Home > algebraic geometry > A coherent sheaf with connection is locally free

A coherent sheaf with connection is locally free

This post will be rather technical, but I think it’s worthwhile to write down the following “folklore” proof, which doesn’t seem to be laid out carefully in the literature. Let X be a smooth variety over \mathbb{C} (I think the proof also goes through on a complex manifold), \mathcal{F} a coherent sheaf on X, and \nabla : \mathcal{F} \to \Omega_X^1 \otimes_{\mathcal{O}_X} \mathcal{F} a connection on \mathcal{F}, not assumed flat. We will prove that \mathcal{F} is then locally free, or in other words the sheaf of sections of a vector bundle on X. This shows that the presence of a connection strongly “rigidifies” the underlying sheaf. The proof is suggested by Malgrange in his article “Regular connections after Deligne,” although he gives no details there. I have seen a more sophisticated proof, although it assumes the connection is flat: then one can view \mathcal{F} as a \mathcal{D}_X-module and apply Kashiwara’s theorem (see Proposition 5.13 in Gaitsgory’s notes on geometric representation theory).

Since \mathcal{F} is coherent, it suffices to show that \mathcal{F} is locally free at any (closed) point x \in X. To this end, choose a minimal set of generators e_1,\cdots,e_m for \mathcal{F}_x as an \mathcal{O}_x-module, and consider a relation

f_1e_1 + \cdots + f_me_m = 0

where f_1,\cdots,f_m \in \mathcal{O}_x. Note that we must have f_1,\cdots,f_m \in \mathfrak{m}_x by the minimality of e_1,\cdots,e_m, and if not all these coefficients are zero there is a unique k \geq 1 such that f_i \in \mathfrak{m}_x^k for all 1 \leq i \leq m but f_{\ell} \notin \mathfrak{m}_x^{k+1} for some \ell. Choose f_1,\cdots,f_m so that k is minimal among all such relations.

Denote by \mathcal{T}_X the tangent sheaf on X and recall that \nabla induces a “covariant derivative”

\mathcal{T}_X \otimes_{\mathbb{C}} \mathcal{F} \to \mathcal{T}_X \otimes_{\mathbb{C}} \Omega_X^1 \otimes_{\mathcal{O}_{X}} \mathcal{F} \to \mathcal{F}

\xi \otimes s \mapsto \nabla_{\xi} s,

and the Leibniz identity for \nabla implies that for all \xi \in \mathcal{T}_X, s \in \mathcal{F}, and f \in \mathcal{O}_X we have

\nabla_{\xi}(f \cdot s) = f \cdot \nabla_{\xi} s + \xi(f) \cdot s.

We assume for the moment a crucial lemma, namely that there is a tangent vector \xi \in T_x(X) such that \xi(f_{\ell}) \in \mathfrak{m}_x^{k-1} (this is automatic) but \xi(f_{\ell}) \notin \mathfrak{m}_x^k. Applying \nabla_{\xi} to our relation yields

f_1 \cdot \nabla_{\xi} e_1 + \cdots + f_m \cdot \nabla_{\xi} e_m + \xi(f_1) \cdot e_1 + \cdots + \xi(f_m) \cdot e_m = 0,

and now if we write \nabla_{\xi}e_i = \sum_j g_{ij} e_j and regroup we obtain

\sum_j \big[ \big( \sum_i f_ig_{ij} \big) + \xi(f_j) \big] e_j = 0.

Now by the assumption on f_1,\cdots,f_m we have \sum_i f_ig_{ij} \in \mathfrak{m}_x^k, and our choice of \xi guarantees that each coefficient of this new relation lies in \mathfrak{m}_x^{k-1}, but when j = \ell not in \mathfrak{m}_x^k. This contradicts the minimality of k.

Finally, we prove the lemma. Choose local coordinates x_1,\cdots,x_n \in \mathfrak{m}_x, meaning these germs represent a basis of \mathfrak{m}_x/\mathfrak{m}_x^2, so by our assumption that X is smooth we have \text{gr } \mathcal{O}_x \cong \mathbb{C}[x_1,\cdots,x_n] as graded algebras. Moreover, if we write \frac{\partial}{\partial x_1},\cdots,\frac{\partial}{\partial x_n} \in T_x(X) for the dual basis, then the map \mathfrak{m}_x^k/\mathfrak{m}_x^{k+1} \to \mathfrak{m}_x^{k-1}/\mathfrak{m}_x^k induced by a tangent vector \xi = \sum_i a_i \frac{\partial}{\partial x_i} is identified with the usual action of partial differential operators on polynomials. Thus we have reduced ourselves to the case that X = \mathbb{A}^n is the affine space, x = (0,\cdots,0) is the origin, and f_{\ell} = f \in \mathbb{C}[x_1,\cdots,x_n] is a homogeneous polynomial of degree k \geq 1. But now since \text{char } \mathbb{C} = 0 the claim is obvious: if x_j appears with nonzero exponent in any monomial in f, then \frac{\partial f}{\partial x_j} has degree k-1.

  1. BC
    3 October 2012 at 4:11 am

    The proof of this “folklore” fact is laid out clearly by dimension induction early in Deligne’s SLN book on regular singular points.. In fact, Deligne’s book does much better, simultaneously proving the relative version for smooth morphisms in the category of complex analytic spaces allowing the base to be arbitrary (no smoothness hypothesis on it!).

    • 4 October 2012 at 1:27 pm

      Thanks for the reference! I once had a copy of this book and lost it while I was on vacation in Florida. Maybe someday I’ll get another one.

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