weight 4 eigenforms with rational coefficients---is it reasonable to expect they all come from Calabi-Yaus? - MathOverflow

weight 4 eigenforms with rational coefficients---is it reasonable to expect they all come from Calabi-Yaus?

2010-04-19T18:32:32Z

A weight 2 modular form which happens to be a normalised cuspidal eigenform with rational coefficients has a natural geometric avatar---namely an elliptic curve over the rationals. It seems to be a subtle question as to how this notion generalises---this question was raised by Tony Scholl in conversation with me the other day. For example I guess I wouldn't expect a weight 3 normalised cuspidal eigenform with rational coefficients to be the $H^2$ of a smooth projective surface, because any such surface worth its salt would have (1,1)-forms coming from a hyperplane section, whereas the Hodge numbers of the motive attached to a weight 3 form are 0 and 2.

But in weight 4 one can again dream. A rigid Calabi-Yau 3-fold defined over the rationals has 2-dimensional $H^3$ and the Hodge numbers match up. Indeed there are many explicit examples of pairs $(X,f)$ with $X$ a rigid Calabi-Yau 3-fold over $\mathbf{Q}$ and $f$ a weight 4 cuspidal modular eigenform, such that the $\ell$-adic Galois representation attached to $f$ is isomorphic to $H^3(X,\mathbf{Q}_\ell)$ for all $\ell$.

The question: Is it reasonable to expect that (the motive attached to) every weight 4 normalised cuspidal eigenform with rational coefficients is associated to the cohomology of a rigid Calabi-Yau 3-fold over $\mathbf{Q}$?

Answer by David Hansen for weight 4 eigenforms with rational coefficients---is it reasonable to expect they all come from Calabi-Yaus?

2010-04-20T16:13:04Z

There is a recent preprint of Paranjape and Ramakrishnan where they discuss such matters. In particular, they realize Ramanujan's delta function in the middle-dimensional cohomology of an 11-dim. Calabi-Yau! Perhaps this is not so surprising - their variety is birational to a Kuga-Sato variety.

Let me also point out something what has mystified me greatly. (This may all be wildly incorrect) Suppose $X/ \mathbf{Q}$ is a rigid Calabi-Yau threefold. As you point out, the $(3,0)$-chunk of $H^3(X)$ gives a weight four modular form $f$. Now, people have conjectured the following various items:

The intermediate Jacobian $J(X)=H^{3,0}(X) / H_3(X,\mathbf{Z})$, an elliptic curve, is defined over $\mathbf{Q}$. It thus gives rise to a weight two form $g$.
The Abel-Jacobi map from the group $Ch(X)^2_0$ of homologically trivial one-cycles on $X$ / rat. equivalence to $J(X)$ is injective and defined over $\mathbf{Q}$.
The rank of $Ch(X)^2_0 / \mathbf{Q}$ is equal to the order of vanishing of $L(s,f)$ at its central point. (Bloch)

On the other hand, if the Abel-Jacobi map were injective and defined over $\mathbf{Q}$, then $J(X)(\mathbf{Q})$ should have rank at least the rank of $Ch(X)^2_0 / \mathbf{Q}$. At the level of L-functions, this should force the L-fn attached to the weight two form of $g$ to vanish to order $\geq$ the order of vanishing of the L-fn of the weight 4 form $f$. How could two modular forms of different weights know about each other in such a way as for the orders of vanishing of their L-functions to be entwined? The only thing that I can imagine is that they satisfy a congruence. Maybe they lie in the same Hida family? Wild speculation!

Answer by Kim Hopkins for weight 4 eigenforms with rational coefficients---is it reasonable to expect they all come from Calabi-Yaus?

