# Can Duflo type map be defined for invariant differential operators ? (In a way compatible with “Harish-Chandra” isomorphism defined by F. Knop)

## Background 1) Knop's theorem

In fundamental paper in Annals 1994 F. Knop proved the following theorem.

Let G be a connected reductive group and X a smooth G-variety.

Theorem: Assume that X is either spherical or affine. Then the center Z(X) of the ring of G-invariant differential operators on X is a polynomial ring. More precisely, Z(X) is isomorphic to the ring of invariants of a finite reflection group.

## Background 2) Duflo's map

M. Duflo defined fundamental construction: for any Lie algebra $g$ a map: $DufloMap: S(g) \to U(g)$, such that it is

1) Restricted to the $S(g)^g$ it will give isomorphism of commutative algebras $S(g)^g \to Z(U(G))$

2) It is isomorphism of graded vector spaces, moreover it is identity map on "principal symbolds"

3) Isomorphism of $g$ modules

D. Calaque, C. Rossi "Lectures on Duflo isomorphisms in Lie algebras and complex geometry"

http://people.mpim-bonn.mpg.de/crossi/LectETHbook.pdf

See Capelli determinant = Duflo ( determinant) - was it known ? , Is the Duflo map for Lie algs. unique ? for some info on Duflo map.

## Question

Let $G$ be a Lie group (reductive may be necessary). $M$ is manifold (may be affine).

Question Is it possible to find such a map from Functions(T*M) to Differential operators on $M$, such that it satisfies requirements similar to the Duflo map, where the role of $S(g)$ is played by $Fun(T^*M)$ and of $U(g)$ is played by $Dif(M)$ ?

Moreover we should require is is compatible with Knop-Harish-Chandra isomorphism.

If center of $Dif(M)^G$ is trivial, then there is no point to ask the question, but if center is non-trivial the requirements seems quite non-trivial.

Let $X=T^*M$, and apply Dolgushev's equivariant formality theorem (see http://arxiv.org/abs/math/0307212) to get a $G$-equivariant $L_\infty$-quasi-isomorphism $$\mathcal U:T_{poly}X \longrightarrow D_{poly}X$$

Let $\pi$ be the Poisson structure on $X$ which comes from the canonical symplectic form on $T^*M$. Then one gets a $G$-equiavriant quasi-isomorphism of complexes $\mathcal U_\pi^1$ from the Poisson cochain complex of $(X,\pi)$ to the Hochschild cochain complex of $(\mathcal O_X,\star_\pi)$, where $\star_\pi$ is the star-product the quantizes $\pi$ through $\mathcal U$.

Moreover, one has a homotopy betwen the cup-product one both sides, and it is $G$-equivariant (compatibility with cup-products is sketched in the original paper of Kontsevich http://arxiv.org/abs/q-alg/9709040, is rigorously proved in http://arxiv.org/abs/math/0106205 by Manchon and Torossian, and the globalization is adressed in http://arxiv.org/abs/0805.3444 - in this last reference you will easily see that if you start with a $G$-invariant connection to perform globalization then your homotopy will be $G$-invariant too).

Conclusion. taking $G$-invariants and restricting yourself to the degree zero part of cohomology (Poisson and Hochschil, respectively), you'll find that the Poisson center of $\mathcal O_X^G$ is isomorphic, as an algebra, to the center of the subalgebra of $G$-invariants elements in the quantization.

Things I have been hiding under the carpet. 1. issues involving $\hbar$. In this case you have to prove that the quantization is actually convergent and can be specialized to $\hbar=1$. 2. You have to prove that the quantization really gives back differential operators... this is probably the more tricky part.

• @Damien Thank you very much ! Can it be done somewhat "explicit" similar like Duflo map for Lie algebras ? – Alexander Chervov Jun 25 '12 at 12:29
• E.g. if V-vector space and G - group acting on it. In particular if V - multiplicity free, one can defined "Capelli" elements. So we might want to ask how to compare Capelli with Duflo. They should be related (mathoverflow.net/questions/92348/…) but actually there is some shift in the formula "-(n-1)Id" on the RHS, which makes everything not quite clear... For determinant we have such a shift, but for other elements (like "immanants") it is not clear what happens... – Alexander Chervov Jun 25 '12 at 12:37
• I haven't thought seriously about an explicit formula. Locally, it should be the "obvious one" (this is just because the Poisson structure being constant in local coordinates, the majority of Kontsevich's graphs will not contribute to the quasi-isomorphism in tangent cohomology). What you have to take care about is: 1. how explicit is your identification with the ring of differential operators. 2. globalization issues (even if you work with an affine space there are globalization issues, as you will use a non-trivial connection to get G-equivariance). – DamienC Jun 25 '12 at 15:36