In a previous post, I cite a dimension condition commonly satisfied by homogeneous spaces and claim that a counterexample must have deficiency at least $3$. For convenience, I reproduce the definition of deficiency in the next paragraph.
Let $G/K$ be a homogeneous space with $K$ connected. The quotient map $G \to G/K$ induces a ring map $H^*(G/K) \to H^*(G)$ in rational cohomology whose image is an exterior algebra $\Lambda \hat P$. The dimension of $\hat P$ is bounded above by $\mathrm{rk \,}G - \mathrm{rk \,}K$, and the deficiency is the difference from this upper bound: $$\mathrm{df}(G/K) := \mathrm{rk \,}G- \mathrm{rk \,}K - \dim_{\mathbb Q} \hat P.$$
Now one thing that is clear about deficiency is that it is additive in the sense that given two homogeneous spaces $G_1/K_1$ and $G_2/K_2$, one has $$\mathrm{df}\Big(\frac{G_1 \times G_2}{K_1 \times K_2}\Big) = \mathrm{df}(G_1/K_1) + \mathrm{df}(G_2/K_2).$$ Morally, this is an uninteresting way of picking up deficiency, and particularly, it is not useful in constructing counterexamples to the dimension condition of the previous post. Call a homogeneous space $G/K$ reducible if it admits a direct product factorization in terms of other homogeneous spaces.
All examples of irreducible homogeneous spaces that I know of from the literature are of deficiency $0$ or $1$. I have been able to construct high-deficiency irreducible examples by modifying reducible ones, but these examples have a feel of cheapness to them and wind up satisfying the dimension condition anyway.
Is anyone aware of any examples in the literature of irreducible homogeneous spaces of deficiency at least $2$?
