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Let $(M,g)$ be a simply connected 2-dimensional Riemannian manifold without boundary, and let $K$ be the Gausssian curvature defined on $M$. If $M$ is compact, then by Gauss-Bonnet Theorem, we have $$\int_M K dA = 4\pi,$$where $dA$ is the area element of $M$ under the metric $g$.

If $M$ is not compact, then the above equality is no longer true. For example, let $M=\mathbb{R}^2$ and we define the conformal metric $g=e^{2u}\delta$, where $u=\ln(sech x)$ and $\delta$ is the Euclidean metric, then one can verify that $K \equiv 1$ on $M$, but the total area of $M$ is $\infty$, not $4\pi$.

Motivated by this example, my question is that, if we assume $(M,g)$ is a noncompact simply connected surface without boundary, and the total area of $M$ is finite, then is it true that
$$\int_M K dA =4\pi?$$ Or is it true at least for the conformal case?

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Consider the metric on $\mathbb R^2$ that is rotationally symmetric metric outside a compact set, namely, it is $dr^2+m(r)^2 d\phi^2$ for $r>R>0$. Here $m$ is a positive function on $[R,\infty)$.

The area form at points with $r>R$ is $dA=m(r)drd\phi$, so the surface has finite area if and only if $m$ is integrable on $[R,\infty)$. The total curvature of the rotationally symmetric end is $$\int_R^\infty -\frac{m^{\prime\prime}}{\!\!\! m} dA=-2\pi\int_R^\infty m^{\prime\prime} dr=2\pi\left(m'(R)-\lim_{r\to\infty} m'(r)\right).$$ It is easy to find examples where the limit on the right hand side does not exist but $m$ is integrable. If the limit exists and $m$ is integrable, the limit is zero, and the total curvature of the end is $2\pi m^\prime(R)$.

The metric is arbitrary if $r<R$, and the total curvature of this region can be computed with the usual Gauss-Bonnet for surfaces with boundary. The geodesic curvature of the boundary is easy to compute from the $r\ge R$ side (I don't remember the answer).

EDIT: As Willie Wong points out by Gauss-Bonnet every smooth metric on the region $\{r<R\}$ will have the same total curvature. So just extend $m$ to a smooth function on $[0,R]$ so that $m(r)=r$ near $0$ and consider the metric $dr^2+m(r)^2d\phi^2$ for all $r>0$. Its metric completion is smooth at the origin. (More generally, the metric is smooth at the origin if and only if $m^\prime(0)=1$ and $m$ extends to an odd smooth function on $\mathbb R$). Now the above computation gives total curvature as $2\pi(m^\prime(0)-m^\prime(\infty))=2\pi(1-m^\prime(\infty))$ and if $m^\prime(\infty)$ exists and the area is finite, the total curvature is $2\pi$.

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  • $\begingroup$ To be lazy, you can set $m'(R) = 0$ and you have that the circle $r = R$ is geodesic. Then the total curvature is exactly $2\pi$. $\endgroup$ – Willie Wong Jun 24 '20 at 0:51
  • $\begingroup$ Right, but it would be nice to compute the general case. Here is a sketch (to be checked !). The vector field $X=\frac{1}{m(R)}\frac{d}{d\phi}$ is a unit vector field tangent to the curve $r=R$. The geodesic curvature is the length of $\nabla_X X=-\frac{m^\prime(R)}{m(R)}$. So it looks like the geodesic curvature of the boundary is $-2\pi \frac{|m^\prime(R)|}{m(R)}$. $\endgroup$ – Igor Belegradek Jun 24 '20 at 1:00
  • $\begingroup$ Thank you very much for this inspirational post. I'm just curious that, does there exist a positive function $m(r)$ such that $m(r)$ is integrable, while $\lim_{r \rightarrow \infty} m'(r) \ne 0$? Also, I think the metric cannot be arbitary if $r< R$, since we require the metric is smooth, right? $\endgroup$ – student Jun 24 '20 at 1:11
  • $\begingroup$ Taking my laziness further: by Gauss Bonnet it doesn't matter what the portion $< R$ is, so might as well be a spherical cap and assume $R < \pi \rho$ where $\rho$ is the radius of the spherical cap. The curvature integral of the cap is $2 \pi (1- \cos(R/\rho))$. Then since $m(r) = \rho \sin(r/ \rho)$ we have $m'(R)$ is simply $\cos(R)$. So in fact even in the general case you will have that that any surface that you constructed will have total curvature integral exactly $2\pi$. $\endgroup$ – Willie Wong Jun 24 '20 at 1:22
  • $\begingroup$ @student: if the limit $\lim m'(r)$ exists, it must be zero (otherwise $m(r)$ is unbounded). The limit may fail to exist (imagine of $m(r)$ containing a train of skinnier and skinnier bumps). // Sufficient regularity of the metric (for Gauss Bonnet to apply) is implicitly assumed. You just need the metric to be piecewise $C^2$ (IIRC) for Gauss-Bonnet to work. $\endgroup$ – Willie Wong Jun 24 '20 at 1:30
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If $M$ has finite total curvature, then $$\int_M KdA\leq 2\pi$$ by the Cohn-Vossen inequality. The $2\pi$ arises since the Euler characteristic of a noncompact simply-connected surface without boundary is $1$.

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    $\begingroup$ I don't think this is right. A simple example is to consider $\mathbb{R}^2, \frac{4}{1+x^2+y^2}\delta$. This is simply-connected without boundary, and it is not compact (we don't compatify it at $\infty$), then the total curvature is $4\pi$. I checked th eCohn-VOssen inequality, and it requires that the manifold is complete. $\endgroup$ – student Jun 24 '20 at 0:29
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    $\begingroup$ For Cohn-Vossen you need to assume that the surface is complete. $\endgroup$ – Willie Wong Jun 24 '20 at 0:30
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Since you asked about the conformal case: consider the metric $g = e^{2\phi} \delta$. The area element is $e^{2\phi} ~dx$. The Gauss curvature is $$ K = - e^{-2\phi} \Delta \phi $$ and so the curvature integral is equal to $$ - \int_{\mathbb{R}^2} \Delta \phi ~dx $$

Suppose now that $\phi = \phi(|x|)$ is radial. Then the total integral can be evaluated using Gauss-Green as $$ - \lim_{R\to\infty} 2\pi R\phi'(R). $$ Consider the case that $\phi$ is a smooth function such that for all $|x|$ sufficiently large we have $\phi(|x|) = - \kappa \ln(|x|)$. Notice that when $\kappa > 1$ we have that $M$ has finite total area. Observe also that by a direct computation that the total curvature integral can be evaluated to equal exactly $2\pi \kappa$.

And hence we have that in the conformal case the valid range of values of the total curvature integral contains at least the full range $(2\pi, \infty]$. (The $\infty$ endpoint is attained for, e.g., $\phi = - |x|^2$.)

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  • $\begingroup$ Thank you very much! $\endgroup$ – student Jun 24 '20 at 2:57

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