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For $C$ a compact convex set of $\mathbb{R}^n$ $(n\geq 1)$ whose interior is nonempty and whose boundary $\partial C$ is a smooth manifold, is $\partial C$ homeomorphic to the sphere $\mathbb{S}^{n-1}$?

Gregoire Panel
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Yes. You can suppose that $B_1(0)\subset C$. Then in the halfline of positive scalar multiples of every unit vector $v$ there is only one point of $\partial C$, because if $\lambda>\mu>0$, $\mu v$ is in the convex envelope of $B_1(0)$ and $\lambda v$, which is open, and so it can't belong to $\partial C$.

So the continuous function $\rho:{\mathbf R}^n\setminus 0 \to {\mathbf S}^{n-1}$ restricts to a bijective continuous function $\partial C\to {\mathbf S}^{n-1}$. Since $\partial C$ is compact and ${\mathbf S}^{n-1}$ is Hausdorff's, by a famous lemma of topology it is a homeomorphism.

By extending it linearly on every ray, it can even been proved that $C$ is homeomorphic to $\overline{B_1}(0)$.

Romulus Augustulus
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    Thank you, is it possible to prove that $\partial C$ diffeomorphic to the sphere $\mathbb{S}^{n-1}$? – Gregoire Panel Dec 21 '20 at 11:24
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    It depends on the regularity. If $\partial C$ is a $C^k$ hypersurface, it is $C^k$-diffeomorphic to $S^{n-1}$. It is because we already know a homeomorphism, which is the restriction of a $C^k$ function. To prove that the inverse is $C^k$, by the inverse function theorem, you only need to prove that the differential of $\rho:\partial C\to S^{n-1}$ is invertible everywhere. It is the restriction of the differential of $\rho:\mathbf{R}^n\setminus{0}\to S^{n-1}$, which kernel at every point is spanned by the radial vector. (continue) – Romulus Augustulus Dec 21 '20 at 12:22
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    So to prove the invertibility one has to prove that the tangent space of $\partial C$ at $p$ doesn't contain that vector. This can be proved observing that $C$ contains a cone with vertex $p$ (the one that joines $p$ with the points of $B_1(0))$. – Romulus Augustulus Dec 21 '20 at 12:25