Proof of the Theorem

In the following, denotes a primitive cubic root of 1 (Note that as , and if , if ).

Proof. First we prove that acts irreducibly on . Let be a subspace of invariant under the action of , and let . Then and therefore is invariant under the action of . We may assume that the last two coordinates of are not both zero (otherwise is such a vector of ). Then , , and . If , ; whereas if , . Thus we may assume . Since , , and are four independent vectors in , it follows .

Now we prove that acts primitively on . First assume with (), be a decomposition of into 1-dimensional subspaces, invariant under . Since and , acts reducibly on , if fixes all the 's. Thus has a unique fixed point in its action on , say . We may assume , and . Moreover is an eigenvector of on , hence either (with not both zero), or where and . In the former case , . However both and have the last coordinate equal to zero, a contradiction. In the latter case , , , are dependent, a contradiction. Now assume be a decomposition of into 2-dimensional subspaces invariant under . Since , fixes both and and therefore permutes them. Also fixes both and . Let . If , then with not both zero, and . Therefore, by the symmetry of and we may assume . Let . This forces , . But , a contradiction. So , and . But then , again a contradiction.

Proof. In order to show that cannot fix a linear congruence consisting of isotropic lines, we make use of the orthogonal representation of (see 2.2). A linear congruence as above, is represented by a non-degenerate quadric lying in a 3-subspace of , being the complex of the isotropic lines. Denote by , the images of , in , and by , their linear preimages. Then, computing and with respect to the basis of , and restricting to , it is readily seen that, choosing the frame for , , can be represented by the matrices We need to show that cannot fix the quadric . Assume the contrary. Then, since contains a frame of , leaves invariant. Since is a hyperplane of , and both leave invariant iff and acting on the dual space of have a common eigenvector. It is readily seen that, if , the eigenspaces of in the dual action are and , whereas the eigenspaces of are , , and (possibly over ). Therefore any eigenspace of has trivial intersection with any eigenspace of , and we are done.

Proof. Let be the symmetric matrix defining a quadric . is fixed by iff and for some . It is easy to show that (if ) this leads to . For: This gives

If , then by (1), by (2) and by (3) (note that the discriminant of (3) is ), hence . If and , then by (2) and . Finally, if , (1) gives and Now consider the action of . If , hence . If , the equation gives and again .

Proof. acts on a 5-dimensional vector space . Let be a direct decomposition into 1-dimensional subspaces , , invariant under the action of . Since has order 2, fixes one of the points , say . The eigenspaces of on are , and whereas the eigenspaces of are and (possibly over ) , . It follows that have no common eigenvectors. Hence is not fixed by . Let , , so that and . Assume . Then it turns out that
and

It is easily checked that, since , and cannot be both eigenvectors of . Now assume . In this case,
and
Since , both these vectors are moved by . Hence and interchanges and . Assume first . Then . But this last point is not invariant under , a contradiction. Now assume . Then , which is again not invariant under , a contradiction.

Similarly, if we assume , we get a contradiction.

Proof. Assume . Let a twisted cubic in be invariant under . Then, since has points, fixes at least two points on . Let be one of these fixed points and suppose first is an eigenvector of of shape . Then , , and all lie on . These are four distinct points lying on the plane . Since four coplanar points cannot lie on a twisted cubic, we get a contradiction. Hence we may assume that fixes the points , on , where both and are fixed by , i.e. have shape . Again we find that are coplanar. Thus at least two of them must coincide. This forces, up to scalar multiples, and . Computing up to six distinct points in the orbit of containing and , it turns out that the parametric equation of has to be the following: where

Now has to be symplectic, since two cubics are associated to the same symplectic form iff they can be transformed into one another by an element of . This implies . We conclude that, if (and this choice for is clearly always possible), does not fix a cubic in . (Note: if , may actually fix a cubic. E.g. if , , is actually a fixing a cubic.)

Now assume . If leaves invariant a cubic of parametric equation with , then does not necessarily have fixed points on . But we may consider the ``extended" cubic in i.e. the cubic having parametric equation with . will act on , and, since , will have fixed points on . Thus the above argument will apply.

Proof of the Theorem. Let be the projective image of in . cannot be contained in a maximal subgroup of type 1 or 2 (Proposition 1), nor in a subgroup of type 3(a)i (Proposition 2), 3(a)ii (Proposition 3), 3c (Proposition 4) or 3(d)i (Proposition 5). Thus we are left to show that is not contained in a subgroup of type 3b or 3(d)ii. As already mentioned in 2.3, the subgroups in 3(b)i form a unique conjugacy class in , while the subgroups in 3(b)ii form a unique conjugacy class in . Since is invariant under conjugation in and , it follows that is not contained in any subgroup of type 3(b)i whenever . Next observe that, being the characteristic polynomial of , the characteristic polynomial of () is . Thus has all its coefficients in iff . Hence is not contained in a subgroup of type 3(b)ii whenever . The latter certainly occurs if .

As for the exceptional subgroups (type 3(d)ii), it is enough to notice that .