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 $\int \text{blah} dp$?


1 Answer 1


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 $$ \frac{\partial^2 h}{\partial p^2}\equiv 0\,. $$

  • $\begingroup$ Thank you, Kurt. I’m sorry for all the typos, I was switching between notations as I was writing it up! Can you explain your intuition for taking the derivative again after (2)? $\endgroup$ Feb 12 at 19:58
  • 1
    $\begingroup$ @cheeseboardqueen . Hm intuition. We want to know something about $\partial^2h/\partial p^2\,.$ So I thought "why not giving that a go?" $\endgroup$
    – Kurt G.
    Feb 12 at 21:30
  • $\begingroup$ @cheeseboardqueen Out of all the answers that you received on your 40 questions in more than three years you seem to have accepted only two. Why? This means those questions will be pushed back to the active queue despite possibly having good answers. $\endgroup$
    – Kurt G.
    Feb 13 at 6:54
  • $\begingroup$ I didn't realize that was a function - this is purely a user error on my part. I'd be happy to go back and correct that for the other questions at some point! $\endgroup$ Feb 14 at 19:56
  • $\begingroup$ @cheeseboardqueen Thanks. Another advantage of ticking those answers will be that they can be used as duplicates to close same questions coming in. $\endgroup$
    – Kurt G.
    Feb 14 at 20:32

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