Spin representations, Clifford action, 3D, 1D
Irreducibly Representing 2 Dimensional Matrix Groups
{$V_j=\langle x_j,y_j \rangle$}
{$C(V_j)=\langle 1,x_j,y_j,x_jy_j \rangle$}
{$a_j=\frac{1}{2}(x_j + iy_j)$}
{$\bar{a}_j=\frac{1}{2}(x_j - iy_j)$}
{$x_j=a_j+\bar{a}_j$}
{$y_j=i(\bar{a}_j-a_j)$}
{$W_j\oplus W_j^*=\langle a_j, \bar{a}_j \rangle$}
Representation given by action on {$\bar{a}_j$}
{$W^*_j=\langle \bar{a}_j \rangle $}
{$S'_j = \langle 1, \bar{a}_j \rangle $}
{$C(V_j)=\langle 1,x_j,y_j,x_jy_j \rangle\cong \textrm{End}(\langle 1, \bar{a}_j \rangle)$}
{$1\rightarrow \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}$}
{$a_j\rightarrow \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix}$}
{$\bar{a}_j\rightarrow \begin{pmatrix} 0 & 0 \\ 1 & 0 \\ \end{pmatrix}$}
{$a_j\bar{a}_j\rightarrow \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix}$}
{$\bar{a}_ja_j\rightarrow \begin{pmatrix} 0 & 0 \\ 0 & 1 \\ \end{pmatrix}$}
{$a_j^2=\bar{a}_j^2=0$}
{$x_j\rightarrow \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$}
{$y_j\rightarrow \begin{pmatrix} 0 & -i \\ i & 0 \\ \end{pmatrix}$}
The Lie algebra is given by the bivectors {$k_jx_jy_j, k\in\mathbb{C}$}.
{$x_jy_j\rightarrow \begin{pmatrix} i & 0 \\ 0 & -i \\ \end{pmatrix}$}
{$y_jx_j\rightarrow \begin{pmatrix} -i & 0 \\ 0 & i \\ \end{pmatrix}$}
Note that the Lie algebra is abelian. Thus it is not semisimple. The Lie algebra equals its Cartan subalgebra.
The Cartan subalgebra of the Lie algebra for the spin group can be generated by {$\bar{a}_j\wedge a_j$} which is mapped to {$\frac{1}{4}[\bar{a}_j,a_j]=\frac{1}{4}\bar{a}_ja_j - \frac{1}{4}a_j\bar{a}_j$}. Acting on an arbitrary element {$c_j+d_j\bar{a}_j\in S'$}, and noting the factor {$\sqrt{2}$} in applying a vector, we have {$(\frac{1}{4}\bar{a}_ja_j - \frac{1}{4}a_j\bar{a}_j)(c_j+d_j\bar{a}_j)=(\frac{1}{4}d_j\bar{a}_j - \frac{1}{4}c_j)2$}. This means the action is given by
{$\begin{pmatrix} -\frac{1}{2} & 0 \\ 0 & \frac{1}{2} \\ \end{pmatrix}\begin{pmatrix} c_j \\ d_j \end{pmatrix} = \begin{pmatrix} -\frac{1}{2}c_j \\ \frac{1}{2}d_j \end{pmatrix}$}
The weight spaces are {$S'_+ = \langle 1 \rangle$} with weight {$\lambda_1(h)=-\frac{1}{2}$} and {$ S'_- = \langle \bar{a}_j\rangle $} with weight {$\lambda_{a_j}(h)=\frac{1}{2}$}. The Cartan subalgebra element {$\bar{a}_j\wedge a_j$} has eigenvalue {$-\frac{1}{2}$} on {$\langle 1 \rangle$}, which does not contain {$\bar{a}_j$}, and eigenvalue {$\frac{1}{2}$} on {$\langle \bar{a}_j \rangle$}, which does contain {$\bar{a}_j$}.
