# Analogue Bialgeras vs Lie bialgebras

I once thought that the analogue of bialgebras and Lie bialegras is similar to that of (associative) algebras and Lie algebras, but it seems not that trivial.

Recall the definitions: a) bialgebra $A$ is a algebra $A$ with a comultiplication $\delta: A \to A\otimes A$ such that $\delta$ is coassociative and a morphism of algebras. b)Lie bialgebra $\mathfrak{g}$ is a Lie algebra $\mathfrak{g}$ with a cobracket $\Delta:\mathfrak{g} \to \mathfrak{g}\otimes\mathfrak{g}$ such that $\Delta$ subjects to co-Jacobi identity and $\Delta$ is a cocycle, where the action of $\mathfrak{g}$ on $\mathfrak{g}\otimes \mathfrak{g}$ is by adjoints.

Naïvely, we may expect the cobracket $\Delta$ to be a Lie algebra morphism but not a cocycle. Why so? This is the first part of my question, the other part: Is it possible to build a Lie bialgebra out of a bialgebra via alternating? Thanks in advance.

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There is of course a much more profound "analogy" between bialgebras and Lie bialgebras: the quantization and dequantization functors of Etingof and Kazhdan. From this point of view, Lie bialgebras appear as first order terms of a deformation theory which ultimately assigns to every Lie bialgebra a bialgebra by deformation (quantization). If I recall correctly, the construction uses a Drinfel'd associator, so it probably works only in characteristic zero or evenover the complex numbers. You can find this with many details in a sequel of papers (4?).

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I wanted only to add that while Drinfeld associators were originally defined somewhat analytically, and so seem to be transcendental and dependent on C, it is known that they may be also defined over Q. See, eg, impan.pl/~burgunde/WSBC09/Drinfeld_Associator_Brochier.pdf arxiv.org/pdf/1002.2331 and references therein. So probably just char zero suffices for EK quantization. – David Jordan Mar 21 '11 at 0:39
@David Yes, the first associators from the KZ equations where transcendental and it took quite some effort (Drinfel'd I think) to show that there are rational ones as well. Unfortunately, this is just an abstract existence proof, so no explicit rational associators are known. Still, you're right. It should work as soon as your field (or even ring) contains $\mathbb{Q}$... – Stefan Waldmann Mar 21 '11 at 8:06

Lie bialgebras and bialgebras cannot be analogous. The tensor product of two Lie algebras is not a Lie algebra. We can check this, let $L_1$ and $L_2$ be two Lie algebras, so for $a,b,c \in L_1$ and $x,y,z \in L_2$ we have $$[a,[b,c]] = [[a,b], c] + (-1)^{|a||b|} [b, [a,c]]$$ $$[x, [y,z]] = [[x,y], z] + (-1)^{|x||y|} [y, [x,z]].$$ We use this version of the Jacobi identity-that $[,]$ is a derivation of itself- to keep track of signs, although it won't be that important.

So we can try to define a Lie bracket on $L_1 \otimes L_2$ by $[a\otimes b , x \otimes y]= [a,b] \otimes [x,y].$ But because the Jacobi identity has three terms (and not two like in the associative identity), we get

$$[a \otimes x, [b \otimes y, c \otimes z]] = [a \otimes x , [b,c] \otimes [y,z]]$$ $$= [a, [b,c] \otimes [x,[y,z]]$$ $$= \left([[a,b], c] + (-1)^{|a||b|} [b, [a,c]] \right) \otimes \left( [[x,y], z] + (-1)^{|x||y|} [y, [x,z]] \right).$$

When we expand this out, we get four terms, and not two terms as needed. So $L \otimes L$ is not a Lie algebra, and we can't ask $\Delta$ to be a Lie algebra map.

Edit: There seems to be some related notions. One could ask that $\Delta: L \rightarrow L \otimes L$ be a map of Lie modules. The analogous thing with associative algebras is an open (non-commutative) Frobenius algebra. This is an algebra $A$ with multiplication and comultiplication $\Delta: A \rightarrow A \otimes A$, such that $\Delta$ is a map of $A$ modules.

One could also ask that $\Delta: L \rightarrow L \otimes L$ be a derivation of $L$ with values in the bimodule $L \otimes L$. This is the cocycle condition. The analogous thing for associative algebras is called an infinitesimal bialgebra. Aguiar discusses these objects in the paper "On the associative analog of Lie bialgebras." The idea of symmetrizing the associative product of an infinitesimal bialgebra to obtain a Lie bialgebra is discussed (this process does not always yield a Lie bialgebra).

In "Skein quantization of Poisson algebras of loops on surfaces," Turaev describes a Lie bialgebra. He also discusses a Hopf algebra which quantizes the Lie bialgebra. Since a Hopf algebra is a bialgebra with an antipode, maybe there is a connection to be said between Lie bialgebras and bialgebras.

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It's not clear to me that there aren't other brackets you could try on the tensor product. If I wanted the tensor product of $\mathfrak{gl}(n)$ and $\mathfrak{gl}(m)$ to naturally be a subalgebra of $\mathfrak{gl}(nm)$, then this isn't the bracket I would use. – Qiaochu Yuan Mar 14 '11 at 19:10
You're right, there might be another bracket. But since the definition of a bialgebra used the fact that $A \otimes A$ is an associative algebra with naive combination of products, we would try to do the same thing with a Lie version of a bialgebra, if such a thing could exist. – Micah Miller Mar 14 '11 at 21:01
cont'd: I'm not sure how you would define another kind of bracket, if you start with two abstract Lie algebras. Perhaps, with certain examples like $gl(n)$, you can use extra knowledge about the Lie algebra to try something different. But if all you have is $L_i \otimes L_i \rightarrow L_i$, for $i=1,2$, then what else can you do with $(L_1 \otimes L_2)^{\otimes 2} \rightarrow L_1 \otimes L_2$ besides combining the brackets? – Micah Miller Mar 15 '11 at 2:43
@Qiaochu: One way to say Micah's point is that the category of Lie algebras does not have a functorial monoidal structure covering tensor product of underlying vector spaces. This is equivalent to saying that the operad Lie is not an operad of coalgebras. You can see this immediately: Lie(2) is one-dimensional, and the symmetric group acts by the sign representation. But the sign permutation does not have a nontrivial map to its tensor square. So just at the level of the anticommutativity you're hosed. – Theo Johnson-Freyd Mar 15 '11 at 14:18
@Theo: that's a better way to put it. @Qiaochu: Let $A$ be an associative algebra and $[A]$ be the Lie algebra obtained by symmetrizing the product. Then $A \otimes A$ is an algebra, and $[A \otimes A]$ is a Lie algebra obtained by symmetrizing the product on $A \otimes A$. This is note the same as the bracket on $[A \otimes A]$, which won't satisfy Jacobi. I think this would cover your example. – Micah Miller Mar 15 '11 at 20:04