Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 302 / 4

For simple roots ${\alpha }_i$, the Cartan matrix isIts diagonal entries are two, while its off-diagonal entries encode the angles and relative lengths in the Dynkin diagram.

The fundamental weights ${\omega }_i$ are the basis dual to the simple coroots:

The Dynkin labels of a weight $\lambda$ areso $\lambda ={\sum }_i{\lambda }^i{\omega }_i$.

A dominant weight has nonnegative Dynkin labels. A dominant integral weight additionally lies in the weight lattice, so all those labels are nonnegative integers.

The highest weight of a representation is the weight $\Lambda$ of a nonzero weight vector annihilated by every positive-root operator. Every other weight is obtained from $\Lambda$ by subtracting a nonnegative integer combination of simple roots. A finite-dimensional irreducible representation of a complex semisimple Lie algebra is determined by its dominant integral highest weight.

Pairing ${\alpha }_i={\sum }_j{c}_{ij}{\omega }_j$ with ${\alpha }_k^{\vee }$ gives $c_{ik}=A_{ik}$. Hence the relation between the two bases is

With $A_{ij}=2({\alpha }_i,{\alpha }_j)/({\alpha }_j,{\alpha }_j)$ and roots labeled from left to right, the three Cartan matrices are

All roots of $A_3$ have the same length, so $|{\alpha }_2|/|{\alpha }_3|=1$. In $B_3$, ${\alpha }_2$ is long and ${\alpha }_3$ short, giving $\sqrt{2}$. In $C_3$ their roles are reversed, giving $1/\sqrt{2}$. These ratios also follow from $A_{23}/A_{32}=|{\alpha }_2{|}^2/|{\alpha }_3{|}^2$.

Disconnected nodes in a Dynkin diagram represent orthogonal roots, so $\angle ({\alpha }_1,{\alpha }_3)=90^{\circ }$ in all three cases. A single edge gives $120^{\circ }$, hence $\angle ({\alpha }_1,{\alpha }_2)=120^{\circ }$ and, for $A_3$, also $\angle ({\alpha }_2,{\alpha }_3)=120^{\circ }$. The double edge in both $B_3$ and $C_3$ gives $\angle ({\alpha }_2,{\alpha }_3)=135^{\circ }$.

Starting from $[1,0,0]$ and applying the simple-root lowerings gives the four weightsEach has weight multiplicity one, so $d_1$ is the four-dimensional defining representation of $A_3\cong {\mathfrak{sl}}_4(\mathbb{C})$.

The representation with highest weight $[0,0,1]$ is the dual representation of $d_1$. Its weights are the negatives of those of $d_1$:Thus $\dim d_2=4$.

For $B_3\cong {\mathfrak{so}}_7(\mathbb{C})$, the highest weight $[0,0,1]$ is the spinor representation. Its eight weights, all of multiplicity one, areEquivalently, in an orthonormal basis they are $(\pm e_1\pm e_2\pm e_3)/2$ with independent signs. Hence $\dim d_3=8$.

The tensor product of the defining representation and its dual is the endomorphism representation. Splitting an endomorphism into its matrix trace and traceless part givesIn Dynkin labels, the two irreducible summands have highest weights $[0,0,0]$ and $[1,0,1]$.

Applying $P[a,b,c]=[b,a,b+c]$ to the eight weights of $d_3$ givesThis is exactly the union of the weight sets of $d_1$ and $d_2$. It expresses the branching rule for the standard inclusion $D_3\cong A_3\subset B_3$: the eight-dimensional spinor representation of ${\mathfrak{so}}_7$ restricts to the two four-dimensional chiral spinor representations of ${\mathfrak{so}}_6\cong {\mathfrak{sl}}_4$,