Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
    
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. –  Qiaochu Yuan Jan 19 '11 at 22:58
1  
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. –  Deane Yang Jan 19 '11 at 23:02
1  
@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. –  Willie Wong Jan 20 '11 at 0:11
    
Willie, that's right. I just noticed that your answer is the same as Igor's below with "far-away" removed. –  Deane Yang Jan 20 '11 at 4:21

2 Answers 2

up vote 3 down vote accepted

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.

share|improve this answer
    
Igor, you're much kinder person than me. –  Deane Yang Jan 20 '11 at 3:43
    
Igor, why does the base point have to be far away? Doesn't the argument work for any base point? –  Deane Yang Jan 20 '11 at 3:45
2  
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. –  Igor Rivin Jan 20 '11 at 4:15
    
Igor, that's amazing. Is there a picture or description somewhere? Also, can you name names? –  Deane Yang Jan 20 '11 at 4:20
1  
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. –  Ramsay Feb 18 '11 at 9:15

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

share|improve this answer
    
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. –  Todd Trimble Jul 18 at 14:17

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.