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It's "well known" that, for any weight $k$ and level $N$, the space $S_k(\Gamma_1(N))$ of cusp forms of that weight and level has a basis in which all the Hecke operators act by matrices with entries in $\mathbb{Z}$; consequently all the Hecke eigenvalues are algebraic numbers (indeed algebraic integers).

I was reflecting on how to prove this while teaching an undergraduate course on modular forms. For $k \ge 2$ it's not hard: there's the Eichler-Shimura machinery which relates it to a question about cohomology, and the cohomology with $\mathbb{Z}$ coefficients does the job. Alternatively, and more or less equivalently, you use the pairing with modular symbols. Both of these methods break down for $k = 1$; the only argument I know that works in this case is to use the fact that $X_1(N)$ has a model as an algebraic variety, and weight $k$ modular forms correspond to sections of the $k$-th power of a line bundle that has a purely algebraic definition. But that's not really something I can stand up and explain to a class of undergraduate students!

For cusp forms of weight $k = 1$, can the algebraicity of the Hecke eigenvalues be proved without quoting heavy machinery from arithmetic geometry?

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up vote 10 down vote accepted

Let $S = S_{\mathbf{Q}} = M_{13}(\Gamma_1(N),\mathbf{Q})$, and $S_{\mathbf{C}} = S \otimes \mathbf{C}$ denote the corresponding space of modular forms over $\mathbf{C}$.

Let $V \subset S \times S$ be the subspace cut out by pairs of forms $(A,B)$ satisfying the following equation:

$$A \cdot E_{12} = B \cdot \Delta $$

As equations in the Fourier coefficients of $A$ and $B$ these are linear equations with coefficients in $\mathbf{Q}$. Since, by the $q$-expansion principle, a modular form can be recovered from some finite number of Fourier coefficients, $V$ is determined by the null space of some finite matrix with coefficients in $\mathbf{Q}$. Since a linear system over $\mathbf{Q}$ has the same rank over $\mathbf{C}$, it follows that $V_{\mathbf{C}} = V \otimes \mathbf{C}$, where $V_{\mathbf{C}}$ is the set of solutions in $S_{\mathbf{C}} \times S_{\mathbf{C}}$ of the same equations.

On the other hand, there is an isomorphism: $$V_{\mathbf{C}} \rightarrow M_{1}(\Gamma_1(N),\mathbf{C})$$ given by $$(A,B) \mapsto \frac{A}{\Delta} = \frac{B}{E_{12}}$$ The point is that $E_{12}$ and $\Delta$ do not have any common zeros, so the image of this map clearly consists of holomorphic forms. Hence the map is well defined. However, if $F$ has weight one, then $(A,B) = (F \cdot \Delta, F \cdot E_{12})$ maps to $F$, so the map is surjective. It is clearly injective, so it is an isomorphism.

It follows that the image of $V$ under this map gives a rational basis for $M_1(\Gamma_1(N))$. Since $V$ is then preserved by Hecke operators (as is obvious on $q$-expansions), the result follows.

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Of course! Very pretty. This one could certainly use in an undergrad course. –  David Loeffler May 9 '13 at 6:12
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Here is another proof: Deligne and Serre have proven that the corresponding L-function equal the Artin L-function of a Galois representation. Deligne had proven similar facts known for weight $k \geq 2$ modular forms before that, and their proof essentially relies on the former results. This implies algebraicity and also is the only approach I know for the same result for Maass Hecke cusp forms of Laplace eigenvalue $1/4$, where the same algebraicity result is unknown.

This is probably even harder than what you suggest, or equivalent(?). Quoting that result of Deligne and Serre would be a reasonable choice, and deducing algebraicity from it. The question is pretty old and didn't receive an answer so far, so I guess the class isn't running anymore anyways.

There is no trace formula for Hecke eigenvalues of weight one forms available, since the limit of discrete series representations opposed to the discrete series representations are not square integrable, and have no pseudo-matrix coefficients.

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I think filling in all the steps here is actually quite a lot harder than the argument I mentioned above. Nonetheless, if you assume Deligne--Serre it's certainly a nice short argument! –  David Loeffler May 8 '13 at 10:58
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Dear Marc, Rebecca Black is correct that this is circular. As noted in another recent comment I made on a question of yours, you can't begin to to do the Deligne--Serre argument without congruences between wt. 1 forms and higher wt. forms, and such congruences don't make sense without knowing that wt. 1 eigenforms have algebraic integer $q$-expansion coefficients (or some equivalent algebro-geometric statement of the kind that the OP was trying to avoid). Regards, Matthew –  Emerton May 10 '13 at 23:32
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