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Theorem. (Reeb) If $M$ is a compact manifold and $f$ is a differentiable function on $M$ with only two critical points, both of which are nondegenerate, then $M$ is homeomorphic to a sphere.

Proof) The two critical points must be the minimum and maximum points. Say that $f(p)=0$ is the minimum and $f(q)=1$ is the maximum. If $\epsilon$ is small enough then the sets $f^{-1}[0,\epsilon]$ and $f^{-1}[1-\epsilon,1]$ are closed $n$-cells by the Morse lemma.

What I can't understand is the last sentence. The Morse lemma is stated below. Since $f(p)=0$ is the minimum, we must have $f=(y^1)^2+\cdots+(y^n)^2$ near $p$. For small enough $\epsilon$, we will have $\{y\in U: (y^1)^2+\cdots+(y^n)^2\leq \epsilon \} \subset f^{-1}[0,\epsilon] $, and the former set is the closed $n$-disk. But how can we show the reverse inclusion $\{y\in U: (y^1)^2+\cdots+(y^n)^2\leq \epsilon \} \supset f^{-1}[0,\epsilon]$ holds for small $\epsilon$?

Morse Lemma. Let $p$ be a nondegenerate critical point of $f$. Then there is a local coordinate system $(y^1,\dots,y^n)$ in a neighborhood $U$ of $p$ with $y^i(p)=0$ for all $i$ and such that the identity $f=f(p)-(y^1)^2-\cdots-(y^\lambda)^2+(y^{\lambda+1})^2+\cdots+(y^n)^2$ holds throughout $U$, where $\lambda$ is the index of $f$ at $p$.

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2 Answers 2

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Let us look closer at the implications the normal form for $f$ given by Morse Lemma. Centered at $p$, being the minimum point (in particular a local minimum), there is a chart $(y,U)$ with $y(r)=\big(y^1(r),\ldots, y^n(r)\big)$ for $r\in U$, such that $$f\circ y^{-1}\big(y^1(r),\ldots, y^n(r)\big)=y^1(r)^2+\ldots+y^n(r)^2,$$ for any $r\in U$. This is the precise meaning of $f=(y^1)^2+\ldots+(y^n)^2$ near $p$.
Now, take $\epsilon>0$ such that $y(U)\supset \mathbb{D}^n_{\epsilon}$, where $\mathbb{D}^n_{\epsilon}$ is the closed ball in $\mathbb{R}^n$ with center $0_{\mathbb{R}^n}$ and radius $\epsilon^{\frac{1}{2}}$. You can do this, since $y(U)$ is an open neighborhood of $y(p)=0_{\mathbb{R}^n}$. Clearly, \begin{align} f^{-1}[0, \epsilon]&\supset\{r\in U\ |\ 0\leq f(r)\leq\epsilon\}\\ &= \{r\in U\ |\ 0\leq f\circ y^{-1}\big(y(r)\big)\leq\epsilon\}\\ &= \{r\in U\ |\ 0\leq y^1(r)^2+\ldots+y^n(r)^2\leq\epsilon\}. \end{align} For the other inclusion, put a Riemannian metric $\langle, \rangle$ on $M$ and define the gradient of $f$, $\nabla f$, by $$\langle \nabla f_p, X_p \rangle=df_p(X_p).$$ Since $M$ is compact, $\nabla f$ is a complete vector field. Denote its flow by $\phi:M\times\mathbb{R}\rightarrow M$.
Assume there is $r\in M\setminus U$ such that $f(r)=\epsilon_0\in [0,\epsilon]$.
For any $t\in\mathbb{R}$, \begin{align} \frac{d}{dt}f\phi(t,r)&=df_{\phi(t,r)}\big(\frac{\partial}{\partial t}\phi(t,r)\big)\\ &=\langle \frac{\partial}{\partial t}\phi(t,r), \nabla f_{\phi(t,r)}\rangle\\ &=|\nabla f_{\phi(t,r)}|^2>0, \end{align} since $r$ cannot be in the orbits of $p$, $q$, being constant the integral curves of $\nabla f$ through these points. Hence, the map $\mathbb{R}\rightarrow [0,1]$, $t\mapsto f\phi(t,r)$ strictly increasing.
Let $$[0,\infty)\ni\delta=\inf_{(-\infty,0]} |\nabla f_{\phi(t,r)}|^2.$$ If $\delta>0$, then \begin{align} f\phi(0,r)-f\phi(t,r)&=\int^0_t \frac{d}{ds}f\phi(s,r)\\ &=\int^0_t |\nabla f_{\phi(s,r)}|^2 ds\\ &\geq \int^0_t \delta ds = -\delta t. \end{align} Making $t\to -\infty$ gives a contradiction, since the LHS is a number in $[0,1]$.
So, $\delta=0$ and we have a sequence $\{t_n\}_n\subset (-\infty,0]$ such that $$\lim_{n\to\infty}|\nabla f_{\phi(t_n,r)}|=0.$$ If $\{t_n\}_n$ had a bounded subsequence, we could rename it to $\{t_n\}_n$ and assume w.l.o.g that $t_n\to t_0\in (-\infty,0]$. Hence, $\nabla f_{\phi(t_0,r)}=0$, and this is absurd, as we said before.
Thus $t_n\to -\infty$. Again, since $M$ is compact we can assume w.l.o.g that $$\lim_{n\to\infty}\phi(t_n,r)=r_0\in M,$$ and then $r_0=p$. This shows that $$\lim_{t\to -\infty}\phi(t,r)=p.$$ In particular, the curve $t\mapsto \phi(t,r)$ intersects the $(n-1)$-dimensional sphere $f^{-1}(\epsilon_0)\cap U$, for some $t<0$. Of course this cannot happen, as $f$ strictly increases along the integral curve $t\mapsto \phi(t,r)$.
Thus, such an $r$ does not exist, and we get the other inclusion.

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    $\begingroup$ I can't see why the statement "since $f:M\to \Bbb R$ is continuous, $p\in U$, $U$ is open,and $f(p)=0$, there is an $\epsilon>0$ such that $f^{-1}(-\epsilon,\epsilon)\subset U$. This is not true if $M=\Bbb R$, $f\equiv 0$, $p=0$ and $U=(-1,1)$ (which is an example that does not satisfying the assumptions of the theorem, though) $\endgroup$
    – blancket
    Sep 25, 2020 at 3:26
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If this is wrong, we have $\epsilon_n\to 0$ such that $f^{-1}[0,\epsilon_n]$ doesn't included in $U$. But then we would have $x_n\notin U$ that $f(x_n)\leq \epsilon_n$. Since M is compact, we can select a convergent subsequence, assume it converge to a point $x_0$, then $f(x_0)=0$, so $x_0$ is the minimum point, which is a contradiction.

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