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Let $(E,O)$ be an elliptic curve over $\mathrm{Spec}(\mathbb{C})$. Then a 1-dimenstional represenation of $\pi_1(E^{\mathrm{an}}) = \mathbb{Z} \times \mathbb{Z}$ over $\mathbb{C}$ is just a pair of elements $(a, b)$ of $\mathbb{C}^\times$. Under a standard equivalence of categories, the representation corresponding to a pair $(a, b)$ corresponds to an analytic connection on $E^{\mathrm{an}}$ which we'll denote by $L^{\mathrm{an}}_{(a,b)}$ (see, e.g., theorem 4.2.4 in the book "D-modules, perverse sheaves and representation theory" in the book by Hotta et al). One description of $L^{\mathrm{an}}_{(a,b)}$ is given in Mariano's answer below.

Now we have a equivalence of categories between analytic connections on $E^{\mathrm{an}}$ and algebraic connections on $E$ (see, e.g., corollary 5.3.9 in Hotta et al.).

I'd like to know an explicit description of the algebraic connection $L_{(a,b)}$ on $E$ corresponding to the analytic connection $L^{\mathrm{an}}_{(a,b)}$ on $E^{\mathrm{an}}$ under this equivalence.

The first step would be to determine the underlying invertible $\mathscr{O}_E$-module, which should be of degree 0 (see, e.g., Atiyah's "Complex Analytic Connections in Fiber Bundles," proposition 19). But $\mathrm{Pic}^0(E) = E(\mathbb{C})$ since $E$ is an elliptic curve. In other words, we get a map $$(\mathbb{C}^\times)^2 \to E(\mathbb{C}),$$ taking $(a,b) \in (\mathbb{C}^\times)^2$ to the closed point $P_{(a,b)} \in E(\mathbb{C})$ such that $\mathscr{O}_E(P_{(a,b)}-O)$ is the line bundle underlying the connection $L_{(a,b)}$. What is $P_{(a,b)}$ in terms of $(a, b)$? Given $P \in E(\mathbb{C})$, what are the $(a,b) \in (\mathbb{C}^\times)^2$ such that $P = P_{(a,b)}$?

How can we describe the connection on $L_{(a,b)}$ in terms of $(a, b)$? We know that $\Omega^1_E$ has a global nowhere-vanishing section $\omega$. Let $\partial$ be the derivation of $\mathscr{O}_E$ dual to $\omega$. We know that $L_{(a,b)}$ is trivial on $E \setminus \{O, P_{(a,b)}\}$, so the connection on this open is given by a differential equation of the form $\partial - f$ for some $f \in \Gamma(U, \mathscr{O}_E)$. Can we explicitly write down such a $f$?

chthonian
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1 Answers1

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$\def\ZZ{\mathbb{Z}}\def\CC{\mathbb{C}}$Let $X=\CC\times\CC$, fix $\tau$ in the open upper halfspace, and let $a$ and $b$ be in $\CC^\times$. There is on $X$ an action of $\ZZ\times\ZZ$ such that we have $$(x,y)\cdot(z,v)=(z+x+\tau y,a^xb^yv)$$ whenever $(x,y)\in\ZZ\times\ZZ$ and $(z,v)\in E$. The first projection $\pi:X\to\CC$ is equivariant for the usual action of $\ZZ\times\ZZ$ on $\CC$, so by passing to the quotient we end up with a map $\bar\pi:V=X/(\ZZ\times\ZZ)\to E$, with $E$ an elliptic curve, and this is easily seen to be a line bundle on $E$. There is a connection on $V$ whose horizonal sections on the open subsets of $E$ well covered by the projec tion $\CC\to E$ are precisely the constant sections of $\pi$. This means that the Christoffel symbols of the connection in charts induced from the covering $\CC\to E$ are zero.

Mariano Suárez-Álvarez
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    This works always. If $V$ is a representation of $\pi_1$ of a manifold $M$ and $\tilde M$ is the universal covering, then $\tilde M\times V\to\tilde M$ is a trivial vector bundle on $\tilde M$, which passes to the quotient to give an usually nontrivial vector bundle $\tilde M\times_{\pi_1}V\to \tilde M/\pi_1=M$, and there is a connection on this last vector bundle whose horizontal sections (on well-covered open sets of $M$) "are" the constant sections of the trivial bundle on $\tilde M$. – Mariano Suárez-Álvarez Jan 13 '17 at 20:49
  • This is helpful, but it's still a little abstract: I was hoping for something more explicit still. For example, which line bundle on $E$ is it? One of the nice things about elliptic curves is that we have a classification of line bundles on $E$ in terms of $E$ itself, as I had indicated in my question. We also know that $\Omega^1_{E/\mathbb{C}}$ is generated by the invariant differential, which should make describing the connection very explicitly possible also? – chthonian Jan 13 '17 at 20:59
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    Well, I explicitly constructed the line bundle! It does not get more concrete than that. As you know the classification of line bundles, you should be able to pass from my description to Pic :-) – Mariano Suárez-Álvarez Jan 13 '17 at 21:03