3
$\begingroup$

I am struggling to see how the symplectic action 1-form $\alpha_H$ on the loop space of a symplectic manifold $(M,\omega)$ is closed. It is defined by

$\alpha_H(Y) = \int_0^1 \omega(\dot{x}(t)-X_H,Y)dt$

with $Y \in T_{x(t)}M$ and $X_H$ a Hamiltonian vector field on $M$. Going the other way and obtaining it as the differential of the symplectic action is fine (and so I understand why it is closed from this point of view), but I am not sure how to calculate $d\alpha_H$ explicitly and have not seen it done anywhere. To be honest, calculations on loop space are confusing me a bit and it would be useful to know of a source where this is all worked out in detail.

Improve this question
$\endgroup$
1
  • $\begingroup$ Have you looked at various proofs of the Poincare Lemma? Typically it involves taking the derivative of an integral of the contraction of a form. This is precisely what you are interested in. $\endgroup$
    – Ryan Budney
    Apr 8, 2015 at 17:58

2 Answers 2

Reset to default
1
$\begingroup$

A very thorough introduction to the symplectic geometry of the loop space is given by Wurzbacher in his 1995 paper.

Improve this answer
$\endgroup$
1
$\begingroup$

It is an exact form and hence closed. Actually, this 1-form is the differential of the Symplectic action functional defined on the loop space. For a complete proof see for example Proposition 6.3.3 in "Morse Theory and Floer Homology" by Michèle Audin and Damian Mihai.

Improve this answer
$\endgroup$
1
  • $\begingroup$ This form is not exact in general, I think Audin and Damian only consider Symplectically aspherical manifolds, which it is the case there. But in more general cases, this action functional is still closed. For example in some cases it can be realized as differential of some circle valued action functional. (Lectures on Floer homology, Dietmar Salamon) $\endgroup$
    – Saman Habibi Esfahani
    Sep 25, 2019 at 19:55

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .