Why is the trace of matrices in any power a Casimir function for the Poisson structure on $\mathfrak{gl}(N)$?

Question

Let's assume we have a Poisson bracket $$ \{S_{ab}, S_{cd}\} = S_{cb} \delta_{ad} - S_{ad} \delta_{cb} \,. $$ It's obvious that $C_k = \mathrm{tr}(S)$ (where $S = \sum_{ij} S_{ij} E_{ij}$) is a Casimir function for it: \begin{align*} \{f,C_k\} &= \sum_{abcd} \{S_{ab}, S_{cd}\} \frac{\partial f}{\partial S_{ab}} \frac{\partial C_k}{\partial S_{cd}} \\ &= \sum_{abcd} (S_{cb} \delta_{ad} - S_{ad} \delta_{cb}) \frac{\partial f}{\partial S_{ab}} \mathrm{tr}\Biggl( \frac{\partial \sum S_{ij}E_{ij}}{\partial S_{cd}} \Biggr) \\ &= \sum_{abcd} (S_{cb} \delta_{ad} - S_{ad} \delta_{cb}) \frac{\partial f}{\partial S_{ab}} \mathrm{tr}(E_{cd}) \\ &= \sum_{abcd} \biggl( S_{cb} \delta_{ad} \frac{\partial f}{\partial S_{ab}} \delta_{cd} - S_{ad} \delta_{cb} \frac{\partial f}{\partial S_{ab}} \delta_{cd} \biggr) \\ &= \sum_{abcd} \biggl( S_{ab} \frac{\partial f}{\partial S_{ab}} - S_{ab} \frac{\partial f}{\partial S_{ab}} \biggr) \\ &= 0 \,. \end{align*} But I have no idea why it also works for $\mathrm{tr}(S^k)$. I need an idea for the derivation.

Answer 1

I will not give all the details, but here is an outline of one way to see this.

(Step 1) Show that for any two functions $f, g \in C^\infty(\mathfrak{gl}(N))$, the Poisson bracket is given by $$ \{f,g\}(S) = \mathrm{tr}\Big(\left[ S, \, \nabla f(S) \right] \cdot \nabla g(S) \Big) $$ On the right-hand side, $\mathrm{tr}(\cdot)$ is the trace, $[-,-]$ is the matrix commutator, and the "gradient" $\nabla f$ is the transpose of the matrix of partial derivatives. That is, the $(i,j)$-entry of $\nabla f$ is $\frac{\partial f}{\partial S_{ji}}$. For example the gradients of the coordinate functions are $\nabla S_{ab} = E_{ba}$.

I think you can (probably) prove this formula using a calculation similar to the one in your original post.

(Step 2) Show that for $f(S) = \mathrm{tr}(S^k)$, we have $\nabla f(S) = k \, S^{k-1}$.

(Step 3) Since $S$ (and all powers of $S$) commutes with itself, we have $[S, S^{k-1}] = 0$. Now use Step 1 to conclude that $\{\mathrm{tr}(S^k), g\} = 0$ for all $g$.