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Let $P$ be an analytic linear differential operator defined on some open interval $X=(a,b)$ and $\mathcal{M}=\mathcal{D}_X / \mathcal{D}_X \bullet P$ the corresponding $\mathcal{D}$-module. I'm trying to understand what the correct definition of a Green's function is for $P$ in this context.

For starters I'm not sure what kind of algebraic object it is. Should it be a...

  1. A function on $X \times X - \triangle$?
  2. A $1$-form on $X \times X - \triangle$?
  3. A class in local cohomology $H^1_{\triangle}(X \times X, \Omega^1_{X \times X})$?
  4. A $\mathcal{D}_{X \times X}$- module?
  5. None of the above... :( ?

Obviously I'm phrasing everything in the most simple terms but ideally I'm looking for the answer that will generalize easily and immediately to general case.

If there is a standard reliable reference for these kind of questions I'd be happy to know about it.

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    $\begingroup$ The article Differential Equations and Integral Geometry by Goncharov seems to contain something in that direction. $\endgroup$
    – Michael Bächtold
    Commented May 23, 2017 at 21:29
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    $\begingroup$ For those who can't be bothered to read the article, Goncharov defines a "Green class" for a $D$-module $M$ as the class in $H^{dim(X)} (DR(M^* \otimes_{O_X} M))$ induced by the identity map in $Hom_{D_X}(M,M)$ (here $M^*$ is the $D_X$-dual of $M$ and $DR$ is the De Rham complex). Then goes on to tell you how this thing behaves like a Green function. $\endgroup$
    – Ketil Tveiten
    Commented May 24, 2017 at 8:24
  • $\begingroup$ Alternately, the direct translation of the definition of a Green's function for $P$ at $x\in X$ would be a solution of $I_x\cdot P$, where $I_x$ is the ideal of the point $x$. This isn't very useful unfortunately. $\endgroup$
    – Ketil Tveiten
    Commented May 24, 2017 at 8:37

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