I am trying to understand the 2012 blog post What is a symplectic manifold, really?

It says (with correction of a typo in the second point):

  • If $f: M \to \mathbb{R}$ is a smooth compactly supported function, then there is a time evolution $a_f:M\times R \to M$ such that $a_f(a_f(x,u),t)=a_f(x,u+t)$. Physically, we think of this as the energy function specifying how the system evolves over time.
  • Conservation of energy: $f(a_f(x,t))=f(x)$.
  • ...
  • The assignment from energy functions to flows is equivariant under any of the flows: $a_{f(a_g(-,t))}(x,u)=a_f(a_g(x,u),t)$.

All of these are hopefully intuitive properties for a physically system to have.

I do not understand the last bullet point ("equivariance").

First of all, when $g=f$ it seems to contradict the first two points.

Secondly, calculating in the plane $\mathbb{R}^2$ with coordinates $x=(q,p)$: for $f=\tfrac{1}{2}q^2$ and $g=\tfrac{1}{2}p^2$ (okay, not compactly supported, but I doubt that is the biggest issue here), I get the Hamiltonian vector fields $X_f=-q\frac{\partial}{\partial p}$ and $X_g=p\frac{\partial}{\partial q}$ which integrate to $a_f((q,p),t) = (q,p-tq)$ and $a_g((q,p),u) = (q+up, p)$. The right-hand side of the desired equality is $a_f(a_g(x,u),t)=(q+up,p-t(q+up))$. The left-hand side is the evolution of the Hamiltonian vector field associated to $h:=f(a_g((q,p),t))=\tfrac{1}{2}(q+tp)^2$, i.e. of the vector field $X_h=t(q+tp)\frac{\partial}{\partial q}-(q+tp)\frac{\partial}{\partial p}$, hence it is $a_h((q,p),u)=(q+ut(q+tp),p-u(q+tp))$, which is not the same as the right-hand side.

What am I doing wrong and/or what is the correct statement?

It should be related to the Jacobi identity, but that seems more like an infinitesimal version of this.


The condition I was trying to write in that last bullet point is that the symplectic form itself is invariant under these flows. Indeed, this is equivalent to the Poisson bracket satisfying the Jacobi identity.

For fixed $u$, we can think about $a_f(a_g(x,u))$ as integrating the vector field $X_f$ (defined by $\langle X_f,\omega\rangle=df$) starting at the point $a_g(x,u)$. This is equivalent to considering the vector field $X’$ which is the pushforward of $X_f$ by the map $a_g(-,-u)$, flowing by that starting at $x$, and then applying $a_g(-,u)$. The point of invariance is that $X’$ is uniquely defined by $\langle X’, a_g(-,u)^*\omega\rangle= df(a_g(-,u))$; by the invariance of the symplectic form (by Cartan’s magic formula, $L_{X_g}\omega=d^2g+\iota_{X_g}d\omega=0$), this says that $X’=X_{f(a_g(-,u))}$.

So, at least one correct version of the statement is $a_g(a_{f(a_g(-,u))}(x,t),u)=a_f(a_g(x,u),t).$ The Jacobi identity for Poisson bracket is obtained by taking $\frac{\partial^2}{\partial u \partial t}h$ for some function $h$; the RHS gives $\{\{ h, f\},g\}$ and the left hand side gives $\{\{h,g\},f\}+\{h,\{f,g\}\}$.

  • $\begingroup$ Thanks a lot! I understand now about the pushforward by the flow in the opposite direction. Is $X'$ also Hamiltonian when the Poisson structure is degenerate? $\endgroup$ – Ricardo Buring Feb 18 '20 at 23:59
  • $\begingroup$ @RicardoBuring Yes, everything I wrote remains true for Poisson manifolds. Just take the equations I wrote with the symplectic form and rewrite them with the Poisson tensor on the other side of the equation. $\endgroup$ – Ben Webster Feb 20 '20 at 1:42
  • $\begingroup$ Thanks! I may be a bit dense so I asked for the details on Math.SE: Pushforward of Hamiltonian vector field by (reverse) Hamiltonian flow is Hamiltonian $\endgroup$ – Ricardo Buring Feb 20 '20 at 17:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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