Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 302 / 3

A Cartan subalgebra $\mathfrak{h}$ of a complex semisimple Lie algebra is a maximal commuting subalgebra consisting of semisimple elements. A nonzero functional $\alpha \in {\mathfrak{h}}^{\ast }$ is a root when its space in the root-space decompositionis nonzero. A Cartan-Weyl basis combines a basis of $\mathfrak{h}$ with root vectors $E_{\alpha }\in {\mathfrak{g}}_{\alpha }$.

A choice of regular hyperplane divides roots into positive and negative roots. The simple roots are the positive roots that are not sums of two positive roots; they form a basis of the real root span. The Cartan matrix isup to the equivalent transposed indexing convention.

Let $\beta$ be a positive root. If it is not simple, it is a sum of two positive roots. Repeating this decomposition terminates because the height with respect to a regular positive functional strictly decreases, and it writeswith at least one $n_i>0$. Equivalently, one may repeatedly choose a simple root ${\alpha }_i$ with $(\beta ,{\alpha }_i)>0$ and use the root string to replace $\beta$ by the root $\beta -{\alpha }_i$.

For uniqueness, the simple roots are linearly independent. Therefore two such expansions have identical coefficients. Here “positive integer coefficients” must allow zero coefficients: a simple root itself has coefficient one on its own basis vector and zero on the others.

The matrix is the Cartan matrix of type A3. The positive roots areTogether with their negatives they formso $|\Phi |=12$.

In Dynkin label coordinates, subtracting ${\alpha }_i$ subtracts row $i$ of the stated Cartan matrix. Starting from $(1,0,0)$ gives the weight chainThese are the four weights of the defining representation of $A_3\cong {\mathfrak{sl}}_4$.

The highest weight $(0,1,0)$ gives the six-dimensional second exterior power of the defining representation. Adding each unordered pair of the four defining weights from part i givesThere are no multiplicities, in agreement with the assumption in the question.

For equal-length candidate roots, the required inner products are

Set A is invalid because ${\alpha }_1\cdot {\alpha }_3=-1$ rather than zero; indeed its three vectors sum to zero and are not linearly independent. Set B has all the displayed inner products and is linearly independent, so it is valid. Set C is the standard realizationof the A3 root system and is also valid.