# Hecke operators on universal elliptic curves

Suppose that $$\tau \in \mathbf{H}$$ belongs to the complex upper half plane. The quotient $$\mathbf{C}/(\mathbf{Z}+\mathbf{Z}\tau)$$ gives an elliptic curve over $$\mathbf{C}$$. Write this elliptic curve as $$E_{\tau}$$. We can write $$E_{\tau}$$ as follows:

$$E_{\tau}: y^2 = 4x^3 - \left( \dfrac{27 j(\tau)}{j(\tau)-1728} \right)x - \left( \dfrac{27 j(\tau)}{j(\tau)-1728} \right),$$

where $$j(\tau)$$ is the $$j$$-invariant function. This is a model for the universal elliptic curve $$E \to Y_1(N)$$ over a modular curve $$Y_1(N)$$.

In very vague terms, my question is: suppose we have an eigen-cuspform $$f \in S_1(\Gamma_1(N))$$. Given the differential $$\omega = dx/y$$ on $$E$$, how do we compute the "$$f$$-part" of $$\omega$$? That is, how do we compute the direct summand of $$\omega$$ where the Hecke operators act via the Hecke eigenvalues of $$f$$?

More rigorously: Let $$H^0(E, \Omega_{E/Y_1(N)})$$ be the space of holomorphic one-forms on $$E$$ over $$Y_1(N)$$ and let $$f \in S_1(\Gamma_1(N))$$ be an eigen-cuspform. Given the differential $$\omega = dx/y \in H^0(E, \Omega_{E/Y_1(N)})$$, how do we explicitly compute the $$f$$-isotypical component of $$\omega$$ under the action of the Hecke operators?

• I think the eigenvalues will rather come from modular forms of weight $1$, since $H^0(E, \Omega_{E/Y_1(N)}$ is the space of modular forms of weight $1$ and level $N$. Oct 1 at 15:44
• edited the question, thanks for the correction. Oct 1 at 17:21
• @WillSawin A follow up on this: in what sense is $H^0(E, \Omega_{E/Y_1(N)})$ isomorphic to $M_1(Y_1(N))$? The former space is differential forms on $E$, whereas the latter space is differential forms on $Y_1(N)$. How do you get a differential form on $E$ and produce from it a differential form on $Y_1(N)$? Oct 1 at 21:50
• Please use a high-level tag like "nt.number-theory". I added this tag now. Oct 2 at 2:45

The universal elliptic curve can be written $$y'^2 = 4x'^3- g_2 x - g_3$$ where $$g_2$$ and $$g_3$$ are Eisenstein series. Given any differential on $$E$$, simply change coordinates to this family, then divide by $$dx'/y'$$ to obtain a modular form of weight $$1$$.
The point is that $$g_2$$ is modular of weight $$4$$ and $$g_3$$ is modular of weight $$6$$, so if $$y$$ and $$x$$ are coordinates in a $$\Gamma_1(N)$$-invariant presentation of the universal family, then $$x/x'$$ is a modular function of weight $$2$$ and $$y/y'$$ is a modular function of weight $$3$$. Thus $$dx/y$$ is equal to $$dx'/y'$$ times a modular function of weight $$2-3=-1$$. Dividing the $$\Gamma_1(N)$$-invariant equation for your differential form by a modular function of weight $$-1$$ produces a modular form of weight $$1$$.
I'm not sure the presentation you gave necessarily works for this, because I think it is only the correct one up to possible quadratic twist. The correct model is the one that actually has a section of order $$N$$.
Modular forms of weight $$1$$ certainly don't correspond to differentials on the modular curve. Rather these are modular forms of weight $$2$$. The relation between differentials on $$E$$ and differentials on modular curves is simply that they satisfy similar transformation laws under $$SL_2(\mathbb Z)$$, or, algebro-geometrically, that the line bundle of relative differentials squares to the line bundle of differentials of the base.
• Thanks for the very helpful answer! A clarification, what exactly do you mean by a "$\Gamma_1(N)$-invariant presentation of the universal family"? Explicitly, under what change of variables for $x,y$ would this presentation have to be invariant under? Oct 1 at 23:25
• @AdithyaChakravarthy I mean a presentation like yours where there is no change of variables for $x,y$ when $\tau$ is changed by an element of $\Gamma_1(N)$. Oct 1 at 23:38
• I see. So given a differential on $E$, how would one go about computing its $f$-isotypical component? Oct 2 at 20:17
• @AdithyaChakravarthy In terms of $q$-expansions, I would find all the eigenforms, write a system of linear equations writing the differential as a sum of eigenforms, and solve. It may be possible to also do this using the Petersen inner product, taking advantage of the orthogonality of eigenforms to write the answer as the inner product of the differential with $f$, times $f$, divided by the inner product of $f$ with itself. Oct 2 at 21:03