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For the real Grassmannian Gr$(N,k)$ we have the well-known isomorphism $$ \text{Gr}(N,k) = O(N)/(O(k) \times O(N-k)) $$ For the complex case, we have $$ \text{Gr}(N,k) = U(N)/(U(k) \times U(N-k)) $$ I would like to know if anything like this holds in the finite field setting, ie can the finite field Grassmannians be described as a homogeneous space of an algebraic group over a finite field, or something like this?

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Yes. It's a homogeneous space over GL of the base field, for all fields. – Ben Webster Mar 31 '12 at 14:55
Do you have a reference for this? What's the stabilizer subgroup, and how does it act on GL? – Ago Szekeres Mar 31 '12 at 15:41
Wikipedia:… Which it's considered good form to check before asking MO questions. – Ben Webster Mar 31 '12 at 16:38
Duly noted, thanks – Ago Szekeres Mar 31 '12 at 16:45
Perhaps it should be noted that over finite fields there are "outer forms" of some Grassmannians, e.g., the Grassmannian of $k$-dimensional subspaces of a vector space of dimension $2k$. These are not homogeneous spaces for the "split form" of $\textbf{GL}_{2k}$. Also, I believe the OP mistakenly wrote $O(1) \times O(N-1)$ instead of $O(k) times O(N-k)$. – Jason Starr Apr 2 '12 at 11:31

There is a description of these homogeneous spaces over finite fields in terms of graphs in "Distance-Regular Graphs" by Brouwer, Cohen and Neumaier (Springer, 1989) and in "Algebraic combinatorics I: Association schemes" by E.Bannai and T.Ito (Benjamin/Cummings, 1984).

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