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I am currently reading a paper on symplectic geometry: Periodic orbits for Hamiltonian systems in cotangent bundles by Christopher Golé.

It deals with the question how the stability properties of a sequence of (periodic) points or fixed points can be related to the second derivative of the action functional. So the underlying question that is covered in this theorem is how stability/instability of a point relates to minimizers/maximizers of the action functional.

The proof of this is partly done in the appendix on p.22, but I don't really see how equation A.2 is derived.

So the situation is something like this (it is a reduced situation of the original question):

We are given a symplectomorphism $F$ with a fixed point $x = (q_0,p_0).$ In this situation, this means that $\hat{F}^{2N}(x)=x.$ Notice, that $F$ here is actually a composition of $2N$ many maps. (see page 10)

So, we can think of $F$ as being $F= \hat{F} \circ...\circ \hat{F}$ $2N$ times, where $\hat{F}$ is a symplectomorphism.

Furthermore, $(q_i,p_i)_i$ is an orbit under $\hat{F},$ so $\hat{F}^i(q_0,p_0) = (q_i,p_i).$Thus, the Euler-Lagrange equation is satisfied which is here the equation between $A.1$ and $A.2$.

What I now really don't understand is, how they get equation $A.2$?

If you have any questions, please let me know.

Does anybody see where it comes from? Maybe I am also misinterpreting their notion of a tangent orbit, but to me it is just an arbitrary tangent vector $\delta x_0 := (\delta q_0, \delta p_0)$ and then they iterate $d\hat{F}^i(\delta x_0).$

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    $\begingroup$ Asking for clarification of points in papers is okay but given the way you've written your question, it's doubtful many people will click on your link. You should perhaps try to motivate your question with more particulars, about the context in the paper as well as your own motivations. $\endgroup$
    – Ryan Budney
    Aug 5, 2015 at 18:03
  • $\begingroup$ @RyanBudney thanks for the hint, I updated my question a bit. $\endgroup$
    – Zlatan12
    Aug 5, 2015 at 18:26
  • $\begingroup$ The question would be easier to follow if you made it more self-contained. As is, any reader will have to compare the PDF with what you've written several times. $\endgroup$
    – j.c.
    Aug 6, 2015 at 2:51
  • $\begingroup$ Crossposted to math.stackexchange.com/q/1385993/11127 $\endgroup$
    – Qmechanic
    Aug 6, 2015 at 19:54

1 Answer 1

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It just looks like a basic application of the chain rule to the immediately preceding equation. Maybe it's the notation that's confusing you; the previous equation has the form $$ (\partial_2F)(x,y) + (\partial_1 G)(y,z) = 0. $$ Considering the left hand side as a function of three independent variables $(x,y,z)$, and differentiating in the direction $(\delta x, \delta y, \delta z)$, gives $$ \partial_1\partial_2 F\, \delta x + (\partial_2\partial_2F + \partial_1\partial_1G)\,\delta y + \partial_2\partial_1 G\,\delta z = 0, $$ which is equation $(A.2)$. This gives a constraint that must be satisfied by the three variations $(\delta x, \delta y, \delta z)$.

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