Coadjoint orbits coincide with symplectic leaves, Dufour's proof

Question

Edit: I have included a proof of what I'm asking here. I would appreciate it if anyone could check if it is correct.

In Poisson Structures and Their Normal Forms by Dufour and Zung I found the following proof:

I don't know how Leibniz rule proves what I have underlined in red. I know that, for any point $x\in \mathfrak{g}^*$, the tangent space is given by $\text{span}\{X_f(p):f\in C^\infty( \mathfrak{g}^*)\}$, but I don't see how to obtain any $X_f(p)$ using vector fields of linear functions.

Thanks in advance.

Answer 1

The Leibniz rule tells you that $X_f(p)$ only depends on the value $df(p)$ for the following reason: Taylor's theorem tells us that we may write $f(p+v)= f(p) +\langle df(p), v\rangle+||v||^2h(v)$ where $h(v) \to 0$ as $v\to 0$. In the following, we write $df(p)$ for the map $v\mapsto \langle df(p), v\rangle$. $df(p)$ yields a map $\mathfrak{g}^*\to \mathbb{R}$ since the identification $T^*_p\mathfrak{g}^*\cong (\mathfrak{g}^*)^*\cong \mathfrak{g}$ (assuming $G$ is finite dimensional or is modeled on a reflexive Banach space) allows us to think of $df(p)\in T^*_p\mathfrak{g}^*$ as an element of $(\mathfrak{g}^*)^*$. We then have \begin{align} X_f(p) & =X_{f(p)}(p) +X_{df(p)}(0) +X_{||v||^2 h(v)}(0)\\ &=X_{df(p)}(0)+h(0)X_{||v||^2} +||0||^2 X_{h}(0)\\ &=X_{df(p)}(0)\end{align} so the linear function $g(p+v)=\langle df(p), v\rangle$ has $X_g(p) =X_f(p)$.

In a general, one can complete this procedure locally for any Poisson manifold. Whenever we have an assignment $\delta:C^{\infty}(M)\to V$ a derivation based at $p\in M$, i.e. $\delta$ is linear and $\delta(fg)=\delta(f)g(p)+f(p)\delta(g)$, this yields a map $\delta|_{U}: C^\infty(U)\to V$ for $U\subset M$ open. The preceding argument tells us that the value of $\delta(f)$ depends only on $df(p)$. Since $T^*_pM$ is spanned by $df(p)$ for all functions $f$, $\delta$ factors through the map$f\mapsto df(p)$ and hence descends to a map $\tilde{\delta}$ which makes the following diagram commute

Then, if we have a family of derivations $\delta_p: C^\infty(M)\to V_p$ for each $p\in M$ with the $\delta_p$ and $V_p$ varying smoothly, they glue together to yield a global derivation $\Delta: C^\infty(M)\to \Gamma(\mathcal{V})$ Where $\mathcal{V}$ is the vector bundle gotten by gluing the $V_p$'s together which factors through a bundle map $\hat{\Delta}:T^*M\to \mathcal{V}$. In the case of a Poisson structure, we yield a bundle map $T^*M\to TM$ and the tangent space to the foliation is the image of this bundle map. We can think of the data of $\hat{\Delta}: T^*M\to TM$ as a section of $TM\otimes TM$ and skew symmetry means that this is a section $\Pi\in \Gamma(\bigwedge^2 TM)$. This object $\Pi$ is the Poisson bivector of the Poisson structure. Returning to the case of $\mathfrak{g}^*$, one can create such a bivector by using the Lie bracket and the natural pairing, i.e. $\Pi_{\alpha}(\xi, \eta):=\pm \langle \alpha, [\xi, \eta]\rangle$.

Answer 2

The Lie-Poisson bracket can be written in terms of the structure constants of the algebra as $$\{f,g\}=\sum_{i,j,k=1}^nc_k^{ij}x^k\frac{\partial f}{\partial x^i}\frac{\partial g}{\partial x^j}. $$ (the coordinates $x^i$ are global and given by $x^i(\alpha)=\alpha(e_i) $ for any $\alpha\in\mathfrak{g}^*$ and each element $e_i$ of the base of $\mathfrak{g}$). Let's consider some $f\in C^\infty(\mathfrak{g}^*)$, then, the corresponding vector field is $$X_f=\displaystyle\sum_{i,j,k=1}^nc_k^{ij}x^k\frac{\partial f}{\partial x^i}\frac{\partial }{\partial x^j}=\sum_{i=1}^nX_{x_i}\frac{\partial f}{\partial x^i}.$$ Therefore, for any point $p\in \mathfrak{g}^*$, we have $X_f(p)=X_{a_1\cdot x_1+\cdots+a_n\cdot x_n}(p) $, where $a_i=\displaystyle\left.\frac{\partial f}{\partial x^i}\right|_p$. Therefore, $g=\displaystyle\sum_{i=1}^na_i\cdot x_i $ is the required linear function.