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 $k-1$.