2010-04-20T23:59:21Z

Apologies if I'm misunderstanding your question (which I probably am), but in http://arxiv.org/abs/0904.1141v1, I construct a map for a Hecke eigenform $f\in S_k^-(N)$ (set k=4 for example) by integrating $\int_{i\infty}^\tau f(z) (az^2 + bz + c) dz$ where $\tau:=\frac{-b+\sqrt{D}}{2a}$ is a Heegner point of level N. The period integrals $$L_f:= { \int_{i\infty}^{\gamma(i\infty)} f(z) z^m dz, \gamma \in \Gamma_0(N), 0\leq m\leq 2}$$ form a lattice, so this gives a well defined map on Heegner points mod $\Gamma_0(N)$ to an elliptic curve $\mathbb{C}/L_f$. This map on Heegner divisors turns out to be equivalent to the $f$-component of the Abel-Jacobi map for the Kuga-Sato variety (on Heegner cycles) into its intermediate Jacobian. In this sense, this gives an association between the elliptic curve $\mathbb{C}/L_f$ and the piece $H^{3,0}(X)/H_3(X,\mathbb{Z})$ you mentioned above. I've computed the $j$-invariant for some examples and the elliptic curves so far seem to not be defined over $\mathbb{Q}$ or any number field for that matter. Is this relevant at all?

Answer by Junkie for weight 4 eigenforms with rational coefficients---is it reasonable to expect they all come from Calabi-Yaus?

2010-04-21T12:34:47Z

"At the level of L-functions, this should force the L-fn attached to the weight two form of g to vanish to order the order of vanishing of the L-fn of the weight 4 form f. How could two modular forms of different weights know about each other in such a way as for the orders of vanishing of their L-functions to be entwined?"

A paper of Dummigan has congruences with vanishing, but not how you say.

http://neil-dummigan.staff.shef.ac.uk/dsw_13.dvi

A paper of Schoen gives CM examples for the Abel-Jacobi map.

http://www.jstor.org/stable/2154210

A paper of Villegas has a [3,0] type of conductor $59^2$ that has vanishing order 2 (page 437).

http://www.math.utexas.edu/~villegas/publications/square-root-2.pdf

William Stein mentions the j-invariant calculation in his thesis (page 68).

http://wstein.org/papers/thesis/stein-thesis.pdf

Answer by Kim Hopkins for weight 4 eigenforms with rational coefficients---is it reasonable to expect they all come from Calabi-Yaus?

2010-04-23T00:51:57Z

@David: Not quite I don't think. Let's see if I can avoid butchering this... In weight 4, the intermediate Jacobian of the Kuga-Sato variety W is $$J^2(W)= \frac{(H^{3,0} \oplus H^{2,1})^\vee}{H_3(W,\mathbb{Z})^\vee}.$$ We care about the holomorphic piece, $H^{3,0}$ which is in bijection with $S_4$. Let $\Lambda$ be the corresponding sub-lattice in $H_3(W,\mathbb{Z})^\vee$. If we take a Heegner cycle $z_\tau$, which is more-or-less the graph of multiplication by a Heegner point $\tau$ in $W$, then the Abel-Jacobi map on it is $$AJ(z_\tau)(\omega) = \int\limits_{\Delta_\tau} \omega \mod H_3(W,\mathbb{Z})^\vee$$ where $\Delta_\tau$ is a $3$-chain bounded by $z_\tau$. If we choose a weight 4 form $f$ with $\omega_f:=f(\tau)d\tau dz_1dz_2$, then one can show that $$AJ(z_\tau)(\omega_f) = \int\limits_{i\infty}^\tau f(z) (az^2 + bz + c)dz \bmod \Lambda'$$ where $\Lambda'$ is a slightly larger lattice which contains $\Lambda$ with finite index. Moreover $L_f\subset \Lambda'$. This is what I meant when I said that the map I construct above evaluated on Heegner points is equal to the Abel-Jacobi map on Heegner cycles.

@Kevin: I don't think I know the answer to your question but would also like to hear an answer.

The paper Complex multiplication cycles on elliptic modular threefolds' and the preprintGeneralized Heegner cycles and p-adic Rankin L-series' by Bertolini, Darmon, and Prasanna describe this well.

Answer by Remke Kloosterman for weight 4 eigenforms with rational coefficients---is it reasonable to expect they all come from Calabi-Yaus?

2010-08-23T13:56:29Z

I am not that sure that your question can be answered positively, but the following is merely speculation, so you should not pin me down on it. The basic idea is that there might be "more" weight 4 forms than rigid CY 3-folds.

