# Determinant as a Hamiltonian

Are there two symplectic structures $\omega_1, \omega_2$ on $M_{2n}(\mathbb{R})$ such that the function $Det:M_{2n}(\mathbb{R})\to \mathbb{R}$ is completely integrable with respect to $\omega_{1}$ but is not completely integrable with respect to $\omega_2$

Note that we do not limit the symplectic structures to structures with constant coefficients.

I had already asked a weaker version of this question in the following link but I confess that I did not understand the details(How does the action of orthonormal group guarantee's the integrability?)

Correction of the above Mathforum link: For $n=1$, the appropriate first integral showing complete integrability of $Det =xw-yz$ is $xz+yw$. It works on $M_{2}(\mathbb{R})\setminus \{0\}$. At the origin(the zero matrix) we loose the independence of differentials of two first integrals.
• @KonstantinosKanakoglou Sorry for my delay. I mean the standard definition: existence of n independent first integral $f_1,f_2,\ldots, f_n$ with $\omega(X_{f_i},X_{f_j})=0$. – Ali Taghavi Jul 27 '18 at 22:13