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There was this question, whether a torus in dimension n, $T^n$, can carry a riemannian metric with positive sectional curvature.

A read a proof, which goes as follows:

$T^n$ is complete, because it's compact. Assume $sec_x>0$ for all $x\in T^n$. Then his universal covering has also positive sectional curvature, hence is compact after the Bonnet-Myers theorem. So the fundamental group of $T^n$ is finite, which is a contradiction.

I think this is no complete proof. The Theorem of Bonnet-Myers needs an unifome bound on the curvature, so strictly positive curvature everywhere is not sufficient. Here's mny counterexample: Take the upper hyperboloid in the three dimensional euclidian space $M:=\{x\in\mathbb{R}^3|x_1^2+x_2^2-x_3^2=-1,x_3>0\}$ with the induced metric from $R^3$. This has positive sectional curvature evreywhere, but is not compact. (Because as $|x|$ goes to infinity, the sectional curvature of $x$ goes to zero.)

What do I make wrong? Or am I right and this proof is incomplete? If so, how can I get an uniformly bound on the sectional curvature? Using that $T^n$ is compact? How can I make this precise?

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SInce $T$ is compact, its sectional curvature is uniformly bounded. The same is true then of its universal covering space, with the same bounds, since the section curvatures of the latter are exacty the same as those of the former! –  Mariano Suárez-Alvarez Mar 22 '13 at 16:07
    
@MarianoSuárez-Alvarez, we finally struggled through to your point in comments below. I should point out that, while i have had my morning tea, I have not yet had breakfast. How is your giant matrix over the field with two elements doing? –  Will Jagy Mar 22 '13 at 16:16
    
I brute forced it with the help of a random number generator and patience, and got it down to a 5 equation incompatible system, which is small enough :-) –  Mariano Suárez-Alvarez Mar 22 '13 at 16:18
    
@MarianoSuárez-Alvarez, good. At that point you actually could check all subsets of four rows and confirm that those are compatible, should that be important, which was my impression, a confirmed minimum. –  Will Jagy Mar 22 '13 at 16:22
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A continuous function on a compact set achieves its minimum. –  Will Jagy Mar 22 '13 at 17:03

1 Answer 1

Your hyperboloid has negative sectional curvature everywhere, indeed constant $-1.$ This is one of the standard models of the hyperbolic plane. http://en.wikipedia.org/wiki/Hyperboloid_model

In case this is the source of the confusion, the restriction of the Minkowski form to the hyperboloid becomes a positive metric, that is, Riemannian.

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@archipelago, what is the source you are quoting? You are correct about taking the other induced metric. However, the hyperboloid cannot be the Riemannian covering space of the ordinary torus. For one thing, the integral of the sectional (Gauss) curvature on the torus is $0.$ –  Will Jagy Mar 22 '13 at 15:53
    
As an isolated fact, yes, you need a positive lower bound. However, your quote/translation from your book is also isolated, somewhere before this there should be a more refined discussion of the universal covering space. –  Will Jagy Mar 22 '13 at 16:00
    
@archipelago Good. Local isometry! The torus itself is compact, positive lower bound on the sectional curvature, same bound in the covering space! –  Will Jagy Mar 22 '13 at 16:14

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