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I'm having a little trouble here and would appreciate some hints.

Let $M$ be a compact manifold without boundary and let $X$ be a smooth vector field on $M$ with only isolated zeros. Let $\theta_t$ denote the flow of $X$.

I would like to show that the following is true:

For small enough $t > 0$ the only fixed points of $\theta_t$ are the zeros of $X$.

I need this, since I was trying to write down a proof of Poincaré-Hopf by showing that for the index $\iota(X)$ of $X$ we have $$\iota(X) = L(\theta_t)$$ where $L(\theta_t)$ denotes the Lefschetz fixed point number of $\theta_t$. From the above equality it would immediately follow that $\iota(X) = L(\theta_t) = L(id) = \chi(M)$.

However I got stuck in proving that we can choose $t$ sufficiently small to ensure that no unwanted fixed points of $\theta_t$ turn up.

I think the statement should be true (I'm guessing compactness will be key). At least, it seems right to me.

Thanks!

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I can't answer the actual question, but compactness definitely is key: Consider the vector field on $S^1\times\mathbb{R}$ given by $V(\theta, t) = t \frac{\partial}{\partial\theta}$. It's easy to see that for any fixed time $t_0$, the point $(\theta, t_0/2\pi)$ is sent to itself under this flow. –  Jason DeVito Jun 27 '11 at 22:01
    
@Jason: Thanks for the counterexample to the general case! –  Sam Jun 28 '11 at 15:50
    
I can remove the topology from the question. The right question to ask is: Given a vector field on $\mathbb{R}^n$, with an isolated fixed point at $0$, are there positive numbers $\delta$ and $\epsilon$ such that, within a ball of radius $\delta$ around $0$, there are no cycles of period $<\epsilon$? If so, then choose such a $\delta$ ball around each fixed point, cover the rest of your manifold by open sets for which there is no cycle of length $< \epsilon$, and conclude by compactness that there is some finite $\epsilon$ so there are no cycles $\< \epsilon$ anywhere on your manifold. –  David Speyer Jun 28 '11 at 16:42
    
Quite right, yes. I have also thought about this. However, there might be some approach which takes into account the global structure instead of arguing locally (there might also not be such an approach of course, since any such global argument seems to be a stronger assertion than a proof of your observation)..? –  Sam Jun 28 '11 at 16:47
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My guess is that the answer to my question above is "yes". Sketchy argument: Suppose there were a sequence of cycles with period $\epsilon_i \to 0$. Suppose that these cycles were circles, of radii $\rho_i \to 0$, parameterized at constant speed $2 \pi \rho_i/\epsilon_i$. Between two opposite points of each cycle, the velocity changes by $2 (2 \pi \rho_i)/\epsilon_i$, over a distance of $2 \rho_i$. So the spatial derivative of your vector field is as large as $2 \pi/\epsilon_i $. This is unbounded, so your vector field can't be smooth. –  David Speyer Jun 28 '11 at 16:55

2 Answers 2

up vote 12 down vote accepted

I think I have found a more elementary approach to the problem, so I'll post it for anyone who might be interested (and maybe to check whether I haven't made some silly mistake).

The idea is actually quite simple: I approximate the flow to the first order and use this to get a lower bound on the periods of nonfixed points.

Proposition: Let $X$ be a smooth vector field on $\mathbb R^n$ such that $|X|$ and $|dX|$ are bounded. Then there is a $\tau >0$ such that for all $0<t<\tau$: $$\theta(t,p) = p \quad \iff\quad X(p) = 0$$

Proof: By Taylor expansion we have

$$\theta(t,p) = p + tX(p) + \int_0^t (t-\tau) \; dX\left(\theta\left(\tau,p\right)\right) X\left(\theta\left(\tau,p\right)\right) \; d\tau$$

By choosing $t_0$ small enough, we may assume

$$\left|X\left(\theta\left(\tau,p\right)\right)\right| \le 2\left|X(p)\right|$$

for all $p\in \mathbb R^n$ and $0\le \tau \le t < t_0$.

Edit: As has been pointed out by David Speyer in the comments, the existence of such a $t_0$ isn't as clear as I had initially thought. To see that such $t_0$ exists, we assume $|dX|<M$ for some $M>0$ and $|X| < \tilde M$. By Taylorexpansion we have

$$\left|X\left(\theta\left(\tau,p\right)\right)\right| \le |X(p)| + \tau M |X\left(\theta\left(s,p\right)\right)|$$

where $s \in [0,\tau]$ is chosen to maximize $|X\left(\theta\left(s,p\right)\right)|$. Now let $t_0 := 1/(2M)$. By iterating the same argument with $|X\left(\theta\left(s,p\right)\right)|$ we get the estimate

\begin{align*} \left|X\left(\theta\left(\tau,p\right)\right)\right| &\le |X(p)| + \tau M \; \big(\; |X(p)| + \tau M |X\left(\theta\left(s,p\right)\right)|\; \big) \\ &\vdots \\ &\le \sum_{k=0}^\infty \left(\tau M\right)^k |X(p)| + \lim_{k\to \infty} (\tau M)^k\tilde M \\ &\le 2|X(p)| \end{align*}

Now let us define $$\Phi(t,p) = p + t X(p)$$ From the above and the properties of $X$, there is some $C>0$ and $t_0>0$ such that for $0<t<t_0$

$$|\theta(t,p) - \Phi(t,p) | = \left|\int_0^t (t-\tau) \; dX\left(\theta\left(\tau,p\right)\right) X\left(\theta\left(\tau,p\right)\right) \; d\tau\right| \le C|X(p)|t^2$$

for all $p$. But then

\begin{align*} |\theta(t,p) - \theta(0,p)| &\ge |\Phi(t,p) - \theta(0,p)| - |\theta(t,p) - \Phi(t,p)| \\ &\ge t|X(p)| - t^2C|X(p)| \\ &= t|X(p)| (1 - Ct) \end{align*}

So if $p$ is a point such that $\theta(T,p) = \theta(0,p)=p$ it follows that either $X(p)=0$ or $T \ge C^{-1}$. Proving the proposition.

