Is there a bound on the length of the longest Morse trajectory?

Given a compact Riemannian manifold (with a fixed metric) and a Morse function on it (also fixed). Is there a bound (depending on the metric and the Morse function) on the length of the Morse trajectories? (You can assume the Morse-Smale condition if helpful.)

EDIT (In response to Dick's answer): The Morse function and the metric are fixed. I am just looking for something like $\int_{-\infty}^{\infty}\| \nabla f(\phi_t(p))\|dt\leq C$ and $C=C(f,g)$ where $f$ is the Morse function in question, $g$ stands for the metric and $\phi$ denotes the flow of $-\nabla f$. Note here that the constant is independent of the starting point $p$, as it is easy to see that such a constant additionally depending on $p$ exists (you use the hyperbolicity of $\nabla f$ to deduce exponential convergence towards a critical point). Furthermore it is also easy to see that the above integral is bounded if you include a $2$ in the exponent of the norm (a.k.a. $L^2$), as $\| \nabla f(\phi_t(p))\|^2=-\frac{d}{dt}f(\phi_t(p)).$

EDIT2 (In response to Bill's answer): I changed the statement to make it abundantly clear that the bound may depend on the metric and the Morse function. Maybe I was somewhat unclear in my formulation - sorry for that.

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the point of the answers given below is that you are not completely clear about what you meant by $C = C(f,g)$. Do you actually want a practical bound? Or just a statement that the length is finite? The way you stated the question sounds like you want the former. In that case, what is interesting is what does $C$ depend on. And Thurston's answer shows that $C$ cannot just depend on, say, the $C^1$ norm of the Morse function. – Willie Wong Oct 4 '10 at 17:14
Incidentally, the compactification of the space of flow lines is used in proving the existence of Morse homology: en.wikipedia.org/wiki/Morse_homology – Ian Agol Oct 4 '10 at 18:35
@Agol: The function {point \mapsto length of the flow line through that point} is not continuous, as you can easily convince yourself. Hence I do not know how to use the compactification of flow lines. (The compactification, by the way, uses L^2 bounds, not L^1 bounds!) – Orbicular Oct 4 '10 at 18:46
@ Willie: The integral I wrote above popped up in my research (f and g are auxiliary data and hence fixed). I just needed a bound for it. This bound can depend on this auxiliary data. That's what I mean by $C=C(f,g)$ Maybe you can suggest a "good way" (i.e. understandable without disambiguity) of posing the question. – Orbicular Oct 4 '10 at 18:49
@Orbicular: you're correct, but flowlines are compactified by broken flow-lines. So the space of broken flow-lines is compact, and the length is continuous. Incidentally, there is no compactness in the space of circle-valued Morse functions, which leads to the necessity of introducing Novikov rings. – Ian Agol Oct 4 '10 at 19:59

4 Answers

As explained by Bill Thurston, on a given Riemann manifold $(M,g)$ you may really have Morse functions with gradient lines of any length. However, as now you are asking for a bound in terms of the Morse function $f$, the following argument shows why lengths of all Morse trajectories (MT) are bounded, and how to get a quantitative bound provided you are able to evaluate simpler geometric quantities.

Thanks to the hyperbolic structure of the flow near critical points, any critical point $x$ has a nbd $U_x$ such that any flow line crosses $U_x$ in an arc interval, and the length of the arc interval is bounded above by a constant $c=c(f,g).$ In order to do this quantitatively you may use the Morse lemma. Also, if the $U_x$ are taken small enough, and assuming the flow is Morse-Smale, any flow line or MT can possibly meet them only in strictly decreasing order of the relative Morse indices: in particular, it can meet at most $\dim(M)+1$ of them. In conclusion, the contribute to the length of any flow line or MT inside the set $U:=\cup _ {x\in\operatorname{crit}(f)} U _ x$ is bounded by $(\dim(M)+1)c.$

Now consider an arc of gradient line $\gamma$ in $M\setminus U$ of length say $L$.You can (re)parametrize it with respect to arc-length on the interval $[0, L].$ It solves the ODE $\gamma'(s)=-\frac{\nabla f(\gamma(s))}{\|\nabla f(\gamma(s))\|},$ so this time the derivative of $f$ along $\gamma$ wrto $s$ is $\frac{d}{ds}f(\gamma(s))=-\|\nabla f(\gamma(s))\|$ and integrating you get

$f(\gamma(0))-f(\gamma(L))= \int_0^L \|\nabla f(\gamma(s))\|ds \geq L\\ \min_ {M\setminus U} \|\nabla f\|$.

