\twoheadrightarrow & \mathrm{SO}(3)
\\
{\mathbin{\text{\rotatebox[origin=c]{-90}{$=$}}}}& & {\mathbin{\text{\rotatebox[origin=c]{-90}{$\hookrightarrow$}}}}& & {\mathbin{\text{\rotatebox[origin=c]{-90}{$\hookrightarrow$}}}}\\
\{\pm1\} & \hookrightarrow & \mathrm{Pin}_-(3) & \twoheadrightarrow & \mathrm{O}(3)
\\
 & & {\mathbin{\text{\rotatebox[origin=c]{-90}{$\twoheadrightarrow$}}}}& & {\mathbin{\text{\rotatebox[origin=c]{-90}{$\twoheadrightarrow$}}}}\\
&  & \{\pm1\} & = & \{\pm1\}
\end{array}$$ Within the Pin (and Spin) group, one obtains the fermionic property that a $4\pi$ rotation equals 1. The subgroups are known as double point groups, ${\hat{G}\subset\mathrm{Pin}_-(3)}$, such that ${\hat{G}/{\mathbb{Z}}_2=G}$. However, when taking into account the projective nature of the Hilbert space and constructing the spinor representation, this amounts to setting a $2\pi$ rotation to be ${(\hat{c}_n)^n=-1}$, and a double inversion to be ${\hat{I}^2=1}$. This is referred to as the spinor representation, $\Pi(G)$. To clarify things [@Ando2015Topological], we note that:\
(i) In $G$ a $2\pi$ rotation is $1$.\
(ii) In $\hat{G}$ a $2\pi$ rotation squares to $1$.\
(iii) In $\Pi(G)$ a $2\pi$ rotation is $-1$.

For brevity, we shall use $\hat{c},\hat{s},\hat{\sigma}$ for the generators of both $\hat{G}$ and $\Pi(G)$. Moreover, one finds that a reflection satisfies ${\hat{\sigma}^2=I^2\hat{c}_2^2=-1}$, and that since the product of orthogonal reflections $\hat\sigma_h\hat\sigma_v$ is a $\pi$ rotation, they satisfy ${(\hat\sigma_h\hat\sigma_v)^2=-1}$. The rotoreflections, however, depend on the parity, as ${(\hat s_{2n})^{2n}=\hat c_{2n}^{2n}(\hat\sigma_h^2)^n=-(-1)^n}$ (see Fig. \[fig:PiG\]).

The point group symmetries each correspond to one of the abstract groups 