Let be the Frobenius map which raises elements to the q-th power, and let it act entrywise on G. The set of fixed points under F in G is of orderLet W be the 16-dimensional vector space of forms modulo cubics we have been analyzing, but now defined over , so that . We want to consider an orbit Gf say of the form f. We assume here that f has coefficients fixed by F; in other words, they lie in , which is the fixed field of F. For f as in Table 1, they are in the ground field . We will try to analyze the set Y of elements in Gf which are fixed by F, that is, . We are using the two actions of F on G and also on V. Let . If gf has coefficients fixed by F, so does hgf for each . This means that the elements of H act on Y and form orbits under this action. As , the cardinality of Y is finite.
Let R be the stabilizer in G of f and S its connected component. Then is finite, say of size n, so there are coset representatives of S in R. We assume that q has been chosen so they are fixed by F. Assume is a complete set of representatives for the conjugacy classes of . We prove the following: Theorem. There are exactly r orbits of H acting on Y. The orbits are precisely , , where with . Moreover, if denotes the subgroup of R consisting of all elements for which , the number of elements in is .
We proceed in five steps.
Notice first that , as f has coefficients in . It follows that if , then , and so so for and . Suppose on the other hand that for some i, and s in S; then . This means that all elements in Y are of the form gf for such a g. B. For each i, there is a for which .
Lang's theorem applies to G acting on itself as G is connected. The fixed points of the automorphism F are finite and so there is a for which .
If , then .
To see the converse suppose satisfies . Then and so . This means the coset is exactly the set of elements g of G for which . There is one orbit of H in Y given by .
D. Suppose that satisfies with and for some i. Then .
We will show that there exists satisfying . Then by C.
Notice that , and so we want to find with or, equivalently,Let be conjugation by t acting on S by . The condition for is
Let . Note that since , the Frobenius map F stabilizes R and so S as well. By Lang's theorem, all elements in the coset FG are conjugate in . Thus fixes only finitely many points (indeed |H| points) on G and so also on S. By Lang's theorem, applied to acting on S, there is such an . E. For , we have if and only if and are conjugate in .
Suppose and are conjugate in by an element . This means for . Then . From D., it follows that . Note that and so the H-orbits of and of are the same.
Suppose on the other hand that and are in the same H-orbit. We may assume for some and so for one of the coset representatives t and . Now and so . But under F, the coset tS is mapped to tS as S is mapped to S and so mod S, and are conjugate. This means we need only consider the orbits .
F. The only remaining part is to show that the number of elements in the orbit of is as indicated. The stabilizer in H of is exactly . But, for , the conjugate is in H if and only if it is fixed by F. This gives which is equivalent to as given, that is, . Corollary. Suppose the stabilizer R splits as an extension over S (so that we can choose the coset representatives of S in R in such a way that they form a group themselves fixed by F). Then
As above, let be the set of elements r in R satisfying , so that . Suppose . Write with . Then Because we have chosen the to be fixed by F, this means . But this implies that and commute modulo S and so commute as the form a group. This implies . Thus, where is the centralizer of in . By the theorem, we now have . But is the size of the conjugacy class and so the result follows. Corollary. If R is finite, then |Y|=|H|.
|S|=1 in the above.
Corollary. If (in particular if R=S), then (where are the just the fixed points of F on S).
Any induces an inner automorphism on S. It follows by Lang's theorem, that is conjugate to F by an element of S and so has the same number of fixed points as F on S.
We now add the contributions to from the orbits of forms , . Note that the stabilizers in G of these forms are given in .2. Here h=|H| is as in the beginning of this section. For , we shall denote by the set .
The stabilizer of is finite and so, by Corollary 6.3, the number of points in is h.
The stabilizer of is where T is a one-dimensional torus and is the cyclic group of order 2 generated by the matrix . There are two orbits, each of size , so .
The stabilizer of is connected and so the number of orbits is one and the size is .
Now adding the contributions for these three orbits and subtracting from gives a polynomial of degree 13.
We now quote a result by [LW] which shows there are no more orbits of dimension 14 or higher.
Theorem. [LW]. Let X be a d-dimensional subvariety of an n-dimensional projective space defined over . There exists a constant c (depending only on X, d and n - in fact, not on X explicitly, but on the degree of X in the projective space)
Note this result also works for subvarieties of affine space. For one direction, embed the affine variety in projective space and consider the closure; then where is the number of -points of the closure and use the upper bound for that. For the other bound, choose an open subset O contained in the original variety. Then the number of -points on the closure of O and the complement of O in can be estimated, and so also on O; this gives the appropriate lower bound for . Corollary. Theorem holds.
If satisfies , then, by Theorem 6.5, the number of -points of Gf (the only dense orbit of ) is at least (for some constant c), which contradicts the fact that the number of -points of is a polynomial in q of degree 13. This shows there are no further orbits of dimension 14 or higher. In particular, all other orbits lie in . In , any form in was shown to be listed in Table 1. Our work is over the algebraic closure of . However arguments in [GLMS] show the same applies over any algebraically closed field of characteristic 3.