In conclusion, for any gradient line or MT $\Gamma$, summing over the components of $\Gamma\setminus U$ and $\Gamma\cap U$ one gets this bound on the length $$\mathrm{length}(\Gamma)\leq (\dim(M)+1)c + \frac{\max_M(f)-\min_M(f)}{ \min_ {M\setminus U} \|\nabla f\| }$$

(BTW, note that, as a consequence, you can reparametrize on $[0,1]$ all Morse trajectories so as to be Lipschitz of constant $C(f,g)$, and that by Ascoli-Arzelà these are therefore a compact set in the uniform distance).

 A more realistic variant, from a quantitative point of view. If the flow is not asssumed to be Morse-Smale one can simply bound the first term (the contribute to the length inside $U$) with a bound $N$ on the number of critical points, times the constant $c$. This way the choice of the nbd's $U_x$ is only subjected to local properties (on the contrary, telling how small they have to be in order to ensure the mentioned monotonicity of the Morse indices, requires informations on global properties of the flow).

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Thanks, Pietro. The proof will generalize to the case of a fixed Morse function f on a Hilbert manifold (with fixed complete metric) assuming the PS condition if one fixes the endpoints, right? Because the f-values on the Morse trajectory stay in a finite windows. Hence the number of possible critical points it comes close to is finite. Furthermore dim(M) in your equation should be replaced by the relative Morse index. – Orbicular Oct 4 '10 at 19:02
Yes, that's correct. – Pietro Majer Oct 4 '10 at 19:17
Excuse me, just a small question: how does one obtain the neighborhood $U_x$ and the bound $c=c(f,g)$? – Orbicular Nov 29 '10 at 17:50

When a function is very close to flat, the trajectories tend to be very erratic and wandering. The answer is NO, even if linear bounds are specified on the derivatives of the function.

To be more specific: in the unit sphere, draw any smooth path $\gamma$ starts from the south pole, bottom, ends at the north pole, and weaves and wanders as much as you like, for as long as you like. Now define a function $f$ on a neighbhood of $\gamma$ that is 0 at the bottom, $1$ at the top, and for which $\gamma$ is a trajectory of the gradient, and the north pole and south pole are critical points. You can do this by using closest-point projection in a regular neighborhood. Smooth functions on a regular neighborhood can be extended to $C^\infty$ funtions on $S^2$, and a smooth extension can be perturbed away from $\gamma$ to be a Morse function, so the particular curve is a gradient line of a Morse function with arbitrary length.

Funtions that have a path like $\gamma$ in their gradient flow are obviously very inefficient. If you want functions with more efficiency, you could look at linear combinations of eigenfunctions of the Laplacian with small eigenvalue. In those cases, I think you can get reasonable inequalities concerning the average length of gradient flow lines; this is related to the known and widely used Cheeger type relationships between diameter of manifolds (or graphs), size of separators, and eigenvalues of the Laplacaian. I'm not sure what you can conclude about the maximum length of gradient flow lines, but I suspect something could be done, and may well be known.

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Maybe I'm misunderstanding something, but once your $\gamma$ is fixed there should be a bound on the length of the longest Morse trajectory, right? Obviously there is no bound independent of the function and the metric (as already pointed out by Dick Palais). – Orbicular Oct 4 '10 at 15:45
Yes, of course, but I was answering an earlier version of the question that I understood to ask for a bound depending only on the metric. – Bill Thurston Oct 4 '10 at 23:10

I think you will need to be more precise and say in terms of what you want to bound the lengths. Of course there cannot be an absolute bound, since to multiply the lengths of all curves (and in particular of all the Morse trajectories) by $a$ you can just scale the metric by $a$

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If everything in sight is real-analytic you can use \Lojasiewicz's results to bound the Morse trajectories. See e.g. http://front.math.ucdavis.edu/9703.5224

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