Let $f \in S_0(N)$ be a normalized Hecke eigenform. It is well known that its coefficients are algebraic integers, and $f^\sigma$ lies in $S_0(N)$ for $\sigma \in G_{\mathbb{Q}}$. At CM points $z \in \mathbb{H}$, $f$ takes an algebraic value as well.

Can anything be said about the relation between $f(z)$ and $f^\sigma(z)$ or $f(z^{\sigma})$?

I am aware that there is a nice relation in the $p$-adic setting, due to the $p$-adic convergence of cusp forms, and that this is behind the remarkable usefulness of the Tate curve, but I am more interested in the global behavior.

Perhaps this is wishful thinking, but are there any ($p$-adic or global) results for non-cusp forms as well?

  • $\begingroup$ "..I am aware that there is a nice relation in the p-adic setting, due to the p-adic convergence of cusp forms, and that this is behind the remarkable usefulness of the Tate curve,.." can you elaborate what you mean here? $\endgroup$
    – DBS
    Commented Jul 10, 2013 at 4:42
  • 1
    $\begingroup$ For an elliptic curve $E/K$, $K$ a p-adic field, there is a uniformization, as Galois modules, $E(\overline{K}) \cong \overline{K}^*/q^{\mathbb{Z}},$ for some $q \in K^*$. It is a Galois isomorphism, because for the p-adic power series which define it, the Galois action is continuous, so you can apply it to every term in the sequence, the same as applying to the infinite sum. $\endgroup$
    – Jesse
    Commented Jul 10, 2013 at 14:54
  • $\begingroup$ Just to note that the accepted answer (and the entire premise of the question) is completely wrong. Both the question and the answer claim that if $f$ is a normalized Hecke eigenform then the value of $f$ at a CM point is an algebraic value, and this is the basis of the question. But this is false. It's so false, in fact, that the values of $f$ at such points are never algebraic unless they either vanish or $f$ has weight $0$ and so is a constant. $\endgroup$
    – user297024
    Commented May 25 at 10:06

1 Answer 1


When working with modular forms as analytic objects, the Galois structure is somewhat invisible and has to be rediscovered using the Hecke algebra. However, there is a purely algebraic notion of modular form, due to Katz, which makes the Galois structure (among other things) a lot more transparent. Katz's construction works well for level structures for which the corresponding moduli functor is representable, and unfortunately this excludes the case of $\Gamma_0(N)$. However, for level $\Gamma = \Gamma_1(N)$ or $\Gamma = \Gamma(N)$ (with $N>4$), the moduli problem is representable by an affine curve $Y_\Gamma$ over $\mathbf Q$, and using Katz's theory, the following result becomes almost tautological:

  1. Let $f$ be a modular form of weight $k$ and level $\Gamma$. Suppose that the $q$-expansion of $f$ has coefficients in $\overline{\mathbf Q}$. Let $\sigma \in G_{\mathbf Q}$. Then $f^\sigma$, obtained by applying $\sigma$ to the $q$-expansion of $f$, is a modular form of the same weight and level as $f$.

  2. Let $p \in Y_\Gamma(\overline{\mathbf Q})$ be a $\overline{\mathbf Q}$-point of $Y_\Gamma$. Let $f$, as in (1), be a modular form with coefficients in $\overline{\mathbf Q}$. Then $f(p) \in \overline{\mathbf Q}$ and $$f^\sigma(p) = f(p^\sigma) = f(p)^\sigma.$$

From the theory of complex multiplication, it is known that a CM point $p \in Y_\Gamma(\mathbf C) = \mathbf H/\Gamma$ is in fact a $\overline{\mathbf Q}$-point of $Y_\Gamma$ (more precisely, it is defined over the Hilbert class field of the imaginary quadratic field to which it belongs). (Warning: just because a point of the upper-half plane is algebraic, does not mean that it corresponds to an algebraic point of $Y_\Gamma$. The identification of $\mathbf H/\Gamma$ with $Y_\Gamma(\mathbf C)$ is transcendental. It is a special property of CM points that they are algebraic "on both sides", but in the present context, this is more of a pitfall than anything else.) It follows from this and from (2) that:

Theorem. Let $f$ be a modular form of weight $k$ and level $\Gamma$ whose $q$-expansion has coefficients in $\overline{\mathbf Q}$. Let $p$ be a CM point in the upper-half plane. Then $f(p)$ is algebraic, and $f^\sigma(p) = f(p)^\sigma$ for every $\sigma \in G_{\mathbf Q}$.

However, keep the above-stated pitfall in mind: the equality with $f(p^\sigma)$ makes no sense from the classical point of view because the Galois action on $p$ takes place in $Y_\Gamma(\overline{\mathbf Q})$, not in the upper half-plane. In fact, in the naive sense, either $p^\sigma=p$ or $p^\sigma = \overline{p}$ does not even lie on the upper half-plane, so the expression $f(p^\sigma)$ is meaningless from the classical point of view.

Over $\Gamma_0(N)$, the same results are true, but to prove it in this way, we have to pull everything back to $\Gamma(N)$ and push everything back down at the end.

  • $\begingroup$ This answer is completely wrong (see comment on the question above). $\endgroup$
    – user297024
    Commented May 26 at 0:46

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