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I'm reading a paper, in which we have $M^n$ an n-dimensional compact hypersurface embedded in $\mathbb{R}^{n+1}$. We take the scalar cuvature $R$ to be the elementary symmetric polynomial of degree 2 in the principal curvatures of $M$. We know that $R$ is constant.

The author then says "As $M$ has one elliptic point, $R$ is a positive constant and the mean curvature is positive somewhere".

I'm lost here - Why does $M$ have an elliptic point? And how does this affect $R$ and the mean curvature?

Thanks for any help.

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  • $\begingroup$ A generic coordinate function on M has a single maximum. My geometric intuition says that such a maximum is an elliptic point, but I don't actually know any Riemannian geometry. $\endgroup$
    – Qiaochu Yuan
    Commented Jan 19, 2011 at 22:58
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    $\begingroup$ I believe this is explained, at least for a compact surface in $R^3$, in any basic differential geometry textbook or at least those that talk about surfaces in $R^3$. I suggest you look there first. Or ask on math.stackexchange.com. Or there's probably someone a lot nicer than a grumpy old differential geometer like me who's willing to tell you the answer. $\endgroup$
    – Deane Yang
    Commented Jan 19, 2011 at 23:02
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    $\begingroup$ @Qiaochu: it is easier to consider the maximum point(s) on $M$ of the smooth function $f(x) = |x|^2$, I think. Then you don't have to worry about genericity and actually get local convexity. $\endgroup$
    – Willie Wong
    Commented Jan 20, 2011 at 0:11
  • $\begingroup$ Willie, that's right. I just noticed that your answer is the same as Igor's below with "far-away" removed. $\endgroup$
    – Deane Yang
    Commented Jan 20, 2011 at 4:21

3 Answers 3

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Elliptic point is, by definition, a point where all the principal curvatures are positive, hence $R$ is positive. A point of maximal distance from some far-away basepoint is elliptic.

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  • $\begingroup$ Igor, you're much kinder person than me. $\endgroup$
    – Deane Yang
    Commented Jan 20, 2011 at 3:43
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    $\begingroup$ Igor, why does the base point have to be far away? Doesn't the argument work for any base point? $\endgroup$
    – Deane Yang
    Commented Jan 20, 2011 at 3:45
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    $\begingroup$ Yes, of course, where there is a sphere, there is a way. For some reason I always think of the basepoint as being somewhere far. An amusing fact, rediscovered several times (the last in a Rutgers PhD thesis about five years ago) is that this does not work for PL surfaces (more precisely, there are compact embedded polyhedral surfaces in $R^3$ whose curvature (in the PL/cone sense) is nowhere positive. $\endgroup$
    – Igor Rivin
    Commented Jan 20, 2011 at 4:15
  • $\begingroup$ Igor, that's amazing. Is there a picture or description somewhere? Also, can you name names? $\endgroup$
    – Deane Yang
    Commented Jan 20, 2011 at 4:20
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    $\begingroup$ In a response to his own question here: mathoverflow.net/questions/31222/… Joseph O'Rourke linked to a paper by Zalgaller: springerlink.com/content/hu76g212137g2864 which describes a PL embedding of a flat torus in $R^3$. I am curious to know where that paper sits in the line of successive rediscoveries. $\endgroup$
    – Ramsay
    Commented Feb 18, 2011 at 9:15
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You can refer to this more general Proposition, mentioned by Jose Ivan in Daczer's book.

Let $f:M^n\rightarrow R^{n+p}$ be an isometric immersion of a compact manifold. Then there is a point $x_0$ in $M^n$ and a normal vector $u\in T_{x_0}M^{\perp}$ such that the second fundamental form $A_u$ is positive definite.

The proof is simple, just consider smooth function $\frac{1}{2}|f(x)|^2$ on compact manifold, then the consider second fundamental form at maximum point $x_0$ and it is not hard to see that $<A_uX,X>\geq|X|^2$ for $X\in T_{x_0}M$.

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See Marcos Daczer, Submanifolds and isometric immersions, proprosition 1.3.

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  • $\begingroup$ This text appears in Google Books, but I can't seem to access specific pages there or through Amazon. It would help if you could either state the proposition or provide a link to the page, preferably both. Thanks. $\endgroup$
    – Todd Trimble
    Commented Jul 18, 2014 at 14:17

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