Some comments on the conjectures of Ivanov and Putman-Wieland

Posted by: matheuscmss | April 24, 2015

Some comments on the conjectures of Ivanov and Putman-Wieland

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 ${SL(2,\mathbb{R})}$ -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 ${S}$ be a ramified finite cover of the two-torus ${T}$ (say branched at only one point ${0\in T}$ ). Denote by ${H}$ the subspace of ${H_1(S,\mathbb{Q})}$ generated by the homology classes of all simple closed loops on ${S}$ covering such a curve on ${T}$ .

Question 1. Is it true that one always has ${H=H_1(S,\mathbb{Q})}$ 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 ${g\geq 3}$ and ${n\geq0}$ . Consider ${\Gamma}$ a finite-index subgroup of the mapping-class group ${\textrm{Mod}_{g,n}}$ of isotopy classes of homeomorphisms of a genus ${g}$ surface ${S}$ fixing pointwise a set ${\Sigma=\{x_1,\dots, x_n\}}$ of marked points. Then, there is no surjective homomorphism from ${\Gamma}$ to ${\mathbb{Z}}$ .

Remark 1 This 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 ${\mathbb{Z}}$ because they satisfy Kazhdan property (T). Nevertheless, Jorgen Andersen recently proved that the mapping-class groups ${\textrm{Mod}_{g,n}}$ do not have Kazhdan property (T) when ${g\geq 2}$ .

Remark 2 It was proved by John McCarthy and Feraydoun Taherkhani that the analog for ${g=2}$ 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 ${g\geq 2}$ and ${n\geq 0}$ . Given a finite-index characteristic subgroup ${K}$ of the fundamental group ${\pi_1(S_{g,n}, x_{n+1})}$ of a surface ${S=S_{g,n}}$ of genus ${g}$ with ${n}$ punctures ${x_1, \dots, x_n}$ , denote by ${S_K\rightarrow S}$ the associated finite cover, and let ${\overline{S_K}}$ be the compact surface obtained from ${S_K}$ by filling its punctures.

Then, the natural action on ${H_1(\overline{S_K},\mathbb{Q})-\{0\}}$ of the group of lifts to ${\overline{S_K}}$ of isotopy classes of diffeomorphisms of ${S_{g,n}}$ fixing ${x_1,\dots, x_{n+1}}$ pointwise has no finite orbits.

Remark 3 This 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 ${S\rightarrow T}$ .

Remark 4 The analog of Putman-Wieland conjecture in genus ${g=1}$ 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 5 In the context of Putman-Wieland conjecture, one has a representation ${\textrm{Mod}_{g,n+1}\rightarrow \textrm{Aut}(H_1(\overline{S_K}, \mathbb{Q}))}$ (induced by the lifts of elements of ${\textrm{Mod}_{g,n+1}}$ to ${\overline{S_K}}$ ). This representation is called a higher Prym representation by Putman-Wieland. In this language, Putman-Wieland conjecture asserts that higher Prym representations have no non-trivial finite orbits when ${g\geq 2}$ and ${n\geq 0}$ .

Secondly, they proved that:

Theorem 1 (Putman-Wieland) Fix ${g\geq 3}$ and ${n\geq 0}$ .

(a) If Putman-Wieland conjecture holds for every finite-index characteristic subgroup ${K}$ of ${\pi_1(S_{g-1, n+1}, x_{n+2})}$ , then Ivanov conjecture is true for any finite-index sugroup ${\Gamma}$ of ${\textrm{Mod}_{g,n}}$ .

(b) If Ivanov conjecture holds for every finite-index subgroup of ${\textrm{Mod}_{g,n+2}}$ , then Putman-Wieland conjecture is true for any finite-index characteristic subgroup ${K}$ of ${\pi_1(S_{g,n+1}, x_{n+2})}$ .

Moreover, if Ivanov conjecture is true for all finite-index subgroups of ${\textrm{Mod}_{g,n}}$ for all ${n\geq0}$ , then it is also true for all finite-index subgroups of ${\textrm{Mod}_{G,m}}$ with ${G\geq g}$ , ${m\geq 0}$ .

