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 $x$?

  • 2
    $\begingroup$ @Addendum: see Prop. 2. 16, 2.21 in "Notes on crystalline cohomology", by Berthelot-Ogus (Princeton Univ. Press) $\endgroup$ Nov 19, 2011 at 13:07
  • $\begingroup$ (late comment) Not that you have to assume that ${\rm char}(k)=0$. $\endgroup$ Nov 27, 2011 at 7:37

1 Answer 1


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

  • $\begingroup$ @Damian: this is exactly what I needed! Thanks a lot for your answer! $\endgroup$
    – Veen
    Nov 19, 2011 at 13:56
  • 6
    $\begingroup$ Actually, Katz's quoted proposition assumes that the connection is integrable. The hypothesis is not necessary : see Corollary 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 $\endgroup$
    – ACL
    Nov 20, 2011 at 10:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.