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[x,[y,z]] = [[x,y],z] + [y,[x,z]]   for all x,y,z in L

Remark. What is being defined above is a left Leibniz algebra. There is also the notion of a right Leibniz algebra where the Leibniz identity now says that it is the right multiplication [-,x] which is a derivation, instead of the left multiplication [x,-] as in the equation above.

Since the bracket is no longer alternating, left- and right-multiplications are no longer related simply by a sign as in the case of Lie algebras, and this means that representations in general admit two actions of L: one on the left and one on the right, satisfying some identities which are explained, for example, in a paper of Loday (who seems to have introduced the concept) and Pirashvili (Math. Ann. 296 (1993) pp. 139–158). In that paper they also define the universal enveloping algebra U(L) of a Leibniz algebra L and show that the there is a categorical equivalence between representations of L and left modules over U(L). (Right modules of U(L) correspond to the notion of corepresentation.) Also notice that in that paper they work with right Leibniz algebras, so everything there is the mirror image to what I'm saying here. One difference with the case of a Lie algebra is that U(L) is a quotient of the tensor algebra of L\oplus L, to take into account the two actions of L on a representation.

My question is whether there is a Hopf algebra structure on U(L).

My interest in this question is that in some recent work on the deformation theory of n-Leibniz algebras, I studied cohomology with values in a representation M of a Leibniz algebra L and also with values on End(M). The action of L on End(M) follows from the formalism and one can check that it is indeed a representation, but it does not follow in any obvious way from the action of L on M. In Lie theory, we are used to the fact that if M is a (finite-dimensional) representation of a Lie algebra G, then we have an isomorphism End(M) = M \otimes M^* as representations of G, where to determine the action of G on M \otimes M^* we use the Hopf algebra structure on U(G). Hence my question.

EDIT: I am adding more details about U(L), as requested in the comment below by Theo Johnson-Freyd.

To motivate it, let us first define a representation M of a (left) Leibniz algebra L to be a vector space admitting two actions of L:

  (x,m) \mapsto [x,m]  and  (m,x) \mapsto [m,x] for all m in M and x in L

satisfying three identities, which are obtained from the Leibniz identity above by replacing x,y,z in turn by m; that is,

           [m,[x,y]] = [[m,x],y] + [x,[m,y]]

           [x,[m,y]] = [[x,m],y] + [m,[x,y]]

           [x,[y,m]] = [[x,y],m] + [y,[x,m]]

To define U(L) we start with the tensor algebra T(L+L) of L \oplus L. Let l_x = (x,0) and r_x = (0,x) in L \oplus L. Then U(L) is the quotient of T(L+L) by the two-sided ideal generated by the following elements (which can be read off from the conditions defining a representation):

r_[x,y] - r_y r_x - l_x r_y

l_x r_y - r_y l_x - r_[x,y]

l_x l_y - l_y l_x - l_[x,y]

for all x and y in L, and where I have omitted the \otimes's.

Notice that adding the first two, we can substitute one of them by the simpler

r_y (l_x + r_x) = 0

$$ [x,[y,z]] = [[x,y],z] + [y,[x,z]] $$ for all $x,y,z \in L$

Remark. What is being defined above is a left Leibniz algebra. There is also the notion of a right Leibniz algebra where the Leibniz identity now says that it is the right multiplication $[-,x]$ which is a derivation, instead of the left multiplication $[x,-]$ as in the equation above.

Since the bracket is no longer alternating, left- and right-multiplications are no longer related simply by a sign as in the case of Lie algebras, and this means that representations in general admit two actions of $L$: one on the left and one on the right, satisfying some identities which are explained, for example, in a paper of Loday (who seems to have introduced the concept) and Pirashvili (Math. Ann. 296 (1993) pp. 139–158). In that paper they also define the universal enveloping algebra $U(L)$ of a Leibniz algebra $L$ and show that the there is a categorical equivalence between representations of $L$ and left modules over $U(L)$. (Right modules of $U(L)$ correspond to the notion of corepresentation.) Also notice that in that paper they work with right Leibniz algebras, so everything there is the mirror image to what I'm saying here. One difference with the case of a Lie algebra is that $U(L)$ is a quotient of the tensor algebra of $L\oplus L$, to take into account the two actions of $L$ on a representation.

