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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.

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2 Answers 2

up vote 2 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.

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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

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.

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