During the graduate workshop on moduli of curves (organized by Samuel Grushevsky, Robert Lazarsfeld, and Eduard Looijenga last July 2014), Alex Wright gave a minicourse on the -orbits on moduli spaces of translation surfaces (the videos of the lectures and the corresponding lecture notes are available here and here).

These lectures by Alex Wright made Eduard Looijenga ask if some “remarkable” translation surfaces could help in solving the following question.

Let be a ramified finite cover of the two-torus (say branched at only one point ). Denote by the subspace of generated by the homology classes of all simple closed loops on covering such a curve on .

**Question 1.** Is it true that one *always* has in this setting?

By following Alex Wright’s advice, Eduard Looijenga wrote me asking if I knew the answer to this question. I replied to him that my old friend *Eierlegende Wollmilchsau* provided a *negative* answer to his question, and I directed him to the papers of Forni (from 2006), Herrlich-Schmithüsen (from 2008) and our joint paper with Yoccoz (from 2010) for detailed explanations.

In a subsequent email, Eduard told me that my answer was a good indication that notable translation surfaces could be interesting for his purposes: indeed, the Eierlegende Wollmilchsau is *precisely* the example described in the appendix of a paper by Andrew Putman and Ben Wieland from 2013 where Question 1 was originally solved.

After more exchanges of emails, I learned from Eduard that his question was motivated by the attempts of Putman-Wieland (in the paper quoted above) to attack the following conjecture of Nikolai Ivanov (circa 1991):

**Conjecture (Ivanov).** Let and . Consider a finite-index subgroup of the mapping-class group of isotopy classes of homeomorphisms of a genus surface fixing pointwise a set of marked points. Then, there is no surjective homomorphism from to .

Remark 1This conjecture came from the belief that mapping-class groups should behave in many aspects like lattices in higher-rank Lie groups and it is known that such lattices do not surject on because they satisfy Kazhdan property (T). Nevertheless, Jorgen Andersen recently proved that the mapping-class groups do not have Kazhdan property (T) when .

Remark 2It was proved by John McCarthy and Feraydoun Taherkhani that the analog for of Ivanov’s conjecture fails.

In fact, Putman-Wieland proposed the following strategy to study Ivanov’s conjecture. First, they introduced the following conjecture:

**Conjecture (Putman-Wieland).** Fix and . Given a finite-index characteristic subgroup of the fundamental group of a surface of genus with punctures , denote by the associated finite cover, and let be the compact surface obtained from by filling its punctures.

Then, the natural action on of the group of lifts to of isotopy classes of diffeomorphisms of fixing pointwise has *no* finite orbits.

Remark 3This conjecture is closely related (for reasons that we will not explain in this post) to a natural generalization of Question 1 to general ramified finite covers .

Remark 4The analog of Putman-Wieland conjecture in genus is false: the same counterexample to Question 1 (namely, the Eierlegende Wollmilchsau) serves to answer negatively this genus 1 version of Putman-Wieland conjecture.

Remark 5In the context of Putman-Wieland conjecture, one has a representation (induced by the lifts of elements of to ). This representation is called ahigher Prym representationby Putman-Wieland. In this language, Putman-Wieland conjecture asserts that higher Prym representations have no non-trivial finite orbits when and .

Secondly, they proved that:

Theorem 1 (Putman-Wieland)Fix and .

- (a) If Putman-Wieland conjecture holds for every finite-index characteristic subgroup of , then Ivanov conjecture is true for any finite-index sugroup of .
- (b) If Ivanov conjecture holds for every finite-index subgroup of , then Putman-Wieland conjecture is true for any finite-index characteristic subgroup of .

Moreover, if Ivanov conjecture is true for all finite-index subgroups of for all , then it is also true for all finite-index subgroups of with , .

In other words, Putman-Wieland proposed to approach an algebraic problem (Ivanov conjecture) via the study of a geometric problem (Putman-Wieland conjecture) because these two problems are “essentially” equivalent.

