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Let $g^*$ be the dual vector space of a vector space $g$. Suppose that $g^*$ is a Lie algebra and $[,]_{g^*}: \Lambda^2 g^* \to g^*$ satisfies the Jacobi identity. Let $\delta: g \to \Lambda^2 g$ be the dual map of $[,]_{g^*}: \Lambda^2 g^* \to g^*$. Do we have: $\delta$ satisfies the coJacobi identity (the coJacobi identity is: $\text{Alt}(\delta \otimes 1)(\delta)(a) = 0$ for all $a \in g$, where $\text{Alt}(a \otimes b \otimes c) = a \otimes b \otimes c + b \otimes c \otimes a + c \otimes a \otimes b$)? If this is true, how to show that $\delta$ satisfies the coJacobi identity? Any help will be greatly appreciated!

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  • $\begingroup$ If it deserves the name, the co-Jacobi identity should just be the dual of the Jacobi identity. $\endgroup$ – Qiaochu Yuan Feb 12 '16 at 16:25
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We have $\delta: g \to g \otimes g$. It has a dual map $\delta^*: g^* \otimes g^* \to g^*$. The bracket on $g^*$ is given by $\delta^*$: $[x,y]_{g^*} = \delta^*(x \otimes y)$. Therefore for $x, y, z \in g^*$, we have \begin{align} [[x,y]_{g^*}, z]_{g^*} = \delta^*([x,y]_{g^*} \otimes z) = \delta^*(\delta^*(x \otimes y) \otimes z) = \delta^*(\delta^* \otimes 1)(x \otimes y \otimes z). \end{align} It follows that \begin{align} & [[x,y]_{g^*}, z]_{g^*} + [[y,z]_{g^*}, x]_{g^*} + [[z,x]_{g^*}, y]_{g^*} \\ & = \delta^*(\delta^* \otimes 1)( x \otimes y \otimes z + y \otimes z \otimes x + z \otimes x \otimes y ). \end{align} Therefore $g^*$ satisfies the Jacobi identity is equivalent to \begin{align} \delta^*(\delta^* \otimes 1)( x \otimes y \otimes z + y \otimes z \otimes x + z \otimes x \otimes y ) = 0 \end{align} for all $x, y, z \in g^*$. That is $ \delta^*(\delta^* \otimes 1)\text{Alt}: g^* \otimes g^* \otimes g^* \to g^*$ is a zero map. The dual of a zero map is also zero. Therefore the dual map $\text{Alt} (\delta \otimes 1)\delta: g \to g \otimes g \otimes g$ is zero.

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