Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 18 / 4 / d

Choose positive rootsThe corresponding simple roots are ${\epsilon }_1-{\epsilon }_2$ and ${\epsilon }_2$. The vector $v_{1,+}\wedge v_{2,+}$ has weight $\lambda ={\epsilon }_1+{\epsilon }_2$. Adding any positive root to this weight produces a weight absent from part (a), so every positive root vector annihilates it. It is consequently a highest-weight vector with highest weight $\lambda$.

A representation of a compact Lie group is completely reducible: Haar averaging makes an inner product invariant, and orthogonal complements of invariant subspaces remain invariant. Therefore the exterior square contains an irreducible summand $L_{\lambda }$ of this highest weight. This can also be seen by projecting the displayed highest-weight vector into irreducible summands; a nonzero projection remains annihilated by all positive root vectors and has the same weight. Its multiplicity is one because the weight space at $\lambda$ is one-dimensional.

Use the Weyl dimension formula, expressly allowed in this part:Here $L_{\lambda }$ is the irreducible highest-weight representation, ${\Phi }^{+}$ is the chosen positive root system, $\rho$ is half its sum, and the pairing is the invariant Euclidean inner product on the real weight plane, with ${\epsilon }_1,{\epsilon }_2$ orthonormal. Using coroots instead gives the same ratio because the length factor cancels for each root. In this convention,The four factors, in the listed positive-root order, are respectivelyTheir product is $10$. Since $\dim {\Lambda }^2{\mathbb{C}}^5=\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10$, this irreducible summand occupies the entire exterior square. HenceThe proof uses the dimension formula only where it is allowed, not in the symmetric-power argument of Question 2. Its B2 adjoint representation weight diagram is exactly the eight roots and the double zero weight obtained above.