# Differential operators on M/G are factor of invariant differential operators on M ? (Quantization commute with reduction ) What is state of art ?

Naively we may expect: Diff(M/G) = Diff(M) // G, where Diff(M) // G = Diff(M)^G /I, where "I" is two-sided ideal in Diff(M)^G of operators which act by zero on invariant functions.

This is example of quantization commute with reduction ideology, since Diff(N) - quantization of T*N.

My questions is what is the state of art ? I would expect it is known to be true if action of M on G is free, M - smooth oriented manifold. If yes what is the reference ?

There is paper by B. Fedosov:

Non-Abelian Reduction in Deformation Quantization, Lett. Math. Phys. 1998 Volume 43, Number 2, 137-154,

As far as I understand his result should imply the positive answer for the compact group G. But I am not sure about the details, may be he assume some compactness, or some other assumptions...

If any one would be so kind to send me this paper, it would be very kind of him.

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In the algebraic situation one could do it as follows: First of all, you have a canonical map $Diff(M) // G\rightarrow Diff(M/G)$. Indeed if $f$ is a function on the quotient, $D\in Diff(M)^G$, then $D(\pi^* f)$ is $G$-invariant, so it descents to the quotient.

Now lets assume that the action on $M$ is free and proper. Then $\pi:M\rightarrow M/G$ is a principal $G$-bundle. So for checking whether our canonical map is an isomorphism we can restrict to the product situation $G\times X\rightarrow X$. In this case we compute: $$Diff(M) // G = Diff(G\times X)^G /I=(U(\mathfrak g)\otimes Diff(X))/(U(\mathfrak g)\otimes 1) =Diff(X)=Diff(M/G)$$

So here we used $Diff(X\times Y)=Diff(X)\otimes Diff(Y)$, in the algebraic situation this is literally true. In the analytic situation I would guess it is true if one interpretes the $\otimes$ in the right way... Maybe someone more experienced with these things can tell whether the above computation can be translated into the smooth manifold situation?

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What means "proper" ? Why we get M/G - principal G-bundle, but not G/H bundle for some subgroup H ? why we can trivialize bundle ? Diff(G)^G = U(g) may be requires some simply-conectedness of G ? –  Alexander Chervov Oct 31 '11 at 9:50
What if G - is finite group ? –  Alexander Chervov Oct 31 '11 at 9:52
So far, $G$ could be anything. For Diff(G)^G = U(g) you don't need any assumptions on $G$ (other than Lie group/algebrac group). There is a theorem, that if a Lie group acts free and proper on a manifold, than the quotient is again a manifold. Furthermore $M\rightarrow M/G$ is locally trivial in this situation. A definition of proper action and proofs of such theorems can be found in most books on differential geometry. You can also find these things online, for example here: ma.huji.ac.il/~karshon/monograph/app-actions.ps If $G$ is finite and the action is free everything works! –  Jan Weidner Oct 31 '11 at 10:25
"Proper" usually means preimage of compact set is compact. Consider SO(3) acting by rotations on R^3/0. Action is free. And seems proper. But fiber is S^2, not SO(3). Seems contradiction ? –  Alexander Chervov Oct 31 '11 at 11:27
For finite group there is no Lie algebra, so your argument is not applied literally. –  Alexander Chervov Oct 31 '11 at 11:28
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