$\newcommand{\g}{\mathfrak{g}} \newcommand{\Cinf}[1]{\mathcal{C}^\infty (#1)} \renewcommand{\d}{\mathrm{d}} \newcommand{\R}{\mathbb{R}} $

Let $\g$ be a Lie algebra and consider in $\g^*$ the following Poisson structure: $$\{f, g \} (\xi):= [f_{*,\xi},g_{*,\xi}](\xi), \qquad f,g \in \Cinf{\g^*},$$ where $f_{*,\xi},g_{*,\xi}: T_\xi \g^* = \g^* \to \R$ are considered as elements in $\g^{**}= \g$.

Given a Poisson map $\mu: M \to \g^*$ where $(M, \omega)$ is a symplectic manifold (therefore also a Poisson manifold) and $\g^*$ has the previous Poisson structure, consider also $\mu$ also as the map $\mu : \g \to \Cinf M, v \to \mu_v $ (linear in $v$), with $$\mu (x)(v)= \mu_v (x)$$

I want to show that $$\mu_{[u,v]}= \{\mu_u, \mu_v \}$$

Any idea, suggestion? This is part of a bigger problem but actually I am stuck here. Thanks.


Do you mean to say that $\mu: M\to \mathfrak{g}^\ast$ is a moment map? In fact, what you want to show is only true if the map $\mu$ is equivariant. If $\mu$ is a moment map, then $$d\mu([u,v])=\iota_{[u,v]}\omega=-L_u\iota_{v}\omega +\iota_{v}L_u\omega=-L_u\iota_{v}\omega.$$ You can show that this is equal to $d\{\mu(u),\mu(v)\}$.

It turns out that the constant $\mu([u,v])-\{\mu(u),\mu(v)\}$ measures exactly the equivariance of $\mu$. In particular, it is zero if $\mu$ is equivariant.

  • $\begingroup$ I need to prove that there's an action of $\mathfrak{g}$ on $M$ such taht $(M, \omega, \mu )$ is a $\mathfrak{g}$-Hamiltonian space . I wrote all the hypothesis I have, but... what is equivariant talking about an infinitesimal action of a Lie algebra on $M$? I know what equivariance is but in Lie groups. Thanks!! $\endgroup$
    – Minkowski
    Oct 3 '17 at 14:00
  • $\begingroup$ By the way, how do you do $i_v \omega$ if $v$ is an element of the Lie algebra? $\endgroup$
    – Minkowski
    Oct 3 '17 at 14:39
  • $\begingroup$ I am abusing notation and letting $v$ denote both an element of $\mathfrak{g}$ and its infinitesimal generator. The map $\mu:M\to\mathfrak{g^\ast}$ is equivariant if $\mu(g\cdot x)=g\cdot \mu(x)$. Differentiating this, (i.e. put $g=\exp(tv)$ and differentiate at $t=0$) gives infinitesimal equivariance. The derivative is exactly $\mu([u,v])-\{\mu(u),\mu(v)\}$. $\endgroup$
    – JonHerman
    Oct 3 '17 at 15:20
  • $\begingroup$ In regards to your first point, see here $\endgroup$
    – JonHerman
    Oct 3 '17 at 15:22
  • $\begingroup$ Thanks for your answers! The point of them that I haven't catched yet it that in order to speak about equivariance you need a Lie group don't you? When you wrote $\mu (g \cdot x)= g \cdot \mu (x)$ I understand that there's a Lie group $G$ acting on $M$ and $\mathfrak{g}^*$ (and $\mathfrak{g}$ is indeed its Lie algebra). But I don't have any Lie group in my problem, so I don't know how to read the definition of equivariance you told me. Thanks again. $\endgroup$
    – Minkowski
    Oct 3 '17 at 15:47

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