Pin and Spin Groups and Their Representations: Examples and Counterexamples

In ${\mathrm{C}\mathrm{l}}_{1,0}$, $e^2=1$ and the unit-vector-generated Pin group is $\{1,-1,e,-e\}$; its reflection lift $e$ has square $+1$. In ${\mathrm{C}\mathrm{l}}_{0,1}$, $f^2=-1$ and Pin is $\{1,-1,f,-f\}\cong C_4$; its reflection lift has square $-1$. In both cases the even subgroup is $\{1,-1\}$.

In ${\mathrm{C}\mathrm{l}}_{2,0}$ put $E=e_1{e}_2$. Then $E^2=-1$, and by [F3] the even algebra is exactly $\mathbb{R}\oplus \mathbb{R}E\cong \mathbb{C}$. For $a^2+b^2=1$, $a+bE$ is a product of two unit vectors: $e_1(a{e}_1+b{e}_2)=a+bE$. Thus every unit $a+bE$ belongs to Spin, and conversely every even unit product has norm one, so Spin$(2)=\{a+bE:a^2+b^2=1\}$.

The matrices in the Example satisfy $A^2=B^2=I$ and $AB=-BA=\mathrm{diag}(-i,i)$. Hence $(z_{1}A+z_{2}B{)}^2=(z_1^2+z_2^2)I$ for all $z_1,z_2\in \mathbb{C}$. Applying [F4] to the complexification of the positive plane defines a unital complex Clifford action on $S$. Its image contains $P_{+}=(I+iAB)\mathrm{/}2=\mathrm{diag}(1,0)$ and $P_{-}=(I-iAB)\mathrm{/}2=\mathrm{diag}(0,1)$, as well as $P_{+}A$ and $P_{-}A$, which are the other two matrix units. The image is therefore all of $M_2(\mathbb{C})$; [F3] gives complex dimension $4$ for the source, so this map is an algebra isomorphism. Every nonzero invariant subspace contains both coordinate vectors by applying these matrix units to a nonzero vector, proving irreducibility of the complex dimension-$2$ Clifford module.

For $\theta \in \mathbb{R}$, let $g_{\theta }=\cos (\theta \mathrm{/}2)-E\sin (\theta \mathrm{/}2)$. Since $g_{\theta }^{-1}=\cos (\theta \mathrm{/}2)+E\sin (\theta \mathrm{/}2)$, direct multiplication using $E{e}_1=-e_2$ and $e_{1}E=e_2$ gives ${\rho }_{g_{\theta }}(e_1)=\cos \theta e_1+\sin \theta e_2$ and ${\rho }_{g_{\theta }}(e_2)=-\sin \theta e_1+\cos \theta e_2$. Therefore the angle doubles; $g_{2\pi }=-1$ and $g_{4\pi }=1$.

Since $c(E)=AB=\mathrm{diag}(-i,i)$, the operator $c(\omega )=iAB$ has the displayed eigenspaces $S^{\pm }$, each a complex line. Restricting the Clifford action to the circle in step 1.2 gives on these lines the scalars $a-ib$ and $a+ib$, of absolute value one. The scalar functions are continuous in $(a,b)$, and multiplicativity of the Clifford action makes them group characters; the one-dimensional modules are irreducible. They are inequivalent because the Spin element $E$ acts by $-i$ on $S^{+}$ and $i$ on $S^{-}$. Substituting $a=\cos (\theta \mathrm{/}2)$ and $b=-\sin (\theta \mathrm{/}2)$ yields the two asserted half-angle actions. In particular the kernel element $g_{2\pi }=-1$ acts by $-1$ and $g_{4\pi }=1$ acts by $1$, consistently with the vector double-angle map in step 2.1.

From the polarized relation, $e_1{e}_2+e_2{e}_1=0$. Thus ${\rho }_{e_1}(e_1)=-e_1{e}_1{e}_1=-e_1$ and ${\rho }_{e_1}(e_2)=-e_1{e}_2{e}_1=e_2$. Linearity gives $(x,y)\mapsto (-x,y)$, the reflection in $e_1^{\perp }$.

By [F1], every Pin element acting trivially is a nonzero real scalar $\lambda$. Its norm is $N(\lambda )={\lambda }^2$; Pin requires this norm to be $\pm 1$, so positivity gives ${\lambda }^2=1$ and $\lambda =\pm 1$. Both scalars act trivially, agreeing with [F2].

The polarized Clifford relation gives $e_{+}{e}_{-}=-e_{-}{e}_{+}$ and hence $E^2=-e_{+}^2{e}_{-}^2=1$. By [F3], $1,E$ form a basis of ${\mathrm{C}\mathrm{l}}_{1,1}^0$. The map $a+bE\mapsto (a+b,a-b)$ preserves products and is bijective, so it identifies this algebra with $\mathbb{R}\oplus \mathbb{R}$.

The idempotents $p_{\pm }=(1\pm E)\mathrm{/}2$ are the two factor identities. Let each factor act on $\mathbb{R}$ by its scalar coordinate and the other factor by zero. Each module is one-dimensional and hence irreducible; they are inequivalent because $E$ acts by opposite scalars. By [F1], invariance under full-even Spin is equivalent to invariance under its linear span ${\mathrm{C}\mathrm{l}}^0$, so these are precisely the two split-row Clifford spinor types in this example.

The Clifford relations give $e_{-}{e}_{+}=-e_{+}{e}_{-}$, so $E^2=e_{+}{e}_{-}{e}_{+}{e}_{-}=-e_{+}^2{e}_{-}^2=1$. It is a product of two normalized vectors and therefore belongs to full-even Spin. By [F1], $N(E)=(-Q(e_{+}))(-Q(e_{-}))=(-1)(1)=-1$.

The reflection from $e_{+}$ negates $e_{+}$ and fixes $e_{-}$; the reflection from $e_{-}$ fixes $e_{+}$ and negates $e_{-}$. Their product is ${\rho }_E=-I$, with determinant $+1$. Its polar block factor is $K_{+}=-1$ on the positive line and $K_{-}=-1$ on the negative line, so [F3] puts it outside $S{O}_0(1,1)$. Hence the full-even lift $E$ is excluded from norm-one Spin although it lies over $SO(1,1)$, as asserted.