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 $\Bbb R^n$?

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    $\begingroup$ The claim that all geodesically convex sets are contractible is clearly false. $\endgroup$ Commented Jul 26, 2016 at 1:44
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    $\begingroup$ Your proof is correct, but it applies only to special convex sets. $\endgroup$ Commented Jul 26, 2016 at 2:50
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    $\begingroup$ The standard definition of geodesic convexity does not require uniqueness, only existence. $\endgroup$ Commented Jul 26, 2016 at 3:33
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    $\begingroup$ @studiosus I own nine books that define it with uniqueness, and one that defines it in the "standard" way. Perhaps you can understand my confusion. $\endgroup$
    – Ryan Unger
    Commented Jul 26, 2016 at 3:47
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    $\begingroup$ Interesting. The definition you are using I know as strong convexity which incidentally is the terminology that you use in your proof. $\endgroup$ Commented Jul 26, 2016 at 4:10

1 Answer 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. $\quad\Box$

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    $\begingroup$ You claim that a strongly convex neighbourhood $U$ is automatically normal about each of its points. At first I was not sure about whether this is true, as besides the unique minimizing geodesic between two points there might be another geodesic which does not minimize lengths. This option however may be ruled out by an argument along the lines of the answer to this question . $\endgroup$
    – Jan Bohr
    Commented Mar 12, 2018 at 9:26
  • $\begingroup$ However, in order to prove the Theorem used in Bott and Tus book, one does not need this implication: The neighbourhood that do Carmo provides on pg. 76 is totally normal for other reasons and one could theoretically argue with sets satisfying both properties (strongly convex and normal about each of their points), since both properties remain true upon taking finite intersections. $\endgroup$
    – Jan Bohr
    Commented Mar 12, 2018 at 9:30

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