In particular, this gives the following concrete route to establish Ivanov conjecture:

- (I) if we want to show that Ivanov conjecture is
*true*for all and , then it suffices to prove Putman-Wieland conjecture for (and all ); indeed, this is so because item (a) of Putman-Wieland theorem would imply that Ivanov conjecture is true for (and all ) in this setting, and, hence, the last paragraph of Putman-Wieland theorem would allow to conclude the validity of Ivanov conjecture in general. - (II) if we want to show that Ivanov conjecture is
*false*for some and , then it suffices to construct a counterexample to Putman-Wieland conjecture for and .

Once we got at this point in our email conversations, Eduard told me that Question 1 was just a warmup towards his main question:

**Question 2. **Are there remarkable translation surfaces giving counterexamples to Putman-Wieland conjecture?

By inspecting my list of “preferred” translation surfaces, I noticed that I knew such an example: in fact, there is exactly one member in a family of translation surfaces that I’m studying with Artur Avila and Jean-Christophe Yoccoz (for other purposes) which is a counterexample to Putman-Wieland conjecture in genus (and ).

In other words, one of the translation surfaces in a forthcoming paper joint with Artur and Jean-Christophe answers Question 2.

Remark 6This shows that Putman-Wieland’s strategy (I) above doesnotwork (because their conjecture is false in genus ). Of course, this doesnotmean that Ivanov conjecture is false: in fact, by Putman-Wieland strategy (II), one needs a counterexample to Putman-Wieland conjecture in genus (rather than in genus ). Here, it is worth to point out that Artur, Jean-Christophe and I havenogood candidates of counterexamples to Putman-Wieland conjecture in genus and/or Ivanov conjecture.

Below the fold, we focus on the case and of Putman-Wieland conjecture.

**Update (September 7, 2015)**: Last June 2015, Eduard gave a talk on algebro-geometrical aspects of mapping-class groups, and he wrote the following summary here where the connections between (a natural generalization of) Question 1 and the conjectures of Ivanov and Putman-Wieland are discussed.

**1. A genus cover of a genus surface **

Let be the genus surface associated to the Riemann surface . The genus surface corresponding to the Riemann surface has the structure of a triple cover given by . Observe that is unramified off the six (Weierstrass) points of located at the five roots of unit , , and the point at infinity .

Recall that the ramified finite cover corresponds to a finite-index subgroup of where is a genus surface with punctures and is a point of located at .

It is possible to check that is *not* a characteristic subgroup of . Nevertheless, we can easily construct a subgroup of such that is a finite-index characteristic subgroup of . Indeed,

is a subgroup of which is characteristic in . Furthermore, has finite-index in because has the same (finite) index of for all and has only finitely many subgroups of a given index (since is finitely generated).

Denote by the compact surface associated to the finite ramified cover of induced by , and let be the mapping-class group of . Since is characteristic, we can lift any element of to a mapping-class of , so that we have a higher Prym representation .

Theorem 2There exists a eight-dimensional subspace such that the orbit of any under the higher Prym action of on is finite. In particular, the Putman-Wieland conjecture is false in the case and .

We will deduce this theorem as a consequence of the following result:

Theorem 3There exist a eight-dimensional subspace and a finite-index subgroup of with the following properties. Any element of lifts to a mapping-class of and the orbit of any under the corresponding representation is finite.

The proof of Theorem 3 relies on the unusual features of the Hodge filtration of .

**2. Proof of Theorem 2 assuming Theorem 3**

Since is a subgroup of , we have that the cover associated to *factors* through the cover (associated to ), that is, we have a cover such that the composition is the cover corresponding to .

Given a eight-dimensional subspace and a finite-index subgroup of as in the statement of Theorem 3, let be the subspace of cohomology cycles projecting to which are also invariant under the whole group of deck transformations of .

By Theorem 3, the natural action of on factors a finite group of matrices. By construction, all orbits of the action of on are finite. Since is a finite-index subgroup of , it follows that all orbits of the action of on are also finite.

