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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 guarantees the integrability?)

http://mathforum.org/kb/thread.jspa?forumID=253&threadID=1653483&messageID=5990478#5990478

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.

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  • $\begingroup$ What is the definition of "complete integrability" you are using ? $\endgroup$
    – Konstantinos Kanakoglou
    Commented Jul 25, 2018 at 13:42
  • $\begingroup$ @KonstantinosKanakoglou Sorry for my delay. I mean the standard definition: existence of n independent first integral $f_1,f_2,\ldots, f_n$ with $\omega(X_{f_i},X_{f_j})=0$. $\endgroup$
    – Ali Taghavi
    Commented Jul 27, 2018 at 22:13
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    $\begingroup$ this is a quite interesting question. I do not know the answer in general, but i think that -since you are asking a scenario involving differerent symplectic structures on the same underlying manifold- that the question transcends hamiltonian mechanics. So, i was thinking of suggesting that maybe the question should be retagged: "Classical Mechanics" seems more suitable -imo- than "hamiltonian mechanics" and maybe "differential geometry" would be better than "hamiltonian vector fields". $\endgroup$
    – Konstantinos Kanakoglou
    Commented Jul 29, 2018 at 18:54
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    $\begingroup$ Don't misunderstand me, i am just trying to think of ways to increase the visibility of the question. $\endgroup$
    – Konstantinos Kanakoglou
    Commented Jul 29, 2018 at 19:03
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    $\begingroup$ @KonstantinosKanakoglou Thank you very much for your interesting suggestion. I revise the tags. $\endgroup$
    – Ali Taghavi
    Commented Jul 30, 2018 at 7:28

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