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I think it is something like a folkore result that a coherent sheaf $\mathcal F$ on a smooth algebraic variety $X$ over $k$, which is equipped with a connection

$\nabla: \mathcal F \rightarrow \mathcal F \otimes \Omega^1_{X/k}$

is already locally free.

Maybe one may weaken the assumptions, but I think the proof wouldn't alter very much.

I thought about how to prove it, but couldn't make it rigorous. In any case one should show that the stalks $\mathcal F_x$ are free, and this somehow must follow from the existence of the connection.

Addendum: It seems that the existence of a connection in some way rigidifies the underlying sheaf. A related question in this respect is: in how much is a horizontal morphism $\phi: \mathcal F \rightarrow \mathcal F$ already rigidified by $\nabla$. E.g. if one knows that $\phi$ is the identity on $\mathcal F(x) \rightarrow \mathcal F(x)$, then is it so already around $x$?

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@Addendum: see Prop. 2. 16, 2.21 in "Notes on crystalline cohomology", by Berthelot-Ogus (Princeton Univ. Press) –  Damian Rössler Nov 19 '11 at 13:07
    
(late comment) Not that you have to assume that ${\rm char}(k)=0$. –  Damian Rössler Nov 27 '11 at 7:37
    
Yes, I realized that, thanks. –  Veen Nov 29 '11 at 8:27

1 Answer 1

up vote 6 down vote accepted

See Prop. 8.8, p. 206 in N. Katz, "Nilpotent connections and the monodromy...", Publications Mathématiques de l'IHES, 39 (1970), p. 175-232.

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@Damian: this is exactly what I needed! Thanks a lot for your answer! –  Veen Nov 19 '11 at 13:56
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Actually, Katz's quoted proposition assumes that the connection is integrable. The hypothesis is not necessary : see Corollary 2.5.2.2 of Yves André's paper, Différentielles non commutatives et théorie de Galois différentielle ou aux différences. Annales scientifiques de l'École Normale Supérieure, Sér. 4, 34 no. 5 (2001), p. 685-739. numdam.org/item?id=ASENS_2001_4_34_5_685_0 –  ACL Nov 20 '11 at 10:24
    
Thanks again, this is very useful! –  Veen Nov 29 '11 at 8:29

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