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 ${g\geq 3}$ and ${n\geq 0}$ , then it suffices to prove Putman-Wieland conjecture for ${g=2}$ (and all ${n\geq 0}$ ); indeed, this is so because item (a) of Putman-Wieland theorem would imply that Ivanov conjecture is true for ${g=3}$ (and all ${n\geq0}$ ) 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 ${g\geq 3}$ and ${m\geq 2}$ , then it suffices to construct a counterexample to Putman-Wieland conjecture for ${g=3}$ and ${n=m-1}$ .

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 ${g=2}$ (and ${n=6}$ ).

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

Remark 6 This shows that Putman-Wieland’s strategy (I) above does not work (because their conjecture is false in genus ${2}$ ). Of course, this does not mean that Ivanov conjecture is false: in fact, by Putman-Wieland strategy (II), one needs a counterexample to Putman-Wieland conjecture in genus ${g\geq 3}$ (rather than in genus ${g=2}$ ). Here, it is worth to point out that Artur, Jean-Christophe and I have no good candidates of counterexamples to Putman-Wieland conjecture in genus ${g\geq 3}$ and/or Ivanov conjecture.

Below the fold, we focus on the case ${g=2}$ and ${n=6}$ 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 ${10}$ cover of a genus ${2}$ surface

Let ${M}$ be the genus ${2}$ surface associated to the Riemann surface ${z^2=x^5-1}$ . The genus ${10}$ surface ${N}$ corresponding to the Riemann surface ${y^6=x^5-1}$ has the structure of a triple cover ${p:N\rightarrow M}$ given by ${p(x,y):=(x,y^3)}$ . Observe that ${p}$ is unramified off the six (Weierstrass) points of ${M}$ located at the five roots of unit ${x=\exp(2\pi ia/5)}$ , ${a=1,\dots, 5}$ , and the point at infinity ${x=\infty}$ .

Recall that the ramified finite cover ${p:N\rightarrow M}$ corresponds to a finite-index subgroup ${H}$ of ${\pi_1(S_{2,6}, 0)}$ where ${S_{2,6} = M-\{6 \textrm{ Weierstrass points}\}}$ is a genus ${g=2}$ surface with ${n=6}$ punctures and ${0}$ is a point of ${M}$ located at ${x=0}$ .

It is possible to check that ${H}$ is not a characteristic subgroup of ${\pi_1(S_{2,6}, 0):=\Pi_{2,6}}$ . Nevertheless, we can easily construct a subgroup ${K}$ of ${H}$ such that ${K}$ is a finite-index characteristic subgroup of ${\Pi_{2,6}}$ . Indeed,

$\displaystyle K:=\bigcap\limits_{\varphi\in\textrm{Aut}(\Pi_{2,6})} \varphi(H)$

is a subgroup of ${H}$ which is characteristic in ${\Pi_{2,6}}$ . Furthermore, ${K}$ has finite-index in ${\Pi_{2,6}}$ because ${\varphi(H)}$ has the same (finite) index of ${H}$ for all ${\varphi\in\textrm{Aut}(\Pi_{2,6})}$ and ${\Pi_{2,6}}$ has only finitely many subgroups of a given index (since ${\Pi_{2,6}}$ is finitely generated).

Denote by ${\overline{S_K}}$ the compact surface associated to the finite ramified cover of ${S_{2,6}}$ induced by ${K}$ , and let ${\textrm{Mod}_{2,7}}$ be the mapping-class group of ${S_{2,7}=S_{2,6}-\{0\}}$ . Since ${K}$ is characteristic, we can lift any element of ${\textrm{Mod}_{2,7}}$ to a mapping-class of ${\overline{S_K}}$ , so that we have a higher Prym representation ${\textrm{Mod}_{2,7}\rightarrow \textrm{Aut}(H_1(\overline{S_K},\mathbb{Q}))}$ .

Theorem 2 There exists a eight-dimensional subspace ${U\subset H_1(\overline{S_K}, \mathbb{Q})}$ such that the orbit of any ${u\in U}$ under the higher Prym action of ${\textrm{Mod}_{2,7}}$ on ${H_1(\overline{S_K}, \mathbb{Q})}$ is finite. In particular, the Putman-Wieland conjecture is false in the case ${g=2}$ and ${n=6}$ .

