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Oct 13, 2015 at 16:57 answer added t3suji timeline score: 1
Jul 27, 2015 at 10:00 comment added Nick Gill You only need to prove the statement for $r=n$, as the rest will follow directly... And the statement is true when $2n=4$ and the field is ${\mathbb F}_3$ - see p.33 of this: math.lsu.edu/~hoffman/papers/spreads4.pdf (Whether that provides any evidence for the statement in general, I couldn't say. The group ${\rm Sp}_4(3)$ is a bit special.)
Jul 15, 2015 at 13:51 answer added user76083 timeline score: 0
Jul 3, 2015 at 13:36 comment added user33209 @ Dear Jason, yes you areright, it is totally isotropic! since it relate to symplectic groups, I used finite-group
Jul 3, 2015 at 13:25 comment added Name by the way in the literature, the subspaces $U$ with $U\subset U^\perp$ are called totally isotropic.
Jul 3, 2015 at 13:22 comment added Name why the tag finite-groups?
Jul 3, 2015 at 12:27 comment added user33209 @Jason, I mean the first!
Jul 3, 2015 at 12:26 comment added Jason Starr When you say "trivial intersection", do you mean that every pairwise intersection is trivial, or do you mean the common intersection of all specified subspaces is trivial?
Jul 3, 2015 at 12:25 comment added user33209 a set of isotropic subspaces with trivial intersection, that covers $V$
Jul 3, 2015 at 12:24 comment added Jason Starr What is an "isotropic spread"?
Jul 3, 2015 at 12:22 history asked user33209 CC BY-SA 3.0