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In this section we show that the only remaining orbits, i.e., those in are , , and . We do this by counting the

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 order

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
.

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
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**.

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