In this way, we can get a simplicity criterion for the Kontsevich-Zorich cocycle over -orbits of origamis based on a “pinching” property and a “twisting” property (see this post). However, since the twisting property was stated in a “strong form” (in SPCS6), before extracting consequences from SPCS6 and SPCS7, we will spend the first half of today’s post with the study of several versions of the twisting properties.
1. Twisting properties
For the discussion of twisting properties, it is convenient to revisit a little bit the features of Noetherian topological spaces.
1.1. Noetherian spaces
Then, for each , we can define an hyperplane section as
Using these hyperplane sections, we have a (“pseudo-Zariski”) topology on by considering the coarsest topology such that the hyperplane sections are closed. Equivalently, we consider the topology on whose closed sets are the (arbitrary) intersections of finite unions of hyperplane sections.
Notice that this topology is coarser than the Zariski topology: for instance, hyperplane sections are defined by degree one (linear) equations while Zariski topology involves taking equations of arbitrary degree (in other words, this topology is a sort of “Zariski topology for dummies”🙂 … by the way, I would be happy to know if this topology has a “standard” name in the literature…). In particular, this topology is not Hausdorff as the same is true for the Zariski topology. As we’re going to see in a moment (see Example 1 below), this topology has the following property described in the definition below:
Definition 1 A topological space is Noetherianif one of the following equivalent conditions is satisfied:
- (i) any decreasing sequence of closed sets is stationary (in the sense that there exists such that for all ).
- (ii) any increasing sequence of open sets is stationary.
- (iii) every intersection of a family of closed sets is the intersection of a finite subfamily .
- (iv) every union of a family of open sets is the union of a finite subfamily .
Observe that any subspace of a Noetherian space is also Noetherian.
Example 1 It is a classical fact that the Zariski topology is Noetherian. In particular, since the topology on associated to hyperplane sections is coarser than the Zariski topology, it is also Noetherian.
Definition 2 A Noetherian (topological) space is irreducible if is not the union of two non-trivial closed sets.
Proof: We will show only the existence part of this statement (leaving the uniqueness as an exercise). We argue by contradiction: otherwise, since is Noetherian, we could find closed set that is not the finite union of irreducible sets and minimal (with respect to the inclusion) for this property. Of course, by construction, is not irreducible. Hence, we can write with . By minimality, and where and are irreducible closed sets. So,
is also a finite union of irreducible closed sets, a contradiction.
Remark 1 This proposition is not true if we replace “Noetherian space” by “Zariski topology”, i.e., the product of Zariski topologies is not the Zariski topology on the product. In particular, this is the reason why one does not insist on the product of Zariski topologies in Algebraic Geometry.
Proof: We illustrate the arguments in the case of the product of two Noetherian spaces , as the general case is similar.
It is not hard to see that our task is to prove that every union of products is the union of a finite subfamily. Suppose that this is not true. Then, for each , such that , for all and for all .
Define and . By construction, . We claim that:
- (a) either there exists an infinite sequence with for all
- (b) or there exists an infinite sequence with for all .
Here, we note that these possibilities are notmutually exclusive, i.e., it may happen that both of them occur. In any event, assuming momentarily that this claim is true, say (a) holds, it follows that is an infinite sequence of open sets such that but , a contradiction with the fact that is Noetherian. In other words, we reduced the proof of the proposition to showing the our claim above.
We prove the claim by recurrence. Put and assume that the sets are already constructed with the properties that for all , and is infinite.
Consider the sets for . We have two possible scenarios:
- there exists such that is infinite: in this case, we can define and advance one step in our argument by recurrence;
- for all the set is finite: in this case, since , one has that, for all , the set contains all but finitely many elements of ; then, we can pick and we can define and it is not hard to see that satisfies item (b) of our claim.
In resume, the recursive argument above says that either we can advance indefinitely in the construction of the sequence fitting item (a) of our claim, or we get “blocked” at some stage but then we can produce in “one shot” the sequence meeting the requirements of item (b) of our claim. Of course, this proves the claim and thus the proof of the proposition is complete.
where are closed irreducible sets.
Proof: By the previous proposition, one can check that closed sets of are finite unions of products of closed sets of ‘s. Thus, the irreducible closed sets of are precisely the products of irreducible closed sets .
