Relationship between the zeros of a vector field and the fixed points of its flow

Question

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!

Answer 1

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.

Answer 2

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.

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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