The universal sub-bundle and the dual of the universal quotient bundle on $\operatorname{Gr}(2,4)$ are specific examples of spinor bundles on a quadric. They arise from an irreducible representation of a semi-simple Lie algebra. In the paper of Ottaviani (1988), a theorem of Ramanan is quoted to argue that spinor bundles are stable.
I was wondering if there is a direct and elementary way to prove the stability of a spinor bundle on the Grassmannian $\operatorname{Gr}(2,4)$, and why is it exceptional.
Let $V$ be a 4-dimensional vector space and let $S$ be the universal subbundle on $\mathrm{Gr}(2,V)$. Let $0 \ne f \in V^*$ and consider it as a global section of $S^*$. Then its zero locus is the sub-Grassmannian $$ \Pi := \mathrm{Gr}(2,V_f) \cong \mathbb{P}^2, $$ where $V_f \subset V$ is the annihilator of $f$. Therefore, there is an exact sequence (Koszul complex) $$ 0 \to \mathcal{O}(-1) \to S \to \mathcal{O} \to \mathcal{O}_\Pi \to 0.\tag{*} $$ Since $\mathcal{O}(-1)$ has no cohomology, while $H^\bullet(\mathcal{O}) = H^\bullet(\mathcal{O}_\Pi) = \Bbbk$ is the base field, it follows that $$ H^\bullet(S) = 0. $$ In particular, the vanishing of $H^0(S)$ implies the stability of $S$.
Tensoring $(*)$ by $S^*$, we obtain $$ 0 \to S \to S \otimes S^* \to S^* \to S^*\vert_\Pi \to 0. $$ Now $S$ has no cohomology, while $$ H^\bullet(S^*) = V^*, \qquad H^\bullet(S^*\vert_\Pi) = V_f^*, $$ and the map is the restriction (the dual to the embedding $V_f \hookrightarrow V$), hence it is surjective and its kernel is 1-dimensional. This shows that $$ H^\bullet(S \otimes S^*) = \Bbbk, $$ hence $S$ is exceptional.