Finally, since the homology group is in a natural duality with the cohomology group , our reduction of Theorem 2 to Theorem 3 is complete.

**3. Proof of Theorem 3**

Let be the automorphism where generating the group of deck transformations of the cover where is the Riemann sphere and .

Note that factors : indeed, where is the natural projection from to the quotient of by its hyperelliptic involution , .

Since has genus , the elements of commute with the hyperelliptic involution : this is a very special property of the genus setting whose proof follows from the results in this paper here (see also page 77 of Farb-Margalit book [while paying attention that our convention differs from them because our mapping-class groups are required to fix pointwise each puncture]).

It follows that the elements with form a finite-index subgroup of such that the lift to of any such commutes with the automorphism (in fact, this is so because projects under to the hyperelliptic involution of ). [**Update (August 24, 2015)**: See also the exchange between Ben Wieland and myself in the comments below.]

By construction, acts on and our task is to show the existence of a eight-dimensional subspace of such that the -orbit of any is finite.

For this sake, we start by analyzing the action of on . Here, the crucial point is that was built in such a way that all of its elements commute with . In particular, the action of preserves each summand of the decomposition

into the eigenspaces associated to the eigenvalues of . (Note that the eigenspace is trivial because ).

Recall that the action of on preserves the intersection form . Since each eigenspace has a Hodge decomposition

and the intersection form is positive definite on the space of holomorphic -forms and negative definite on the space of anti-holomorphic -forms, we have that acts on via a indefinite unitary group of a pseudo-Hermitian form of signature where

In our context, is associated to the curve , so that

is an explicit basis of the space of holomorphic -forms on . From this, we infer that and (and, in general, for each ).

In other words, , , and acts on the eight-dimensional complex subspace

via (a subgroup of) the *compact* group .

Next, we study the action of on . We begin by noticing that the eight-dimensional complex subspace is defined over . In fact, this is a consequence of the following elementary observation (from Galois theory): is the sum of *all* eigenspaces associated to *all* primitive th roots of unity .

Since is defined over , it intersects into a lattice of rank . In particular, acts on via (a subgroup of) the symplectic group because respects the symplectic intersection form on .

In summary, we proved that:

- on one hand, acts on via the compact group ;
- on the other hand, acts on via .

In other terms, acts on through a *compact* subgroup of the *discrete* group , i.e., acts on the eight-dimensional subspace through a *finite* subgroup of symplectic matrices.

It follows that the -orbit of any is finite, so that the proof of Theorem 3 is complete.

I do not understand Theorem 3. I do not see any reason to believe that Γ commutes with T. That T acts at all is only because you chose a special

markedcurve, namely one where the ramification points are the Weierstrass points. If you replace Mod_2,6 with Mod_2 or Mod_0,6, then the monodromy does commute with T and thus does fix an 8-dimensional subspace, but I do not believe that is true for Γ in Mod_2,6. Look at it on the level of moduli spaces: These ramified covers form a bundle of curves over BΓ, which is approximately M_2,6. The hyperelliptic involution acts on the total space of this bundle, but it does not act by bundle maps. It does not act on each individual fiber because it does not act trivially on the base space. The hyperelliptic involution acts on M_2,6 by rearranging the marked points, and similarly on BΓ. Since it transforms the fiber over one point into the fiber over another point, it transforms the homology of the fiber over one point into the homology of the fiber over another point. Since it does not act by automorphisms, it does not have a bundle of eigenspaces over BΓ. Over the fixed locus of the hyperelliptic involution it does act by automorphisms and thus it does have a bundle of eigenspaces on this locus, but there is no reason for monodromy along an arbitrary element of Γ to preserve this bundle, since the element must be represented by a curve which leaves the locus.By:

Ben Wielandon August 20, 2015at 7:00 pm

Dear Ben Wieland,

First of all, thank you very much for your comments!

