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?



What is the universal enveloping algebra?
I have now understood the situation better so my previous post has been replaced
by this.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.