Representation given by action on {$a_j$}
{$W_j=\langle a_j \rangle $}
{$S_j = \langle 1, a_j \rangle $}
{$C(V_j)=\langle 1,x_j,y_j,x_jy_j \rangle\cong \textrm{End}(\langle 1, a_j \rangle)$}
{$1\rightarrow \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}$}
{$a_j\rightarrow \begin{pmatrix} 0 & 0 \\ 1 & 0 \\ \end{pmatrix}$}
{$\bar{a}_j\rightarrow \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix}$}
{$a_j\bar{a}_j\rightarrow \begin{pmatrix} 0 & 0 \\ 0 & 1 \\ \end{pmatrix}$}
{$\bar{a}_ja_j\rightarrow \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix}$}
{$a_j^2=\bar{a}_j^2=0$}
{$x_j\rightarrow \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$}
{$y_j\rightarrow \begin{pmatrix} 0 & i \\ -i & 0 \\ \end{pmatrix}$}
The Lie algebra is given by the bivectors:
{$x_jy_j\rightarrow \begin{pmatrix} -i & 0 \\ 0 & i \\ \end{pmatrix}$}
{$y_jx_j\rightarrow \begin{pmatrix} i & 0 \\ 0 & -i \\ \end{pmatrix}$}
The Lie algebra is abelian, thus not semisimple. It equals its Cartan subalgebra.
The Cartan subalgebra of the Lie algebra for the spin group can be generated by {$\bar{a}_j\wedge a_j$} which is mapped to {$\frac{1}{4}[\bar{a}_j,a_j]=\frac{1}{4}\bar{a}_ja_j - \frac{1}{4}a_j\bar{a}_j$}. Acting on an arbitrary element {$c_j+d_ja_j\in S$}, and noting the factor {$\sqrt{2}$} in applying a vector, we have {$(\frac{1}{4}\bar{a}_ja_j - \frac{1}{4}a_j\bar{a}_j)(c_j+d_ja_j)=(\frac{1}{4}c_j - \frac{1}{4}d_j)2$}. This means the action is given by
{$\begin{pmatrix} \frac{1}{2} & 0 \\ 0 & -\frac{1}{2} \\ \end{pmatrix}\begin{pmatrix} c_j \\ d_j \end{pmatrix} = \begin{pmatrix} \frac{1}{2}c_j \\ -\frac{1}{2}d_j \end{pmatrix}$}
Alternatively, consider the values above for {$\bar{a}_ja_j$} and {$a_j\bar{a}_j$} and calculate {$\bar{a}_ja_j - a_j\bar{a}_j$}. This will give a value that differs by a multiplicative constant. Note that a Lie algebra representation is always isomorphic to its product with a nonzero scalar.
Note that this is also just a complex constant times the value for {$x_jy_j$}.
The weight spaces are {$\langle 1 \rangle$} and {$\langle a_j \rangle$}. The respective weights are {$\lambda_1(h)=\frac{1}{2}$} and {$\lambda_{a_j}(h)=-\frac{1}{2}$} for {$h\in\frak{h}$}. The Cartan subalgebra element {$\bar{a}_j\wedge a_j$} has eigenvalue {$-\frac{1}{2}$} on {$\langle a_j \rangle$}, which contains {$a_j$}, and eigenvalue {$\frac{1}{2}$} on {$\langle 1 \rangle$}, which does not contain {$a_j$}.
Relating the representations
The Lie algebra for {$\textrm{Spin}(2)$} has a single generator, the bivector {$x_jy_j$}.
The representation {$\bar{\phi}$} based on {$S'=\langle 1,\bar{a}_j\rangle$} sends
{$x_jy_j\rightarrow \begin{pmatrix} i & 0 \\ 0 & -i \\ \end{pmatrix}$}
The representation {$\phi$} based on {$S=\langle 1, a_j \rangle$} sends
{$x_jy_j\rightarrow \begin{pmatrix} -i & 0 \\ 0 & i \\ \end{pmatrix}$}
Thus these are dual representations. {$\phi(x_jy_j)=-\bar{\phi}(x_jy_j)^T$} and {$\bar{\phi}(x_jy_j)^T=-\phi(x_jy_j)$}.
Note also that the values of the dual representation is given by the value on {$y_jx_j$}, which is to say, the value in the opposite Clifford algebra.
The weight {$\lambda_1(h) = \frac{1}{2}$} of {$S_+=\langle 1 \rangle$} is the negative of the weight {$-\lambda_1(h)=\lambda_{a_j}(h) = -\frac{1}{2}$} of {$S_-=\langle a_j \rangle$}.