Secondly, concerning Theorem 3, let me try to phrase the construction of in a slightly different (but equivalent) way.

As you pointed out, the curve is “special” because my punctures are placed exactly at Weierstrass points. In particular, the elements of commute with the hyperelliptic involution (whose fixed points are the 6 Weierstrass points).

Thus, descends to an element on the sphere with 7 punctures (obtained from the quotient of by the hyperelliptic involution after adding an extra puncture at the “origin” ). In other terms, we can think of elements of as lifts of elements of in our special situation.

Now, we observe that the automorphism is the generator of the deck group of the finite ramified cover of the sphere.

Therefore, we can take a finite-index subgroup of such that the elements of can be lifted to into mapping-classes that commute with .

Finally, we set to be the finite-index subgroup of “corresponding” to the finite-index subgroup of . In this way, if we start with an element , we “descend” it to and then we lift and to a mapping-class on commuting with .

In summary, we begin with a mapping-class in , then we try to lift it to (while commuting with ) by looking at the mapping-class on associated to (i.e., is the quotient of by the hyperelliptic involution), and studying how to lift to (while commuting with ).

Best,

Matheus

By:

matheuscmsson August 21, 2015at 11:16 am

Where are you using something special about this representation? If you use nothing special about this representation, it seems that you make a general claim that when one considers a special marked curve as basepoint that the hyperelliptic involution commutes with Mod_2,6. In particular, what about the universal case, acting on the curve S_2,6 itself? Do the two groups commute? Do you claim that the hyperelliptic involution is a central element of Mod_2,6?

I do not know what you mean by “descend” and “lift.” Do you claim that there is a natural homomorphism from Mod_2,1 to Mod_0,7 covering the map from Mod_2 to Mod_0,6? But the kernels are completely different: a genus 2 surface group and a free group on 5 generators. I do not think that moving to Mod_2,7 helps. The seventh point must avoid the first six, but it need not avoid their images downstairs. If they stay as Weierstrass points, there is only one preimage, so it avoids the image, but Mod_2,7 allows the points to move around.

But if there were a map Mod_2,7 to Mod_0,7, I do not see what good it does. Indeed, there is a map Mod_2,6 to Mod_0,6. The map has large kernel, almost the same as the kernel from Mod_2,6 to Mod_2. Thinking of these elements as lifts of the identity in the braid group does not seem to me to be of any use. I think you are making a leap from the surjection Mod_2,6 to Mod_2 virtually splitting to the kernel acting trivially, but it does not follow.

By:

Ben Wielandon August 23, 2015at 2:12 am

The convention adopted for mapping-class groups here is the same from your paper with Putman: Mod_n,p is the group of orientation-preserving homeomorphisms fixing pointwise the punctures modulo isotopies fixing pointwise the punctures.

Note that this convention is slightly different from Farb-Margalit’s book and from some papers in the Algebraic Geometry literature (for example) where the punctures can move around. In fact, Farb and Margalit take a special care of referring to “our” Mod_n,p as pure mapping class group (and denoting it by PMod(S_g,n)) in order to avoid confusion.

~~If you take the special curve S_{2,6} with the 6 punctures at Weierstrass points, then the argument at page 80 of this version of Farb-Margalit’s book here http://www.maths.ed.ac.uk/~aar/papers/farbmarg.pdf implies that the elements of (our version of) Mod_2,6 commute with the hyperelliptic involution since a generating set of Dehn twists do so. (Observe that half-twists are out of this picture because the punctures are not allowed to move around)~~By “descending” f from M to the sphere, I mean that the quotient of f by the hyperelliptic involution (=generator of the deck group of the cover from M to the sphere) is a well-defined element g on the sphere.

Similarly, by “lifting” I mean what is written at page 3 of your joint paper with Putman http://www.math.rice.edu/~andyp/papers/HigherPrym.pdf . Moreover, the paragraphs in this page also explain the reason for marking an extra point (say at the “origin” 0), i.e., for discussing Mod_2,7: if we don’t add an extra point, the “lifting” would be defined only modulo a deck group.

