Computing Poisson brackets Suppose I have Hamiltonian :H = $\frac{1}{2}(S_1^2+S_2^2+\beta S_3^2)+R_1$,
and two first integrals: $f_1 = R_1^2+R_2^2+R_3^2$,$f_2 = R_1S_1+R_2S_2+R_3S_3$.
And also i know how to compute Poisson bracket: first i write down $w^{ij} = $ {$x^i,x^j$},$(x_1,x_2,x_3,x_4,x_5,x_6)=(S_1,S_2,S_3,R_1,R_2,R_3)$ and {$S_i,S_j$}=$\epsilon_{ijk}S_k$,
{$R_i,S_j$}=$\epsilon_{ijk}R_k$, {$R_i,R_j$}=0
and to compute {f,g} = $\sum_{ij} w^{ij} \dfrac{f}{x^i} \dfrac{g}{x^j}$.So my problem is when i compute {$f_1,f_2$} i get {$f_1,f_2$} = $2R_1R_2R_3$, but i know that this functions are Kasimir functions , so there Poisson bracket must be zero.
 A: Maybe this is just a typo in your post, but your formula for $\{f,g\}$ is not quite right. It should be $\frac{\partial f}{\partial x_i}$ rather than $\frac{f}{x_i}$, and similar for $g$:
$$ \{f,g\} = \sum_{i,j} w^{ij} \frac{\partial f}{\partial x_i} \frac{\partial g}{\partial x_j} $$
Anyways, let's finish the calculation. If you write the matrix $W = (w^{ij})_{i,j=1}^n$ (this is just the Poisson tensor in coordinates), then this formula is equivalent to the matrix-vector product $\{f,g\} = (df)^\top W (dg)$, where $df$ and $dg$ are the vectors of partial derivatives, i.e. $(df)^\top = \left( \frac{\partial f}{\partial x_1}, \dots, \frac{\partial f}{\partial x_n} \right)$. In your case, we have
$$
W = \left( \begin{array}{ccc|ccc}
0 & S_3 & -S_2 & 0 & R_3 & -R_2 \\
-S_3 & 0 & S_1 & -R_3 & 0 & R_1 \\
S_2 & -S_1 & 0 & R_2 & -R_1 & 0 \\ \hline
0 & R_3 & -R_2 & 0 & 0 & 0 \\
-R_3 & 0 & R_1 & 0 & 0 & 0 \\
R_2 & -R_1 & 0 & 0 & 0 & 0
\end{array} \right)
$$
$$ (df_1)^\top = 2 \, \left( 0, 0, 0, R_1, R_2, R_3 \right), \quad \quad
df_2 = \begin{pmatrix} R_1 \\ R_2 \\ R_3 \\ S_1 \\ S_2 \\ S_3 \end{pmatrix} $$
And you are right that both $f_1$ and $f_2$ are Casimir functions, which can easily be checked by seeing that both matrix products are zero:
$$ (df_1)^\top W = 0 \quad \text{ and } \quad W df_2 = 0 $$
