Fix $n$ and a field $k$ of characteristic zero. Let $G$ be the pro-algebraic group of automorphims of $k[[x_1,...x_n]]$. Let $G_0$ be the subgroup of automorphisms preserving the closed point (note that for general $T$, $G_0(T)$ can be a proper subgroup of $G(T)$). Let $X$ be a regular variety over $k$ and let $P$ be the principal $G$ bundle of formal coordinate systems, naturally a $G$ torsor over $X$. I hear that there is a connection between $P$ and $D_X$-modules. what is this connection?
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1$\begingroup$ @Sergiy: you removed a pair of brackets, turning $k[[x_1.\dots.x_n]]$ into $k[x_1, \dots, x_n]$, which is a rather differet thing! You should undo that change. $\endgroup$ – Mariano Suárez-Álvarez Oct 10 '13 at 16:31
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Assume X is n-dimensional and regular. Then there is a functor from G-modules V to DX-modules, given by an associated bundle construction. Take the trivial (ind-)bundle on P with fiber V, and quotient by the action of G on P and V. If you replace G with G0 and P with the canonical G0-torsor, the same construction yields an OX-module. The extra structure of a G-action lets you identify infinitesimally nearby fibers.
answered Oct 14 '09 at 22:34
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$\begingroup$ I'm confused. Shouldn't you need the the action of a Harish-Chandra pair (roughly since k[[x_1,...,x_n]] has derivations that don't integrate)? Group actions shouldn't be able to glue infinitesimally close points. Or is this some subtle thing where thinking about the algebraic group encapsulates the Lie algebra bit? $\endgroup$ – Ben Webster♦ Oct 14 '09 at 22:49
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1$\begingroup$ You can formally exponentiate a Harish-Chandra pair. G_0 is a proalgebraic group, but G is a formal group with a big proalgebraic subgroup. You can think of it as a group ind-scheme. $\endgroup$ – S. Carnahan♦ Oct 14 '09 at 23:01
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$\begingroup$ It's sort of a yes for both questions, but you change "algebraic" to "formal" in the second. You can think of G as G_0 with extra fuzz. $\endgroup$ – S. Carnahan♦ Oct 15 '09 at 15:31
