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
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:
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:
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 .
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
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
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 .