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Let $A$ be an abelian variety over a field $k$ of characteristic $0$. How do I prove, without using transcendental methods, that if $\nabla$ is an integrable connection on a vector bundle $L$ on $A$ then $(L, \nabla)$ comes from a $G$-bundle with connection for some abelian algebraic subgroup $G \subset GL(n)$? And the corollary there exists a $\nabla$-stable flag$$L = L_n \supset L_{n-1} \supset \dots \supset L_1 \supset 0$$with $\text{rank }L_i = i$? It is easy to prove for $k = \mathbb{C}$ by looking at the monodromy of $(L, \nabla)$, and the general case follows by the Lefschetz principle, but I want to find a different proof which does not use $\mathbb{C}$ at all.

Any help would be greatly appreciated!

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2 Answers 2

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Here is one approach using the Fourier transform for $D$-modules on an abelian variety due to Laumon.

Let $A^\flat$ be the moduli space of rank one local systems on $A$, it is the universal extension of the dual abelian variety $A^\vee$ by a vector space. Note that L is a D-module such that the underlying O-module is coherent. This means that its Fourier transform $\hat{L}\in D^b(A^\flat)$ has the property that its pushforward to $A^\vee$ is coherent (or more precisely all of its cohomology are coherent). This is equivalent to the $supp(\hat{L})$ being finite over $A^\vee$. Thus, $supp(\hat{L})$ is complete (i.e., proper over $k$).

The claim we are trying to prove is equivalent to finiteness of $supp(\hat{L})$. Thus, we are trying to prove that $A^\flat$ contains no non-trivial complete sub-varieties. This is very easy analytically (it is analytically equivalent to an affine variety), but there is also an algebraic proof.

Indeed, it suffices to check that any map from a projective curve $X$ to $A^\flat$ is constant. Such a map is simply a line bundle on $X\times A$ together with a flat connection along $A$. Now looking at it as an $A$-family of line bundles on $X$, we see that it is an exterior product of two line bundles, and it is easy to see that the connection must be constant as well.

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  • $\begingroup$ This is a good argument, but I think it only gives a proof for $G=GL(n)$. $\endgroup$ Sep 29, 2015 at 20:04
  • $\begingroup$ The question was for $GL(n)$ :) It also seems that once you prove it for $GL(n)$, it extends to other groups automatically (basically by Tanakian formalism). Maybe there is a shorter way? $\endgroup$
    – t3suji
    Sep 29, 2015 at 20:21
  • $\begingroup$ Actually, I am not sure about the last part: since the reduction to Borel is not canonical I don't see how to use Tannakian formalism. But I agree that the question was for $GL(n)$ (I missed that part). $\endgroup$ Sep 29, 2015 at 20:45
  • $\begingroup$ (Sorry, I keep misreading things and having to delete comments. Hopefully this makes sense:) Consider the category of vector bundles with connections on A. It is Tannakian with the fiber functor, say, fiber at 0. By the Fourier transform, it is equivalent to the category of torsion sheaves on A♭. We now see that it is actually a category of representations of a commutative pro-algebraic group H (which explicitly looks like something horrible: the Cartier dual of the group which is a disjoint union of completions of A♭ at all points)... $\endgroup$
    – t3suji
    Sep 30, 2015 at 5:17
  • $\begingroup$ By the Tannakian formalism, a G-connection on A is the same as a morphism H→G. Since H is commutative, the morphism factors through a commutative subgroup of G. $\endgroup$
    – t3suji
    Sep 30, 2015 at 5:19
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Here is an outline of a different approach from that of t3suji.

First, define an algebraic group structure on the group $G$ of pairs $(a, \tau)$, where $a \in A$ and $\tau$ is a lift of the transformation automorphism $T_a: A \to A$ to an automorphism of $L$, then interpret $\nabla$ as an abelian Lie subalgebra of $\text{Lie}\,G$, and show that for any algebraic group $G$, an abelian Lie subalgebra of $\text{Lie}\,G$ is contained in the Lie algebra of an abelian algebraic subgroup of $G$.

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