Let $\mathfrak g$ be a finite-dimensional Lie algebra over $\mathbb C$. Define $\mathcal Z(\mathfrak g)$ to be the center of the universal enveloping algebra $\mathcal U\mathfrak g$, and define $(\mathcal S\mathfrak g)^{\mathfrak g}$ to be the ring of invariant elements of the symmetric algebra $\mathcal S\mathfrak g$ under the induced adjoint action of $\mathfrak g$. (Clearly $\mathcal Z(\mathfrak g) = (\mathcal U\mathfrak g)^{\mathfrak g}$ via the adjoint action.) The Duflo isomorphism is an isomorphism of algebras $\mathcal Z(\mathfrak g) \cong (\mathcal S\mathfrak g)^{\mathfrak g}$. At the level of vector spaces, the trick is to realize that the PBW map $\mathcal U\mathfrak g \to \mathcal S \mathfrak g$ is an isomorphism of $\mathfrak g$-modules. For the isomorphism of algebras in the semisimple case, see for example my unedited notes on the class by V. Serganova.

(I read here that this isomorphism can be realized as a composition of the PBW vector-space isomorphism $\mathcal U\mathfrak g \to \mathcal S\mathfrak g$ with the map $\mathcal S\mathfrak g \to \mathcal S\mathfrak g$ given by $x \mapsto \sinh(x/2)/(x/2)$. But it's not at all obvious that this composition is even well-defined or linear when restricted to $\mathcal Z(\mathfrak g)$. I should mention that $\mathcal Z$ is not a functor, I think. The PBW isomorphism is non-canonical, although a canonical version can be given via the symmetrization map, and I guess on the center it is canonical.)

When $\mathfrak g$ is semisimple of rank $n$, at least, one can further show that $(\mathcal S\mathfrak g)^{\mathfrak g} \cong \mathbb C[x_1,\dots,x_n]$, although you have some choice about how to make this isomorphism. Thus, at least when $\mathfrak g$ is semisimple, $\mathcal Z(\mathfrak g)$ is a polynomial ring.

But any polynomial ring can be given a Hopf structure. By choosing an isomorphism with $\mathbb C[x_1,\dots,x_n]$, we can take the Hopf structure generated by $\Delta: x_i \mapsto x_i \otimes 1 + 1 \otimes x_i$. In fact, this structure doesn't depend quite on the full choice of isomorphism. A Hopf structure on a commutative algebra $R$ is by definition the same as an algebraic group structure on $\text{Spec}(R)$. But $\text{Spec}(\mathbb C[x_1,\dots,x_n])$ is $n$-dimensional affine space — the algebra isomorphisms of $\mathbb C[x_1,\dots,x_n])$ are precisely the affine maps — so picking a commutative group structure is the same as picking an origin. (For certain values of $n$ there are also non-commutative group structures on affine $n$-space, and so non-cocommutative Hopf structures on the polynomial ring. For example, the group of upper-triangular matrices with $1$s on the diagonal is affine.)

My question is whether this Hopf structure can be picked out canonically.

Question: If $\mathfrak g$ is a finite-dimensional Lie algebra over $\mathbb C$, can the center $\mathcal Z(\mathfrak g)$ of the universal enveloping algebra be given a canonical (cocommutative) Hopf algebra structure? If no, how much extra structure on $\mathfrak g$ is needed?

Here by "canonical" I of course don't mean that there is a unique one, so you may make choices once and for all. But there should be some definition/construction that does not require the user to make any choices to implement it. By "extra structure" I mean either extra structure (an invariant metric, for example) or extra properties (semisimplicity, for example).

My suspicion is that the answer is "yes" for a metric Lie algebra, which is a Lie algebra $\mathfrak g$ along with a choice of an invariant nondegenerate metric, i.e. a chosen isomorphism of $\mathfrak g$-modules $\mathfrak g \cong \mathfrak g^*$. Metric Lie algebras include the semisimples and the abelians, and certain extensions of these (in fact, I believe that there is a structure theorem that any metric Lie algebra is a metric extension of semisimples and abelians, but don't quote me), but generally there are many choices of metric (e.g. any metric on an abelian Lie algebra $\mathfrak a$ is invariant, so there are $\mathfrak{gl}(\dim \mathfrak a)$ many choices).

The motivation for my question is this: by studying Vassiliev invariant and/or perturbative Chern-Simons theory, Bar Natan and others have defined a certain commutative and cocommutative Hopf algebra $A$ of "diagrams". Any choice of metric Lie algebra $\mathfrak g$ determines an algebra homomorphism $A \to \mathcal Z(\mathfrak g)$. I would like to know if this can be made into a Hopf algebra homomorphism.

