We begin with a quick revision of the second part of SPCS8. Let be a symplectic
matrix with integer coefficients. Denote by
the characteristic polynomial of
and let
be the set of roots of
. Since
is symplectic,
is a reciprocal polynomial, i.e.,
, so that
whenever
. Then, we put
, and we define
. Also, for each
, we let:
the set of subsets of
of cardinality
,
the set of subsets of
of cardinality
, and
.
Next, we suppose that the Galois group of
is the largest possible, i.e.,
. In other words, by dividing
into
pairs
, we have that
contains all permutations of these
pairs (and this is the
factor), and for each pair
, we may or may not permute
and
independently of other pairs (and this is the
).
Now, assuming that is the largest possible, we fix a choice of basis
consisting of eigenvectors of
that is coherent with the action of the Galois group
, i.e.,
and
for all
.
Then, we consider a matrix such that
and
don’t share a common invariant subspace, and we want to prove that
Theorem 1 Under the assumptions above, adequate products of powers of
and
are twisting with respect to the pinching matrix
.
Indeed, this is a main step towards the application (in the next and final post of this series) of the simplicity criterion for the Kontsevich-Zorich cocycle over -orbits of origamis to concrete cases of genus
origamis because it permits to reduce the verification of the twisting condition to an algebraic condition on Galois groups of characteristic polynomials of matrices in
.
Moreover, by the end of SPCS8, we saw that given , it is twisting with respect to
if and only if
for all
for
, where
is the coefficient of the matrix
acting on
. Here, we used that
is generated by
,
, so that morally
corresponds to an “isotropy condition”. Furthermore, we constructed a graph
whose set of vertices is
and whose set of arrows is
. Thus, in this language
is twisting with respect to
if and only if
is a complete graph. Finally, we mentioned that checking directly that
is complete is not obvious, and we proved that:
Proposition 2 If
is mixing, then an adequate product
of powers of
and
is
-twisting with respect to
(in the sense that
twists
-dimensional isotropic
-invariant subspaces).
So, after this brief revision of SPCS8, we will spend today’s post with the presentation of some elements of the proof of Theorem 1.
1. Strategy of proof of Theorem 1
The outline of the proof of Theorem 1 is:
- Step 0: We will show that the graphs
are always non-trivial, i.e., there is at least one arrow starting at each of its vertices.
- Step 1: Starting from
and
as above, we will show that
is mixing and hence, by Proposition 2, there exists
twisting
-dimensional (isotropic)
-invariant subspaces.
- By Step 1, the treatment of the case
is complete, so that we have to consider
. Unfortunately, there is no “unified” argument to deal with all cases and we are obliged to separate the case
from
.
- Step 2: In the case
, we will show that
(with
as in Step 1) is mixing for all
. Hence, by Proposition 2, we can find
twisting
-dimensional isotropic
-invariant subspaces for all
. Then, we will prove that
is mixing and, by Proposition 2, we have
twisting with respect to
, so that this completes the argument in this case.
- Step 3: In the special case
, we will show that either
or a closely related graph
are mixing and we will see that this is sufficient to construct
twisting
-dimensional isotropic
-invariant subspaces.
In the sequel, the following easy remarks will be repeatedly used:
Remark 1 If
, then the graph
is invariant under the action of Galois group
on the set
(parametrizing all possible arrows of
). In particular, since the Galois group
is the largest possible, whenever an arrow
belongs to
, the inverse arrow
also belongs to
. Consequently,
always contains loop of even length.
Remark 2 A connected graph
is not mixing if and only if there exists an integer
such that the lengths of all of its loops are multiples of
.
2. Step 0: is nontrivial
Lemma 3 Let
. Then, each
is the start of at least one arrow of
.
Remark 3 Notice that we allow symplectic matrices with real (not necessarily integer) coefficients in this lemma. However, the fact that
is symplectic is important here and the analogous lemma for general invertible (i.e.,
) matrices is false.
Proof: For , since every
-dimensional subspace is isotropic,
and the lemma follows in this case from the fact that
is invertible. So, let’s assume that
(and, in particular,
is a proper subset of
). Since
is invertible, for each
, there exists
with
Of course, one may have a priori that , i.e.,
, and, in this case, our task is to “convert”
into some
with
.