After this little digression on Noetherian spaces, we can pass to twisting properties.
1.2. Twisting monoids
Let be a monoid acting on a Noetherian space by homeomorphisms. Here, of course, our main example is:
Example 2 Given a countable family of matrices , , with as in (1), we can consider the natural action of the monoid generated by acting on the Grassmanian equipped with the Noetherian topology associated to hyperplane sections.
Proof: Otherwise, would be a strictly decreasing infinite sequence of closed sets of the Noetherian space .
Proof: This follows from the uniqueness part of Proposition 3.
- (i) there exists no non-trivial closed invariant set by
- (ii) for every and open set, there exists such that
- (iii) for every , and open set, there exists such that for all
- (iii)’ for every , and open sets, there exists such that for all
Proof: The equivalence between and follows from the fact that the monoid also acts (diagonally) on the Noetherian space (see Proposition 4).
Also, it is clear that , so that it remains only to prove that implies .
Suppose that holds but is not true. Then, we can find , a point and with a non-trivial open set of such that the (diagonal) action of on satisfies:
for all .
Observe that, by Proposition 4, is a (non-trivial) open set of . So, is a closed set such that
In particular, is a closed set such that for all . By Proposition 6, it follows that for all .
Now, we write the decomposition of into irreducible closed sets . By Proposition 5,
where are irreducible closed subsets of . Since for all , we have that, by Proposition 7, every permutes the irreducible closed sets .
Here, we used that to ensure . Because acts by homeomorphisms and permutes the irreducible closed subsets , any is sent by into some , and hence
for every . On the other hand, since is irreducible (by hypothesis), , a contradiction with the statement of item .
In view of Example 2 and the discussion in SPCS6 (related to Avila-Viana simplicity criterion), it is natural to call a (“strong form” of) twisting condition.
Remark 2 The twisting condition (i.e., or ) is satisfied for the monoid if and only if it is satisfied for the group generated by (e.g., this follows immediately from the statement of item ).
1.3. Twisting with respect to pinching matrices
Coming back to the context of Example 2 (and using the notations from SPCS6), assume that we have a word such that is simple (i.e., pinching).
for every -invariant and (that is, and ).
For later use, we will abbreviate the second statement in this proposition by saying that is twisting with respect to .
Remark 3 It can be checked that isotropic is -invariant (i.e., ) if and only if is generated by the sum of eigenvectors of with eigenvalues such that and for every . In particular, there are only finitely many with , and hence the proposition above says that, in the presence of a pinching matrix , the (strong form of) twisting condition (an “infinitary” statement about all and ) can be reduced to the property of twisting with respect to (a “finitary” statement as it concerns only the finitely many [isotropic and coisotropic] -invariant subspaces).
Proof: Of course, “twisting with respect to ” is a necessary condition. Conversely, given and , we are searching for a matrix such that for all . We affirm that it suffices to take for sufficiently large: indeed, as , we have
- converges (in the usual topology) to some with ;
- converges (also in the usual topology) to with .
It follows that, by assumption, for (as and are -invariant). Since these transversality conditions are open, one gets that for sufficiently large (depending on and ), that is, for sufficiently large.
At this stage, we are ready to state our main result towards the verification (based on Galois theory) of the pinching and twisting conditions for the Kontsevich-Zorich cocycle over the -orbits of square-tiled surfaces.
2. Sufficient conditions for pinching and twisting in the case of origamis
- (i) the eigenvalues of are real
- (ii) the Galois group of the characteristic polynomial of is the largest possible, that is,
- (iii) and don’t share a common non-trivial invariant subspace
Then, the pinching and twisting conditions are satisfied.
Remark 4 Concerning the Galois group (see item ), since is symplectic, is a reciprocal polynomial (i.e., ), so that its roots have the form . Thus, by saying that is the largest possible, we want to be able make all permutations of roots of respecting the symplectic constraint/symmetry, that is, we want to permute all pairs , (this gives the factor of ), and for each given pair we want to permute and without touching the other pairs (this gives the factor of ).
