Let ${\mathfrak g}$ be a Lie algebra in a symmetric monoidal category enriched over $K$-vector spaces, i.e., in particular, hom-s are $K$-vector spaces (where $K$ is a field of characteristic zero). What is its universal enveloping algebra?

As one can talk about associative and Lie algebras there, I can imagine the definition in terms of the universal property but I am interested in its existence, a construction, if you may. Completing the category appropriately (direct sums and direct summands) could give familiar tensor and symmetric algebras $T({\mathfrak g})$ and $S({\mathfrak g})$ (i.e. they are objects in a certain completion of the original category). Is there a way to quotient $T({\mathfrak g})$ or to deform $S({\mathfrak g})$ at this point?

  • 3
    $\begingroup$ Exactly what kind of category are we talking about? If it is abelian you can construct the enveloping algebra as a quotient as in the usual case. If not how do you construct the symmetric algebra? $\endgroup$ May 17, 2010 at 16:14
  • 1
    $\begingroup$ No, not abelian, Torsten, just symmetric monoidal, with homs as vector spaces. To construct $S({\mathfrak g})$, one needs to complete by direct summands (or idempotents, maybe, Karoubian completion - my terminology is wonky). The symmetric group $S_n$ acts on the tensor power $T^n ({\mathfrak g})$, then its group algebra acts and $S^n ({\mathfrak g})$ is a direct summand of $T^n ({\mathfrak g})$ corresponding to the trivial idempotent... I will correct the question. $\endgroup$
    – Bugs Bunny
    May 17, 2010 at 16:54

4 Answers 4


I have now understood the situation better so my previous post has been replaced by this. (The only thing that was in the original but will not be here are some explicit formulas but Theo has given a reference for that.)

As I understand the question the poser wanted a construction of the enveloping algebra of a Lie algebra in a symmetric monoidal pseudoabelian (i.e., idempotents have kernels) $K$-category $\mathcal C$ with arbitrary sums over a field $K$ of characteristic zero. This means that for any $K[\Sigma_n]$-module $M$ and any object $V\in\mathcal C$ we can define $M\bigotimes_{\Sigma_n}V^{\otimes n}$ and for any $\Sigma$-module $M_\bullet$ (i.e., a collection $(M_n)$ of $K[\Sigma_n]$-modules) we can define $M(V):=\bigoplus_nM_n\bigotimes_{\Sigma_n}V^{\otimes n}$ which is an endofunctor of $\mathcal C$. Furthermore, a map $M \to N$ of $\Sigma$-modules gives a natural transformation of functors $M(V) \to N(V)$. In the particular case when $\mathcal C$ is the category of $K$-vector spaces such a natural transformation comes from a unique map of $\Sigma$-modules. The idea is to do what we know to do for $K$-vector spaces, interpret it as a set of natural transformations, get the corresponding maps of $\Sigma$-modules and use them to induce natural transformations for a general $\mathcal C$.

Let thus $S(V)$ be the symmetric algebra on the $K$-vector space $V$, $T(V)$ the tensor algebra on $V$ and $L(V)$ the free Lie algebra on $V$. Symmetrisation gives an isomorphism $S(L(V)) \to U(L(V))=T(V)$, where $U(-)$ is the enveloping algebra of Lie algebras. As $T(V)$ is a $T$-algebra (i.e., an associative algebra) and we can use this isomorphism to give $S(L(V))$ a $T$-algebra structure (i.e., a natural transformation $T(S(L(V))) \to S(L(V))$ fulfilling the appropriate conditions with respect to the monad structure on $T(-)$). Furthermore, if $\mathfrak g$ is a Lie algebra, then the $T$-algebra structure on $S(\mathfrak g)$ induced by the isomorphism $S(\mathfrak g) \to U(\mathfrak g)$ is given as the composite of $T(S(\mathfrak g)) \to T(S(L(\mathfrak g)))$ induced by the inclusion $\mathfrak g \to L(\mathfrak g)$, the map $T(S(L(\mathfrak g))) \to S(L(\mathfrak g))$ given by the $T$-module structure on $S(L(V))$ above and the map $S(L(\mathfrak g)) S(\mathfrak g)$ induced by the structure map $L(\mathfrak g) \to \mathfrak g$

