Let $E$ be an elliptic curve and $x,y \in H^1(E, \mathbb{Q})$ be a basis for the first rational cohomology group of $E$. There is an action of the linear group $SL_2(\mathbb{Q})$ on $H^*(E,\mathbb{Q})$ as follows: for $g=\left(\begin{array}{cc}a & b \\\ c & d \\ \end{array}\right)$ we set $$g \cdot x=ax+cy, \qquad g \cdot y=bx+dy.$$ Question: Is there a modular interpretation of this action? Does it come from an action of the group $SL_2(\mathbb{Q})$ on the elliptic curve $E$ or on the moduli space of elliptic curves?


As asked, I think that the answer to your latter two questions are both no.

A few points:

1) The action you write down isn't well-defined, as it depends on the choice of a basis of the cohomology.

2) A "random" elliptic curve (one without CM) only has has multiplication by elements of $\mathbb{Z}$ as endomorphisms, so you're not going to find an interesting action of the group $SL_2(\mathbb{Q})$ on $E$ in the first place.

3) There also isn't a meaningful action of $SL_2(\mathbb{Q})$ on the moduli space of elliptic curves. Think complex-analytically: this moduli space (using the term loosely) consists of equivalence classes of $\tau$ in the upper-half plane $\mathbb{H}$ modulo the action of $SL_2(\mathbb{Z})$. Try to cook up a well-defined action of $SL_2(\mathbb{Q})$ here. The obvious choice doesn't work because $SL_2(\mathbb{Z})$ isn't normal in $SL_2(\mathbb{Q})$. (Also, even if you did have an action on the moduli space, such a thing wouldn't beget an action on the cohomology of particular elliptic curves anyway).

Regarding my point (1) - there may be a way to salvage your action by looking at a moduli space of elliptic curves together with a choice of basis of $H^1(\mathbb{Q})$. Over $\mathbb{C}$, I think that you can cook up such a thing pretty easily by considering a diagonal action of $SL(\mathbb{Z})$ on $\mathbb{H}\times \mathbb{Q}^2$ (though I think that you may have to twist the action on the second component slightly to get the right thing) and passing to the quotient.

Perhaps something in this picture gives you what you're looking for?


As mentioned in the answer by Ramsey, there is no natural action of $SL_2(\mathbf{Q})$ on an elliptic curve, but there is an action of $SL_2(\mathbf{Z})$, which can be thought of as the monodromy of the universal elliptic curve. Let me describe how to get an action of a finite index torsion free subgroup $\Gamma\subset SL_2(\mathbf{Z})$ in this way.

Let $\mathbf{H}$ stand for the upper half-plane. Consider the quotient $U$ of $\mathbf{C}\times\mathbf{H}$ by the following action of $\mathbf{Z}\times\mathbf{Z}:$

$$(a,b)\cdot (z,\tau)=(z+a+b\tau,\tau)$$ where $z\in\mathbf{C}, a,b\in\mathbf{Z}$ and $\tau\in\mathbf{H}$. This quotient is fibered over $\mathbf{H}$ with fiber over $\tau$ being the elliptic curve $\mathbf{C}/\langle 1,\tau\rangle$. $SL_2(\mathbf{Z})$ acts on $U$ by $$\left(\begin{array}{cc} a&b\\ c& d\end{array}\right)\cdot[z,\tau]=\left[\frac{z}{c\tau+d},\frac{a\tau+c}{c\tau+d}\right].$$

If we take the quotient $U/\Gamma$, we get a fiber bundle over $\mathbf{H}/\Gamma$ with fiber an elliptic curve. This fiber bundle is manifestly flat, so the fundamental group of the base, which is $\Gamma$, acts on the fiber. This action is just the restriction of the standard $SL_2(\mathbf{Z})$-action on the 2-torus.

Getting the action of the whole of $SL_2(\mathbf{Z})$ in this way is possible, I believe, but a bit trickier, the problem being, of course, that the action of $SL_2(\mathbf{Z})$ on $\mathbf{H}$ can have stabilizers. So I would say instead of the topological quotient one should consider the stack quotient $[U/SL_2(\mathbf{Z})]$. The first derived pushforward of the constant sheaf on $U$ under the projection $U\to [U/SL_2(\mathbf{Z})]$ should be a local system and the standard 2-dimensional $SL_2(\mathbf{Z})$-module should be the resulting monodromy. I do not know whether or where this has been worked out in detail.

  • 1
    $\begingroup$ Aha! I was thinking in the context of algebraic (or complex-analytic) geometry, in which case the situation is no better for $SL_2(\mathbb{Z})$. However, the action that turns up in algori's answer is smooth, but not complex-analytic. This is, of course, sufficient to induce an action on $H^1(E,\mathbb{Z})$! Moreover, this action is basically what the OP writes down (restricted to $SL_2(\mathbb{Z})$). $\endgroup$ – Ramsey Feb 12 '11 at 16:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.