Remark 5 The proof of this result is somewhat long (as it involves considering several cases), and hence the general plan for the next two lectures (March 7 and 14) by J.-C. Yoccoz is the following: for the next lecture (March 7), we will see some ideas of the proof of the theorem above and for the final lecture (March 14) we will see some applications of this result to the case of origamis in the stratum (of genus square-tiled surfaces with a single zero of order ) where we will take and we will consider matrices on .
The rest of this post concerns some preparatory facts towards the proof of this theorem.
We begin by noticing that the matrix verifies the pinching condition: indeed, since the Galois group of is the largest possible, we have that is irreducible, and thus its roots are simple. By the assumption , all roots , , of are real, so that the pinching condition is violated by precisely when there are such that . However, this is impossible because is the largest possible: for instance, since , we have an element of fixing and exchanging and ; applying this element to the relation , we would get that and , so that a contradiction with the fact that is irreducible.
Remark 6 Concerning the applications of this theorem to the case of origamis, we observe that item is satisfied whenever the splitting fields and of the characteristic polynomials of and are disjointas extensions of , i.e., . Indeed, if is invariant by and , one has that
- is generated by eigenvectors of (as is pinching, i.e., has simple spectrum), so that is defined over , and
- is invariant by , so that is also defined over .
Since and are disjoint, it follows that is defined over . But this is impossible as doesn’t have rational invariant subspaces (by and ).
Once we know that the matrix satisfies the pinching condition, the proof of Theorem 10 is reduced to checking the twisting condition, and, by Proposition 9, it suffices to check the condition of twisting with respect to the pinching matrix . Keeping this goal in mind, we introduced the following notations.
We denote by the set of roots of the polynomial (so that ), for each , we put , and we define (so that ).
Given , let , resp. be the set whose elements are subsets of , resp. with elements, and let be the set whose elements are subsets of with elements such that is injective. In other words, consist of those such that if , then .
Next, we make a choice of basis of as follows. For each , we select an eigenvector of associated to , i.e., . In particular, is defined over ). Then, we assume that the choices of ‘s are coherent with the action of the Galois group , i.e., (and thus ) for each . In this way, for each , we can associated a multivector (using the natural order of the elements of ).
By definition, where .
From our assumptions and on , we have that:
- , , is a basis of
- the subspace generated by is isotropic if and only if
Also, by an elementary (linear algebra) computation, it is not hard to check that a matrix is twisting with respect to if and only if
for all , where are the coefficients of the matrix in the basis .
In order to organize our discussions, we observe that the condition (2) can be used to define an oriented graph as follows. Its vertices are , and we have an arrow from to if and only if . In this language, (2) corresponds to the fact that is a complete graph.
Of course, the verification of the completeness of is not simple in general, and hence it could be interesting to look for more soft properties of ensuring completeness of for some matrix constructed as a product of powers of and . Here, we take our inspiration from Dynamical Systems and we introduce the following classical notion:
Here, we note that it is important in this definition that we can connect two arbitrary vertices by a path of length exactly (and not of length ). For instance, the figure below shows a connected graph that is not mixing because all paths connecting to have odd length while all paths connecting to have even length.
As the reader can guess by now, mixing is a soft property ensuring completeness of a “related” graph. This is the content of the following proposition:
satisfies (2) for all sufficiently large .
We close this post with the proof of this proposition.
Proof: To alleviate the notation, let’s put . By definition,
where and .
Our goal is to prove that . Of course, even though was expressed as a sum of exponentials with non-vanishing coefficients, there is a risk of getting non-trivial cancelations so that the resulting expression vanishes. The idea is to show that can be chosen suitably to avoid such cancelations, and the heart of this argument is the observation that, for , the linear forms and are distinct. Indeed, given and , , we claim that the following coefficients of and differ:
Otherwise, we would have a relation
But, since , we have that if then . In particular, by taking an element , and by considering an element of the Galois group with and otherwise, one would get on one hand that
but, on the other hand,
so that or , a contradiction in any event (as is real and by the pinching conditions on ).
Now, we define . Since and are distinct linear forms for , it follows that is a hyperplane. Because there are only finitely many paths of length on , the collection of corresponds to a finite family of hyperplanes .
Finally, we complete the proof by noticing that if , then
for sufficiently large because the coefficients are mutually distinct.