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

Theorem 3 There exist a eight-dimensional subspace ${V\subset H^1(N,\mathbb{Q})}$ and a finite-index subgroup ${\Gamma}$ of ${Mod_{2,7}}$ with the following properties. Any element of ${\Gamma}$ lifts to a mapping-class of ${N}$ and the orbit of any ${v\in V}$ under the corresponding representation ${\Gamma\rightarrow Aut(H^1(N),\mathbb{Q})}$ is finite.

The proof of Theorem 3 relies on the unusual features of the Hodge filtration of ${N}$ .

2. Proof of Theorem 2 assuming Theorem 3

Since ${K}$ is a subgroup of ${H}$ , we have that the cover ${\overline{S_K}\rightarrow M}$ associated to ${K}$ factors through the cover ${p:N\rightarrow M}$ (associated to ${H}$ ), that is, we have a cover ${q:\overline{S_K}\rightarrow N}$ such that the composition ${q\circ p}$ is the cover ${\overline{S_K}\rightarrow M}$ corresponding to ${K}$ .

Given a eight-dimensional subspace ${V\subset H^1(N,\mathbb{Q})}$ and a finite-index subgroup ${\Gamma}$ of ${\textrm{Mod}_{2,7}}$ as in the statement of Theorem 3, let ${U\subset H^1(\overline{S_K},\mathbb{Q})}$ be the subspace of cohomology cycles projecting to ${V}$ which are also invariant under the whole group of deck transformations of ${q}$ .

By Theorem 3, the natural action of ${\Gamma}$ on ${V}$ factors a finite group of matrices. By construction, all orbits of the action of ${\Gamma}$ on ${U}$ are finite. Since ${\Gamma}$ is a finite-index subgroup of ${\textrm{Mod}_{2,7}}$ , it follows that all orbits of the action of ${\textrm{Mod}_{2,7}}$ on ${U}$ are also finite.

Finally, since the homology group ${H_1(\overline{S_K},\mathbb{Q})}$ is in a natural duality with the cohomology group ${H^1(\overline{S_K},\mathbb{Q})}$ , our reduction of Theorem 2 to Theorem 3 is complete.

3. Proof of Theorem 3

Let ${T:N\rightarrow N}$ be the automorphism ${T(x,y) = (x,\varepsilon y)}$ where ${\varepsilon=\exp(2\pi i/6)}$ generating the group of deck transformations of the cover ${r:N\rightarrow S}$ where ${S=\overline{\mathbb{C}}}$ is the Riemann sphere and ${r(x,y)=x}$ .

Note that ${r:N\rightarrow S}$ factors ${p:N\rightarrow M}$ : indeed, ${r=s\circ p}$ where ${s:M\rightarrow S}$ is the natural projection from ${M}$ to the quotient ${S=M/\langle s\rangle}$ of ${M}$ by its hyperelliptic involution ${\iota:M\rightarrow M}$ , ${\iota(x,z)=(x,-z)}$ .

Since ${M}$ has genus ${2}$ , the elements of ${\textrm{Mod}_{2,7}}$ commute with the hyperelliptic involution ${\iota}$ : this is a very special property of the genus ${2}$ 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 ${f\in \textrm{Mod}_{2,7}}$ with ${f_*(H)\subset H}$ form a finite-index subgroup ${\Gamma}$ of ${\textrm{Mod}_{2,7}}$ such that the lift to ${N}$ of any such ${f}$ commutes with the automorphism ${T:N\rightarrow N}$ (in fact, this is so because ${T}$ projects under ${p:N\rightarrow M}$ to the hyperelliptic involution ${\iota}$ of ${M}$ ). [Update (August 24, 2015): See also the exchange between Ben Wieland and myself in the comments below.]

By construction, ${\Gamma}$ acts on ${H^1(N,\mathbb{Q})}$ and our task is to show the existence of a eight-dimensional subspace ${V}$ of ${H^1(N,\mathbb{Q})}$ such that the ${\Gamma}$ -orbit of any ${v\in V}$ is finite.