My question is whether there is a Hopf algebra structure on $U(L)$.

My interest in this question is that in some recent work on the deformation theory of n-Leibniz algebras, I studied cohomology with values in a representation $M $of a Leibniz algebra L and also with values on $End(M)$. The action of $L$ on $End(M)$ follows from the formalism and one can check that it is indeed a representation, but it does not follow in any obvious way from the action of $L$ on $M$. In Lie theory, we are used to the fact that if $M$ is a (finite-dimensional) representation of a Lie algebra $G$, then we have an isomorphism $End(M) \cong M \otimes M^*$ as representations of $G$, where to determine the action of $G$ on $M \otimes M^*$ we use the Hopf algebra structure on $U(G)$. Hence my question.

EDIT: I am adding more details about $U(L)$, as requested in the comment below by Theo Johnson-Freyd.

To motivate it, let us first define a representation $M$ of a (left) Leibniz algebra $L$ to be a vector space admitting two actions of $L$: $$ (x,m) \mapsto [x,m] \textrm{ and } (m,x) \mapsto [m,x], \forall m \in M \textrm{ and } x \in L $$

satisfying three identities, which are obtained from the Leibniz identity above by replacing $x,y,z$ in turn by $m$; that is, $$ [m,[x,y]] = [[m,x],y] + [x,[m,y]] \\ [x,[m,y]] = [[x,m],y] + [m,[x,y]] \\ [x,[y,m]] = [[x,y],m] + [y,[x,m]] $$

To define $U(L)$ we start with the tensor algebra $T(L\oplus L)$ of $L \oplus L$. Let $l_x = (x,0)$ and $r_x = (0,x) \in L \oplus L$. Then $U(L)$ is the quotient of $T(L+L)$ by the two-sided ideal generated by the following elements (which can be read off from the conditions defining a representation): $$ r_{[x,y]} - r_y r_x - l_x r_y \\ l_x r_y - r_y l_x - r_{[x,y]} \\ l_x l_y - l_y l_x - l_{[x,y]} $$ for all $x, y \in L$, and where I have omitted the $\otimes$'s.

Notice that adding the first two, we can substitute one of them by the simpler $$ r_y (l_x + r_x) = 0 $$

EDIT: I am adding more details about U(L), as requested in the comment below by Theo Johnson-Freyd.

To motivate it, let us first define a representation M of a (left) Leibniz algebra L to be a vector space admitting two actions of L:

  (x,m) \mapsto [x,m]  and  (m,x) \mapsto [m,x] for all m in M and x in L

satisfying three identities, which are obtained from the Leibniz identity above by replacing x,y,z in turn by m; that is,

           [m,[x,y]] = [[m,x],y] + [x,[m,y]]

           [x,[m,y]] = [[x,m],y] + [m,[x,y]]

           [x,[y,m]] = [[x,y],m] + [y,[x,m]]

To define U(L) we start with the tensor algebra T(L+L) of L \oplus L. Let l_x = (x,0) and r_x = (0,x) in L \oplus L. Then U(L) is the quotient of T(L+L) by the two-sided ideal generated by the following elements (which can be read off from the conditions defining a representation):

r_[x,y] - r_y r_x - l_x r_y

l_x r_y - r_y l_x - r_[x,y]

l_x l_y - l_y l_x - l_[x,y]

for all x and y in L, and where I have omitted the \otimes's.

Notice that adding the first two, we can substitute one of them by the simpler

r_y (l_x + r_x) = 0

I don't know what the coalgebra structure is, though. That's part of the original question.