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This is for all that are interested in classical groups and their representations. We are investigating the following situation:

Let $V$ be $d$-dimensional $k$-vector space (where $k$ is a finite field) equipped with a non-degenerate quadratic form $Q$, so that $V$ is an orthogonal space. Let $GO$ be the group of similarities (which are a bit more general than isometries; we use the language of Kleidman's and Liebeck's book 'The Subgroup Structure of the Finite Classical Groups') and fix $g_1,g_2\in GO$. Suppose we have the following setting:

There are non-degenerate subspaces $U_i$ which are irreducible as $\langle g_i\rangle$-modules and which have large dimension $e_i>d/2$ for $i=1,2$. (we call $g_i$ 'fat' in this case, a generalization of ppd-elements as described in A. Niemeyer and C. Praeger 'A recognition algorithm for classical groups over finite fields'; our project actually deals with finding the proportions of such elements within the classical groups)

What we are interested in here is a third subspace:

Let $W$ be an irreducible $\langle g_1,g_2\rangle$-module which intersects trivially with $U_i$ and assume that $W$ is totally singular i.e. $Q_W=0$. Is it possible that $W$ - as a subspace of $U_i^{\perp}$ - is maximal totally singular in $U_1^{\perp}$ or $U_2^{\perp}$, which is means that the dimension of $W$ equals the Witt index of $U_i^{\perp}$?

We can control the case where $W$ is non-degenerate or generated by a non-singular element $v$ (i.e. $Q(v)\neq 0$). The bad guys are such $W$ as above, as the group $\Omega\le GO$ as described in (Kleidman and Liebeck) does not act transitvely on maximal totally singular subspaces in general. The best result would be if such subspaces cannot exist in the above setting.

P.S. For simplicity we could assume that the characteristic of $k$ is odd.

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I am confused. First you say that $W$ intersects trivially with $U_i$, then you talk about $W$ being a subspace of $U_i$! –  Derek Holt May 4 '12 at 22:19
    
@Derek, you are absolutely right, $W$ is a subspace of $U_i^{\perp} - sorry for that. –  Natalie May 6 '12 at 21:17
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