# Symplectomorphism is linear in second component

I've got a problem (da Silva homework 3, question 3) that I have rewritten as the following:

Let $$g:M\to M$$ be a symplectomorphism such that it takes $$(q,p)\to (q',p')= (f(q,p), h(q,p))$$.

I want to show that $$\partial f/\partial p=0$$ and $$\partial^2h/\partial p^2=0$$.

I have by the assumption that $$g^{*}\alpha=\alpha$$ that $$p'dq'=pdq$$ i.e. $$h(q,p)dq'=pdq$$ [$$\alpha$$ is the tautological 1-form].

I have that $$dq'=\frac{\partial f}{\partial p}dp+\frac{\partial f}{\partial q}dp.$$ Therefore $$h(q,p)dq'=pdq$$ implies that

$$h\left(\frac{\partial f}{\partial p}dp+\frac{\partial f}{\partial q}dp\right)=pdq$$ If it is the case that $$h(q,p)=0$$, then we have that $$h$$ is the zero function so $$h(q,\lambda p)=h(q,p)=0$$ and $$\frac{\partial^2 h}{\partial q^2}=0$$.

So I'm stuck when $$h\neq 0$$. This tells us that $$\frac{\partial f}{\partial p}=0$$. Then I know $$g=g(q)$$ and it remains to show that $$\frac{\partial^2h}{\partial q^2}=0$$.

I tried doing IBP by showing that $$\int h(q,p)\frac{\partial f}{\partial q}dq=\int pdq$$ implies $$h(q,p)f(q,p)-\int \frac{\partial h}{\partial q}f(q,p)dq=qp-\int qdp$$.

I'm not sure what to do at this point or if this is even a reasonable approach. I also wonder if there is a way to do IBP that can swap the $$dq$$ and $$dp$$? By this I mean if there is a way to write $$\int \frac{\partial h}{\partial q}f(q,p)dq$$ as something with $$\int \text{blah} dp$$?

Since $$q'=f$$ your $$dq'$$ should be $$dq'=\frac{\partial\color{red}f}{\partial p}\,dp+\frac{\partial\color{red}f}{\partial q}\,d\color{red}q\,.$$ From $$p'\,dq'=h\,dq=p\,dq$$ it then follows that $$\tag{1} \frac{\partial\color{red} f}{\partial p}\equiv 0\,.$$
Likewise, from $$dp'=\frac{\partial h}{\partial p}\,dp+\frac{\partial h}{\partial q}\,dq$$ and $$dp'\wedge dq'=dp\wedge dq$$ it follows that $$dp\wedge dq=\frac{\partial h}{\partial p}\frac{\partial f}{\partial q}\,dp\wedge dq+\underbrace{\frac{\partial h}{\partial q}\frac{\partial f}{\partial p}\,dq\wedge dp}_{0}\,.$$ Therefore $$\tag{2} \frac{\partial h}{\partial p}\frac{\partial f}{\partial q}\equiv 1\,.$$ Taking a derivative gives $$\frac{\partial^2 h}{\partial p^2}\frac{\partial f}{\partial q}+\frac{\partial h}{\partial p}\!\!\!\!\!\underbrace{\frac{\partial^2 f}{\partial p\,\partial q}}_{=0\text{ because of }(1)}\equiv 0\,.$$ By (2) it is not possible that $$\frac{\partial f}{\partial q}$$ vanishes anywhere. Therefore $$\frac{\partial^2 h}{\partial p^2}\equiv 0\,.$$
• @cheeseboardqueen . Hm intuition. We want to know something about $\partial^2h/\partial p^2\,.$ So I thought "why not giving that a go?" Feb 12 at 21:30