Spin representations, 1D, 2D, 3D
Irreducibly Representing {$n$}-Dimensional Matrix Groups
Focusing, for now, on the spin representations of the special orthogonal group {$SO(n)$} and its Lie algebra {$\frak{so}$}{$(n)$}. See also the tensorial representations of the classical Lie groups.
Orthogonal group {$O(n)$}
The orthogonal group {$O(n)$} consists of isometries (distance-preserving transformations) of an {$n$}-dimensional Euclidean space that preserve the origin. An orthogonal matrix {$Q$} preserves the length of any vector {$v$}, namely {$\langle v,v\rangle=v^Tv=(Qv)^T(Qv)=v^TQ^TQv$}, which is true if and only if {$Q^TQ=I$}.
More generally, given a symmetric bilinear form or a quadratic form, the orthogonal group of the form is the group of invertible linear maps that preserve the form. {$\langle v,v\rangle=v^TBv=(Qv)^TB(Qv)=v^TQ^TBQv$}, which is true if and only if {$Q^TBQ=I$}, in other words, {$QQ^T=B^{-1}$}, and again {$Q^TQ=Q^{-1}B^{-1}Q$}.
Special orthogonal lie algebra {$\frak{so}$}{$(n)$}
For a matrix Lie group {$G$}, the exponential map {$\textrm{exp}:\frak{g}\rightarrow G$} can be defined concretely as {$\textrm{exp}(X)=\sum_{k=0}^\infty \frac{X^k}{k!}$}. The Lie algebra for the orthogonal group, and likewise the special orthogonal group and the spin group, is given by {$n\times n$} matrices {$X$} such that {$e^x\in O(n)$}, which means {$(e^X)^T=(e^X)^{-1}$}. We have {$X^T=-X$}. These are the skew-symmetric matrices {$X$}.
The skew-symmetric matrices form a Lie algebra. Suppose {$X,Y$} are skew-symmetric. Then
{$ [X,Y]^T=(XY-YX)^T=(XY)^T-(YX)^T=Y^TX^T-X^TY^T=(-Y)(-X)-(-X)(-Y)=YX-XY=[Y,X]=-[X,Y] $}.
Thus {$ [X,Y] $} is likewise skew-symmetric.
Skew-symmetric means that if we pick out two dimensions, then they need to be linearly independent. We are focusing on linear independence.
It means that if we focus on two dimensions, we will have rotation by i.
It means that the Lie algebra expresses the smooth action (rotation) in terms of relationships between discrete dimensions. So it expresses continuity in terms of discreteness.
The Lie algebra relates the Lie group element {$I$} with its tangent plane, which is orthogonal to it. For example, in {$U(1)$} it relates {$1$} and {$it$}.
Basis for {$\frak{so}$}{$(n)$}
A basis for {$\frak{so}$}{$(n)$} is given by matrices of the form {$M(i,j)=\begin{pmatrix} & & & \\ & & 1_{ij} & \\ & -1_{ji} & & \\ & & & \\ \end{pmatrix}$}. Let {$e_j$} be the related basis for {$V$} so that any vector {$v\in V$} can be written as {$v=\sum_{j=1}^n v_je_j$}. Consider how {$M(i,j)$} acts on {$v_{ij}=v_ie_i + v_je_j$} as a ninety-degree rotation, sending it to {$M(i,j)v_{ij}=v_je_i - v_ie_j$}:
{$\begin{pmatrix} & & & \\ & & 1_{ij} & \\ & -1_{ji} & & \\ & & & \\ \end{pmatrix}\begin{pmatrix} \\ v_i \\ v_j \\ \\ \end{pmatrix} = \begin{pmatrix} \\ v_j \\ -v_i \\ \\ \end{pmatrix}$}
This can be understood as multiplication by the imaginary number in the relevant complex plane. Moreover, we can understand this action as sending {$v=\sum_{j=1}^n v_je_j$} to {$v_je_i - v_ie_j = \langle e_j,v\rangle e_i - \langle e_i,v\rangle e_j$}.
