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The usual modular parametrization says that if one takes a modular form $f$ which is a newform for $\Gamma_0(N)$ (i.e. a form in the new subspace which is a normalized eigenfunction for the Hecke operators), then one can associate to $f$ an abelian variety $A_f$ of dimension $d$, where $d$ is the degree of $K$ over $\mathbb Q$, where $K = \mathbb Q[a_n]$, being $a_n$ the Fourier coefficients of $f$ at the infinity cusp. The construction goes as follows, to $f$ one attachs the module (over $\mathbb Z$)

$$I_f = (T \in Hecke_{\mathbb Z} \quad : \quad Tf =0)$$ (sorry but the usual latex bracket didn't work in the previous formula)

Then take the quotient of $J_0(N)$ by the image of $I_f$ (i.e. $J_0(N)/I_fJ_0(N)$). My question is the following: if one considers $\mathbb Z_K$ the ring of integers of $K$ and defines $I_f$ in the same way but looks at the module generated over $\mathbb Z_F$ (which distinguishes between $f$ and its Galois conjugates), and then take the quotient of $J_0(N)$ by this ideal, do one gets an elliptic curve over $K$?

If not, I have a second question, if one considers the image of the Abel-Jacobi map for $f$, i.e. integration over all the homology of $X_0(N)$ of the form $f(z)dz$, is the image of such map a lattice in $\mathbb C$? (I would expect to get the lattice which coincides with the first question if true). I know that if you consider all the conjugates of $f$ then you get an abelian variety over $\mathbb Q$, and I wonder if it is the restriction of scalars of an elliptic curve over $K$.

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2 Answers 2

up vote 7 down vote accepted

No, the abelian variety $A_f$ is typically absolutely simple, and hence does not factor (even up to isogeny, and even over $\overline{\mathbb Q}$ or $\mathbb C$) as a product of elliptic curves. What will happen is that if you integrate $f(z)dz$ over the homology of $X_0(N)$, you will get a finitely generated, but non-discrete, $\mathbb Z$-submodule of $\mathbb C$. If $f$ has coefficients in $\mathbb Q$, then miraculously this submodule will actually be a lattice. If $f$ has cofficients over $K$ of degree $d$, and you integrate all the conjugates of $f$ to get a $\mathbb Z$-submodule of $\mathbb C^d$, then again you miraculously get a lattice!

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The idea I had in mind was to compute the action of the Hecke algebra in the homology and "diagonalize" it (only for primes prime to $N$). If you do diagonalize them, you will get something of rank 2 in the image (since an eigenvector in the homology with different eigenvalue than that of $f$ will integrate zero). BUT my mistake was that when you diagonalize you need to go to the field of coefficients of $f$, so the image will actually be generated by $2g$ elements (which works ok if you end in $\mathbb C^g$). Do you have a reference of the fact that $A_f$ is absolutely simple? –  A. Pacetti Jun 3 '11 at 0:15

This is an answer to the question A. Pacetti asked in his comment to Emerton's answer.

The modular variety $A_f$ does not have to be geometrically simple. William Stein and I computed many examples of this years ago: we were looking for modular abelian surfaces with Quaternionic Multiplication and it turns out that often when you actually compute the Hilbert symbol you find that your quaternion endomorphism algebra is $M_2(\mathbb{Q})$, i.e., the surface geometrically splits as $E \times E$.

However, when the level $N$ is squarefree $A_f$ is always geometrically simple. This follows from a 1975 theorem of Ribet: when $N$ is squarefree, the $\mathbb{Q}$-rational endomorphism algebra of $J_0(N)$ is equal to the geometric endomorphism algebra. The former is just the algebra generated by the Hecke operators and is a product of totally real fields (multiplicity one!). Indeed, what he actually shows is that all the endomorphisms of a semistable abelian variety over a number field $K$ are defined over an everywhere unramified extension of $K$. This means that the geometric endomorphism algebra of any given $A_f$ is just the Fourier field. Since this is a division algebra, $A_f$ is geometrically simple in this case.

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Dear Pete, Thanks for this cogent answer. Best wishes, Matt –  Emerton Jun 3 '11 at 1:37
    
Thanks Pete, do you have the examples in your web page? I will look at Ribet's paper. –  A. Pacetti Jun 3 '11 at 2:27
    
@Ariel: no, they are lost in the sands of time. But if you contacted William Stein, I'll bet he could reproduce them without much difficulty. –  Pete L. Clark Jun 3 '11 at 6:01

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