Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

As I'm trying to work my way through Dietmar Salamon's "Notes on Floer Homology", I'm having trouble with the very first exercise.

Let $(M, \omega)$ be a compact symplectic manifold. Let $H$ be a real function on $M$, let $X$ be the vector field associated to $dH$ under the isomorphism $TM \simeq T^{*}M$ induced by $\omega$.

Let $x \in M$ be a critical point for $H$, ie. $dH(x) = 0$. If $\phi _{t}: M \times \mathbb{R} \rightarrow M$ is the flow associated to $X$ (that is, $\frac{\partial}{\partial t} \phi _{t} = X \circ \phi_{t}$), then $x$ is a fixed point of this flow. In particular $\phi_{1}(x) = x$. We will say that $x$ is non-degenerate if $det(id - d\phi_{1}(x)) \neq 0$. This happens exactly when $d\phi_{1}(x)$ has no fixed vectors. The problem is as follows.

Show that under these assumptions $x$ is also a non-degenerate critical point for $H$. One possible way to state that is that in any local coordinate system $x_{i}$ around $x$, the matrix $(\frac{\partial^{2}}{\partial x_{i} \partial x_{j}} H) _{i, j}$ is non-singular.

I tried to reason as follows. Since $X(x) = 0$, infinitesimally at time $0$ and around $x$ points are (up to first order) not moved by $\phi$ at all. However, as I let the flow run for one unit of time I perform some form of "integration" that makes the second order changes by $\phi$ "go up an order" and become visible in $d\phi_{1}$. Thus, I should interpret the condition that $d\phi_{1}$ has no fixed vectors as a sing that $X = dH$ moves all vectors around $x$ in a second-order change, ie. $dH$ is non-degenerate.

I was trying to formalize the "implication"

$(\frac{\partial}{\partial t} \phi_{t} = X \circ \phi_{t}) \ \Rightarrow \ \phi_{1} = \int\limits_{0}^{1} X \circ \phi_{t} \ dt$

so that later I could differentiate both sides with respect to $x$, but I quickly run into trouble because I am not working in the euclidean space, where - up till now - I have always "gotten by" using coordinate-free descriptions, since the problem I worked on were rather simple-minded. I would be very interested in a coordinate-free and proof with coordinates, but it would be most useful if it was formal, because I think my problems stem from a large lack of experiance with these kinds of problems.

Feel free to retag the question as needed.

share|improve this question
add comment

2 Answers

up vote 1 down vote accepted

Your idea was good and here is a formal proof. However, it does use local charts and I would be also interested in seeing a coordinate-free solution.

Choose Darboux coordinates around the critical point $x_0$, so that its neighbourhood is identified with $\mathbb{R}^{n}$, the symplectic form $\omega$ is locally represented by a constant $n\times n$ antisymmetric matrix and $X = \omega^{-1} \nabla H$. For fixed $x$ from a small neighbourhood of $x_0$ we have

$\psi_1(x) - x = \psi_1(x) - \psi_0(x) = \int_0^1 \frac{d}{dt}\left( \psi_t(x) \right) dt = \int_0^1 X(\psi_t(x)) dt = \int_0^1 \omega^{-1}\nabla H (\psi_t(x)).$

Differentiating both sides with respect to $x$ (note that $\omega$ is constant), we obtain

$d \psi_1(x) - I = \int_0^1 \omega^{-1} \text{Hess}(H)(\psi_t(x)) \frac{d \psi_t(x)}{dx} dt,$

where $\text{Hess}(H)$ denotes the Hessian matrix of $H$. Now put $x = x_0$. As it is a critical point, $\psi_t(x_0) = x_0$ for all $t$ and we get

$ d \psi_1(x_0) - I = \text{Hess}(H)(x_0) \left( \int_0^1 \frac{d \psi_t(x_0)}{dx} dt \right),$

which proves that the matrix $\text{Hess}(H)(x_0)$ is non-singular.

share|improve this answer
    
It's nice, thank you! I accept it, of course, although I would also like to see a coordinate-free solution. –  Piotr Pstrągowski Apr 9 '13 at 15:16
add comment

An alternative way of stating the non-degeneracy condition is that the following self-adjoint unbounded operator on $L^2(S^1, \gamma^*TM)$ does not have $0$ in the spectrum. (This formulation is actually what one uses in the proofs where the non-degeneracy assumption is used.)

Let $\nabla$ be an arbitrary torsion-free connection on $M$. Then, define the differential operator by \[ \cdot \mapsto -J(\nabla_t \cdot + \nabla_{\cdot} X_H). \] If I haven't made a mistake, this should be just $-J L_{\cdot} X_H$ (and thus doesn't depend on the connection). (I learned this formulation from Siefring's paper on asymptotics, but I don't believe it is an original observation from him.)

In particular, if we don't impose the condition that this operator act on periodic sections of $\gamma^*TM$, the linearized flow acting on any vector i.e.~$d\phi_t \cdot v$ for any $v$, is in the kernel of this operator. The kernel is then non-trivial if and only if there is some vector such that $d\phi_t \cdot v$ is of period 1 in $t$, which is precisely equivalent to having an eigenvector of eigenvalue $1$.

With this formulation, it becomes immediate: if $x_0 = \gamma(t)$ is a constant solution, then this operator becomes \[ h \mapsto -J (\nabla_t h + \nabla_h X_H. \] However, we know that $X_H = J \nabla H$ and $X_H(x_0) = 0$, so $\nabla_h (J \nabla H) = (\nabla_h J) \nabla H(x_0) + J \nabla_h \nabla H = J \nabla_h \nabla H$. This means the differential operator just simplifies to \[ -J \nabla_t + \nabla \nabla H. \] Since the eigenvalues of the Hessian are real, this operator has non-trivial kernel if and only if $\nabla^2 H$ has a zero eigenvalue.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.