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Every smooth embedding of $S^2$ into $\mathbb{R}^3$ has at least one umbilic point (in fact, the recent proof of the Caratheodory conjecture yields two such points). The usual proof of this is to use the Hopf index lemma. Alternatively, one can appeal to the hairy ball theorem. These proofs don't seem to generalize to higher dimensional hypersuraces, which motivates my question:

Question: Does every topologically spherical hypersurface of $\mathbb{R}^{n+1}$ have an umbilic point? If not, are there any known conditions which guarantee the existence of such a point?

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    $\begingroup$ In higher dimensions, do you want to define 'umbilic point' to be a point where two of the eigenvalues are equal or a point where all of the eigenvalues are equal? $\endgroup$
    – Robert Bryant
    Commented Feb 25, 2016 at 16:22
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    $\begingroup$ I'm looking for points where all curvatures are equal. I guess that is too much to ask? $\endgroup$
    – User0.9999999.....
    Commented Feb 26, 2016 at 0:29
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    $\begingroup$ Yes, it's too much to ask that all the curvatures be equal. When $n>2$, the general ellipsoid in $\mathbb{R}^{n+1}$ will have no points where all the principal curvatures are equal. (Just do the computation for $n=3$ and you will see why.) Only for $n=1$, $3$, or $7$ is there any chance of having an $n$-sphere (convex or not) in $\mathbb{R}^{n+1}$ with $n$ distinct principal curvatures at each point. (For $n=1$, this is trivial, of course.) There are examples (not convex and not embedded) of $3$-spheres immersed in $\mathbb{R}^4$ that have $3$ distinct principal curvatures at each point. $\endgroup$
    – Robert Bryant
    Commented Feb 28, 2016 at 16:39

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Because people have asked for it, I thought I would supply an example of what I mentioned in my comment above, an immersion of the $3$-sphere into $\mathbb{R}^4$ that has three distinct principal curvatures at every point. I'm sure that this example is well-known, but I don't know, off-hand, an explicit place where it is written up. [When I get back home, I imagine that I'll be able to find such a reference, but it's difficult to do while I'm traveling.]

The simplest example that I know is the following one: Consider $\mathbb{R}^5$ as the space of symmetric $3$-by-$3$ matrices with real entries and trace zero, and let $A\in \mathrm{SO}(3)$ act on $m\in\mathbb{R}^5$ by $A\cdot m = AmA^T$.

If $m_0\in\mathbb{R}^5$ has three distinct eigenvalues $\lambda_1<\lambda_2<\lambda_3$ such that ${\lambda_1}^2+{\lambda_2}^2+{\lambda_3}^2= 1 = \mathrm{tr}({m_0}^2)$, then the orbit $\mathrm{SO}(3)\cdot m_0\subset\mathbb{R}^5$ is a hypersurface in $S^4\subset\mathbb{R}^5$ that has three distinct principal curvatures at each point (as a hypersurface in $S^4$). Since it is also a quotient of $\mathrm{SO}(3)$ by a finite subgroup (of order $4$), and since $\mathrm{SO}(3)$ is, itself, double-covered by $S^3 = \mathrm{Spin}(3) = \mathrm{SU}(2)$, it follows that this hypersurface is the image of an immersion of $S^3$ into $S^4$ with three distinct (constant) principal curvatures.

Now, $S^4$ is conformally flat, and after removing a single point $p\in S^4$ that is not in $\mathrm{SO}(3)\cdot m_0\subset\mathbb{R}^5$, one can find a conformal diffeomorphism $\phi:S^4\setminus \{p\}\to \mathbb{R}^4$. The submanifold $\phi\bigl(\mathrm{SO}(3)\cdot m_0\bigr)\subset\mathbb{R}^4$ is then the image of an immersion of $S^3$ into $\mathbb{R}^4$ that has three distinct principal curvatures at each point.

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