# Assignment of energy functions to flows is “equivariant”?

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.

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 $$\{\{h,g\},f\}+\{h,\{f,g\}\}$$.
• 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? – Ricardo Buring Feb 18 '20 at 23:59