It seems that (but Kevin probably knows this better than me) that it is still an open question whether there are finitely or infinitely many weight 4 eigenforms up to twisting.

Suppose there were infinitely many weight 4 eigenforms up to twisting. To realize every weight 4 eigenform we then need infinitely many $\overline{\mathbb{Q}}$-isomorphism classes of rigid CY 3-folds defined over $\mathbb{Q}$. All hodge numbers of a rigid CY 3-fold are a priori fixed, except for $h^{1,1}$ and $h^{2,2}$, which coincide. The Euler characteristic of a rigid CY 3-fold is $2h^{1,1}$ and rigid CY 3-folds do not admit deformation, hence in order to realize every weight 4-form we either find a Hodge diamond $D$ such that

{ $X | X$ smooth projective complex variety with Hodge diamand $D$}/deformations

is infinite, or the absolute value of the Euler characteristic of a CY 3-folds can be arbitrarily large. The first conclusion would be quite remarkable, the second would solve an open problem (as far as I know).

Suppose now there were only finitely many weight 4 eigenforms up to twisting. If you want to avoid the above mentioned problems you need that every eigenform $f$ is realized by a CY 3-fold $Y_f$ admitting an involution, so that a twist of $f$ is realized by a twist of $Y_f$.

Still it is not clear whether there are enough rigid CYs to realize every eigenform. Some computational evidence can be found in the book of Christian Meyer, Modular Calabi-Yau Threefolds. He realizes close to 100 eigenforms (up to twisting). The corresponding list at http://www.fields.utoronto.ca/publications/supplements/weight4.pdf contains much more forms. The smallest level that he could not realize is 7.

If you allow $h^{2,1}$ to be nonzero, i.e., you allow the motive of the form to be a factor of $H^3$ or if you are happy to work with quasi-projective varieties $Y$ such that its completions is a CY 3-fold then you are in a much better position.

Answer by Matthias Schuett for weight 4 eigenforms with rational coefficients---is it reasonable to expect they all come from Calabi-Yaus?

2010-09-03T14:16:21Z

To follow up the question of the intermediate Jacobian, there is indeed a later survey by Noriko Yui (Arithmetic of Calabi-Yau varieties. Mathematisches Institut, Georg-August-Universität Göttingen: Seminars Summer Term 2004, 9--29) where she does conjecture it to be defined over $\mathbb Q$ for any (modular) rigid CY3 over $\mathbb Q$.

Moreover, she refers to a joint paper with X. Xarles in preparation (still unpublished) which claims to settle this in the CM case:

Let $X$ be a rigid CY3 of CM type (i.e. with commutative Hodge group) over some number field $F$. Then the intermediate jacobian $J^2(X)$ is an elliptic curve with CM by an order in an imaginary quadratic field (understood: the same field, since there will be a relation of Hecke characters over some extension), and it has a model over $F$.

Let's apply this to rigid CY3's over $\mathbb Q$ and assume that there is one for each newform of weight 4 with rational coefficients. Pick one of the weight 4 newforms with rational coefficients and CM of class number 3, i.e. induced by a Hecke character for an imaginary quadratic field $K$ of class number 3 such as $\mathbb Q(\sqrt{-23})$ (or generally of class group exponent 3). By assumption there is an associated CY3 $X$ over $\mathbb Q$ (which ought to have CM). But then, by the above result, its intermediate Jacobian is an elliptic curve over $\mathbb Q$ with CM in $K$, contradiction.

Of course, one can still ask whether all non-CM newforms can be realized in some CY3's over $\mathbb Q$, but I would find it surprising if this would only fail at certain CM forms. And after all, I would be glad to allow non-rigid CY3's over $\mathbb Q$ admitting the right submotive, but this might still not be sufficient.

Answer by christian-meyer for weight 4 eigenforms with rational coefficients---is it reasonable to expect they all come from Calabi-Yaus?

2011-06-15T12:48:42Z

@David and others that have been discussing intermediate Jacobians: I still have a printed copy of a draft by Helena Verrill entitled "The Intermediate Jacobian of the Six Kuga-Sato Rigid Calabi-Yau Threefolds", containing numerical computations. It seems to be almost ten years old and not to have been published.