Is there a quick way to prove the following statement, if possible without resorting to the classification of simple Lie groups?
Let $G$ be a simple Lie group of non-compact Hermitian type of rank $r$. Let $Y = G/K$ be the associated Hermitian symmetric space, normalized so that the minimal holomorphic sectional curvature is $-1$. Suppose that there is a complex $2$-plane inside the holomorphic tangent space at some point, $\mathcal{T}Y$, sitting inside the locus having maximal holomorphic sectional curvature $-\frac{1}{r}$. Then in fact $G \cong \text{SU}(p,q)$, with $p \geq 2q$ or $q \geq 2p$.
I have an explicit proof of this fact through some unpleasant matrix computations (also, this proof somehow misses the cases $SO^*(10)$ and $SO^*(14)$, if anybody is interested in the reason for this I have posted an unfruitful question on Math SE: https://math.stackexchange.com/questions/830719/linearly-independent-skew-symmetric-complex-matrices-having-the-least-eigenvalue ). But I sense that the question has already been investigated, and I would like something shorter an nicer, if possible.
I think that if one puts in the condition that this 2-space is tangent to a totally geodesic submanifold the result gets easier (since a complete classification of such manifolds is available, by [Ihara, 1967]), but again, I would like to avoid classification arguments, for something that should be rather easy.
