Timeline for A complete surface with $K\to -\infty$
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Sep 21, 2013 at 14:11 | comment | added | Ted Shifrin | @Robert: Yup! That one I know :) [In fact, I believe I answered a question regarding the history of that estimate here over a year ago.] But with $K\to -\infty$ it's pretty clear, then, that there's no uniform upper bound :) | |
| Sep 21, 2013 at 14:05 | comment | added | Robert Bryant | @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 $p$. | |
| Sep 21, 2013 at 13:06 | comment | added | Ted Shifrin | @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! | |
| Sep 21, 2013 at 13:00 | comment | added | Benoît Kloeckner | @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. | |
| Sep 21, 2013 at 12:25 | comment | added | Ted Shifrin | 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!) | |
| Sep 21, 2013 at 10:27 | comment | added | Robert Bryant | 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 $$f(r) = e^{r^2/2}\ \int_0^r e^{-\rho^2}\ d\rho $$. | |
| Sep 21, 2013 at 0:34 | review | First posts | |||
| Sep 21, 2013 at 0:51 | |||||
| Sep 21, 2013 at 0:31 | vote | accept | Ted Shifrin | ||
| Sep 21, 2013 at 0:22 | answer | added | Igor Rivin | timeline score: 13 | |
| Sep 21, 2013 at 0:15 | history | asked | Ted Shifrin | CC BY-SA 3.0 |