Note that {$M(i,j)$} makes sense only if {$i\neq j$} because we can't have the same entry be {$1_{ii}$} and {$-1_{ii}$}. Note also that {$M(i,j)=-M(j,i)$} but we should realize that the example above suppose {$i\leq j$}. This suggests we define {$M(i,i)=0$}. We see that we have a bijection between bivectors {$e_i\wedge e_j\in \wedge^2 V$} and our basis.
{$e_i\wedge e_j\rightarrow (\phi_{e_i\wedge e_j}:v\rightarrow \langle e_j,v\rangle e_i - \langle e_i,v\rangle e_j)$}
{$\phi_{e_i\wedge e_j} = M(i,j)$}
Lie bracket
{$ [\phi_{e_i\wedge e_j}, \phi_{e_k\wedge e_l}] = \begin{pmatrix} & & & \\ & & 1_{ij} & \\ & -1_{ji} & & \\ & & & \\ \end{pmatrix}\begin{pmatrix} & & & \\ & & 1_{kl} & \\ & -1_{lk} & & \\ & & & \\ \end{pmatrix} - \begin{pmatrix} & & & \\ & & 1_{kl} & \\ & -1_{lk} & & \\ & & & \\ \end{pmatrix}\begin{pmatrix} & & & \\ & & 1_{ij} & \\ & -1_{ji} & & \\ & & & \\ \end{pmatrix}$}
{$ [\phi_{e_i\wedge e_j}, \phi_{e_i\wedge e_j}] = 0 $}
{$ [\phi_{e_i\wedge e_j}, \phi_{e_k\wedge e_l}] = 0 $} if {$i\neq k,l$} and {$j\neq kl$}
{$ [\phi_{e_i\wedge e_j}, \phi_{e_j\wedge e_l}] = 1_{ij}1_{jl} - ((-1_{ji})(-1_{lj})) = 1_{il} - 1_{li} = \phi_{e_i\wedge e_l} $}
We up basis elements {$j,\bar{j}$} as {$\phi_{e_j\wedge e_\bar{j}}$}. We want the pairs disjoint so that we have an abelian algebra. The irreducible representations of an abelian algebra must be one-dimensional. We want a maximal abelian subalgebra of our Lie algebra. This is the Cartan subalgebra.
We can choose {$\bar{i}=m+i,\bar{j}=m+j$}
{$ \begin{pmatrix} & 1_{ij} & & \\-1_{ji} & & &\\ & & & 1_{i\bar{j}}\\ & & -1_{j\bar{i}} & \end{pmatrix}$}
Quadratic form {$Q$} and bilinear form {$B$}
Define the {$n$}-dimensional complex vector space {$V$}. Pair up its basis elements, with an extra basis element {$u$} if {$n=2m+1$} is odd, and {$u=0$} is {$n=2m$} is even.
{$V=\langle x_1,y_1,\cdots,x_m,y_m,u \rangle$}
Define the quadratic form {$Q$} on {$V$} with {$Q(x_j)=1,Q(y_j)=1, Q(u)=0$}, for all {$j, 1\leq j\leq m$}.
The polarization identities associate to {$Q$} a bilinear form {$\langle\;,\;\rangle$}
| {$\langle v, w\rangle = \frac{1}{2}(Q(v+w)-Q(v)-Q(w))$} | {$=\frac{1}{2}(Q(v)+Q(w)-Q(v-w))$} | {$=\frac{1}{4}(Q(v+w)-Q(v-w))$} |
If we want {$\langle x_j, y_k\rangle=0$}, then {$Q(x_j+y_k)=Q(x_j)+Q(y_k)=2$}. Similarily, for {$j\neq k$}, we have {$Q(x_j+x_k)=Q(x_j)+Q(x_k)=2$}. This means that the Pythagorean theorem holds for any two basis elements {$e_j, e_k$}: the square of the sum equals the sum of the squares. Then the matrix {$B_{ij}=\langle e_j, e_k \rangle$} of the bilinear form is given by the identity matrix {$B=I$}, namely {$\langle u,v\rangle = u^TIv=u^Tv$}.
