Past exam of the groups and representations course of the University of Oxford / 2019 / 4

For a differentiable curve $Q(t)\in SO(5)$ with $Q(0)=I_5$, differentiate $Q(t{)}^{T}Q(t)=I_5$. Its tangent matrix $X=Q^{\prime }(0)$ is real and satisfies $X^T+X=0$. Conversely the matrix exponential of any real antisymmetric matrix is orthogonal, and $\det e^{tX}=e^{t\mathrm{tr}X}=1$ because $\mathrm{tr}X=0$. Hence the real Special orthogonal Lie algebra isLet $E_{ab}$ denote a matrix unit. A simple real basis isEach independent entry above the diagonal supplies one real parameter, giving $\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$ parameters. The commutator of two basis matrices iswhere $M_{ba}=-M_{ab}$ and $M_{aa}=0$.

A Cartan subalgebra is spanned by the commuting matrices $M_{12},M_{34}$, which generate independent rotations in two orthogonal planes. A generic $h=a{M}_{12}+b{M}_{34}$, with $ab(a^2-b^2)\ne 0$, has distinct complex eigenvalues $ia,-ia,ib,-ib,0$. A real antisymmetric matrix commuting with it preserves those two planes and the remaining line; it is a multiple of the rotation in each plane and is zero on the line. Thus the commuting Lie subalgebra has dimension two and is maximal toral. Therefore

Use the convention $A_{ij}=⟨{\alpha }_i,{\alpha }_j^{\vee }⟩$, so the rows of the supplied Cartan matrix are the simple roots in Dynkin label coordinates. Realize the B2 root system in an orthonormal plane byThus the first simple root is long and the second is short. The Dynkin diagram has two nodes joined by a double bond, with its arrow pointing to the short node:A Euclidean weight $w=x{e}_1+y{e}_2$ has Dynkin labelsIn particular the fundamental weights are ${\omega }_1=e_1$ and ${\omega }_2=(e_1+e_2)/2$. This long-first ordering is essential: sources that put the short simple root first interchange the two Dynkin labels.

The highest weight $(0,1)$ is the spinor representation. Its Euclidean weights are $(\pm \frac{1}{2},\pm \frac{1}{2})$, whose Dynkin labels areA lowering chain subtracts ${\alpha }_2=(-1,2)$, then ${\alpha }_1=(2,-2)$, then ${\alpha }_2$ again, passing through these four weights in the displayed order. All weight multiplicities are one.

The highest weight $(1,0)$ is the defining vector representation, with Euclidean weights $e_1,e_2,0,-e_2,-e_1$. ThusHere a lowering chain subtracts ${\alpha }_1,{\alpha }_2,{\alpha }_2,{\alpha }_1$. Each weight multiplicity is again one.

Finally $(0,2)$ is the highest root $e_1+e_2={\alpha }_1+2{\alpha }_2$, giving the Adjoint representation. Its nonzero weights are the eight roots $\pm e_1,\pm e_2,\pm (e_1-e_2),\pm (e_1+e_2)$. The Cartan subalgebra supplies zero with weight multiplicity two, soThe dimensions are $4,5,10$. Independently the long-first Weyl dimension formula iswhich gives those same values. These are Lie algebra representations; the four-dimensional spinor representation integrates to the Spin group $\mathrm{Spin}(5)$, and does not descend to the special orthogonal group $SO(5)$. A concrete obstruction is a $2\pi$ rotation: it is the identity in $SO(5)$ but acts as $-I_4$ on the spinor representation, as verified in solution.

Let $r=V(0,1)$ be the four-dimensional spinor representation. The tensor product representation splits into its symmetric square and exterior square, with dimensions ten and six. If $v$ is a highest-weight vector of $r$, then $v\otimes v$ is a highest-weight vector of weight $(0,2)$ in the symmetric square. The irreducible representation with that highest weight has dimension ten by solution, exhausting the symmetric square:For the exterior square, add each pair of distinct weights from solution. The resulting multiset of weights isIts maximal weight $(1,0)$ is the highest weight of a five-dimensional irreducible representation. It accounts for the four nonzero weights and one zero weight, leaving a single zero weight. By complete reducibility of semisimple Lie algebra representations, that remaining one-dimensional summand is the trivial representation. Thus ${\Lambda }^{2}r\cong 5\oplus 1$, and the spinor tensor square for so5 isThe dimension check is $16=1+5+10$. The trivial representation lies in the exterior square, reflecting the invariant nondegenerate alternating bilinear form of this pseudoreal representation. Although $r$ itself does not descend to $SO(5)$, its tensor square does: the central element $-1\in \mathrm{Spin}(5)$ acts as $(-I_4)\otimes (-I_4)=I_{16}$.

Use the basis matrices $M_{ab}=E_{ab}-E_{ba}$ from solution. Define the Pauli matrices and the two-by-two identity byThe following five Hermitian four-by-four matrices, written as Kronecker products, realize a Euclidean Clifford algebra:Define the Lie algebra representation on ${\mathbb{C}}^4$ byThese are anti-Hermitian matrices, appropriate to the real compact Lie algebra. For complete explicitness, all ten representation matrices areEach table entry is the image under $\rho$ of the basis matrix immediately to its left. Since the two-by-two factors are specified above, these Kronecker products specify every entry of each four-by-four matrix.

To verify the representation law, the Clifford algebra relations implyApply this identity to $\rho (M_{cd})=\frac{1}{4}[{\gamma }_c,{\gamma }_d]$ and use the commutator product rule. The result isexactly the Lie bracket relation in solution. Thus $\rho$ is a Lie algebra representation.

It remains to identify its highest weight, rather than just its dimension. The commuting Hermitian Cartan subalgebra operators areTheir simultaneous eigenvalues are $(x,y)=(\pm \frac{1}{2},\pm \frac{1}{2})$. Converting with $(a,b)=(x-y,2y)$ gives exactly the four weights of $V(0,1)$ in solution. A state with $(x,y)=(\frac{1}{2},\frac{1}{2})$ has highest weight $(0,1)$, so this Clifford realization of the so5 spin representation is the requested $r$.

Finally, $\rho (M_{12}{)}^2=-I_4/4$, soThis explicitly exhibits the distinction between the spinor representation of $\mathfrak{so}(5)$, or of $\mathrm{Spin}(5)$, and a single-valued group representation of $SO(5)$.