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.
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 $}.
{$ \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$} flips its values across the anti-diagonal.
Can we identify {$\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 {$\bar{i}y_j$}