Evidently, in order to accomplish this task it suffices to show that if and
, then there exists
with
and
. Keeping this goal in mind, we observe that
implies that we can write
with
. Also, the fact that
is equivalent to say that the
minor of
associated to
and
is invertible, and hence, by writing
, we can find
such that
and
In other words, we can make a change of basis to convert the invertible minor of into the
identity matrix.
Now, denoting by the symplectic form, we observe that
because
, i.e., the span of
is an isotropic subspace, and
. On the other hand, since
is symplectic, we get that
Since (as
), it follows that there exists
with
and
.
Then we define . We have that
. Furthermore, the minor
of
associated to
and
is obtained from the minor
of
associated to
and
by removing the line associated to
and replacing it by the line associated to
. By looking in the basis
, this means that the minor
differs from the identity minor
by the fact that the line associated to
was replaced by the line associated to
. In other words, in the basis
, one of the entries
of
was replaced by the coefficient
. Thus, we conclude that the determinant
of the the minor
is
Therefore, satisfies the desired properties and the argument is complete.
3. Step 1: is mixing
For , the set
consists of exactly one pair
, so that the possible Galois invariant graphs are:
In the first case, by definition, we have that (and
), so that
and
share a common subspace, a contradiction with our hypothesis in Theorem 1.
In the second case, by definition, we have that and
, so that
and thus
and
share a common subspace, a contradiction with our assumptions in Theorem 1.
Finally, in the third case, we have that the graph is complete, and hence
is
-twisting with respect to
.
Now, after this “warm up”, we pass to the general case . Firstly, suppose that the sole arrows in
are of the form
. Then,
, and, since
, the subspace
is non-trivial. In particular, in this case,
and
share a common non-trivial subspace, a contradiction. Of course, this arguments breaks up for
(and this is why we had a separate argument for this case).
Therefore, we may assume that has some arrow
with
. Because the Galois group
is the largest possible and
is invariant under the action of
(see Remark 1), we have that all arrows of this type belong to
. In view of Remarks 1 and 2, it suffices to construct a loop of odd length in
.
Since we dispose of all arrows with
, if
, we can easily construct a loop of length
:
On the other hand, for , we have two possibilities. If
is the non-mixing graph invariant under the Galois group:
we get that and
, so that
and
share a common invariant subspace, a contradiction.
So, we have some extra arrow in the previous picture, say:
In this case, it is not hard to see that the addition of any extra arrow allows to build up loops of lenght , so that, by Remarks 1 and 2, the argument is complete.
Therefore, in any event, we proved that is mixing.
4. Step 2: For ,
is mixing for
, and
is mixing
Given twisting
-dimensional
-invariant subspaces, we wish to prove that
is mixing for all
whenever
. Since
is invariant under the Galois group
(see Remark 1), we start by considering the orbits of the action of
on
.
Proposition 4 The orbits of the action of
on
are
We leave the proof of this proposition as an exercise to the reader. This proposition says that the orbits of the action on
are naturalized parametrized by
$latex \displaystyle \widetilde{I}=\{(\widetilde{\ell}, \ell) \textrm{ satisfying } (1)\}&fg=000000$
In particular, since is
-invariant, we can write
for some
, where
is the graph whose vertices are
and whose arrows are
Proposition 5 The graph
is not mixing if and only if
- either
and
- or
and
Proof: Let for
or
for
. Then, one can see that, since
,
is not mixing simply because it is not connected! For the proof of the converse statement, due to the usual space-time limitations, we’re going only to say a few words on (referring to the forthcoming article by M. Möller, J.-C. Yoccoz and C. M. for formal arguments). Essentially, one starts by converting pairs
into a single point
, so that
becomes a new graph
. Then, one proves that, if
, then
is connected. Using that
is connected, it is possible to prove that
is connected and from this one can construct loops of odd length, thus getting the mixing property.
Coming back to the study of ,
,
, we set
. By the previous proposition, if
is not mixing, then
for
or
for
. For sake of concreteness, we will deal “only” with the case
(leaving the particular case
when
as an exercise to the reader). In this situation, we have an arrow
of
with
. This means that
, and hence we can find
such that
and
In other words, as we also did in Step 0, we can use to “convert” the minor of
associated to
into the identity.
We claim that if , then
for all
. Indeed, by the same discussion around minors and replacement of lines, if this were not true, say
, we could find an arrow from
to
. Since
, we have that
, so that, for some
, one has
, a contradiction showing that the claim is true.
