EricHogle
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The cohomology of $C_2$-equivariant Grassmannians

A database of known values of $H^{\ast,\ast}(\text{Gr}_k(\mathbb{R}^{p,q});\underline{\mathbb F}_2)$, expressed as a module over $\mathbb{M}_2:=H^{\ast,\ast}(pt;\underline{\mathbb F}_2)$. All $RO(C_2)$-grading is motivic, with bidegree $(x,y)$ indicating a $x$-dimensional representation containing $y$ copies of the sign representation (and hence $x-y$ copies of the trivial rep.) The majority of these calculations are produced by the AutoKron, using techniques outlined here.

Note that $\text{Gr}_k(\mathbb{R}^{p,p-q})\cong \text{Gr}_k(\mathbb{R}^{p,q})$ and $\text{Gr}_{p-k}(\mathbb{R}^{p,q})\cong \text{Gr}_k(\mathbb{R}^{p,q})$ so we'll just worry about when $k,q\le \frac p2$.

When $k=1$

These are the projective spaces $\mathbb{P}(\text{Gr}_k\mathbb{R}^{p,q})$. Kronholm defines "twisted" projective space $\mathbb{RP}^n_{tw}=\mathbb{P}(\mathbb{R}^{n+1,\lfloor\frac {n+1}2\rfloor})$.

$k=1$ $q=1$ $q=2$ $q=3$ $q=4$ $q=5$ $q=6$
$p=2$ $\mathbb{RP}^1_{tw}$
$p=3$ $\mathbb{RP}^2_{tw}$
$p=4$ $\mathbb{P}(\mathbb{R}^{4,1})$ $\mathbb{RP}^3_{tw}$
$p=5$ $\mathbb{P}(\mathbb{R}^{5,1})$ $\mathbb{RP}^4_{tw}$
$p=6$ $\mathbb{P}(\mathbb{R}^{6,1})$ $\mathbb{P}(\mathbb{R}^{6,2})$ $\mathbb{RP}^5_{tw}$
$p=7$ $\mathbb{P}(\mathbb{R}^{7,1})$ $\mathbb{P}(\mathbb{R}^{7,2})$ $\mathbb{RP}^6_{tw}$
$p=8$ $\mathbb{P}(\mathbb{R}^{8,1})$ $\mathbb{P}(\mathbb{R}^{8,2})$ $\mathbb{P}(\mathbb{R}^{8,3})$ $\mathbb{RP}^7_{tw}$
$p=9$ $\mathbb{P}(\mathbb{R}^{9,1})$ $\mathbb{P}(\mathbb{R}^{9,2})$ $\mathbb{P}(\mathbb{R}^{9,3})$ $\mathbb{RP}^8_{tw}$
$p=10$ $\mathbb{P}(\mathbb{R}^{10,1})$ $\mathbb{P}(\mathbb{R}^{10,2})$ $\mathbb{P}(\mathbb{R}^{10,3})$ $\mathbb{P}(\mathbb{R}^{10,4})$ $\mathbb{RP}^9_{tw}$
$p=11$ $\mathbb{P}(\mathbb{R}^{11,1})$ $\mathbb{P}(\mathbb{R}^{11,2})$ $\mathbb{P}(\mathbb{R}^{11,3})$ $\mathbb{P}(\mathbb{R}^{11,4})$ $\mathbb{RP}^{10}_{tw}$
$p=12$ $\mathbb{P}(\mathbb{R}^{12,1})$ $\mathbb{P}(\mathbb{R}^{12,2})$ $\mathbb{P}(\mathbb{R}^{12,3})$ $\mathbb{P}(\mathbb{R}^{12,4})$ $\mathbb{P}(\mathbb{R}^{12,5})$ $\mathbb{RP}^{11}_{tw}$
$p=13$ $\mathbb{P}(\mathbb{R}^{13,1})$ $\mathbb{P}(\mathbb{R}^{13,2})$ $\mathbb{P}(\mathbb{R}^{13,3})$ $\mathbb{P}(\mathbb{R}^{13,4})$ $\mathbb{P}(\mathbb{R}^{13,5})$ $\mathbb{RP}^{12}_{tw}$
A pattern emerges here. Kronholm gives $$H^{\ast,\ast}(\mathbb{RP}^n_{tw})=\bigoplus_{k=0}^n\Sigma^{k,\lceil\frac k2\rceil}\mathbb M_2$$ and more generally we can select constructions whose filtration spectral sequences have no nonzero differentials, for example if for $q\le \frac p2$ we identify $V=\mathbb{R}^{p,q}\cong (\mathbb{R}^{+-})^{\oplus q}\oplus (\mathbb{R}^+)^{\oplus (p-2q)}$ then as $\mathbb{M}_2$ modules $$H^{\ast,\ast}(\mathbb{P}(\mathbb{R}^{p,q}))\cong E_1(V)=\bigoplus_{k=0}^p\Sigma^{k,\min(q,\lceil\frac k2\rceil)}\mathbb M_2.$$

