Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 102 / 3

Write an element of the diagonal Cartan subalgebra asThe root system of this Special orthogonal Lie algebra isThus the maps requested are $H(t)\mapsto \pm t_i\pm t_j$, with the two signs independent. This is the D3 root system, with twelve roots.

For a direct check, the defining matrix identity gives the block form $X=\left(\begin{array}{cc}A& B\\ C& -A^T\end{array}\right)$ with $B^T=-B$ and $C^T=-C$. If $E_{ij}$ denotes a matrix unit, the root space for ${\epsilon }_i-{\epsilon }_j$ is spanned by $E_{ij}-E_{j+3,i+3}$; those for ${\epsilon }_i+{\epsilon }_j$ and $-{\epsilon }_i-{\epsilon }_j$ are spanned respectively by $E_{i,j+3}-E_{j,i+3}$ and $E_{i+3,j}-E_{j+3,i}$. Commuting these with $H(t)$ gives exactly the displayed maps.

Choose the root basisIn the usual invariant inner product, each simple root has squared length $2$, while $({\alpha }_1,{\alpha }_2)=({\alpha }_1,{\alpha }_3)=-1$ and $({\alpha }_2,{\alpha }_3)=0$. The labeled Dynkin diagram is therefore the following chain, with a single bond between adjacent nodes:It is type $D_3$, equivalently the A3 root system. To check the root basis explicitly, the corresponding positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_1+{\alpha }_2,{\alpha }_1+{\alpha }_3,{\alpha }_1+{\alpha }_2+{\alpha }_3$; these and their negatives are all twelve roots.

Let $s_i=w_{{\alpha }_i}$. The root reflection formula is $s_i(\gamma )=\gamma -⟨\gamma ,{\alpha }_i^{\vee }⟩{\alpha }_i$. Applying the Cartan integers from part (b), the images of the chosen simple roots areIn particular, the two end-node Weyl reflections fix the opposite end node.

An automorphism of a root system outside the Weyl group is the sign changeIt preserves the set $\{\pm {\epsilon }_i\pm {\epsilon }_j\}$, fixes ${\alpha }_1$, and interchanges ${\alpha }_2$ and ${\alpha }_3$. It therefore realizes the nontrivial symmetry of the Dynkin diagram while preserving the chosen root basis.

To verify that $\tau \notin W$, note that $s_1$ swaps ${\epsilon }_1,{\epsilon }_2$, $s_2$ swaps ${\epsilon }_2,{\epsilon }_3$, and $s_3$ swaps ${\epsilon }_2,{\epsilon }_3$ while changing both their signs. Every product of these generators is a signed permutation with an even number of sign changes, whereas $\tau$ has one sign change. Thus it cannot be a Weyl group element. The parity here concerns sign changes, rather than the determinant of the permutation.

The Dynkin diagram in part (b) is a three-node chain, hence $D_3=A_3$. By the classification of complex semisimple Lie algebras by their root systems, this gives the Isomorphism between so6 and sl4:There is also a concrete realization. On $U={\mathbb{C}}^4$, choose a nonzero alternating four-form $\mathrm{vol}$. The six-dimensional exterior square ${\Lambda }^{2}U$ carries the nondegenerate symmetric bilinear form $B$ specified byIt is symmetric because exchanging two pairs entails four transpositions, and it is nondegenerate because each basis bivector pairs nontrivially with its complementary pair. The special linear Lie algebra acts by $A(u\wedge v)=Au\wedge v+u\wedge Av$. Its action on ${\Lambda }^{4}U$ is multiplication by $\mathrm{tr}A=0$, so it preserves $B$. We obtain a nonzero Lie algebra homomorphism ${\mathfrak{sl}}_4\to \mathfrak{so}({\Lambda }^{2}U,B)$. Its kernel is an ideal of a Lie algebra, hence zero because ${\mathfrak{sl}}_4$ is a simple Lie algebra. Both sides have dimension $15$, so this homomorphism is a Lie algebra isomorphism.