Why are geodesically convex sets diffeomorphic to $\Bbb R^n$?

In the construction of a good cover for a manifold $M^n$, Bott & Tu use the fact that each point in $M$ is contained in a geodesically convex set (after picking a Riemannian metric). They then claim that the intersection of geodesically convex sets is a geodesically convex set, and that makes sense. They further claim that any geodesically convex set is diffeomorphic to $\Bbb R^n$. For this, they reference Spivak, who does not actually show the last bit.

Perhaps one way to do this is to show that the convex set is star-shaped, so the inverse image of it under the exponential map is convex, and somehow diffeomorphic to a ball, and thus to $\Bbb R^n$?

Solution 1:

We provide a careful proof of Theorem 5.1 from Bott & Tu. The main body of the proof is taken from this MO post.

Let $M^n$ be a smooth manifold. An open cover $\{U_\alpha\}$ is good if each nonempty finite intersection $U_{\alpha_1}\cap\cdots\cap U_{\alpha_k}$ is diffeomorphic to $\Bbb R^n$. The goal here is to prove the following

Theorem 1. Every manifold has a good cover.

Equip $M$ with a Riemannian metric $g$. An open set $U\subset M$ is said to be strongly convex if $p,q\in U$ can be joined by a unique minimizing geodesic contained totally in $U$.

Lemma 2. Each point in $M$ is contained in a strongly convex neighborhood.

Proof. See do Carmo, Riemannian Geometry, page 76. $\quad\Box$

A strongly convex neighborhood $U$ is normal about each point in $p\in U$, that is, $\exp_p:T_pM\to M$ is always a local diffeomorphism.

Proposition 3. $\exp_p^{-1}U$ is star-shaped about $0\in T_pM$.

Proof. $p$ can be connected to any $q\in U$ by a unique geodesic $\gamma$. Using normal coordinates, we see that $0$ can be connected to $\exp_p^{-1}q$ by the straight line $\exp_p^{-1}\gamma$.$\quad\Box$

The difficult part of this proof lies in the following technical lemma:

Lemma 4. Let $\Omega\subset\Bbb R^n$ be open and star-shaped about $0$. Then $\Omega\approx \Bbb R^n$.

Proof. The set $F=\Bbb R^n-\Omega$ is closed, so me may find a smooth map $\phi:\Bbb R^n\to \Bbb R_{\ge0}$ such that $F=\phi^{-1}(0)$. (See Lee, Introduction to Smooth Manifolds, page 47.) Define $\lambda:\Omega\to\Bbb R$ by $$\lambda(x)=1+\left(\int_0^{||x||}\frac{\mathrm d t}{\phi(t\hat x)}\right)^2.$$ Since $\phi$ is nonnegative, $\lambda(t\hat x)$ is an increasing function of $t$. Clearly $\lambda$ is smooth. Define a smooth function $f:\Omega\to\Bbb R^n,x\mapsto \lambda(x)x$. Let $A(x)=\sup\{t>0\mid t\hat x\in\Omega\}$. Since $\lambda$ increases along $t\hat x$, $f$ maps the segment $[0,A(x))\hat x$ injectively into $\Bbb R_{\ge 0}\hat x$. Note that $f(0)=0$.

We now want to show that $f(\Omega)=\Bbb R^n$. We need to investigate the behavior of $f$ close to the boundary of $\Omega$. Thus, consider the limit $$L(x)=\lim_{r\to A(x)}||f(r\hat x)||=\lambda(A(x))A(x).$$ If $A(x)=\infty$, then $L(x)=\infty$. The set $[0,A(x))\hat x$ is bounded if $A(x)<\infty$, so it contained in some compact set. Then $D\phi$ is bounded on this compact set, and we may apply Taylor's Theorem with remainder. Namely, since $\phi(A(x)\hat x)=0$, we have $$\phi(r\hat x)=||\phi(A(x)\hat x+r\hat x-A(x)\hat x)-\phi(A(x)\hat x)||\le M||(r-A(x))\hat x||=M(A(x)-r),$$ where $M>0$ is some constant. Thus $$\int_0^{A(x)}\frac{\mathrm d t}{\phi(t\hat x)}\ge \int_0^{A(x)}\frac{\mathrm d t}{M(A(x)-t)}=\infty,$$ so $L(x)=\infty$ in this case too. Thus $f$ is surjective.

We will show that $f$ has no critical points. Suppose that $Df_x(h)=0$ for some $h,x\in\Bbb R^n$. From the definition of $f$, we obtain $Df_x(h)=\lambda(x)h+D\lambda_x(h)x=0$. Since $\lambda(0)=1$, this implies $x\ne 0$. Thus $h=\mu x$, for some $\mu\ne 0$. This implies $\lambda(x)+D\lambda_x(x)=0$. The function $g(t)=\lambda(tx)$ is increasing, so $g'(1)=D\lambda_x(x)>0$. This contradicts $D\lambda_x(x)=-\lambda(x)\le -1$. Thus $f$ is regular on $\Omega$. By the inverse function theorem, $f$ is a local diffeomorphism. A bijective local diffeomorphism is a global diffeomorphism. Thus $f:\Omega\to\Bbb R^n$ is the desired diffeomorphism. $\quad\Box$

Any strongly convex neighborhood is diffeomorphic to a star-shaped open set in $T_pM$, thus diffeomorphic to a star-shaped open set in $\Bbb R^n$, and finally diffeomorphic to $\Bbb R^n$ itself.

Theorem 5. A geodesically convex open set is diffeomorphic to $\Bbb R^n$.

Lemma 6. If $U_1$ and $U_2$ are strongly convex, then so is $U_1\cap U_2$, assuming it is nonempty.

Proof. $U_1\cap U_2$ is open. Let Let $p,q\in U_1\cap U_2$. Let $\gamma\subset U_1$ be the geodesic starting at $p$ going to $q$. In $U_1$, set up normal coordinates at $p$, so that $\gamma$ is a straight line going to $q$. We may also do this in $U_2$, so $\gamma$ is the geodesic connecting $p$ and $q$ in $U_2$ as well. Thus $\gamma\subset U_1\cap U_2$, and $U_1\cap U_2$ is geodesically convex.$\quad\Box$

Proof of the Theorem. By Lemma 2, we may cover $M$ with geodesically convex open sets. Then each set in the cover is diffeomorphic to $\Bbb R^n$, and by Lemma 6, each finite nonempty intersection is as well. $\quad\Box$