For this sake, we start by analyzing the action of ${\Gamma}$ on ${H^1(N,\mathbb{C})}$ . Here, the crucial point is that ${\Gamma}$ was built in such a way that all of its elements commute with ${T}$ . In particular, the action of ${\Gamma}$ preserves each summand of the decomposition

$\displaystyle H^1(N,\mathbb{C}) = \bigoplus\limits_{l=1}^5 H^1(N)_{\varepsilon^l}$

into the eigenspaces ${H^1(N)_{\varepsilon^l}}$ associated to the eigenvalues ${\varepsilon^l}$ of ${T_*}$ . (Note that the eigenspace ${H^1(N)_{\varepsilon^0}}$ is trivial because ${H^1(N)_{\varepsilon^0} \simeq H^1(S,\mathbb{C}) = \{0\}}$ ).

Recall that the action of ${\Gamma}$ on ${H^1(N,\mathbb{C})}$ preserves the intersection form ${\langle\alpha,\beta\rangle := \frac{i}{2}\int \alpha\wedge \overline{\beta}}$ . Since each eigenspace ${H^1(N)_q}$ has a Hodge decomposition

$\displaystyle H^1(N)_q = (H^1(N)_q\cap H^{1,0})\oplus (H^1(N)_q\cap H^{0,1}):= H^{1,0}(N)_q\oplus H^{0,1}(N)_q,$

and the intersection form ${\langle.,.\rangle}$ is positive definite on the space ${H^{1,0}}$ of holomorphic ${1}$ -forms and negative definite on the space ${H^{0,1}}$ of anti-holomorphic ${1}$ -forms, we have that ${\Gamma}$ acts on ${H^1(N)_q}$ via a indefinite unitary group ${U_{\mathbb{C}}(n_q, m_q)}$ of a pseudo-Hermitian form of signature ${(n_q, m_q)}$ where

$\displaystyle n_q:=\textrm{dim}_{\mathbb{C}} H^{1,0}(N)_q \quad \textrm{ and } \quad m_q:=\textrm{dim}_{\mathbb{C}} H^{0,1}(N)_q.$

In our context, ${N}$ is associated to the curve ${y^6=x^5-1}$ , so that

$\displaystyle \left\{\frac{x^ndx}{y^m}: 0\leq n\leq m-2\leq 3\right\}$

is an explicit basis of the space ${H^{1,0}(N)}$ of holomorphic ${1}$ -forms on ${N}$ . From this, we infer that ${(n_{\varepsilon}, m_{\varepsilon}) = (4,0)}$ and ${(n_{\varepsilon^5}, m_{\varepsilon^5}) = (0,4)}$ (and, in general, ${(n_{\varepsilon^k}, m_{\varepsilon^k}) = (5-k, k-1)}$ for each ${1\leq k\leq 5}$ ).

In other words, ${H^1(N)_{\varepsilon} = H^{1,0}(N)_{\varepsilon}}$ , ${H^1(N)_{\varepsilon^5} = H^{0,1}(N)_{\varepsilon^5}}$ , and ${\Gamma}$ acts on the eight-dimensional complex subspace

$\displaystyle V_{\mathbb{C}} := H^1(N)_{\varepsilon}\oplus H^1(N)_{\varepsilon^5}$

via (a subgroup of) the compact group ${U_{\mathbb{C}}(4,0)\times U_{\mathbb{C}}(0,4)}$ .

Next, we study the action of ${\Gamma}$ on ${H^1(N,\mathbb{Q})}$ . We begin by noticing that the eight-dimensional complex subspace ${V_{\mathbb{C}}}$ is defined over ${\mathbb{Q}}$ . In fact, this is a consequence of the following elementary observation (from Galois theory): ${V_{\mathbb{C}}}$ is the sum of all eigenspaces ${H^1(N)_q}$ associated to all primitive ${6}$ th roots of unity ${q}$ .

