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2 added 214 characters in body

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?

1

# Is this an identity in Lie bialgebras?

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 $(1 \otimes \beta \otimes 1) \circ (\delta \otimes \delta)$. In particular, for the calculation to come out right, I'd like for this map to vanish. Does it?