Isotropic basis
Note that {$Q(x_j+iy_j)=Q(x_j)+i^2Q(y_j)=Q(x_j)-Q(y_j)=1-1=0$} because {$x_j, y_j$} are distinct basis elements.
We want to define elements such that {$\langle k(x_s+iy_s), k(x_t-iy_t) \rangle = \delta_{s,t}$}. When {$s\neq t$}, then {$\langle k(x_s+iy_s), k(x_t-iy_t) \rangle = \frac{1}{2}(Q(x_s+x_t+iy_s-iy_t)-Q(x_s+iy_s)-Q(x_t-iy_t))=\frac{1}{2}(Q(x_s)+Q(x_t)+i^2Q(y_s)+i^2Q(y_t))=0$}. When {$s=t$}, then {$\langle k(x_s+iy_s), k(x_s-iy_s) \rangle = \frac{1}{2}(Q(2x_s)-Q(x_s+iy_s)-Q(x_s-iy_s))=\frac{1}{2}(4)=2$}. So we need to set {$k=\frac{\sqrt 2}{2}$}.
{$a_j=\frac{\sqrt 2}{2}(x_j + iy_j)$}
{$\bar{a}_j=\frac{\sqrt 2}{2}(x_j - iy_j)$}
{$x_j=\frac{\sqrt 2}{2}(a_j+\bar{a}_j)$}
{$y_j=\frac{\sqrt 2}{2}i(\bar{a}_j-a_j)$}
Let us focus on the case {$n=2m$}, which is to say, let us momentarily ignore the basis element {$u$}, {$Q(u)=0$}, which is only relevant in the odd case.
{$\langle a_j,\bar{a}_k\rangle =\delta_{j,k}$}. We can write the bilinear form in terms of the basis {$\{a_1,\cdots a_m,\bar{a}_1,\cdots,\bar{a}_m\}$}. If we order them in this manner, then the bilinear form is {$\langle u,v\rangle=u^TBv$} where
{$B=\begin{pmatrix} 0 & I \\ I & 0 \\ \end{pmatrix}$}
Note that a different ordering of the basis elements would yield a permutation of {$B$}. Kac uses {$\{a_1,\cdots a_m,\bar{a}_m,\cdots,\bar{a}_1\}$} and thus for him, {$B$} is the matrix with {$1$}s on the anti-diagonal. That can be notationally convenient in unifying the even case {$n=2m$} and the odd case {$n=2m+1$} in that {$u$} appears in the center and, so to speak, pairs with itself.
We can show the relationship between the bases {$\{f_1,\cdots, f_{2m}\}=\{x_1,\cdots,x_m,y_1,\cdots,y_m\}$} and {$\{e_1,\cdots,e_{2m}\}=\{a_1,\cdots,a_m,\bar{a}_1,\cdots,\bar{a}_m\}$} which are related as {$f_j=\sum_k^2m S_{k,j}e_k$}, specifically {$x_j=\frac{\sqrt{2}}{2}e_j +\frac{\sqrt{2}}{2}e_{m+j}, y_j=\frac{\sqrt{2}}{2}ie_{m+j}-\frac{\sqrt{2}}{2}ie_j$}. Thus we have
{$S=\frac{\sqrt{2}}{2}\begin{pmatrix} I & -iI \\ I & iI \\ \end{pmatrix}$}
The change of basis gives a new matrix {$S^TBS$} for the bilinear form, namely {$I$}:
{$\frac{\sqrt{2}}{2}\begin{pmatrix} I & I \\ -iI & iI \\ \end{pmatrix}\begin{pmatrix} 0 & I \\ I & 0 \\ \end{pmatrix}\frac{\sqrt{2}}{2}\begin{pmatrix} I & -iI \\ I & iI \\ \end{pmatrix}=\frac{1}{2}\begin{pmatrix} I & I \\ iI & -iI \\ \end{pmatrix}\begin{pmatrix} I & -iI \\ I & iI \\ \end{pmatrix}=\frac{1}{2}\begin{pmatrix} 2I & 0 \\ 0 & 2I \\ \end{pmatrix}=I$}
And the matrix {$S^{-1}$} enables us to get back from {$I$} to {$B=(S^{-1})^TIS^{-1}$}.
