Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The Duflo map is the map S(g) -> U(g), which known to satisfy the following properties:

1) identity on g

2) isomorphism of g-modules (and in particular vector spaces)

3) restricted to Poisson center on S(g) it is ISOMORPHISM of commutative algebras S(g)^g to ZU(g) (the center of U(g)). (This is highly non-trivial property). It predicts that the centers on the "classical" and "quantum" level are the same. (Kontesevich generalized to arbitrary Poisson variety).

The question: Is it the only map satisfying such properties ? (At least for semisimple Lie algebras) ?

(I think answer is YES, and it should be known, but I have not seen the reference).

========= Example let g-commutative, then it is true:

Since S(g)=U(g) and since the map is required to be identity on g and it is homomorphism, so it is identity map on the S(g).

===

If you read this question I guess you know what S(g) and U(g) means :) But may be nevertheless to keep good spirit of MO it is more polite to include definitions:

S(g) - means commutative algebra - symmetric algebra of g (i.e. just take some basis xi in g and consider polynomial algebra C[x1 ... xn] - this is S(g) as a vector space. Lie algebra g acts on - it simple way - it acts on xi by adjoint action, and continued further by Leibniz rule).

U(g) - means NON-commutative algebra - universal enveloping algebra of g - which is: you take basis xi and consider non-commutative polynomials C[xi] where generators satisfying the relations: [xi, xj] = C_ij^k x_k , where C_ij^k - are structure constants of g.

The Duflo map is for example discussed here:

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

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

=========

Kontsevich mentions that in general situation (of Poisson manifolds) Grothendieck-Teichmuller groups should act on quantizations and in particular on "Duflo" type maps, but he writes that for semisimple algebras this action is trivial. So we should not have problem from this side. Moreover as far as I understand since nobody seen non-trivial example of this action it might be always like this.

share|improve this question
2  
+1 for being polite and keeping the "spirit of MO". –  Theo Johnson-Freyd Nov 4 '11 at 17:16
add comment

2 Answers

up vote 6 down vote accepted

Choose a map $\varphi$ satisfying these properties and make the difference $\psi=\varphi^{-1}\varphi_D$ with the Duflo map. Then $\psi$ is an automorphism of the $\mathfrak g$-module $S(\mathfrak g)$ which is the identity on $\mathfrak g$ and is multiplicative on invariants.

If you want this map to be universal (namely it should only involve universal formulae in terms of the Lie bracket) then it is very likely to be unique.

But if you want the statement to be true independantly for every single Lie algebra, I would believe the answer is "no". Namely, consider the $2$-dimensional solvable Lie algebra $\mathfrak g:={\bf k}x\oplus {\bf k}y$ with $[x,y]=y$. Now we have that $S(\mathfrak g)={\bf k}[x,y]$ and that $S(\mathfrak g)^{\mathfrak g}={\bf k}$. Therefore any non-trivial automorphism of the $\mathfrak g$-module $S^{\geq2}(\mathfrak g)$ (e.g. a non-trivial multiple of the identity) gives a counter-example. .

share|improve this answer
1  
Aside: I would call the 2-dimensional Lie algebra "solvable", not "nilpotent", because $[x,-]$ is not acting nilpotently. –  Theo Johnson-Freyd Nov 4 '11 at 17:22
    
God, it's a shame. I have fixed it. Thank you. –  DamienC Nov 4 '11 at 22:15
    
@Damien Thank You very much ! What about semisimple case ? –  Alexander Chervov Nov 5 '11 at 18:14
    
I don't know. I'll think about it. –  DamienC Nov 7 '11 at 7:52
    
If you take $\mathfrak{sl}_2$ then invariant polynomials form a polynomial algebra in one variable $k[x]$, where $x=\sigma_2=tr(ad^2)$. In particular there are no invariant polynomials of odd degree. So you have a lot of flexibility. You could e.g. take the identity map on $S^k(\mathfrak{g})$ for $k\neq 3$ and multiply by $2$ on $S^3(\mathfrak{g})$. This would give you a $\psi$ which is not the identity. –  DamienC Nov 24 '11 at 16:28
add comment

Well, a look at These notes of Calaque and Rossi would suggest (Remark 1.3) that the answer is no: $e^{\mathrm{tr}(ad)}$ is an automorphism of $S(\mathfrak{g})$, precomposing with this automorphism you get a different Duflo isomorphism. They call it modified Duflo element. We just need to check that the automorphism is the identity on $\mathfrak{g}$, which seems to be true if $\mathfrak{g}$ is unimodular (in particular if $\mathfrak{g}$ is semisimple).

Edit: oops, the notes I'm attaching are also mentioned in your post, so I might be wrong cause you obviously read them. Ahh I see, it seems that the whole $e^{tr(ad)}$ is the identity in the unimodular case.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.