Past exam of the groups and representations course of the University of Oxford / 2019 / 4 / c / 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 $(-I_4)\otimes (-I_4)=I_{16}$.