Now, the functors $S(-)$, $L(-)$ and $T(-)$ are associated to $\Sigma$-modules which will be denoted by the same letters (instead of the standard $Com$, $Lie$ and $Ass$). Furthermore, composition of functors correspond to the plethysm $\circ$. Hence we get that $S\circ L$ is a $T$-module, i.e., we have a map $T\circ S\circ L \to S\circ L$ compatible with the operad structure on $T$. Consider now the case of a general $\mathcal C$. Each of $S$, $L$ and $T$ give endofunctors on $\mathcal C$ and $\circ$ again corresponds to composition. Let $\mathfrak g$ be a Lie algebra in $\mathcal C$ and define a $T$-algebra structure (i.e., the structure of associative algebra) on $S(\mathfrak g)$ as the composite $$ T(S(\mathfrak g)) \to T(S(L(\mathfrak g))) \to S(L(\mathfrak g)) \to S(\mathfrak g) $$ as above. The verification that this does indeed give a $T$-algebra structure is just a question of unwinding the definitions. The fact that $S$ is an operad gives us a natural transformation $V \to S(V)$ which applied to $\mathfrak g$ gives a morphism $\mathfrak g \to S(\mathfrak g)$ which we now want to show is a Lie algebra homomorphism. Here the Lie algebra structure on $S(\mathfrak g)$ is induced by its $T$-algebra structure and the operad map $L \to T$. Again unwinding definitions shows that it is indeed a Lie algebra morphism.

Finally assuming that $\mathfrak g \to A$ is a Lie algebra homomorphism where $A$ is an associative algebra with $L \to T$ inducing its Lie algebra structure. Note that we have an isomorphism (now going back to vector spaces) $S(L(V))\to T(V)$ and hence an isomorphism is $\Sigma$-modules $S\circ L=T$. This gives us a map $S(\mathfrak g)\to S(L(\mathfrak g))=T(\mathfrak g) \to T(A) \to A$ and it is easy to see that this is an algebra morphism.

  • $\begingroup$ Thank you very much. This is excellent! However, I do not see how one can reduce proving associativity and universality to vector spaces. Suppose we have defined a multiplication $\mu:S({\mathfrak g})\otimes S({\mathfrak g})\rightarrow S({\mathfrak g})$. Can we actually apply the functor $hom(I, )$ to this map? Is this functor tensor? $\endgroup$
    – Bugs Bunny
    May 18, 2010 at 9:44
  • $\begingroup$ This construction does give an associative algebra object in any category over $\mathbb Q$, by the Deligne-Morgan reference above. The trick is to read the formulas like the one Torsten gave above for $u\odot v \odot w$ not as formulas for actual elements (in which case it would be incorrect e.g. in super vector spaces), but as ("multilinear") maps in your category. This means, in particular, that you are not allowed to duplicate or delete variables in your formulas, a restriction that Baez calls "quantum". Once you restrict yourself, whatever symbolic argument you want to run you can. $\endgroup$ May 18, 2010 at 15:47
  • $\begingroup$ I am still thinking but still remain in the dark, at least in my head. I can write quite universal categories of the sort I am asking by generators and relations. They will have functors to abelian categories but how do I pull info back? Deligne-Morgan is helpful as a formula but the proof is still "abelian". Essentially, I think that one needs "diagrammatic" proof of associativity and universality, akin to Bar-Natan's proof of Duflo Conjecture... Ain't I a stinker? $\endgroup$
    – Bugs Bunny
    May 21, 2010 at 9:02
  • $\begingroup$ To celebrate one month's anniversary I am accepting this as an answer, although I still believe it is not general enough... $\endgroup$
    – Bugs Bunny
    Jun 15, 2010 at 10:44

As you say, given a symmetric monoidal category $\mathcal C$ enriched in abelian groups, the words "Lie algebra object in $\mathcal C$" and "associative algebra object in $\mathcal C$" make sense. (Actually, the latter does not depend on the symmetric structure nor the ab-gp enrichment.) In particular, there are natural categories $\text{LieAlg}_{\mathcal C}$ and $\text{AssocAlg}_{\mathcal C}$ — it makes sense to say whether an arrow in $\mathcal C$ between Lie/associative algebra objects is a homomorphism — and there is a natural "forgetful" functor from associative algebra objects to Lie algebra objects. If this functor has a left adjoint, said adjoint deserves to be called "free" or "universal enveloping" (but see below).

Of course, you are not guaranteed such an adjoint. For example, in the category of finite-dimensional vector spaces you cannot build (most) UEAs. You can see this very explicitly: working over characteristic $0$, the Lie algebra $\mathfrak{sl}(2)$ acts faithfully and transitively on representations of arbitrary dimension, and so $U(\mathfrak{sl}(2))$ cannot be finite-dimensional.

The minimum extra structure that I know of to guarantee the existence of a left-adjoint to $\text{Forget}: \text{AssocAlg}_{\mathcal C} \to \text{LieAlg}_{\mathcal C}$ is:

  1. Existence of arbitrary countable direct sums in $\mathcal C$.
  2. Existence of cokernels in $\mathcal C$.

If you have these, then you can do the usual construction to define $U\mathfrak g$.

