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I am reading Etingof's Lectures on Quantum Groups, Chapter 2. There, the authors define a Lie bialgebra structure on $T_eG$ for $G$ a Lie-Poisson group. Questions:

  1. First, they define a Lie algebra structure on $T_x^*M$ for any Poisson manifold $M$ such that $\Pi(x)=0$. The construction is as follows.
    First, the Poisson structure $\{-,-\}$ on $M$ defines a Poisson structure $$ \{-,-\}:\mathcal{O}_{M,x}\otimes\mathcal{O}_{M,x}\to\mathcal{O}_{M,x} $$ on the stalk of germs $\mathcal{O}_{M,x}$.
    Second, notice that $\{f,g\}(x)=0$, since $\Pi(x)=0$, so that the Poisson structure maps to the maximal ideal of $\mathcal{O}_{M,x}$, i.e., $\{-,-\}:\mathcal{O}_{M,x}\otimes\mathcal{O}_{M,x}\to\mathcal{m}_{M,x}$, which gives a Poisson structure to $\mathcal{m}_{M,x}$.
    Finally, they notice that if $f\in\mathcal{m}_{M,x}$ and $g\in\mathcal{m}_{M,x}^2$, then $\{f,g\}\in\mathcal{m}_{M,x}^2$, which implies that $\mathcal{m}_{M,x}^2$ is a Lie ideal of $\mathcal{m}_{M,x}$ and hence $T_x^*M=\mathcal{m}_{M,x}/\mathcal{m}_{M,x}^2$ has a Lie algebra structure.
    I can check that $\mathcal{m}_{M,x}^2$ is a Lie ideal of $\mathcal{m}_{M,x}$ using local coordinates, but I don't really understand what is happening here. Is there an intuitive explanation for this fact?
  2. They then "take the dual" of the commutator $[-,-]$ on $T_e^*G$ for $G$ a Lie-Poisson group obtained using the above construction to get a Lie co-algebra structure $\delta:T_eG\to\Lambda^2T_eG$ on $T_eG$. I do not understand what they mean by "taking the dual" here. Specifically, I don't understand how to get a map $\delta:T_eG\to\Lambda^2T_eG$ from a map $[-,-]:\Lambda^2T_e^*G\to T_e^*G$, even if we have an available Poisson bivector $\Pi\in\Gamma(\Lambda^2TM)$.
    The authors explain: "let us use left translations $\lambda_g$ to identify $T_gG$ with $\mathcal{g}$. This allows us to view the Poisson bivector as a map $\Pi: G\to\Lambda^2\mathcal{g}$. Then $\delta=d\Pi:\mathcal{g}\to\Lambda^2\mathcal{g}$."
    I understand that $\lambda_g$ allows us to identify $T_gG$ with $\mathcal{g}=T_eG$. But I don't understand how that allows us to view $\Pi$ as a map $\Pi: G\to\Lambda^2\mathcal{g}$.
    Also, is there a more algebraic way to get $\delta: \mathcal{g}\to\Lambda^2\mathcal{g}$ from $[-,-]:\Lambda^2\mathcal{g}^*\to\mathcal{g}^*$? If $\Pi$ were non-degenerate, I presume $\delta$ could be defined using some standard identifications $\Pi:T_e^*G\cong T_eG$.

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