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I am reading an article (reference: http://www.jstor.org/stable/1971139?seq=1#page_scan_tab_contents), and in the proof of the main theorem, the author states that "it is a fact that complete, simply-connected surfaces in $\mathbb{R}^3$ of non-positive Gaussian curvature have infinite area" without any reference. I was not able to prove this so far. How can one see this result? Is this fact known as a theorem?

user1234
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  • I assume "surface" means at least $C^2$, since Nash-Kuiper lets you $C^1$-isometrically embed your favorite surface of nonpositive curvature (e.g. the flat torus) and it will clearly have finite surface area. – user7530 May 26 '16 at 19:01
  • You are correct, I assume indeed that the surface is at least $C^2$. – user1234 May 26 '16 at 19:04

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Let $(M, g)$ be a complete surface of non-positive (Gaussian) curvature embedded in Euclidean $3$-space, equipped with the induced metric. By the Cartan-Hadamard theorem, the universal cover $\widetilde{M}$ of $M$ is simply-connected, and the exponential map from an arbitrary point of $\widetilde{M}$ is a diffeomorphism.

If $M$ is simply-connected, then $(M, g)$ is its own universal cover, and the exponential map from an arbitrary point is a diffeomorphism. The fact the curvature is non-positive implies the exponential map does not decrease area. (Intuitively, geodesics move apart at least as fast as rays in the Euclidean plane; use geodesic normal coordinates to calculate the areas of disks in $(M, g)$, concluding that a disk of radius $r$ in $(M, g)$ has area at last $\pi r^{2}$.)

Andrew D. Hwang
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  • Thanks! I think this might do the trick. So $M$ is diffeomorphic to $\mathbb{R}^2$ and since the exponential map does not decrease area, $M$ has infinite area. – user1234 Jun 08 '16 at 12:56
  • Yes, that's the claim. (It may be possible to avoid Cartan-Hadamard in your situation, though I haven't tried to work out the details.) – Andrew D. Hwang Jun 08 '16 at 21:52
  • Would you kindly explain how you would measure the area of disks in (M,g)? – prikarsartam Nov 12 '21 at 09:30
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    @prikarsartam Fix a point $p$ and consider geodesic normal coordinates at $p$. Qualitatively, the circumference of the geodesic circle of radius $r$ grows no slower than $2\pi r$ if $K \leq 0$, so the area of a disk can be estimated by comparison with flat polar coordinates. (I'm guessing you've already searched the site, so if you're seeking details, perhaps post a separate question?) – Andrew D. Hwang Nov 13 '21 at 12:31