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The only point that worries me is the sentence beginning "By choosing $t_0$" small enough. I think it is true, but it isn't obvious to me. A naive continuity argument won't work, because it isn't clear that that '$|X(\theta(\tau, p))|/|X(p)|$' is continuous (as a function of $p$) near a fixed point. –  David Speyer Jun 30 '11 at 10:46
    
@David Speyer: Ah, yes. Thanks for pointing that out. I thought a simple Taylorseries argument would show this. But it actually doesn't seem to follow as easily, as I thought (if it is even true). –  Sam Jun 30 '11 at 18:10
    
@David Speyer: I have fixed it now, I think. –  Sam Jun 30 '11 at 18:50
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Nice, that works! (You have a typo: $t_0$ should be $1/(2M)$, not $M/2$.) This is definitely the right argument. –  David Speyer Jun 30 '11 at 19:08
    
Alright, thanks a bunch for your help and enthusiasm for this question again. :) –  Sam Jun 30 '11 at 19:37

This is true. I needed to use some higher tech than I would have expected to prove it; I'd be curious to hear a simpler solution.

Lemma: Let $M$ be a smooth manifold, $X$ a smooth vector field on $M$, $\theta$ the flow along $X$ and $x$ a point of $M$. Then there is a positive number $\epsilon$, and open sets $U \supset V \ni x$, such that, for $t \in (0, \epsilon)$, the flow $\theta_t$ takes $V$ into $U$, and the only fixed points of $\theta_t$ are the zeroes of $X$.

Proof: Since the statement is local, we may immediately assume that $X = \mathbb{R}^n$. We will fix a Euclidean norm on $\mathbb{R}^n$, so we may talk about arc-lengths, surface areas and so forth.

Take $U$ to be an open ball around $x$ (of finite radius). Choose $\epsilon'$ small enough that flowing from $x$ for time $\epsilon'$ stays within $U$. Choose $V$ a small enough ball around $U$ that flowing by time $\epsilon'$ keeps $V$ within $U$.

Let $K$ be a bound for $|\nabla \times X|$. (If $n=2$ or $3$, you presumably know what this means. In general, I mean to use the inner product on $\mathbb{R}^n$ to turn $X$ into a $1$-form, take $d$ of that $1$-form and then use the induced norm on $\bigwedge^2 \mathbb{R}^n$. I can't tell from your writing whether you are happy with this sort of manipulation -- if not, just think about the curl you are used to.) Our $\epsilon$ will be $\min(\epsilon', 4 \pi K^{-1})$.

Consider a nontrivial closed flow line, $\gamma$, in $U$. Let $T$ be the time taken to transverse $\gamma$; we will show $T>\epsilon$. Let $\sigma$ be the disc of minimal area with boundary $\gamma$. Let $L$ be the length of $\gamma$ and let $A$ be the area of $\sigma$.

By the Cauchy-Schwarz inequality, $$\int_{\gamma} |X|^{-1} ds \cdot \int_{\gamma} |X| ds \geq \left( \int_{\gamma} ds \right)^2 = L^2.$$ Here the integrals are with respect to arc-length.

Now, $\int_{\gamma} |X|^{-1} ds = T$. (The time to travel a path is the integral, over the length of the path, of the inverse speed. If you run one $7$-minute mile, and one $9$-minute mile, it's going to take you $16$ minutes to run two miles.)

Since $\gamma$ is a flow line for $X$, we see $\int_{\gamma} |X| ds = \int X \cdot ds$, the line integral of $X$ along $\gamma$. By Green's theorem, this is the same as $\int_{\sigma} \nabla \times X$, which is $\leq K A$.

Putting it all together, $$T \cdot (KA) \geq L^2.$$

Now, $\sigma$ is the minimal surface with boundary $\gamma$. By a result of Carleman, the isoperimetric inequality $A \leq L^2/(4 \pi)$ holds for $\sigma$ and $\gamma$. The best online refence I could find for the result of Carleman is this paper of Choe; Carleman's result is discussed in the third paragraph.

So $$T \cdot K \cdot L^2/(4 \pi) \geq L^2$$ and we deduce that $$T \geq 4 \pi/K \geq \epsilon$$ as desired. This proves the lemma.


Now that we have the lemma, we can find such a $(V,U, \epsilon)$ for every $x \in M$. We can find finitely many $V$'s with cover $X$ and, taking the minimum of the finitely many $\epsilon$'s, we find that $X$ has no nontrivial cycles with length $< \epsilon$. QED

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Thanks for the answer. I don't really know anything about minimal surfaces, but I'm quite happy with the rest of your argument. =) –  Sam Jun 29 '11 at 15:20

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