Let $\mathfrak{g}$ be a real finite-dimensional Lie algebra, and let $r\in\wedge^2\mathfrak{g}$$r\in\bigwedge^2\mathfrak{g}$ be a solution of the Yang-Baxter equation. The Yang-BaxterYang–Baxter equation states that for all $x\in\mathfrak{g}$, $$\mathrm{ad}_x[r,r]=0,$$$$\operatorname{ad}_x[r,r]=0,$$ where $[r,r]$ represents the Schouten-NijenhuisSchouten–Nijenhuis bracket and the adjoint action $\mathrm{ad}_x$$\operatorname{ad}_x$ is extended to $\wedge^\ast\mathfrak{g}$$\bigwedge^\ast\mathfrak{g}$ as a derivation.
Now, consider a linear map $\delta:\mathfrak{g}\to\wedge^2\mathfrak{g}$$\delta:\mathfrak{g}\to\bigwedge^2\mathfrak{g}$, which is a 1-cocycle, meaning it satisfies the condition $$\delta([x,y])=\mathrm{ad}_x\delta(y)-\mathrm{ad}_y\delta(x).$$$$\delta([x,y])=\operatorname{ad}_x\delta(y)-\operatorname{ad}_y\delta(x).$$
I want to determine the conditions under which the 1-cocycle $\xi=\delta+\mathrm{ad}\,r$$\xi=\delta+\operatorname{ad} r$ defines a Lie bialgebra, i.e., the transpose map $\xi^t:\wedge^2\mathfrak{g}^\ast\to\mathfrak{g}^\ast$$\xi^t:\bigwedge^2\mathfrak{g}^\ast\to\mathfrak{g}^\ast$ defines a Lie bracket on the dual vector space $\mathfrak{g}^\ast$. In particular, is it necessary for $(\mathfrak{g}, \delta)$ to be a Lie bialgebra.?
A particular case I knownknow: if $\delta = \mathrm{ad}\,R$$\delta = \operatorname{ad} R$ is a 1-coboundary with $[R,R]$ ad-invariant, the condition is that the two Poisson tensors associated to $r$ and $R$ have to be compatible, i.e., their Schouten bracket is zero.