By:

matheuscmsson August 23, 2015at 11:26 am

After re-reading your comments and my last reply to it (especially the part indicated with “underlines”), I thought of the following alternative argument.

The proof of Theorem 3 works if the lift h of an element f of Gamma commutes with the automorphism T at the level of the cohomology H^1(N,C).

Now, if we think of f in Gamma as an element of Mod_2, then we see that f is homotopic to F commuting with the involution iota. So, the lift h of f (which is well-defined as f belongs to Gamma) is homotopic to H commuting with T (because T projects to iota by the covering map [and H is an homeomorphism]). Therefore, the actions of h and T on the cohomology H^1(N,C) commute.

Do you agree with this?

By:

matheuscmsson August 23, 2015at 3:53 pm

The deleted paragraph leads directly to the conclusion that the hyperelliptic involution is in the center of Mod_2,6. Any argument that the hyperelliptic involution commutes with Mod_2,6 on a particular representation must explain how it used properties of the particular representation. For if it does not, it applies to all representations and thus concludes that the hyperelliptic involution is in the center. Or, to take a simpler, more analogous example, it should explain why it does not even apply to the standard representation: the homology of the 6-punctured 2-torus. The hyperelliptic involution divides this into invariant homology coming from the 6-puncutured sphere and the -1-eigenspace that survives on the closed surface, but Mod_2,6 does not preserve this direct sum decomposition: it shears the second into the first.

By:

Ben Wielandon August 24, 2015at 4:02 am

Thank you once again for the feedback! If you don’t mind, I will edit my last comment, and I will add a reference in the main post for our comments when discussing the construction of Gamma.

By:

matheuscmsson August 24, 2015at 8:58 am

Dear Ben Wieland,

In relation to your last comment, let me add a remark that you certainly aware of (indeed, I was told by Looijenga that it is known among experts).

The curve N={y^6=x^5-1} of genus 10 also covers the curve L={w^3=x^5-1} of genus 4. In particular, we can repeat the arguments discussed so far to build a higher Prym representation of a *symmetric* mapping-class group in genus 4 (i.e., mapping-classes in genus 4 commuting with a hyperelliptic involution) with non-trivial finite orbits on H^1(N,Q).

However, this observation is not useful to study higher Prym representations *full* mapping-class groups in genus 4 (because typical mapping-classes in higher genus have nothing to do with hyperelliptic involutions), and, in fact, this representation doesn’t respect the eigenspaces of T if I’m not mistaken.

In summary, as it is known to experts, Ivanov’s conjecture and your conjecture with Putman in genus >2 becomes easier to answer in the negative if we replace “mapping-class groups” by “symmetric mapping-class groups”.

Best,

Matheus

By:

matheuscmsson August 24, 2015at 11:35 am

My favorite version of the conjecture is: (S) That the Jacobian of a generic symmetric curve is no more symmetric than the curve itself. By a generic symmetric curve, I mean generic among curves with the given group action; equivalently, that the quotient is generic.

When the quotient has small genus, the symmetry of the quotient introduces an ambiguity in (S) that I had not noticed. I had thought that (S) implied the conjecture that we put in the paper, but your example attempts to squeeze through the ambiguity and makes me concerned about our conjecture in genus 2.

The precise statement of (S) is not about the group of automorphisms of the Jacobian, but about its Mumford-Tate group. For example, an elliptic curve with complex multiplication need not have any more automorphisms than the generic elliptic curve, but it does have extra endomorphisms, which count for this purpose. But I think that a proof for the group of automorphisms could be promoted to a proof of the MT statement.

The strong form of (S) is that if a finite group G acts on a curve with generic quotient, then the MT group of the curve should be the centralizer of G in the symplectic group. The weak form of (S) only considers the case where the group is the full symmetry group. The two forms only differ when the generic quotient has symmetries.

By:

Ben Wielandon August 20, 2015at 7:02 pm