b) - \sin(a - b) e_1 e_2 = \cos(a - b + \pi) + \sin(a - b + \pi) e_1 e_2 = F(a - b + \pi)$$

The full multiplication table is given:

$$E(a) E(b) = F(a - b + \pi)$$ $$E(a) F(b) = E(a - b)$$ $$F(b) E(a) = E(a + b)$$ $$F(a) F(b) = F(a + b)$$

The even elements here are the $F(a)$, whose multiplication is given by the last line, thus $\text{Spin(2)} = U(1)$.

We can use our twisted adjoint representation $\widetilde{\text{Ad}}_x(v) = \alpha(x) v x^{-1}$ on the basis $(e_1, e_2)$ for $\mathbb{R}^2$.

$$\widetilde{\text{Ad}}_{E(a)} = \begin{pmatrix} - \cos (2a) & - \sin(2a) \\ - \sin(2a) & \cos(2a) \end{pmatrix}$$

$$\widetilde{\text{Ad}}_{F(a)} = \begin{pmatrix} \cos(2a) & - \sin(2a) \\ \sin(2a) & \cos(2a) \end{pmatrix}$$

So it is easy to see that the map $\text{Spin}(2) \to SO(2)$ is actually the non-trivial double covering of the circle to itself $z^2: S^1 \to S^1$.

We grade $\text{Pin}^+(2)$ by the homomorphism $\det \widetilde{\text{Ad}}: \text{Pin}^+(2) \to \mathbb{Z}_2$. It is clear that $\ker \det \widetilde{\text{Ad}} = \text{Spin}(2) = U(1)$. Thus we have $$1 \to U(1) \to \text{Pin}^+(2) \to \mathbb{Z}_2 \to 1$$ Notice that this extension is not split. Suppose there was such a section $s: \mathbb{Z}_2 \to \text{Pin}^+(2)$, then it would have to map $s(-1) = E(a)$ for some $a$, since it would have to map $s(1) = F(0) = 1$ but then $E(a)^2 = F(\pi) = -1$. This cannot happen since $s^2(-1) = E^2(a) = -1$ but $s$ is a homomorphism so $s^2(-1) = s(-1)s(-1) = s((-1)^2) = s(1) = 1$.

$\text{Pin}^-(2)$: This case is very similar with altered multiplication table:

$$E(a)E(b) = F(b - a)$$ $$E(a)F(b) = E(a + b)$$ $$F(a)E(b) = E(b - a)$$ $$F(a)F(a) = F(a + b)$$

And this gives a split exact sequence $$1 \to U(1) \to \text{Pin}^-(2) \to \mathbb{Z}_2 \to 1$$ with general split given by $-1 \mapsto E(a)$. Thus $\text{Pin}^-(2) \cong U(1) \rtimes_\alpha \mathbb{Z}_2$ for some action $\alpha$.