How to check that the Poisson bracket of traces of matrix powers is zero? The canonical Lie-Poisson bracket on functions on the space $\mathfrak g$ of n×n square matrices is given by:
$$\{f_1,f_2\}(a)=\langle a, [df_1(a),df_2(a)] \rangle,$$
where $a \in {\mathfrak g}^*$, [,] is the commutator and $\langle, \rangle$ is the canonical linear pairing of $\mathfrak g$ and $\mathfrak g^*$.
A mathematical paper I am reading claims that it is obvious that the Poisson bracket of the two functions $f_k(X)= trace(X^k)$ and $f_m(X)= trace(X^m)$ is zero but I have trouble seeing this for myself.
Could you help me check that $\{f_k, f_m\}=0$?
Thank you very much!
 A: Let $A, B \in \mathfrak{gl}(n)$. Note that $df_k(A)$ (which I'll also denote by $df_k|_{A}$) is a linear functional on $\mathfrak{gl}(n)$ so I guess that when you write $\left[ df_k(A), df_m(A) \right]$, you mean that we should use the pairing $\left< \cdot, \cdot \right>$ to identify $df_k(A), df_m(A)$ with matrices in $\mathfrak{gl}(n)$ and take their commutator.
Let us calculate the differential $df_k|_{A} \colon \mathfrak{g} \rightarrow \mathbb{R}$. Note that we have
$$ (A + tB)^k = A^k + \left( \sum_{i + j = k-1} A^i B A^j \right)t + O(t^2) $$
and so using the fact that $\operatorname{tr}(CD) = \operatorname{tr}(DC)$  we have
$$ f_k \left( A + tB \right) = \operatorname{tr} \left( A^k \right) + k \operatorname{tr}(A^{k-1}B) t+O(t^2)$$
which shows that
$$ df_k|_{A}(B) = \left. \frac{d}{dt} f_k \left( A + tB \right) \right|_{t=0} = k \operatorname{tr} \left( A^{k-1} B \right) = \left<k A^{k-1}, B \right>. $$
This means that under the pairing $\left< \cdot, \cdot \right>$ the differential $df_k|_{A}$ is identified with the matrix $k A^{k-1}$ and so
$$ \left[ df_k(A), df_m(A) \right] = \left[kA^{k-1}, mA^{m-1} \right] = 0. $$
