6
$\begingroup$

Does anyone know an example of a complete surface (certainly not necessarily embedded in $\Bbb R^3$) with negative curvature unbounded below? Any example I have computed so far is either not complete or has curvature bounded below. Any light that might be shed would be welcome. (Sadly, this corner of geometry is far from my expertise.)

$\endgroup$
  • 4
    $\begingroup$ Can't you just take a rotationally symmetric example in polar coordinates? $ds^2 = dr^2 + f(r)^2\,d\theta^2$ where $f$ is a solution of the equation $f''(r)-(1+r^2)f(r) = 0$ with $f(0)=0$ and $f'(0)=1$? This will be complete with $K = -(1+r^2)<0$ (and $r$ is the distance from the pole $r=0$ in this metric). The explicit solution of the equation is $$f(r) = e^{r^2/2}\ \int_0^r e^{-\rho^2}\ d\rho $$. $\endgroup$ – Robert Bryant Sep 21 '13 at 10:27
  • 1
    $\begingroup$ Robert, thanks so much. I was fixating on trying to get a singular metric at the origin, rather than going to infinity, and this was complicating my life a bit. Elegant that the Gaussian emerges yet again :) The question that's motivating this — have you any intuition — is whether as we go off to infinity (with $K\to -\infty$) there is a uniform bound (upper and lower) on $A(R)/R^2$ when $R$ is geodesic distance and $A(R)$ is the area of the geodesic ball of radius $R$. (Happy sabbatical!! More later!) $\endgroup$ – Ted Shifrin Sep 21 '13 at 12:25
  • 1
    $\begingroup$ @Ted Shifrin: You cannot expect to have a upper bound on $A(R)/R^2$: even the hyperbolic plane does not have one since $A(R)$ is exponential! However you do have the lower bound $A(R)/R^2 \ge \pi$ for any simply connected surface of non-positive curvature, by Günther-Bishop inequality. $\endgroup$ – Benoît Kloeckner Sep 21 '13 at 13:00
  • $\begingroup$ @BenoîtKloeckner: Sorry, I phrased it sloppily. I meant for small $R$ only, but I did not know Günther-Bishop. Thanks so much. As I said, I have never been enough of a Riemannian expert :) Much appreciated! $\endgroup$ – Ted Shifrin Sep 21 '13 at 13:06
  • 1
    $\begingroup$ @TedShifrin: As for estimating $A(R)/R^2$ for small $R$, there is the classical expansion $$A(R)/R^2 = \pi\left(1-\frac{K(p)}{12}\ R^2\right) + O(R^4),$$ where $A(R)$ is the area of the disc of radius $R$ centered on $p$. $\endgroup$ – Robert Bryant Sep 21 '13 at 14:05
11
$\begingroup$

Kazdan and Warner (MR0343206 (49 #7950) Reviewed Kazdan, Jerry L.; Warner, F. W. Curvature functions for open 2-manifolds. Ann. of Math. (2) 99 (1974), 203–219. 53C20 (35J05 58G15) )

show that a smooth function on a $\mathbb{R}^2$ is the curvature of a complete metric if and only if $\lim_{r\rightarrow \infty} \inf_{|p| > r} K(p) \leq 0.$ So, presumably this should give plenty of examples of the sort you seek, and maybe if you read the paper it discusses how you might construct such.

$\endgroup$
  • $\begingroup$ Many thanks. I actually knew of some of their results from this era, but didn't think to look here. $\endgroup$ – Ted Shifrin Sep 21 '13 at 0:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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