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We all know the definition in the following

Definition. Let $\frak{g}$ be a Lie algebra. A Lie bialgebra structure on $\frak{g}$ is a skew-symmetric linear map $\delta_{\frak{g}}: \frak{g}\rightarrow \frak{g}\otimes \frak{g},$ called the cocommutator, such that

  1. $\delta_{\frak{g}}^{*}: \frak{g}^{*}\otimes \frak{g}^{*}\rightarrow \frak{g}^{*}$ is a Lie bracket on $\frak{g}^{*}$,
  2. $\delta_{\frak{g}}$ is a $1$-cocycle of $\frak{g}$ with values in $\frak{g}\otimes \frak{g}.$

My question is that ``what is a skew-symmetric linear map'' in the definition?

yuanyuanzhang
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  • See here; $\delta^*$ is skew-symmetric, not $\delta$. See also the Definition in $1.3$ here. – Dietrich Burde Nov 04 '18 at 14:52
  • @Dietrich Burde, dear professor D. Burde, it seems that this definition is also OK. It appears in Char and Pressley's book ``A guide to quantum group'' – Daisy Jul 18 '20 at 01:32
  • @DietrichBurde Unfortunately, the second link doesn't work. – Filippo Apr 13 '22 at 05:55
  • @DietrichBurde Lol - no, I meant that the link didn't (and still doesn't) work on my phone. But now I see that the link does at least work at my computer. Thank you for asking. – Filippo Apr 13 '22 at 17:31

1 Answers1

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The definition you posted is also OK. First, we need a flip map $$\tau:\mathfrak{g} \otimes \mathfrak{g}\rightarrow \mathfrak{g} \otimes \mathfrak{g}, x \otimes y \mapsto y\otimes x, \forall x,y\in \mathfrak{g}.$$ Then the skew symmetry of $\delta_{\mathfrak{g}}$ is given by $$\tau \delta_{\mathfrak{g}}(x)=-\delta_{\mathfrak{g}}(x), \forall x\in \mathfrak{g}.$$

Daisy
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