Since ${V_{\mathbb{C}}}$ is defined over ${\mathbb{Q}}$ , it intersects ${H^1(N,\mathbb{Z})}$ into a lattice ${V_{\mathbb{Z}}}$ of rank ${8 = \textrm{dim}_{\mathbb{C}} V_{\mathbb{C}}}$ . In particular, ${\Gamma}$ acts on ${V:=\mathbb{Q}\otimes V_{\mathbb{Z}}}$ via (a subgroup of) the symplectic group ${Sp(8,\mathbb{Z})}$ because ${\Gamma}$ respects the symplectic intersection form ${(.,.)}$ on ${H^1(N,\mathbb{Z})}$ .

In summary, we proved that:

on one hand, ${\Gamma}$ acts on ${V_{\mathbb{C}}}$ via the compact group ${U_{\mathbb{C}}(4,0)\times U_{\mathbb{C}}(0,4)}$ ;
on the other hand, ${\Gamma}$ acts on ${V}$ via ${Sp(8,\mathbb{Z})}$ .

In other terms, ${\Gamma}$ acts on ${V}$ through a compact subgroup of the discrete group ${Sp(8,\mathbb{Z})}$ , i.e., ${\Gamma}$ acts on the eight-dimensional subspace ${V\subset H^1(N,\mathbb{Q})}$ through a finite subgroup of symplectic matrices.

It follows that the ${\Gamma}$ -orbit of any ${v\in V}$ is finite, so that the proof of Theorem 3 is complete.

Posted in expository, math.AG, math.RT, Mathematics | Tags: Andrew Putman, Artur Avila, Ben Wieland, Eduard Looijenga, Galois theory, Hodge decomposition, Ivanov conjecture, Jean-Christophe Yoccoz, Kazhdan property (T), mapping class group, Nikolai Ivanov, Putman-Wieland conjecture

Responses

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 marked curve, 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 Wieland on August 20, 2015
at 7:00 pm

Reply
- 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 $\Gamma$ in a slightly different (but equivalent) way.
  
  As you pointed out, the curve $S_{2,6}$ is “special” because my punctures are placed exactly at Weierstrass points. In particular, the elements $f$ of $Mod_{2,7}$ commute with the hyperelliptic involution (whose fixed points are the 6 Weierstrass points).
  
  Thus, $f\in Mod_{2,7}$ descends to an element $g\in Mod_{0,7}$ on the sphere with 7 punctures $S_{0,7}$ (obtained from the quotient of $S_{2,6}$ by the hyperelliptic involution after adding an extra puncture at the “origin” $0$ ). In other terms, we can think of elements of $Mod_{2,7}$ as lifts of elements of $Mod_{0,7}$ in our special situation.
  
  Now, we observe that the automorphism $T$ is the generator of the deck group of the finite ramified cover $N$ of the sphere.
  
  Therefore, we can take a finite-index subgroup $\Theta$ of $Mod_{0,7}$ such that the elements $g$ of $\Theta$ can be lifted to $N$ into mapping-classes $h$ that commute with $T$ .
  
  Finally, we set $\Gamma$ to be the finite-index subgroup of $Mod_{2,7}$ “corresponding” to the finite-index subgroup $\Theta$ of $Mod_{0,7}$ . In this way, if we start with an element $f\in \Gamma$ , we “descend” it to $g\in\Theta$ and then we lift $g$ and $f$ to a mapping-class $h$ on $N$ commuting with $T$ .
  
  In summary, we begin with a mapping-class $f$ in $S_{2,7}$ , then we try to lift it to $N$ (while commuting with $T$ ) by looking at the mapping-class $g$ on $S_{0,7}$ associated to $f$ (i.e., $g$ is the quotient of $f$ by the hyperelliptic involution), and studying how to lift $g$ to $N$ (while commuting with $T$ ).
  
  Best,
  
  Matheus
  By: matheuscmss on August 21, 2015
  at 11:16 am
  
  Reply
  - 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 Wieland on August 23, 2015
    at 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: matheuscmss on August 23, 2015
    at 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: matheuscmss on August 23, 2015
    at 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 Wieland on August 24, 2015
    at 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: matheuscmss on August 24, 2015
    at 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: matheuscmss on August 24, 2015
    at 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 Wieland on August 20, 2015
at 7:02 pm

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