11
$\begingroup$

I am trying to get some insight into the Deligne/Scholl construction of the motive of a modular form. First of all I would like to understand the case of weight two, especially when there is complex multiplication. Unfortunately in his paper "Motives for modular forms" Scholl says

"We do not treat here the case $k = 0$, which corresponds to cusp forms of weight 2; the associated motives are then given by the Jacobians of modular curves, and are well understood."

Question 0: Could anyone explain this or give a readable reference?

More precisely, I am interested in the following question. If $E$ is an elliptic curve over a number field $K$, one can look at the Chow motive with rational coefficients $$ h^1(E) $$ cut off from the total motive $(E, \nabla, 0)$ by $\nabla$ minus the two projectors associated to some rational point (over an extension of $K$ if necessary). In general, this motive is indecomposable.

However, when $E$ has complex multiplication by a quadratic imaginary field $F=\mathbb{Q}(\sqrt{-d})$, one gets a decomposition of $h^1(E)_F$ into two rank 1 motives with coefficients in $F$.

On the other hand, one can look to the weight two modular form $f$ attached to $E$ (modularity in the CM case, which is much easier!). Thanks to the construction quoted by Scholl one gets a rank two motive with coefficients in the field generated by the Fourier coefficients of $f$.

Question 1: (General elliptic curve) What is the relation between these two motives? One would like to say there are the same, but the coefficient field are different, aren't?

Question 2 (CM case): Does the geometric construction using Jacobians of modular forms also give the decomposition into two pieces of rank one? If so, how?

$\endgroup$
0

3 Answers 3

4
$\begingroup$

Since this question has come alive again, let me point out that the Hecke operators cannot give a splitting of $h^1(E)$ into two pieces over $F$, since the Hecke correspondences on a modular curve are always defined over $\mathbf{Q}$.

(If you allow some extra stuff like the Atkin--Lehner operator on $X_1(N)$, then you can get some slightly larger fields showing up, but they will still be totally real, and thus not capable of "seeing" the splitting of $h^1(E)$.) (EDIT: Actually I am not sure if this is true, sorry!)

If you take a higher-weight CM cuspform $f$ of weight $> 2$, associated to some Groessencharacter $\Psi$, then the situation is even worse: you can define a motive associated to $f$ using Kuga--Sato varieties, and a motive associated to $\Psi$ using the product of $k-1$ copies of an elliptic curve with CM by $K$. These motives really should be the same, because their $\ell$-adic Galois representations are the same for every $\ell$, but there is (as far as I know) no natural way of writing down a correspondence that gives an isomorphism between them in the category of Chow motives.

$\endgroup$
5
  • $\begingroup$ Allowing the more general Hecke correspondences attached to double cosets $KgK$ in $\mathrm{GL}_2(\mathbb{A}_f)$, we get correspondences defined over the cyclotomic field $\mathbb{Q}(\zeta_N)$ where $N$ is the level of $K$ (e.g. the building blocks of a modular abelian variety $A_f$ are not defined over $\mathbb{Q}$). $\endgroup$ Commented May 19, 2022 at 14:31
  • $\begingroup$ Maybe the reason is that the complex multiplication endomorphisms of $E$ are not defined over an abelian field in general. $\endgroup$ Commented May 19, 2022 at 14:36
  • $\begingroup$ No, this isn't the reason. If $E$ is a CM elliptic curve defined over $\mathbf{Q}$, coming from a CM-field $K$ with class number one, then the CM-endomorphisms are all defied over $K$, which is quadratic and thus certainly abelian over $\mathbf{Q}$. But I think the field of definition of the Hecke correspondences is the real subfield $\mathbf{Q}(\zeta_N)^+$ of $\mathbf{Q}(\zeta_N)$. $\endgroup$ Commented May 19, 2022 at 14:38
  • $\begingroup$ I think it depends on the level $\Gamma \subset \mathrm{GL}_2(\mathbb{A}_f)$: some building blocks in $J_1(N)$ are defined over fields which are not totally real. $\endgroup$ Commented May 19, 2022 at 14:44
  • $\begingroup$ You are right, sorry, I slipped up calculating the field of definition of the components. $\endgroup$ Commented May 19, 2022 at 14:57
3
$\begingroup$

The construction alluded to by Scholl is the Eichler–Shimura construction, and a readable account can be found e.g. in D. Rohrlich, Modular curves, Hecke correspondences, $L$-functions.

If $f$ is the newform of weight 2 associated to a CM elliptic curve $E/K$, then the abelian variety $A_f/\mathbf{Q}$ associated to $f$ by Eichler–Shimura is isogenous over $\overline{\mathbf{Q}}$ to a power of $E$. But some extra work is needed in order to get the minimal field of definition of this isogeny. I don't think it's always true that the isogeny is defined over $K$ — you could imagine replacing $E$ by an elliptic curve which is isomorphic over $\overline{\mathbf{Q}}$, which preserves the CM condition. EDIT : One should also be careful with what is meant by the modular form associated to $E$. It is rather a Grössencharakter over $K$ which is associated to $E/K$. In order to get a classical modular form, I think one should further assume that $K$ is (at most) quadratic.

That being said, if $K'$ is a field of definition of the isogeny $A_f \sim E^n$, then the motive of $A_f$, which is just the restriction of scalars of the motive of $f$, is isomorphic to $h^1(E)^n$ over $K'$.

$\endgroup$
4
  • 1
    $\begingroup$ Thanks for your answer, François! Do you know something about the second question, that is, how to see the two submotives of rank one in $A_f$? $\endgroup$
    – cmotive
    Commented Jun 6, 2013 at 18:48
  • $\begingroup$ This is a good question -- on the Galois side, the Galois representations associated to $f$ and $A_f$ decompose into two rank 1 pieces after restricting to the absolute Galois group of the imaginary quadratic field $F$ by which $E$ has CM. These pieces are cut out by the endomorphisms of $E$ which are not defined over $\mathbf{Q}$, and we may wonder whether these endomorphisms are modular i.e. are induced by some Hecke correspondences on modular curves above $f$. I think this should be true at least in some cases, but I'm not sure and I hope an expert of the CM case can show up. $\endgroup$ Commented Jun 6, 2013 at 19:22
  • $\begingroup$ I once saw a result stating that the Hecke correspondences generate the whole endomorphism algebra of the Jacobian of a modular curve, so the answer should be yes, but I don't recall the reference. $\endgroup$ Commented Jun 6, 2013 at 19:34
  • $\begingroup$ So where is this CM expert? :) $\endgroup$
    – cmotive
    Commented Jun 8, 2013 at 8:03
1
$\begingroup$

Question 1: The field generated by the Fourier coefficients of an elliptic curve associated to a modular form is $\mathbb Q$. (For example, since the Fourier coefficients can be calculated by counting points on the elliptic curve, which gives integer values.)

Question 2: I think the answer is "no".

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .