I am following this paper about momentum maps in Poisson geometry; in particular I am interested in the Poisson Stratification Theorem at page 1273 and the following example 2.5.
One starts with a certain Poisson structure on $\mathbb{C}^{n+1} - \{0\}$. The quotient space given by the action of $\mathbb{C}^*$, namely the complex projective space $\mathbb{C}P(n)$, is a Poisson manifold since the action is free, proper and Poisson.
The action of the $n$-torus on $\mathbb{C}P(n)$ is proper and Poisson but not free, so the quotient space, that can be identified with the $n$-simplex $\Delta^n$, is not a manifold, but it is a Poisson stratified space. It is shown that the strata are the faces of the simplex, so by the Stratification theorem they are Poisson manifolds, the inclusion being a Poisson morphism.
The problem I don't understand the explicit check of this result they do. A $d$-face of an $n$ simplex $\Delta^n \subset \mathbb{R}^{n+1}$ is the collection of points $(x_0, \dots, x_n)$
- being on the simplex, $\sum_{i=0}^n x_i = 1$
- Having $(n-d)$ zero coordinates, $x_{i_1} = \dots = x_{i_{n-d}} = 0 $
- Having the remaining coordinates strictly positive, $x_j > 0$ for $j \notin \{ i_1, \dots, i_{n-d} \} $ (so that, for example, a 1-face is an edge without vertices).
Choose without lost of generality $i_1 = 1, \dots, i_{n-d} = n-d$. Then the corresponding face can be seen as the level set of $F: \mathbb{R}^{n+1} \rightarrow \mathbb{R}^{n+1-d} $, with positive domain, mapping $(x_0, \dots, x_{n}) \mapsto \left( \sum_{i=0}^n x_i, x_1, \dots, x_{n-d} \right)$. The face is precisely $S = F^{-1}(1, 0, \dots, 0)$.
The question To verify the stratification of the simplex by the faces is indeed a Poisson stratification, the authors check that $ \{ x_i, F_i \} _{\Delta} |_S = 0$, where $ \{. , . \} _{\Delta} $ is explicitly known. In other words, they check that the bracket of any function with any component of the constraint $F$ vanishes on the level set of $F$. Why is this so?
In this case you can think of $F$ as a submersion describing the submanifold $S$. So the tangent space of $S$ is given by the kernel of $d F $ restricted to $S$, i.e. $TS =\ker d F|_S$.
The condition $\{ x_i , F_i \}_{\Delta}|_S=0$ implies that all Hamiltonian vector fields $X_i :=\{x_i ,\cdot \}$ are tangent to $S$ and that they are generated by the coordinates describing $S$, i.e. $ X_{F_i}=0 $ for all $i$. Hence the Poisson bracket actually descends to a bracket on $S$, i.e. $ \{f,g\}_{\Delta}|_S=\{f|_S ,g|_S\}_{\Delta}$. Therefore you obtain a well-defined Poisson bracket on $S$ such that the inclusion map is Poisson.