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I want to see the following thing:

$\ \ $If $X$ is a smooth geometrically connected scheme over a field $k$ of characteristic 0, $U\subseteq X$ is a non-empty open, $(E,\nabla)$ is an integrable connection on $X/k$, then $H_{DM}^0(X/k,(E,\nabla))\hookrightarrow H_{DM}^0(U/k,(E,\nabla)|_{U})$ is an isomorphism. Here the De Rham cohomology with respect to an integral connection means I take the hypercohomology of the complex

$$E\xrightarrow{\nabla}E\otimes\Omega^1_{X/k}\to E\otimes\Omega^2_{X/k}\to\cdots$$

$\ \ $ I think it comes from the following "base change property of Guass-Manin sheaf", I don't know the precise formulation for that property, but I guess it is the following (0-th version):

If $S$ and $X$ are smooth geometrically connected schemes over a field $k$ of characteristic 0, $X\to S$ is proper smooth $k$-morphism, if we have a commutative diagramme
$\hspace{100pt}\ Y \xrightarrow{f} X$

$\hspace{100pt}\ |$$a$$\ \ \ \ |b$

$\hspace{100pt}\ T\xrightarrow{g}S$

with the property that $f^*\Omega_{X/S}\to \Omega_{Y/T}$ is an isomorphism, and if $(E,\nabla)$ is an integrable connection on $X/S$, then the canonical map

$g^*b_*E^{\nabla} \to$

$a_*{(f^*E)^{\nabla}}$is an isomorphism. Here $E^{\nabla}$ is the kernel of the $k$-linear map $\nabla: E\to E\otimes\Omega^1_{X/S}$. By definition $b_*E^{\nabla}$ with the canonical connection (the Gauss-Manin connection) on it is $H_{DM}^0(X,(E,\nabla))$. The similar notations for $(f^*E)^{\nabla}$.

Is that true? Is there a reference? If this was true, then we can take $Y/T$ to be $U/k$, this answers my first question. But I think maybe the complete formulation requires the diagramme to be Cartesian.

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You have an arbitrary integrable connection so I don't see what the Gauss-Manin connection has to do with your question. –  ulrich Oct 9 '11 at 12:47
    
You are right, I should just say the base change property for the "Gauss-Manin sheaf" –  unknown Oct 9 '11 at 12:55
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1 Answer

up vote 3 down vote accepted

Here is a proof of your claim:

We want to show that any flat section of $E|_U$ extends to a flat section of $E$. Since the extension is unique if it exists, the question is local on $X$ so we may assume that the bundle $E = \mathcal{O}_X^n$. The connection is then given an $n\times n$ matrix $A$ of $1$-forms on $X$ i.e. for $v$ a section of $E$, $$\nabla v = dv + Av $$ where $d$ is the exterior derivative applied componentwise.

Now any flat section $w$ of $E|_U$ can be viewed as a rational section $v$ of $E$. Since we are in characteristic zero, if any component $f$ of $w$ has a pole of order $r$ along some divisor $D$ in $X \backslash U$, then $df$ has a pole of order $n+1$ along $D$. Since none of the entries of $A$ have poles, it follows that $\nabla w$ cannot be $0$. Thus all components of $w$ are regular functions on $X$ so $w$ is extends to a regular section $v$ of $E$. Since $w$ was a flat section of $E|_U$, it follows that $v$ is a flat section of $E$, proving the desired surjectivity.

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Thank you very much! The answer looks very nice. But I have a stupid question here: Let us assume $X={\rm Spec}(A)$ with an etale map from $X$ to the affine $n$-space. Assume $U={\rm Spec}(A_f)$ with $f\in A$. How do you define the oder of a pole of a component of $w\in A_f$ along a divisor $D\subseteq V(f)$, and how do you define the components of $w$? –  unknown Oct 9 '11 at 14:05
    
In your notation, $w$ is an $n$-tuple $(f_1,f_2,\dots,f_n) \in A_f^n$ and by component I just mean any one of the $f_i$. The local ring of a divisor $D$ in $X \backslash U$ is a discrete valuation ring and by order of pole of an element of $A_f$ is meant the negative of the valuation extended to the quotient field (which contains $A_f$). –  ulrich Oct 9 '11 at 14:30
    
Thanks, ulrich. I think your answer is right and clean, but there is still a point which I am not quite clear yet. You take any divisor $D$ which is the closure of a codimension 1 point $p$ in $X\setminus U$. We assume the bundle is a line bundle, so there is only one component $w$. Let $\pi$ be the uniformizer of $A_p$, then $w_p=a\pi^{−n}$ with $a$ invertible in $A_p$. Assume $n>0$, then $d(w_p)=\pi^{−n}d(a)+na\pi^{-n-1}d(\pi)$, why it has a pole of oder $n+1$? –  unknown Oct 9 '11 at 15:40
    
Let me first clarify what I mean by a pole of order $r$ of a rational $1$-form $w$ along a divisor $D$: In your notation this is defined to be the smallest integer $r$ so that $\pi^{r}w$ is a regular $1$-form at the generic point of the divisor. Since we are in characteristic zero, $na$ is a unit, and $d(a)$ and $d(\pi)$ are regular, so the order must be $n+1$. –  ulrich Oct 9 '11 at 16:42
    
execellent proof, thank you. –  unknown Oct 9 '11 at 20:02
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