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I seek metrics on complete manifolds whose scalar curvature represents a singularity of the form $\frac{h}{\rho^2}$ where $h$ is a continuous function and $\rho$ vanishes on the boundary of some compact set of the manifold.

edited Jul 31 at 11:37

Gjergji Zaimi
37.5k●2●72●149

asked Jul 31 at 11:21

malyou
11●1

A simple example might be to form a surface of revolution by rotating the superellipse $x^m+y^m=1$ about the $y$ axis, and using the induced intrinsic metric on that surface. The $m=2/3$ case, called an astroid, gives a surface of revolution whose Gaussian curvature blows up like $\rho^{-1/2}$, where $\rho$ is the distance from the cusp. I would expect that some $m<2/3$ would give a singularity with the exponent you want. (You didn't define $\rho$, but I assume you intend it to be the distance from the boundary.) – Ben Crowell Jul 31 at 15:43

Anton Petrunin · Answer 1 · 2012-08-01 10:42:46Z UTC

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Take the graph $t=\sqrt{x^2+y^2+z^2}$ with induced intrinsic metric. In $(x,y,z)$-coordinates, the scalar curvature is $$\frac C{x^2+y^2+z^2}.$$

Are you happy?

edited Aug 1 at 10:42

answered Jul 31 at 11:32

Anton Petrunin
16.5k●41●110

Did you mean some other function, perhaps something like $z=1/|x+y|$, rather than $z=|x+y|$? – Ben Crowell Jul 31 at 12:47

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I think, Anton answered question which OP intended to ask, namely, construct a singular Riemannian metric whose curvature behaves like $h/\rho^2$, where $\rho(x)=d(x_0, x)$ and $h(x_0)\ne 0$. Instead, OP asked a question which does not make whole lot of sense, since in his question scalar curvature becomes infinite outside of a bounded region $C$ (assuming that he does not want $h$ to vanish outside of $C$, otherwise one could take, say, $h=-\rho$ and have curvature equal to $-1$ everywhere). – Misha Jul 31 at 13:39

Anton's formula for the scalar curvature doesn't match up correctly with the equation he's given for the surface. The surface $z=|x+y|$ consists of two half-planes joined at the line $x=-y$. The scalar curvature of this surface does not equal $C/|x+y|^2$. The induced intrinsic curvature of this surface is in fact zero everywhere, even on the line $x=-y$. (One way to see that it's zero even on the line is to consider the join between the two planes as the limiting case of a piece of a cylinder with radius $r$ approaching zero; the intrinsic curvature is zero for all $r$.) – Ben Crowell Jul 31 at 14:34

@Anton: Assuming you intended $z=|x+y|^p$ with some other exponent $p\ne 1$, there is still a problem with the answer, because the OP wanted the singularity to occur on the boundary of a compact set, and the line $x=-y$ is not the boundary of a compact set. @Misha: I assume the OP intended $\lim_{\rho\rightarrow0}h\ne 0$. – Ben Crowell Jul 31 at 15:49

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Ben, isn't the one point the boundary of the compact set consisting of that same point? – Deane Yang Jul 31 at 18:45

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Cristi Stoica · Answer 2 · 2012-07-31 14:37:18Z UTC

You can use the warped product of two Riemannian manifolds, with a warping function which vanishes on a compact region.

Let $(B,g_B)$ and $(F,g_F)$ be two Riemannian manifolds, and $f$ a smooth function on $B$. The warped product of $B$ and $F$ with warping function $f$ is the manifold $$B\times_f F:=\big(B\times F, \pi^*_B(g_B) + (f\circ \pi_B)\pi^*_F(g_F)\big),$$ where $\pi_B:B\times F \to B$ and $\pi_F: B \times F \to F$ are the canonical projections. It is customary to call $B$ the base and $F$ the fiber of the warped product $B\times_f F$.

Let $B \times_f F$ be a warped product, with $\dim F>1$. Then, the scalar curvature $s$ of $B \times_f F$ is related to the scalar curvatures $s_B$ and $s_F$ of $B$ and $F$ by $$s = s_B + \frac {s_F}{f^2} + 2\dim F\frac{\Delta f}{f} + \dim F(\dim F - 1)\frac{\langle grad f, grad f\rangle_B}{f^2}.$$

You can play with this formula to obtain what you are looking for - by trying to make $s_B + 2\dim F\frac{\Delta f}{f}$ vanish, or at least to be of the form $\frac{h}{f^2}$. To make it vanish, you may solve the equation $$\Delta f+\frac{s_B}{2\dim F}f=0.$$

More about warped products in O'Neill's "Semi-Riemannian geometry: with applications to relativity", and in this paper.

Ben Crowell · Answer 3 · 2012-07-31 19:48:57Z UTC

There is a description a plane curve called its Cesàro equation, in which the curvature is given as a function $k$ of arc length $s$. Suppose that $k$'s domain is $(0,1]$. Then it should be clear that, given one point and a tangent vector at that point, the corresponding curve $\mathbf{r}(s)$ exists and is unique for $s\in[0,1]$. (Even if the function $k$ misbehaves at $s=0$, the limit $\mathbf{r}(0)$ exists.)

Define the curve C whose Cesàro equation for $s\in(0,1]$ is $k=1/s^2$, with (arbitrarily, but for concreteness) $\mathbf{r}(0)=(1,0)$, and its tangent vector upward at $s=0$. Form a surface of revolution S by revolving C about the $y$ axis. S is compact, and the locus $s=0$ is a unit circle forming a boundary of S.

The Gaussian curvature $\kappa$ is equal in magnitude to the product of the curvatures along the two principal axes. The curvature along the azimuthal axis is 1 for $s=0$, so $|\kappa|\sim 1/s^2$ as $s\rightarrow 0$.

Singular scalar curvature

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