Fix $n$ and a field $k$ of characteristic zero. Let $G$ be the proalgebraic 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?

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 SuárezÁlvarez Oct 10 '13 at 16:31
Assume X is ndimensional and regular. Then there is a functor from Gmodules V to D_{X}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 G_{0} and P with the canonical G_{0}torsor, the same construction yields an O_{X}module. The extra structure of a Gaction lets you identify infinitesimally nearby fibers.

$\begingroup$ I'm confused. Shouldn't you need the the action of a HarishChandra 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

1$\begingroup$ You can formally exponentiate a HarishChandra 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 indscheme. $\endgroup$ – S. Carnahan♦ Oct 14 '09 at 23:01


$\begingroup$ 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