Totally isotropic subspaces {$W$} and {$W^*$}
We define one-dimensional totally isotropic subspaces {$W_j=\langle a_j\rangle, W_j^*=\langle \bar{a}_j\rangle$} and then define maximal totally isotropic subspaces
{$W=\oplus_{j=1}^mW_j, \;\;\; W^*=\oplus_{j=1}^mW^*_j, \;\;\; W\cap W^*=\varnothing$}
In general, from the definition of the bilinear form, when {$B=I$}, for any {$n\times n$} matrix {$A$} we have {$\langle Au,v \rangle=(Au)^TIv=u^TA^Tv=\langle u,A^Tv \rangle$}.
In our case, with {$B=\begin{pmatrix} 0 & I \\ I & 0 \\ \end{pmatrix}$} as above, consider an {$m\times m$} matrix {$A$}, where {$n=2m$}. We want to show that {$\langle Aw,w^*\rangle=\langle w,A^Tw^*\rangle $} for all {$v=w+w^*\in V=W\oplus W^*$}.
{$ \left< \begin{pmatrix} A & 0 \\ 0 & 0 \\ \end{pmatrix}\begin{pmatrix} w & 0 \\ \end{pmatrix}, \begin{pmatrix} 0 \\ w^* \\ \end{pmatrix} \right> = \begin{pmatrix} w & 0 \\ \end{pmatrix}\begin{pmatrix} A^T & 0 \\ 0 & 0 \\ \end{pmatrix}\begin{pmatrix} 0 & I \\ I & 0 \\ \end{pmatrix}\begin{pmatrix} 0 \\ w^* \\ \end{pmatrix} = \begin{pmatrix} w & 0 \\ \end{pmatrix} \begin{pmatrix} 0 & A^T \\ 0 & 0 \\ \end{pmatrix}\begin{pmatrix} 0 \\ w^* \\ \end{pmatrix}$}
{$ = \begin{pmatrix} w & 0 \\ \end{pmatrix} \begin{pmatrix} 0 & I \\ I & 0 \\ \end{pmatrix}\begin{pmatrix} 0 & 0 \\ 0 & A^T \\ \end{pmatrix}\begin{pmatrix} 0 \\ w^* \\ \end{pmatrix} = \left< \begin{pmatrix} w & 0 \\ \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & A^T \\ \end{pmatrix}\begin{pmatrix} 0 \\ w^* \\ \end{pmatrix} \right>$}
Now define {$\rho_A=\begin{pmatrix} A & 0 \\ 0 & -A^T \end{pmatrix}$}. We show below that {$\langle \rho_A u, v \rangle=-\langle u,\rho_A v\rangle $} and thus {$\rho_A$} is skew.
