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This is another naive question that popped up as I'm learning Lie theory.

Given a lie algebra $L$ over a field $k$ of characteristic 0, we can define its universal algebra $UL$ as a suitable quotient of the tensor algebra on the vector space $L$.

This universal algebra $UL$ carries a diagonal map $\Delta : UL\rightarrow UL\otimes_k UL$ which is induced from the usual diagonal map $L\rightarrow L\times L$ given by $x\mapsto (x,x)$, together with the isomorphism $UL\otimes_k UL\cong U(L\times L)$ (an isomorphism which is surprisingly nonobvious, as it seems to use in an essential way the Poincare-Birkhoff-Witt theorem).

In our setting over a field $k$ of characteristic 0, $L$ can be identified with the subspace of $UL$ on which $\Delta(x) = x\otimes 1 + 1\otimes x $.

In general, given any associative algebra $A$, and any map $\Delta : A\rightarrow A\otimes_k A$, the subspace of elements on which $\Delta(x) = x\otimes 1 + 1\otimes x$ is closed under the bracket $[x,y] = xy-yx$, and hence is a Lie subalgebra $L_\Delta$, though I don't see any reason for $A$ to be the universal algebra of $L_\Delta$.

My question is - in general, if $A$ is the universal algebra of some Lie algebra $L$, giving rise to a map $\Delta : A\rightarrow A\otimes_k A$ such that $L = L_\Delta$, then can there exist another $\Delta' : A\rightarrow A\otimes_k A$ such that $A$ is also the universal algebra of $L_{\Delta'}$ and such that $\Delta'$ is the usual diagonal map associated with the identification $A = UL_{\Delta'}$.

Secondly, is every associative algebra the universal algebra of some lie algebra?

EDIT: This second question is certainly false, thanks to Gro-Tsen's comment.

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    $\begingroup$ "Is every associative algebra the universal algebra of some lie algebra?" → Since a Lie algebra embeds in its universal enveloping algebra (henceafter "uea"), a commutative algebra could only be the uea of a zero Lie algebra; but the uea of a zero Lie algebra is a symmetric algebra, i.e., polynomial ring, and not every commutative algebra is a polynomial ring. So, no. $\endgroup$
    – Gro-Tsen
    Commented Sep 3, 2017 at 18:21
  • $\begingroup$ What properties do you want your comultiplication map to have? And do you want to consider the antipode map at all? $\endgroup$
    – user44191
    Commented Sep 3, 2017 at 18:34
  • $\begingroup$ @user44191 I suppose I want the comultiplication map to have any property that the diagonal map coming from a universal algebra has. (Edited now to reflect this). I'm a novice to this area so actually I'm not really familiar with its properties. $\endgroup$
    – Will Chen
    Commented Sep 3, 2017 at 18:39
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    $\begingroup$ You don't need PBW to argue that taking universal enveloping algebras sends direct sums to tensor products. You can argue via universal properties: direct sum and tensor product are both the "commutative coproduct," in that they are the universal thing admitting a map from two other things whose images commute (and the meaning of "commute" in Lie algebras and algebras is compatible). $\endgroup$
    – Qiaochu Yuan
    Commented Sep 3, 2017 at 20:58
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    $\begingroup$ Anyway, for the sake of having the word in this conversation: the structure $U(L)$ acquires from being a universal enveloping algebra is a Hopf algebra structure. The functor $L \mapsto U(L)$, regarded as taking values in Hopf algebras, is fully faithful. Your question is whether it reflects isomorphisms when regarded as taking values in algebras. $\endgroup$
    – Qiaochu Yuan
    Commented Sep 3, 2017 at 21:01

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