# Gauss-Bonnet Theorem on noncompact surface without boundary

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?

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 $$r, and the total curvature of this region can be computed with the usual Gauss-Bonnet for surfaces with boundary. The geodesic curvature of the boundary is easy to compute from the $$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 $$\{r will have the same total curvature. So just extend $$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$$.

• 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 $2\pi$. – Willie Wong Jun 24 '20 at 0:51
• 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 $-2\pi \frac{|m^\prime(R)|}{m(R)}$. – Igor Belegradek Jun 24 '20 at 1:00
• 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? – student Jun 24 '20 at 1:11
• 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 $2\pi$. – Willie Wong Jun 24 '20 at 1:22
• @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. – Willie Wong Jun 24 '20 at 1:30

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 $$1$$.

• 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. – student Jun 24 '20 at 0:29
• For Cohn-Vossen you need to assume that the surface is complete. – Willie Wong Jun 24 '20 at 0:30

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., $$\phi = - |x|^2$$.)

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