{$ \left< \begin{pmatrix} A & 0 \\ 0 & -A^T \\ \end{pmatrix}\begin{pmatrix} u_+ \\ u_- \\ \end{pmatrix}, \begin{pmatrix} v_+ \\ v_- \\ \end{pmatrix} \right> = \begin{pmatrix} u_+ & u_-\\ \end{pmatrix}\begin{pmatrix} A^T & 0 \\ 0 & -A \\ \end{pmatrix}\begin{pmatrix} 0 & I \\ I & 0 \\ \end{pmatrix}\begin{pmatrix} v_+ \\ v_- \\ \end{pmatrix} = \begin{pmatrix} u_+ & u_-\\ \end{pmatrix}\begin{pmatrix} 0 & A^T \\ -A & 0 \\ \end{pmatrix}\begin{pmatrix} v_+ \\ v_- \\ \end{pmatrix} $}
{$= \begin{pmatrix} u_+ & u_-\\ \end{pmatrix}\begin{pmatrix} 0 & I \\ I & 0 \\ \end{pmatrix}\begin{pmatrix} -A & 0 \\ 0 & A^T \\ \end{pmatrix}\begin{pmatrix} v_+ \\ v_- \\ \end{pmatrix} = \left<\begin{pmatrix} u_+ \\ u_- \\ \end{pmatrix}, \begin{pmatrix} -A & 0 \\ 0 & A^T \\ \end{pmatrix}\begin{pmatrix} v_+ \\ v_- \\ \end{pmatrix} \right> $}
We see that passing {$\rho_A$} across this particular matrix {$B$} flips its values across the anti-diagonal.
Now one curious matter is that {$\rho_A$} is skew but is not given by a skew-symmetrix matrix. This is puzzling because skew-symmetry is a basis independent property. To explain this, note that the matrix of the bilinear form is basis dependent: {$\langle u, v \rangle = u^TBv$}. In our case, the equation {$\langle \rho_A u,v\rangle + \langle u,\rho_A v\rangle = 0$} means {$u^T\rho_A^TB v + u^TB\rho_A v=0$}. We have that
{$\rho_A^TB = \begin{pmatrix} A^T & 0 \\ 0 & -A \\ \end{pmatrix}\begin{pmatrix} 0 & I \\ I & 0 \\ \end{pmatrix} = \begin{pmatrix} 0 & A^T \\ -A & 0 \\ \end{pmatrix}$} and {$\begin{pmatrix} 0 & I \\ I & 0 \\ \end{pmatrix}\begin{pmatrix} A & 0 \\ 0 & -A^T \\ \end{pmatrix}=\begin{pmatrix} 0 & -A^T \\ A & 0 \\ \end{pmatrix}=B\rho_A$}
And we can verify that the resulting matrix {$M=\begin{pmatrix} 0 & A^T \\ -A & 0 \\ \end{pmatrix}$} is skew-symmetric. In particular, {$M_{j,m+k}=(A^T)_{j,k}=A_{kj}$} and {$M_{m+k,j}=-A_{kj}=-M_{j,m+k}$}. We also see that we are only getting a subset of the skew-symmetric matrices, namely those whose diagonal blocks are zero in this block form.
In summary, the particular basis we choose for the bilinear form, and the resulting matrix for that bilinear form, is affecting how we visualize the skew-symmetry indicated by the bilinear form. In general, as Kac notes, the special orthogonal Lie algebra is defined as {$\frak{so}$}{$(n)(\mathbb{F})=\{m\in gl_n(\mathbb{F})\;|\;m^TB+Bm=0\}$} and here {$B$} is the matrix of a bilinear form. We can solve using our chosen {$B$} as follows.
{$\begin{pmatrix} M_{11}^T & M_{21}^T \\ M_{12}^T & M_{22}^T \\ \end{pmatrix}\begin{pmatrix} 0 & I \\ I & 0 \\ \end{pmatrix} + \begin{pmatrix} 0 & I \\ I & 0 \\ \end{pmatrix}\begin{pmatrix} M_{11} & M_{12} \\ M_{21} & M_{22} \\ \end{pmatrix} = 0$}
{$\begin{pmatrix} M_{21}^T & M_{11}^T \\ M_{22}^T & M_{12}^T \\ \end{pmatrix} = - \begin{pmatrix} M_{21} & M_{22} \\ M_{11} & M_{12} \\ \end{pmatrix}$}
{$M_{21}^T=-M_{21}, M_{11}^T=-M_{22}, M_{22}^T=-M_{11}, M_{12}^T=-M_{12}$}
{$M=\begin{pmatrix} A & C \\ D & -A^T \\ \end{pmatrix}$} where {$C$} and {$D$} are skew-symmetric.
