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I would like to know if there are any conditions on a compact hypersurface $M^n \subset \mathbb{R}^{n+1}$ that ensure that $M$ is not a minimal hypersurface. The motivation for this question is because I would like to know if there is a generalization of this result and a possible generalization of this result is used on page $786$ of this paper, but the hypotheses for this generalization are not clear for me.

Thanks in advance!

$\textbf{EDIT:}$

Following the hint given by Ted, I tried a proof by finite induction on the dimension of $M^n$:

  1. Base step ($n = 2$):

Is was done in the linked topic.

  1. Induction hypothesis:

a compact hypersurface $M^n \subset \mathbb{R}^{n+1} \Longrightarrow M^n$ is not a minimal hypersurface.

  1. Induction step:

I will prove that induction hypothesis remains for $n+1$ arguing by contradiction. Suppose that exists $p \in M$ such that $H(p) = 0$, then

$$\kappa_1(p) + \cdots + \kappa_{n+1}(p) = 0 \Longrightarrow \kappa_{n+1}(p) = -(\kappa_1(p) + \cdots + \kappa_n(p)) \ (\star)$$

Let be $P_{n+1}$ a hyperplane going through $p$ such that the curvature $\kappa_{n+1}(p)$ of $N^n := M^{n+1} \cap P_{n+1}$ is zero, then I can apply the induction hypothesis to $N^n$ to get

$$\kappa_1(p) + \cdots + \kappa_n(p) \neq 0.$$

This and $(\star)$ imply that $\kappa_{n+1}(p) \neq 0$ on $N^n$, which is a contradiction with the fact that $\kappa_{n+1}(p) = 0$ on $N^n$. $\square$

George
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  • I don’t understand your argument. The $\kappa_i$, the principal curvatures, are curvatures of slices by normal $2$-planes, not Gaussian curvatures. – Ted Shifrin Apr 03 '20 at 04:37

1 Answers1

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No compact hypersurface in Euclidean space can be minimal. There is always a point of positive Gaussian curvature, and the mean curvature at such a point cannot be $0$ (why?).

Ted Shifrin
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  • Thanks for the hint, I sketched a proof by finite induction, but I did not use the compactness in the induction step, only in the base step, can you look and see if it is looks good? – George Apr 03 '20 at 01:56
  • My hint is to give the standard argument that works for surfaces verbatim in higher dimensions. I don’t follow your argument. If you don’t explicitly use compactness, it cannot be correct. – Ted Shifrin Apr 03 '20 at 04:39
  • Can you give me a hint how to give the standard argument in higher dimensions? This was my first thought so much that I linked the topic to the case of surfaces ($M^2 \subset \mathbb{R}^3$), but I really can not see how to apply the same argument in higher dimensions, even if I consider $M^3 \subset \mathbb{R}^4$. I would have $\kappa_1 = - \kappa_2 - \kappa_3$ and $\kappa_1\kappa_2\kappa_3 = - (\kappa_2)^2 \kappa_3 - \kappa_2 (\kappa_3)^2 > 0$ in this case and the inequality would imply $\kappa_3 \kappa_2 < 0$, therefore $\kappa_3 > 0$, $\kappa_2 < 0$ and $\kappa_1 < 0$ – George Apr 03 '20 at 15:48
  • for $\kappa_1 < \kappa_2 < \kappa_3$ – George Apr 03 '20 at 15:49
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    See Proposition 3.5 on pp. 61-62 of my differential geometry text (linked in my profile). In your case, it's not good enough to know that the Gaussian curvature at the point is positive. What you want is the fact that all the principal curvatures are non-zero and have the same sign. (This will follow because each normal slice will be at least as curved as the corresponding slice of the circumscribing sphere.) – Ted Shifrin Apr 03 '20 at 17:34