I'm having trouble with an Exercise (12.3.6) from McDuff's and Solomon's book, "Introduction to Symplectic Topology" (3rd Ed.).
The goal is to prove the monotonicity of the symplectic displacement energy. But in simpler terms, here is the question.
Let $\phi \in Ham_c(\mathbb{R}^{2n},\omega_0)$ be a hamiltonian diffeomorphism with compact support. We define $\phi_\lambda(z) = \lambda \phi(\frac{1}{\lambda}z)$ for $\lambda>0$, and we want to prove that $\phi_\lambda \in Ham_c(\mathbb{R}^{2n}, \omega_0)$.
The book provides a hint: consider the hamiltonian functions $H_\lambda (z) = \lambda^2 H(\lambda^{-1}z)$.
So, following the hint, we consider $H:[0,1]\times \mathbb{R}^{2n}\to\mathbb{R}$ the hamiltonian that generates $\phi$ as the time one map of its flow: $\partial_t \phi_t^H = X_{H_t}\circ \phi_t^{H}$, where $X_{H_t}$ is the hamiltonian vector field of $H_t = H(t,.)$. So, $\phi = \phi_1^H$. Then, given $H$, we consider $H_\lambda$ as defined before.
At this stage, two approaches seem reasonable: either we try to write $X_{H_\lambda}$, or we try to guess an isotopy that generates $\phi_\lambda$.
When I try to compute $X_{H_\lambda}$:
$d_pH_\lambda(v_p) = \lambda^2 d_pH(\lambda^{-1}v_p) \implies d_pH_\lambda(v_p) = \lambda^2\omega_0(X_H(p), \lambda^{-1}v_p) = \omega_0(\lambda^2X_H(p), \lambda v_p)$.
But I am not able to explicitly write $X_{H_\lambda}$ from there!
Also, if I try to guess an isotopy that generates $\phi_\lambda$, I tried to use $\psi_t(z) = \lambda \phi_t^{H}(\lambda^{-1}z)$, but differentiating w.r.t time:
$\partial_t \psi_t(z) = \lambda\partial_t\phi_t^H(\lambda^{-1}z) = \lambda X_{H_t}(\phi_t^H(\lambda^{-1}z)) = \lambda X_{H_t}(\lambda^{-1}\psi_t(z)) \implies \partial_t\lambda^{-1}\psi_t(z) = X_{H_t}(\lambda^{-1}\psi_t(z))$.
But by uniquiness, this would imply that $\lambda^{-1}\psi_t = \phi_t^H$ which makes no sense for me.
I am surely commiting some mistake in the calculations of this derivatives, but I can't seem to find it.
Best Answer
For clarity, I am taking $H_\lambda(z,t) := \lambda^2 H(\lambda^{-1}z,t)$, and suppressing the $t$.
There are a couple of problems in your derivation. First of all the definition of $H_\lambda$ implies $$ d_zH_\lambda(v_z) = \lambda^2 d_{\lambda^{-1}z}H (\lambda^{-1}v_z) $$ (note the different basepoint in the derivative). Also, the next step should be $$ \lambda^2\omega_0(X_H(\lambda^{-1}z),\lambda^{-1}v_z) = \omega_0(\lambda^2 X_H(\lambda^{-1}z),\lambda^{-1}v_z) = \omega_0(\lambda X_H(\lambda^{-1}z),v_z). $$ Here I'm using the fact the $\omega_0$ is bilinear, not linear in $\mathbb{R}^{2n}\oplus\mathbb{R}^{2n}$ as you take it to be. From this, we deduce that $X_{H_\lambda}(z) = \lambda X_H(\lambda^{-1}z)$. It's then not too difficult to show that $\phi^{t,0}_{H_\lambda}$ and $z\mapsto \lambda \phi^{t,0}_H(\lambda^{-1}z)$ are the same diffeomorphism, since they both satisfy the ode $$ \frac{\partial}{\partial t}\psi^t = X_{H_\lambda}\circ \psi^t \quad\text{and} \quad \psi^{0} = \operatorname{id}_{\mathbb{R}^{2n}}. $$ Evaluating at $t=1$, we get $$ \phi^{1,0}_{H_\lambda}(z) = \lambda\phi_H^{1,0}(\lambda^{-1}z) = \lambda\phi(\lambda^{-1}z) =:\phi_\lambda(z). $$ So $\phi_\lambda \in \operatorname{Ham}_c(\mathbb{R}^{2n},\omega_0)$, as claimed.