The weight {$\lambda_{\bar{a}_j}(h)=\frac{1}{2}$} of {$S'_-=\langle \bar{a}_j \rangle$} is the negative of the weight {$\lambda_1(h)=-\frac{1}{2}$} of {$S'_+=\langle 1 \rangle$}.
There is an isomorphism from {$S_+=\langle 1 \rangle$} to {$S'_-=\langle \bar{a}_j \rangle$}, unique up to scale. And there is an isomorphism from {$S_-=\langle a_j \rangle$} to {$S'_+=\langle 1 \rangle$}, unique up to scale.
This can be interpreted to identify the lack of a concept with the existence of a context, and the existence of a concept with the lack of a context.
The isomorphisms can be established by calculating what {$B:S\rightarrow S'$} must be, {$B:\langle 1, a_j \rangle \rightarrow \langle 1, \bar{a}_j \rangle$}. This is a homomorphism if it is {$\frak{spin}(2)$} equivariant so that {$B(X\cdot s)=X\cdot B(s)$} for any {$X\in\frak{spin}(2),$}{$s\in S$}. Let {$X=k_jx_jy_j, s=c_j+d_ja_j$} with {$k_j,c_j,d_j\in\mathbb{C}$}. Then {$B(k_jx_jy_j\cdot(c_j+d_ja_j))=k_jx_jy_j\cdot B(c_j+d_ja_j)$}. Substituting the matrix for {$x_jy_j$} and writing {$B$} as a matrix we have
{$\begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ \end{pmatrix}\begin{pmatrix} k_ji & 0 \\ 0 & -k_ji \\ \end{pmatrix}\begin{pmatrix} c_j \\ d_j \\ \end{pmatrix}=\begin{pmatrix} -k_ji & 0 \\ 0 & k_ji \\ \end{pmatrix}\begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \\ \end{pmatrix}\begin{pmatrix} c_j \\ d_j \\ \end{pmatrix}$}
{$k_ji\begin{pmatrix} b_{11} & -b_{12} \\ b_{21} & -b_{22} \\ \end{pmatrix}\begin{pmatrix} c_j \\ d_j \\ \end{pmatrix}=k_ji\begin{pmatrix} -b_{11} & -b_{12} \\ b_{21} & b_{22} \\ \end{pmatrix}\begin{pmatrix} c_j \\ d_j \\ \end{pmatrix}$}
We get {$b_{11}=0, b_{22}=0$}. Thus we can specify the isomorphism as {$B=\begin{pmatrix} 0 & b_{01} \\ b_{10} & 0 \\ \end{pmatrix}$} with any nonzero {$b_{01},b_{10}\in\mathbb{C}$}.
Nondegenerate bilinear form {$\beta$}
Given {$B$}, we define {$\beta(\phi,\psi)=B(\phi)(\psi)$} on {$\phi,\psi\in S=\langle 1, a_j\rangle$}. Let {$\phi = \phi_0 + \phi_1a_j, \psi= \psi_0+\psi_1a_j$}, then we calculate {$(B\phi)^T\psi = \phi^TB^T\psi$}
{$\begin{pmatrix} \phi_0 & \phi_1 \\ \end{pmatrix}\begin{pmatrix} 0 & b_{10} \\ b_{01} & 0 \\ \end{pmatrix}\begin{pmatrix} \psi_0 \\ \psi_1 \\ \end{pmatrix} = \phi_1b_{01}\psi_0 + \phi_0b_{10}\psi_1$}
We want invariance: {$\beta(\xi\cdot\phi,\psi)+\beta(\phi,\xi\cdot\psi)=0$} where {$\xi=kx_jy_j\in\frak{so}$}{$(2)$}. We have {$\xi= \begin{pmatrix} -ki & 0 \\ 0 & ki \\ \end{pmatrix}$}, {$\xi\cdot\phi= \begin{pmatrix} -ki\phi_0 \\ ki\phi_1 \\ \end{pmatrix}$}, {$\xi\cdot\psi= \begin{pmatrix} -ki\psi_0 \\ ki\psi_1 \\ \end{pmatrix}$}
