Let $X$ be an abelian variety over an algebraically closed field $k$.

Let $L$ be a line bundle on it equipped with an integrable connection $\nabla: L \rightarrow L \otimes \Omega^1_{X/k}$.

Does it then automatically folllow that $L$ is a bundle in $Pic^0(X)$?

And how general can one make such a statement? I mean: Does it also hold for $k$ arbitrary or at least of characteristic zero? And does it hold for abelian schemes over a base $S$, say of characteristic zero?


1 Answer 1


Yes, it is true, though an algebraic proof seems (there may be a simpler proof however) somewhat tricky.

  • Such a line bundle lies in $\mathrm{Pic}^\tau(X)$. This is a general fact as a line bundle lies in $\mathrm{Pic}^\tau(X)$ if its rational Chern classes are trivial (this follows from Riemann-Roch) and the Chern classes of a line bundle with integrable connection are torsion (an algebraic proof is given in Dix exposés sur la cohomologie des schémas). This works without the assumption of $X$ being abelian.
  • It is then a further fact that for abelian varieties $X$ we have that $\mathrm{Pic}^\tau(X)= \mathrm{Pic}^0(X)$ (this I think is an Mumford's abelian varieties somewhere).

This extends to immediately to families by checking fibre by fibre.

Addendum: Sorry forgot to say that this all requires characteristic zero. In characteristic $p$ every $p$'th power line bundle has an integrable connection (even of $p$-curvature $0$) but will in general not lie in $\mathrm{Pic}^0(X)$.

Addendum 1: Lars (in a comment) makes an interesting point about the positive characteristic situation. A module structure over the ring of differential operators (aka a stratification) implies in particular that the line bundle is a $p^n$'th power for each $n$ and as $\mathrm{Pic}(X)/\mathrm{Pic}^\tau(X)$ is a finitely generated group this implies that the line bundle lies in $\mathrm{Pic}^\tau(X)$. The same idea could be applied to the characteristic zero situation if the $p$-curvature of the reduction modulo $p$ for an infinite number of primes $p$ were zero. However that should be true only if the line bundle has finite order so it doesn't give very much.

Addendum 2: Veen is asking about the equality $\mathrm{Pic}^0(A)=\mathrm{Pic}^\tau(A)$ particularly in a family (when $A$ is abelian). The easiest way to answer all of these questions simultaneously is to assume that there is an ample line bundle $\mathcal L$ on $A$ (which is true locally on the base) and then consider the map $A\to \mathrm{Pic}(A)$ given by $a\mapsto \mathcal L_a\bigotimes\mathcal L^{-1}$. To get all equalities needed it is enough to show that the image is $\mathrm{Pic}^\tau(A)$. This is something that can be checked fibrewise and then it can be extracted from Mumford.

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    $\begingroup$ What does $\mathrm{Pic}^\tau$ mean? $\endgroup$ Nov 7, 2011 at 9:03
  • $\begingroup$ Line bundles numerically equivalent to the trivial line bundle. $\endgroup$ Nov 7, 2011 at 9:17
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    $\begingroup$ You should add the hypothesis that ${\rm char}(k)=0$. Otherwise, a counterexample is given by $L^{\otimes p}$, where $L$ is an ample bundle on a scheme $S$, which is smooth over ${\rm F}_p$ ($p$ a prime number). $\endgroup$ Nov 7, 2011 at 9:24
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    $\begingroup$ As a side note: if the base has positive characteristic $p$, and if one replaces the notion of "line bundle with flat connection" by "line bundle with operation of the ring of differential operators of $X/k$", the line bundle will still lie in $Pic^\tau$, at least on smooth $X/k$. $\endgroup$
    – Lars
    Nov 7, 2011 at 22:39
  • $\begingroup$ @Torsten: thanks for your great comment; I just have one question: why does it for an abelian scheme suffice to check it fibre by fibre? I don't see where point here is. $\endgroup$
    – Veen
    Nov 12, 2011 at 22:30

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