2
$\begingroup$

Assume that $n$ is an even number. What is a natural symplectic structure on $M_{n}(\mathbb{R})$, such that for every $1\leq k \leq n$ the manifold of $k$-rank matrices would be invariant under the flow of the Hamiltonian vector field correspond to the Hamiltonian "$H=$ Determinant"? (Or at least the manifold of $n-1$ rank matrices would be invariant under this Hamiltonian vector field)

We consider a contact analogy as follows:

Assume that $n$ is an odd number. Under what contact structures on $M_{n}(\mathbb{R})$, the set of singular matrices is invariant under the flow of corresponding Reeb vector field? Or a refined version: the set of $k$ rank matrices would be invariant under this flow.

$\endgroup$

1 Answer 1

3
$\begingroup$

For $n-1$, doesn't this follow from conservation of energy?

For $k\leq n-2$, the derivative of the Hamiltonian vanishes, so the flow is zero, so it's trivially invariant.

So every symplectic structure will do.

$\endgroup$
4
  • $\begingroup$ Concerning the second sentence: the derivative of the Hamiltonian indeed vanishes in certain directions, but how do you conclude that the associated vector field vanishes as well? (A generic Hamiltonian has vanishing directional derivatives in a subspace of codimension 1 of the tangent space, but that doesn't mean that the flow is zero!) $\endgroup$ Feb 26, 2015 at 19:01
  • $\begingroup$ @TobiasFritz I think thr second sentence is true according to the answer and comment to the following question;mathoverflow.net/questions/197773/… $\endgroup$ Feb 27, 2015 at 12:49
  • $\begingroup$ @WillSawin +1 for your interesting answer. In the literature, are there some result on completly integrability of this Hamiltonian or some ynamical properties of the Hamiltonian vector field correspond to determinant? $\endgroup$ Feb 27, 2015 at 13:03
  • $\begingroup$ @AliTaghavi I don't know anything about the literature, and I also don't know a canonical symplectic structure on $M_n(\mathbb R)$ to evaluate the dynamics in. $\endgroup$
    – Will Sawin
    Mar 4, 2015 at 23:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.