## Length of Floer flow lines

Suppose $(X,\omega)$ is a closed symplectic manifold. Let $H$ denote a time-dependent Hamiltonian, all of whose critical points are non-degenerate, and fix an $\omega$-compatible time dependent family of almost complex structures.

Let $\mathcal{M}$ denote the set of all finite energy Floer flow lines. That is, maps $u:\mathbb{R} \times S^1 \rightarrow X$ that satisfy the Floer equation $\partial_s u + J(u)(\partial_t u - X_{H}(u))=0$, and satisfy $e(u) \lt \infty$, where $e(u)$ is the energy of $u$, defined by $e(u) = \int_{\mathbb{R}} \int_{S^1} |\partial_s u |^2 dsdt$.

Then it's well known that if $u$ is any map satisfying the Floer equation, then $u\in \mathcal{M}$ (i.e. $u$ has finite energy) is equivalent to asking for one of the following two properties:

1. $u \in \mathcal{M}(x,y)$ for two 1-periodic orbits $x,y$ of $H$, that is, $\lim_{s \rightarrow \infty}u(s,t)=x(t)$ and $\lim_{s \rightarrow -\infty}u(s,t)=y(t)$.

2. $u$ decays exponentially - there exist constants $C,\delta$ such that $|\partial_s u(s,t)| \lt Ce^{-\delta |s|}$.

This means that if $u \in \mathcal{M}$ then the length $l(u(\cdot,t)) = \int_{\mathbb{R}} |\partial_s u(s,t)|ds$ is finite for each $t \in S^1$.

Fix two 1-periodic orbits $x,y$. Then if $u \in \mathcal{M}(x,y)$, not only is the energy finite, but there is a uniform bound on the energy $e(u)$ for any such $u$ - namely $\mathcal{A}_{H}(x)-\mathcal{A}_H(y)$ - where $\mathcal{A}_H$ is the action functional.

My question is the following: do there exist uniform bounds (i.e. depending only on $x$ and $y$) on the length $l(u(\cdot,t))$ for every $u \in \mathcal{M}(x,y)$?

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 This should be true for cotangent bundles with quadratic Hamiltonians - length is bounded by energy, which in turn is related to the Hamiltonian action. – Orbicular Feb 15 2011 at 12:54

In your symplectically aspherical setting, bounds on length will indeed exist.

Suppose one has a sequence of solutions $u_n$ to Floer's equation, of bounded energy, and a sequence of points $t_n\in S^1$ with lengths $l(u_n(\cdot,t_n))\to \infty$. Gromov-Floer compactness tells us that after passing to a subsequence, this sequence converges in the Gromov-Floer topology to a broken trajectory plus bubbles. Each cylinder in the domain of the broken trajectory is associated with a time-translation $\sigma_k \colon s\mapsto s+s_k$, and the maps $\sigma_k^\ast u_n$ converge to a limiting cylinder $v_k$ plus bubbles. Outside a long but finite cylinder $[-T,T] \times S^1$, $\sigma_k^\ast u_n$ is exponentially close to $v_k$ (meaning $\leq ce^{-as}$, where $a>0$ depends only on the asymptotic limits $x$ and $y$, and $c$ only on $u_n(s_k,\cdot)$). Similar decay applies to the first derivative of $u_n$. The divergence of lengths must therefore occur inside the finite cylinder.

Inside the finite cylinder but away from finitely many "bad points", $\sigma_k^\ast u_n$ converges to $v_k$, and for any $r$, convergence in $C^r$ is uniform on compact subsets. At the bad points, bubbles form. In the symplectically aspherical case, bubbling is impossible: there are no bad points. Therefore one has uniform $C^1$ bounds on $\sigma_k^\ast u_n$ over the finite cylinder, contradicting the divergence of the lengths.

Maybe it would be interesting to write down a bubbling sequence of holomorphic maps $\mathbb{CP}^1\to \mathbb{CP}^1$ and watch what happens to the lengths.

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 This rather technical answer got three downvotes in quick succession, with no explanation. So maybe there's something drastically wrong with it... or maybe there's not. – Tim Perutz Feb 19 2011 at 14:07 Actually, considering that the question also received downvotes, my guess is that these votes were not comments on the mathematical content of either. – Tim Perutz Feb 20 2011 at 16:31 This question was actually asked on my behalf, so thanks Tim for your answer (and apologies for the delay). Let me also add an answer of my own: in the non-compact case (which is what we were primarily interested in), the existence of such length bounds seems to be essentially equivalent to the existence of $L^{\infty}$ bounds on gradient flows lines. One direction uses Tim's answer - as once $L^{\infty}$ bounds are established one can invoke Gromov compactness as in the closed case. The converse can be made explicit at least in the case of quadratic Hamiltonians on cotangent bundles, say. – Will Merry May 24 2011 at 8:23