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Let $X$ be a smooth variety over $\mathbb{C}$. If $\mathscr{F}$ is a coherent sheaf on $X$ with connection, does it follow that $\mathscr{F}$ is locally free? I can't think of any counterexamples. What if $X$ is a complex analytic manifold?

In his article "Regular connections after Deligne," Malgrange begins to sketch the proof: fix a point $x \in X$ and a minimal system of generators for $\mathscr{F}_x$. Consider a relation in which some coefficient has minimal order at $x$ and differentiate it. How do we finish the proof? I do not see how this calculation implies that the coefficients must vanish (and therefore $\mathscr{F}_x$ is free).

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More details for the proof of Malgrange.

First we easily reduce to the local case. Fix a point $x$ and a minimal system of generators $v_1,\dots, v_n$ of $F_x$. We want to prove it is free. Let $m_x$ be the maximal ideal of $O_{X,x}$, let $$a_1v_1+\dots+a_nv_n=0, \quad a_i\in O_{X,x}$$ and let $r$ be the smallest non-negative integer such that $a_i\in m_x^r$ for all $i$ and $a_i\notin m_x^{r+1}$ for at least one $i$. Using the connection, we are going to find a new vanishing relation as above, but with smaller $r$.

So apply the connection to the above equality, we get $$\sum_i a_i\nabla(v_i)+\sum_i v_i\otimes da_i=0.$$ Note that $a_i\nabla(v_i)\in M\otimes \Omega_{X}^1$ and $da_i\in m_x^{r-1}\Omega_{X}^1\setminus m_x^r\Omega_{X}^1$ (here we use the characteristic $0$). As $\Omega_X^{1}$ is free, considering the above equality components by components, we see that there exist $b_i\in O_{X,x}$ such that $$\sum_i b_iv_i=0$$ at leat one $b_i\notin m_x^{r}$. So we get a new vanishing relation with smaller $r$. Repeating util we get $r=1$, so one of the coefficients is invertible and the system of generators is not minimal. Contradiction.

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These proofs use quite a bit of machinery, and I do not understand them... Surely there is a simple proof of this "folklore" result? – Justin Campbell May 2 '12 at 23:39

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