If you are working in a category in which all hom sets are vector spaces over $\mathbb Q$, then you can also define $U\mathfrak g$ as a deformation of the symmetric algebra $S\mathfrak g$, provided this symmetric algebra exists. Namely, pretend for a moment that our category is just the usual category of $\mathbb K$-vector spaces for $\mathbb K$ a field of characteristic $0$. Then there is a "symmetrization" map $S\mathfrak g \to U\mathfrak g$ given on monomials by $x_1\cdots x_n \mapsto \frac1{n!} \sum_{\sigma \in S_n} x_{\sigma(1)}\cdots x_{\sigma(n)}$, where $S_n$ is the symmetric group in $n$ letters. This is a (filtered) vector space isomorphism (and also a coalgebra isomorphism, and also a $\mathfrak g$-module isomorphism), and so you can use it to pull back the algebra structure on $U\mathfrak g$ to one on $S\mathfrak g$, which you should think of as some sort of "star product".

So do this in $\mathbb K$-vector spaces, and then interpret the formulas on $\mathcal C$. For details, and in particular for an explicit formula for the star product in terms of the usual monomial basis on $S\mathfrak g$, see:

  • Deligne, Pierre; Morgan, John W. Notes on supersymmetry (following Joseph Bernstein). Quantum fields and strings: a course for mathematicians, Vol. 1, 2 (Princeton, NJ, 1996/1997), 41--97, Amer. Math. Soc., Providence, RI, 1999. MR1701597

But I see no conditions weaker than 1–2 above to guarantee the existence of the symmetric algebra.

Finally, I should mention that in general, even if $\text{Forget}$ has a left adjoint $U$, it does not necessarily deserve to be called the "universal enveloping algebra". Namely, simply by being an adjoint, there is a canonical Lie algebra map $\mathfrak g \to U\mathfrak g$. For $U\mathfrak g$ to "envelop" $\mathfrak g$, this map should be a monomorphism in $\mathcal C$.

The following example is due to:

  • Cohn, P. M. A remark on the Birkhoff-Witt theorem. J. London Math. Soc. 38 1963 197--203. MR0148717

Let $\mathbb K$ be a field of characteristic $p \neq 0$, and consider the free associative (noncommutative) algebra $\mathbb K \langle x,y\rangle$. Then $\Lambda_p(x,y) \overset{\rm def}= (x+y)^p - x^p - y^p$ is a non-zero Lie polynomial — it is a sum of compositions of brackets. For example, $\Lambda_2(x,y) = [x,y]$ and $\Lambda_3(x,y) = [x,[x,y]] + [y,[y,x]]$.

Let $R = \mathbb K[\alpha,\beta,\gamma]/(0 = \alpha^p = \beta^p = \gamma^p)$; it is a commutative ring. Let $\mathcal C = R\text{-mod}$ be the category of $R$-modules, with the usual symmetric tensor structure $\otimes_R$. Let $\mathfrak f_3$ be the free Lie algebra in $\mathcal C$, with the generators $x,y,z$, and let $\mathfrak g = \mathfrak f_3 / (\alpha x = \beta y + \gamma z)$.

Then $\Lambda_p(\beta y,\gamma z)$ is non-zero in $\mathfrak g$, but is $0$ in $U\mathfrak g$. Hence, internal to $\mathcal C$, $\mathfrak g$ does not embed into its universal enveloping algebra. (Of course, it does if we were just working over $\mathbb K$, as then the original PBW proof applies. And we always have an embedding in characteristic $0$, as there we can define $U\mathfrak g$ as a deformation of $S\mathfrak g$.)

  • 2
    $\begingroup$ I don't understand your comment about existence of the symmetric algebra. As far as I can see the sum over all n of the symmetric tensors in $V^{\otimes n}$ has a product given by the product in the tensor algebra followed by symmetrisation. This only requires that idempotents have kernels. On the other hand a formula for the star product can not use bases as it is supposed to make sense in an arbitrary symmetric monoidal category. $\endgroup$ May 17, 2010 at 20:27
  • $\begingroup$ Theo, my friend, I have no cokernels. How much easier would the life be if I had cokernels:-))! I doubt one can complete the category by adding cokernels either. Tell me if I am wrong here. Thanks for Deligne-Morgan reference. I will check it and report back... $\endgroup$
    – Bugs Bunny
    May 18, 2010 at 9:13
  • $\begingroup$ @Torsten Ekedahl: That's a good point. In characteristic 0, you win if idempotents have (co)kernels. And, no, the star product above does not use a basis; I was speaking imprecisely, because it's easiest to understand what the construction is doing if you had honest "monomials". $\endgroup$ May 18, 2010 at 15:37
  • 1
    $\begingroup$ @Bugs: Torsten's and my discussions I think prove that if you have access to the symmetric algebra, then you have the universal enveloping algebra, in characteristic 0. (In char=p, all hell breaks loose, but your question was in char=0.) So however you want to get your symmetric algebra, it gives you a UAE too. $\endgroup$ May 18, 2010 at 15:51

