Orbits Under Symplectic Transvections Ii: The Case K = F2

Let V be a not necessarily regular, finite-dimensional symplectic space (with symplectic form . ) over a field K where K\rad(|/) is non-empty. Our previous paper [27] (hereinafter referred to as [I]) began the study of the orbits of K\rad(K) under the action of the group Tv{S) of isometries which are products of transvections from a subset S of V. Let G(S) be the graph with vertex set 5 and an edge between a,b e S if a.b # 0. The main result of [I] was that if K has more than two elements, then the action of Tv(S) on K\rad(K) is transitive if and only if S spans V and G{S) is connected. In this paper we consider the more difficult case where K is the field F2 of two elements, and we assume this throughout. Let P be a basis for V. Then the graph G(P) determines the symplectic form on V. Moreover, if H is a full subgraph of G(P), then the sum of the vertices of H is an element of V. This gives a one-one correspondence between the elements x of V and the full subgraphs x \P of G(P), and enables us to use lattice-theoretic terminology for elements of V. That is, inclusion of subgraphs of G(P) determines a lattice ordering ap on elements of V, with union uP and intersection nP. Let QP be the quadratic form associated to the symplectic form . on V such that QP{a) = 1 for all a e P. Then Tv(P) preserves the form QP. We show that for x e V, QP(x) is the Euler characteristic mod 2 of the graph x |P. Thus this quadratic form is that introduced in [9] for the case where V = Hl(Tg;F2) with the intersection symplectic form. It is easy to prove that if P, R are r-equivalent bases of V, as defined in [I], then QP = QR. Let E6 denote the graph shown in Fig. 1. The vertex v will be called the centre of Eb. A graph G is said to be of orthogonal type if G is a tree and contains £6 as a subgraph. A basis P of V is of orthogonal type if P is r-equivalent to P' where G{P') is a graph of orthogonal type.