Lie bialgebra

Remarks

Example: any Lie algebra with $\delta = 0$ is a Lie bialgebra, since both clauses reduce to $0 = 0$. A richer example is $\mathfrak{sl}_2$ with a cobracket built from its standard basis, which is the classical limit of the quantum group $U_q(\mathfrak{sl}_2)$.

Non-example: let $\mathfrak{g} = \mathfrak{sl}_2$ with basis $e, f, h$ and $[e, f] = h$. Define $\delta(e) = \delta(f) = 0$ and $\delta(h) = e \otimes f - f \otimes e$. Clause 2 for $x = e$, $y = f$ requires $\delta(h) = e \cdot \delta(f) - f \cdot \delta(e) = 0$, but $\delta(h) \neq 0$, so clause 2 fails.

Clause 2 says that $\delta$ is a $1$-cocycle of $\mathfrak{g}$ with values in the module $\mathfrak{g} \otimes \mathfrak{g}$. If $\mathfrak{g}$ is finite-dimensional, the transpose of $\delta$ is a Lie bracket on the dual $\mathfrak{g}^*$, and the transpose of the bracket of $\mathfrak{g}$ is a compatible cobracket on $\mathfrak{g}^*$.