I missed this question but I still want to have my say as I think this deserves to be better known. Perhaps this would make a good topic for a blog post? This is, I think, Poincare's proof of a strong form of the PBW theorem. Birkoff and Witt proved a weaker result a couple of decades later. This story is told in this reference:

MR1793103 (2001f:01039) Ton-That, Tuong ; Tran, Thai-Duong . Poincaré's proof of the so-called Birkhoff-Witt theorem. Rev. Histoire Math. 5 (1999), no. 2, 249--284 (2000).

The set-up is given in Torsten's answer. We have a symmetric monoidal category enriched in the category of vector spaces over a field of characteristic zero. We also assume we can form countable direct sums and that idempotents have images (this is no loss of generality as the original category can be formaly enlarged if necessary). This gives the structure and there is a long list of compatibility conditions most of which should be obvious. I am not sure if we require the tensor product to be distributive over countable direct sums.

Just to be clear I do not assume we have cokernels and I have in mind examples where cokernels do not exist.

However let's start in the category of vector spaces. I was given a copy of notes taken at a talk by Kostant in France in the 1975. The only references I know of are the following (both of which make the construction seem obscure).

MR2301242 (2008d:17015) Durov, Nikolai ; Meljanac, Stjepan ; Samsarov, Andjelo ; Škoda, Zoran . A universal formula for representing Lie algebra generators as formal power series with coefficients in the Weyl algebra. J. Algebra 309 (2007), no. 1, 318--359.


MR1991464 (2004f:17026) Petracci, Emanuela . Universal representations of Lie algebras by coderivations. Bull. Sci. Math. 127 (2003), no. 5, 439--465.

Anyway the basic idea is that we take the symmetric algebra $S(g)$ and define an action of $g$. This then generates the action of $U(g)$ and so constructs $U(g)$. In order to define the action of $g$ it is sufficient to define $x*y^n$ since we obtain the action of $x$ by polarisation. The key is that this is given by:

$$x*y^n = \sum_{j=0}^n \binom{n}{j}B_j ad^j(y)(x)y^{n-j}$$

where the $B_j$ are the Bernoulli numbers with generating function $x/(e^x-1)$.

Once you unwind this you find that you have constructed maps $S^r(g)\otimes S^s(g)\rightarrow S^{r+s-j}(g)$ for $r,s,j\ge 0$ by universal formulae. These formulae then make sense in the abstract setting.

I would be delighted to see a good exposition of this.

Edit: The following reference looks as though it should be relevant but I didn't get much from it.

MR1894038 (2003b:17014) Cortiñas, Guillermo . An explicit formula for PBW quantization. Comm. Algebra 30 (2002), no. 4, 1705--1713.


This is just a short abstract complement to Theo's and Torsten's detailed constructions: if O is any operad in a symmetric monoidal category C with conditions 1 and 2 from Theo's answer, so that tensor algebra is defined (for example, O is $Lie_k$ or $Com_k$, C is $Vec_k$) then "universal enveloping algebra of an algebra over C" exists and can be constructed analogously to U(g). It has the property that the category of L-modules (where L is a C-algebra) is equivalent to the category of U(L)-modules. If my memory serves, this is explained in Ginzburg and Kapranov, Koszul duality for operads.

  • $\begingroup$ That is true but it seems that the general construction of the enveloping algebra requires cokernels and we are not allowed to assume that. $\endgroup$ May 17, 2010 at 20:23
  • $\begingroup$ Why not? Universal enveloping algebra is a quotient of tensor algebra, and in order to be able to form quotients, cokernels are needed. That seems like the natural setting (and is covered by Theo's condition 2). $\endgroup$ May 17, 2010 at 21:53
  • 2
    $\begingroup$ But the posed problem assumed that we wouldn't necessarily have cokernels. I get the impression that that was the whole point of the question and I think I have managed to avoid them in my answer. $\endgroup$ May 17, 2010 at 22:04
  • 2
    $\begingroup$ The category of projective (or flat) modules over a commutative ring with tensor product as monoidal operation. $\endgroup$ May 19, 2010 at 8:13
  • 2
    $\begingroup$ I am by the way not quite convinced that the envelopping algebra can be constructed in the way I suggested for a general operad $O$. What I use is that $U(Lie(V)) = S(Lie(V))$, i.e., that there is a $\Sigma$-module $S$ such that the enveloping algebra of a free algebra is isomorphic to $S$ applied to the free algebra. I don't know if that is true in general. $\endgroup$ May 19, 2010 at 8:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

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