When $k=2$

A pattern emerges for the first two columns and the diagonal, but there's more work to be done.

$k=2$ $q=1$ $q=2$ $q=3$ $q=4$ $q=5$ $q=6$
$p=4$ $\text{Gr}_2(\mathbb{R}^{4,1})$ $\text{Gr}_2(\mathbb{R}^{4,2})$
$p=5$ $\text{Gr}_2(\mathbb{R}^{5,1})$ $\text{Gr}_2(\mathbb{R}^{5,2})$
$p=6$ $\text{Gr}_2(\mathbb{R}^{6,1})$ $\text{Gr}_2(\mathbb{R}^{6,2})$ $\text{Gr}_2(\mathbb{R}^{6,3})$
$p=7$ $\text{Gr}_2(\mathbb{R}^{7,1})$ $\text{Gr}_2(\mathbb{R}^{7,2})$ $\text{Gr}_2(\mathbb{R}^{7,3})$
$p=8$ $\text{Gr}_2(\mathbb{R}^{8,1})$ 2 possible $\text{Gr}_2(\mathbb{R}^{8,3})$ $\text{Gr}_2(\mathbb{R}^{8,4})$
$p=9$ $\text{Gr}_2(\mathbb{R}^{9,1})$ 4 possible 2 possible $\text{Gr}_2(\mathbb{R}^{9,4})$
$p=10$ $\text{Gr}_2(\mathbb{R}^{10,1})$ 8 possible 8 possible 2 possible $\text{Gr}_2(\mathbb{R}^{10,5})$
$p=11$ $\text{Gr}_2(\mathbb{R}^{11,1})$ 24 possible 48 possible 8 possible $\text{Gr}_2(\mathbb{R}^{11,5})$
$p=12$ $\text{Gr}_2(\mathbb{R}^{12,1})$ 72 possible 288 possible 96 possible 4 possible $\text{Gr}_2(\mathbb{R}^{12,6})$
$p=13$ $\text{Gr}_2(\mathbb{R}^{13,1})$ 216 possibilities ?? ?? ?? $\text{Gr}_2(\mathbb{R}^{13,6})$
$\quad\vdots$ $\quad\vdots$ $\quad\vdots$ $\quad\ddots$
$p=n$ $\text{Gr}_2(\mathbb{R}^{n,1})$ $\text{Gr}_2(\mathbb{R}^{n,2})$ $\text{Gr}_2(\mathbb{R}^{n,\lfloor\frac n2\rfloor})$

When $k=3$

$k=3$ $q=1$ $q=2$ $q=3$ $q=4$ $q=5$ $q=6$
$p=6$ $\text{Gr}_3(\mathbb{R}^{6,1})$ $\text{Gr}_3(\mathbb{R}^{6,2})$ $\text{Gr}_3(\mathbb{R}^{6,3})$
$p=7$ $\text{Gr}_3(\mathbb{R}^{7,1})$ 2 possible 14 possible
$p=8$ $\text{Gr}_3(\mathbb{R}^{8,1})$ 4 possible 168 possible ?
$p=9$ 2 possible ? ? ?
$p=10$ 4 possible ? ? ? ?
$p=11$ 8 possible ? ? ? ?
$p=12$ 24 possible ? ? ? ? ?

When $k=4$

$k=4$ $q=1$ $q=2$ $q=3$ $q=4$ $q=5$ $q=6$
$p=8$ $\text{Gr}_4(\mathbb{R}^{8,1})$ 6 possible ? ?
$p=9$ 2 possible ? ? ?
$p=10$ 16 possible ? ? ? ?
$p=11$ 288 possible. ? ? ? ?

When $k=5$

$k=5$ $q=1$ $q=2$ $q=3$ $q=4$ $q=5$ $q=6$
$p=10$ 16 possible ? ? ? ?
$p=11$ 864 possible ? ? ? ?