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Perhaps this will be a trivial question. For this post, everything is over your favorite field of characteristic $0$.

Definitions and notation

Recall that a Lie algebra is a vector space $\mathfrak g$ along with a map $\beta: \mathfrak g^{\wedge 2} \to \mathfrak g$ satisfying the Jacobi identity. One way to write the Jacobi identity is as follows: extend $\beta$ to $\mathfrak g^{\otimes 2} \to \mathfrak g$ via the usual projection $\mathfrak g^{\otimes 2} \to \mathfrak g^{\wedge 2}$, consider the map $\beta \circ (1 \otimes \beta): \mathfrak g^{\otimes 3} \to \mathfrak g$; then the restriction of this map to $\mathfrak g^{\wedge 3} \subseteq \mathfrak g^{\otimes 3}$ vanishes. (Because $\beta$ vanishes on the symmetric product $\mathfrak g^{\vee 2}$, the Jacobi identity is equivalent to $\beta \circ (1 \otimes \beta)$ vanishing on $\mathfrak g^{\vee 3}$.)

A Lie coalgebra is a vector space $\mathfrak g$ with a map $\delta: \mathfrak g \to \mathfrak g^{\wedge 2}$, satisfying the coJacobi identity, which asserts that the map $(\delta \otimes 1) \circ \delta: \mathfrak g \to \mathfrak g^{\wedge 3}$ vanishes. A vector space $\mathfrak g$ that is both a Lie algebra (under $\beta$) and a Lie coalgebra (under $\delta$), is a Lie bialgebra if $\beta$ and $\delta$ satisfy an additional relationship. Namely, let $\sigma: \mathfrak g^{\otimes 2} \to \mathfrak g^{\otimes 2}$ be the usual "flip" map; then the bialgebra identity is that $\delta \circ \beta$ and $(1 \otimes \beta)\circ (\delta \otimes 1) + (\beta \otimes 1) \circ (1 \otimes \delta) + (\beta \otimes 1) \circ (1\otimes \sigma) \circ (\delta \otimes 1) + (1 \otimes \beta) \circ (\sigma \otimes 1) \circ (1 \otimes \delta)$ are equal as maps $\mathfrak g^{\otimes 2} \to \mathfrak g^{\otimes 2}$.

My question

In a calculation I'm doing, I'm led to consider the map $\mathfrak g^{\otimes 2} \to \mathfrak g^{\vee 3}$ given by $(1 \otimes \beta \otimes 1) \circ (\delta \otimes \delta)$. (I mean, $(1 \otimes \beta \otimes 1) \circ (\delta \otimes \delta)$ lands in $\mathfrak g^{\otimes 3}$, but I want the composition with the natural projection $\mathfrak g^{\otimes 3} \to \mathfrak g^{\vee 3}$.) In particular, for the calculation to come out right, I'd like for this map to vanish. Does it?

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A slightly less impressive way to write the compatibility condition is to demand that simply $\delta[a,b]=[a,\delta(b)]+[\delta(a),b]$, where $\mathfrak g$ acts on the left and on the right on $\Lambda^2\mathfrak g$ via the adjoint action. In other words, $\delta$ is a Chevalley-Eilenberg $1$-cocycle for $\mathfrak g$ with values in $\Lambda^2\mathfrak g$ (dually, this is equivalent to $\beta$ being a cocycle for the correspondig cohomology theory for the Lie coalgebra...) –  Mariano Suárez-Alvarez Dec 22 '09 at 22:50
    
@Mariano: Of course. Actually, the best way to define a Lie bialgebra is as follows. In the usual Penrose "birdtrack" notation, draw $\mathfrak g$ as a directed edge, the bracket $\beta$ as a trivalent vertex with two incoming and one outgoing edge, and $\delta$ as a trivalent vertex with one incoming and two outgoing edges. Then Jacobi, coJacobi, and the compatibility condition each asserts that a certain sum over trees with four (directed) leaves vanishes. It is a sum over "all trees" provided you get the signs right. –  Theo Johnson-Freyd Dec 22 '09 at 22:56
    
Heh. "best" for appropriate values of "good", I guess! :P –  Mariano Suárez-Alvarez Dec 22 '09 at 22:59
    
I think in Feynman/Penrose diagrams. See e.g. my notes on Lie bialgebras at math.berkeley.edu/~theojf/GraphicalLanguage.pdf –  Theo Johnson-Freyd Dec 22 '09 at 23:09
    
Oh, incidentally, I thought I had a counterexample, and posted it as an answer, and realized it was wrong, and deleted it. Probably only Mariano saw this anyway. –  Theo Johnson-Freyd Dec 22 '09 at 23:10
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1 Answer 1

up vote 2 down vote accepted

(Hopefully this time I did not mess up the indices in the QYBE :/ )

Let $\mathfrak{sl}\_2$ be spanned by $e$, $f$ and $h$ with $[h,e]=2e$, $[h,f]=-2f$ and $[e,f]=h$, as usual. Let $r=e\wedge f\in\Lambda^2\mathfrak{sl}\_2$ and let $\delta=[\mathord-,r]:\mathfrak g\to\Lambda^2\mathfrak g$ be the inner derivation corresponding to $r$. Of couse $\delta$ is a $1$-cocycle, and one checks by hand that it is a cobracket; explicitely, $\delta(h)=0$, $\delta(e)=e\wedge h$ and $\delta(f)=f\wedge h$. Let $\gamma:\mathfrak g\otimes\mathfrak g\to S^3\mathfrak g$ be the map in question. Then $\gamma(e\otimes f)\neq0$.

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Thanks! My example was bound to fail, because $\gamma$ is antisymmetric in the $\mathfrak g^{\otimes 2}$ part. –  Theo Johnson-Freyd Dec 23 '09 at 2:16
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