Indeed: {$ki[(\phi_1b_{01}\psi_0 - \phi_0b_{10}\psi_1) + (-\phi_1b_{01}\psi_0 + \phi_0b_{10}\psi_1)]=0$}
Antiautomorphism {$\tau$}
Furthermore, we seek an antiautomorphism {$\tau$} of {$Cl_n\mathbb{C}$} such that {$\beta(A\cdot \phi, \psi)=\beta(\phi,\tau(A)\cdot\psi)$} for all {$A\in Cl_n\mathbb{C}=\langle 1, x_j, y_j, x_jy_j\rangle$}. This will force {$b_{01}=b{10}=b$}. Recall that:
{$x_j = \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}, y_j = \begin{pmatrix} 0 & i \\ -i & 0 \\ \end{pmatrix}, x_jy_j = \begin{pmatrix} -i & 0 \\ 0 & i \\ \end{pmatrix}$}
Consequently,
{$x_j\begin{pmatrix} \phi_0 \\ \phi_1 \\ \end{pmatrix} = \begin{pmatrix} \phi_1 \\ \phi_0 \\ \end{pmatrix}, y_j\begin{pmatrix} \phi_0 \\ \phi_1 \\ \end{pmatrix} = \begin{pmatrix} i\phi_1 \\ -i\phi_0 \\ \end{pmatrix}, x_jy_j\begin{pmatrix} \phi_0 \\ \phi_1 \\ \end{pmatrix} = \begin{pmatrix} -i\phi_0 \\ i\phi_1 \\ \end{pmatrix}$}
Then
{$\beta(\phi,\psi) = b(\phi_0\psi_1 + \phi_1\psi_0)$}
{$\beta(x\cdot\phi,\psi) = b(\phi_1\psi_1 + \phi_0\psi_0) = \beta(\phi,x\cdot\psi)$}
{$\beta(y\cdot\phi,\psi) = b(i\phi_1\psi_1 - i\phi_0\psi_0) = \beta(\phi,y\cdot\psi)$}
{$\beta(xy\cdot\phi,\psi) = b(-i\phi_0\psi_1 + i\phi_1\psi_0) = \beta(\phi,yx\cdot\psi)$}
We thus have that {$\tau(x)=x, \tau(y)=y, \tau(xy)=\tau(y)\tau(x)=yx$}.
Symmetry properties of {$\beta$}
Note that {$\beta(\phi,\psi) = b(\phi_0\psi_1 + \phi_1\psi_0) = \beta(\psi,\phi)$}.
Consider for {$A\in\wedge^kV$} what is the sign {$\varepsilon_k$} such that {$\beta(A\cdot\phi,\psi)=\varepsilon_k\beta(A\cdot\psi,\phi)$}.
We have {$\varepsilon_0=1$} because {$\beta(\phi,\psi)=\beta(\psi,\phi)$}
{$\beta(x_j\cdot\phi,\psi)=\beta(\phi,\tau(x_j)\cdot\psi)=\beta(\phi,x_j\cdot\psi)=\beta(x_j\cdot\psi,\phi)$} thus {$\varepsilon_1=0$}.
{$\beta(x_jy_j\cdot\phi,\psi)=\beta(\phi,\tau(x_jy_j)\cdot\psi)=\beta(\phi,y_jx_j\cdot\psi)=\beta(y_jx_j\cdot\psi,\phi)=-\beta(x_jy_j\cdot\psi,\phi)$} thus {$\varepsilon_2=-1$}.
No roots
This Lie algebra is abelian. Thus it has no roots and it has no Weyl group.
Lie group
{$SO(2)$} is abelian. Consequently, the exponential map {$\textrm{exp}:\frak{so}$}{$(2)\rightarrow SO(2)$} is a surjective group homomorphism.
{$SO(2)\cong\{\begin{pmatrix} e^{-ik} & 0 \\ 0 & e^{ik} \\ \end{pmatrix} |\; k\in\mathbb{R}\}$}
Conclusion
The action is defined in terms of adding a concept, going from {$1$} to {$a_i$}, or filling a context, removing it, going from {$\bar{a}_i$} to {$1$}. The product {$x_jy_j$} sets an axis {$x_j$} as an absolute reference and then, with regard to it, rotates away from it in the one direction {$iy_j$} or in the opposite direction {$\bar{i}y_j$}