  • $\begingroup$ Just to comment that the structure theorem for metric Lie algebras to which you refer is due to Alberto Medina and Philippe Revoy and it says that the class of finite-dimensional metric Lie algebras is generated by the simple and one-dimensional Lie algebras under two operations: orthogonal direct sum and "double extension" -- a procedure defined in their paper numdam.org/item?id=ASENS_1985_4_18_3_553_0 (in French). $\endgroup$ Commented Feb 7, 2010 at 1:33
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    $\begingroup$ Was "commutative and commutative" supposed to be "commutative and cocommutative"? $\endgroup$
    – Ben Webster
    Commented Feb 7, 2010 at 1:51
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    $\begingroup$ I just have a brief remark of what you say in brackets about the Duflo isomorphism. The actual Duflo map is a bit different from your guess; see, e.g. the introduction to arxiv.org/abs/math/0506499 for a precise formulation. Also, remarkably it turns out that for metric Lie algebras the Duflo theorem is in a sense trivial - see arxiv.org/abs/0909.3743 (where the term "quadratic" is a synonym for "metric"). Sorry, these were random thoughts, nothing to say about the main question. $\endgroup$ Commented Feb 7, 2010 at 2:44
  • $\begingroup$ @Ben: yes, fixed. $\endgroup$ Commented Feb 7, 2010 at 17:46
  • $\begingroup$ @Theo, if you don’t mind, may I ask a silly question? You mean that when $\mathfrak{g}$ is semisimple, the Hopf algebra structure of $Z(\mathfrak{g})$ induced from the polynomial ring is canonical regardless of the choice of isomorphism? If so, I am wondering that for $X \in Z(\mathfrak{g})$, if $\Delta(X)=\sum X_i’ \otimes X_i’’$ for $X_i’, X_i’’ \in Z(\mathfrak{g})$, then $X(fg)=\sum X_i’(f) X_i’’(g)$ for $f,g \in C^\infty(G)$? $\endgroup$
    – Monty
    Commented Apr 16, 2021 at 14:38

2 Answers 2


In the semisimple case, one has the Harish-Chandra isomorphism between the center ${\mathcal Z}(\mathfrak g)$ and $(S{\mathfrak h})^W$, where ${\mathfrak h}$ is a Cartan subalgebra of ${\mathfrak g}$ and $W$ is the Weyl group. On $S{\mathfrak h}$, there is a natural inner product $(f,g)=f(\partial)g(x)|_{x=0}$ induced by the natural inner product on ${\mathfrak h}$.

Let $I\subset (S{\mathfrak h})^W$ be the augmentation ideal. Let $E$ be the orthogonal complement of the ideal $I^2$ in $(S{\mathfrak h})^W$ (the square of $I$), and let $E_+$ be the positive degree part of $E$. Then $(S{\mathfrak h})^W=SE_+$, so we can define the Hopf algebra structure on $(S{\mathfrak h})^W$ by declaring that $E_+$ consists of primitive elements.

This was explained to me by Kostant. I don't know, however, if it is sufficiently canonical, or helpful in the question related to the Bar-Natan construction.

  • $\begingroup$ So this is great provided I've picked a Cartan subalgebra. And perhaps it doesn't actually depend on that choice, but it's not obvious to me that it does not? $\endgroup$ Commented Feb 7, 2010 at 17:48
  • $\begingroup$ No, it does not depend on the choice. Two choices differ by an element $g$ of the group $G$, so the two primitive subspaces in the center ${\mathcal Z}({\mathfrak g})$ you'll get from the two choices will be conjugate to each other by $g$. But $G$ acts trivially on the center, so they will be the same. $\endgroup$ Commented Feb 7, 2010 at 19:09
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    $\begingroup$ Pasha, I am actually confused by what you say after thinking it over. If we take the orthogonal complement of I^2, would not it contain all polynomials of degree 1 (not all of which are invariant)? I miss something obvious, I am sure, but I need someone to enlighten me. $\endgroup$ Commented Feb 8, 2010 at 22:55
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    $\begingroup$ Sorry, there was a misprint - the superscript $W$ was missing in 4th line in 2 places. I have now corrected this. Thanks! $\endgroup$ Commented Feb 9, 2010 at 1:34
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    $\begingroup$ I am not sure since in general I don't know how to construct a nondegenerate inner product on $(Sg)^g$ (I don't see why the restriction of the inner product from $Sg$ should be nondegenerate if $g$ is not reductive). $\endgroup$ Commented Apr 3, 2010 at 14:21

I would be very surprised if the answer is YES, in general. This would turn the spectrum of Z(g) into an algebraic group, and, in particular, force it to be smooth (if Z(g) is finitely generated). Quick search through Dixmier's book gives a reference to his 1957 paper where he computes Z(g) for nilpotent Lie algebras of dimension up to 5. All of them are polynomial except Z(g_{5,5}). Good luck checking that SPEC(Z(g_{5,5})) is not a 3-dimensional algebraic group.


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