{$MB=\begin{pmatrix} A & C \\ D & -A^T \\ \end{pmatrix}\begin{pmatrix} 0 & I \\ I & 0 \\ \end{pmatrix} = \begin{pmatrix} C & A \\ -A^T & D \\ \end{pmatrix}$} and we see that {$MB$} is a skew-symmetric matrix.
It is also helpful to note another way to think of skew-symmetry of {$M$}, which is that {$\langle Mu,u\rangle=0$} for all {$u\in V$}. Applying this to {$\langle M(u+v),u+v\rangle$} yields {$\langle Mu,v \rangle + \langle Mv,u \rangle=0$} and recall that our bilinear form is symmetric.
Can we identify {$\bar{a}_j$} with the functional {$\langle a_j, \cdot \rangle$}?
{$C(V,Q)=\langle 1,x_1,y_1,\cdots \rangle$} is the {$2^N$}-dimensional complex Clifford algebra on {$V$} with quadratic form {$Q$}
Notes from elsewhere that I'm reworking...
Let {$D$} be the superalgebra {$\mathbb{C}(\varepsilon)$} with {$\varepsilon$} odd and {$\varepsilon^2=1$}.
{$C(V_j)=\langle 1,x_j,y_j,x_jy_j,u,ux_j,uy_j,ux_jy_j \rangle\cong \langle 1, u \rangle\otimes\langle 1, x_j, y_j, x_jy_j \rangle\cong \mathbb{C}(\varepsilon)\otimes\textrm{End}(\langle 1, \bar{a}_j \rangle)\cong\textrm{End}_{\mathbb{C}(\varepsilon)}(\mathbb{C}(\varepsilon)\otimes\langle 1, \bar{a}_j \rangle)$}
Representation given by action on {$\bar{a}_j$}
{$W^*_j=\langle \bar{a}_j \rangle $}
{$S'_j = \langle 1, \bar{a}_j \rangle $}
{$1\rightarrow 1, u \rightarrow \varepsilon$}
{$1\rightarrow \begin{pmatrix} 1 & 0 \\ 0 & 1 \\ \end{pmatrix}$}
{$a_j\rightarrow \begin{pmatrix} 0 & 1 \\ 0 & 0 \\ \end{pmatrix}$}
{$\bar{a}_j\rightarrow \begin{pmatrix} 0 & 0 \\ 1 & 0 \\ \end{pmatrix}$}
{$a_j\bar{a}_j\rightarrow \begin{pmatrix} 1 & 0 \\ 0 & 0 \\ \end{pmatrix}$}
{$\bar{a}_ja_j\rightarrow \begin{pmatrix} 0 & 0 \\ 0 & 1 \\ \end{pmatrix}$}
{$a_j^2=\bar{a}_j^2=0$}
{$x_j\rightarrow \begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$}
{$y_j\rightarrow \begin{pmatrix} 0 & -i \\ i & 0 \\ \end{pmatrix}$}
The Lie algebra is generated by the bivectors {$ux_j, uy_j, x_jy_j$}.
{$ux_j\rightarrow \varepsilon\otimes\begin{pmatrix} 0 & 1 \\ 1 & 0 \\ \end{pmatrix}$}
{$uy_j\rightarrow \varepsilon\otimes\begin{pmatrix} 0 & -i \\ i & 0 \\ \end{pmatrix}$}
{$x_jy_j\rightarrow 1\otimes\begin{pmatrix} i & 0 \\ 0 & -i \\ \end{pmatrix}$}
{$y_jx_j\rightarrow 1\otimes\begin{pmatrix} -i & 0 \\ 0 & i \\ \end{pmatrix}$}
This Lie algebra is not abelian. It is semisimple.
The Cartan subalgebra of this Lie algebra 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 {$\bar{i}y_j$}