From the claim above we deduce that e.g. is a linear combination of
,
, a contradiction with the fact that
twists
-dimensional
-invariant subspaces. In other words, we proved that
is mixing for each
whenever
twists
-dimensional
-invariant subspaces.
By Proposition 2, it follows that we can construct a matrix twisting
-dimensional isotropic
-invariant subspaces for
, and we wish to show that
is mixing. In this direction, we consider the orbits of the action of the Galois group
on
. By Proposition 4, the orbits are
with ,
and
. In particular,
in this case, and the orbits are parametrized by the set
For sake of simplicity, we will denote the orbits of on
by
and we write
where .
It is possible to show (again by the arguments with “minors” we saw above) that if is
-twisting with respect to
, then
contains two consecutive integers say
.
We claim that is mixing whenever
contains two consecutive integers.
Indeed, we start by showing that is connected. Notice that it suffices to connect two vertices
and
with
(as the general case of two general vertices
and
follows by producing a series of vertices
,
,
,
with
,
). Given
and
with
, we select
with
. Then, we consider
obtained from
by replacing the elements of
by their inverses. By definition,
and
(because
). By assumption,
contains
and
, so that we have the arrows
and
in
. Thus, the connectedness of
follows.
Next, we show that is mixing. Since
is invariant under the Galois group, it contains loops of length
(see Remark 1). By Remark 2, it suffices to construct some loop of odd length in
. We fix some arrow
of
. By the construction “
” when
performed in the proof of the connectedness of
, we can connect
to
by a path of length
in
. In this way, we have a loop (based on
) in
of length
.
5. Step 3: Special case
We consider the symplectic form . Since
has dimension
and
is non-degenerate,
has dimension
.
By denoting by the eigenvalues of
, we have the following basis of
,
,
,
;
where
.
In general, given , we can use
to construct a graph
whose vertices are
and
, and whose arrows connect vertices associated to non-zero entries of
.
By definition, , so that
is an eigenvalue of
. In principle, this poses a problem to apply Proposition 12 of SPCS8 (to deduce
-twisting properties of
from
is mixing), but, as it turns out, the fact that the eigenvalue
of
is simple can be exploited to rework the proof of Proposition 12 of SPCS8 to check that
is mixing implies the existence of adequate products
of powers of
and
satisfying the
-twisting condition (i.e.,
twists
-dimensional
-invariant isotropic subspaces).
Therefore, it “remains” to show that either or
is mixing to complete the considerations of this section (and today’s post).
We write with
.
- If
contains
two consecutive integers, then one can check that the arguments of the end of the previous sections work and
is mixing.
- Otherwise, since
(see Step 0), we have
,
or
. As it turns out, the cases
are “symmetric”, as well as the cases
.
For sake of concreteness, we will consider the cases and
(leaving the treatment of their “symmetric” as an exercise). We will show that the case
is impossible while the case
implies that
is mixing.
We begin by . This implies that we have an arrow
with
. Hence, we can find
with
and
Since , the arrows
,
, and
do not belong
. Thus,
. On the other hand, because
is symplectic,
(with
). It follows that
for all
, that is,
preserves the
-subspace spanned by
and
, a contradiction with the fact that
is
-twisting with respect to
.
Now we consider the case and we wish to show that
. We claim that, in this situation, it suffices to construct arrows from the vertex
to
and vice-versa. Notice that the action of the Galois group can’t be used to revert arrows of
involving the vertex
, so that the two previous statements are “independent”. Assuming the claim holds, we can use the Galois action to see that once
contains some arrows from
and some arrows to
, it contains all such arrows. In other words, if the claim is true, we have the following situation:
Thus, we have loops of length (in
), and also loops of length
(based on
), so that
is mixing. In particular, our task is reduced to show the claim above.
The fact that there are arrows from to
follows from the same kind of arguments involving “minors” (i.e., selecting
as above, etc.) and we will not repeat it here.
Instead, we will focus on showing that there are arrows from to
. The proof is by contradiction: otherwise, one has
. Then, we invoke the following elementary lemma (whose proof is a straightforward computation):
Lemma 6 Let
be a symplectic
-plane. Given
a basis of
with
, we define
The bi-vector
is independent of the choice of
as above. Denote by
the symplectic orthogonal of
and put
Then,
is collinear to
if and only if
or
.
Since , from this lemma we obtain that
implies
or
, a contradiction with the fact that
is
-twisting with respect to
.
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