Gr2(R^(9,2))

The AutoKron can't determine the cohomology of $\text{Gr}_2(\mathbb{R}^{9,2})$.

There are 4 possibilities.
Here are their Poincaré polynomials: $$x^{14} y^{4} + x^{13} y^{4} + 2 x^{12} y^{4} + 2 x^{11} y^{4} + 2 x^{10} y^{4} + x^{10} y^{3} + x^{9} y^{4} + 2 x^{9} y^{3} + x^{8} y^{4} + 2 x^{8} y^{3} + x^{8} y^{2} + 2 x^{7} y^{3} + 2 x^{7} y^{2} + 2 x^{6} y^{3} + 2 x^{6} y^{2} + x^{5} y^{3} + 2 x^{5} y^{2} + 3 x^{4} y^{2} + 2 x^{3} y^{2} + x^{2} y^{2} + x^{2} y + x y + 1$$ $$x^{14} y^{4} + x^{13} y^{4} + 2 x^{12} y^{4} + 2 x^{11} y^{4} + 2 x^{10} y^{4} + x^{10} y^{3} + 3 x^{9} y^{3} + x^{8} y^{4} + 3 x^{8} y^{3} + 2 x^{7} y^{3} + 2 x^{7} y^{2} + 2 x^{6} y^{3} + 2 x^{6} y^{2} + x^{5} y^{3} + 2 x^{5} y^{2} + 3 x^{4} y^{2} + 2 x^{3} y^{2} + x^{2} y^{2} + x^{2} y + x y + 1$$ $$x^{14} y^{4} + x^{13} y^{4} + 2 x^{12} y^{4} + 2 x^{11} y^{4} + 2 x^{10} y^{4} + x^{10} y^{3} + x^{9} y^{4} + 2 x^{9} y^{3} + 3 x^{8} y^{3} + x^{8} y^{2} + 3 x^{7} y^{3} + x^{7} y^{2} + 2 x^{6} y^{3} + 2 x^{6} y^{2} + x^{5} y^{3} + 2 x^{5} y^{2} + 3 x^{4} y^{2} + 2 x^{3} y^{2} + x^{2} y^{2} + x^{2} y + x y + 1$$ $$x^{14} y^{4} + x^{13} y^{4} + 2 x^{12} y^{4} + 2 x^{11} y^{4} + 2 x^{10} y^{4} + x^{10} y^{3} + 3 x^{9} y^{3} + 4 x^{8} y^{3} + 3 x^{7} y^{3} + x^{7} y^{2} + 2 x^{6} y^{3} + 2 x^{6} y^{2} + x^{5} y^{3} + 2 x^{5} y^{2} + 3 x^{4} y^{2} + 2 x^{3} y^{2} + x^{2} y^{2} + x^{2} y + x y + 1$$ Here are the corresponding generator grids to these 4 possibilities: