Singular scalar curvature - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T12:49:34Z http://mathoverflow.net/feeds/question/103596 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/103596/singular-scalar-curvature Singular scalar curvature malyou 2012-07-31T11:21:32Z 2012-08-01T10:42:46Z <p>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.</p> http://mathoverflow.net/questions/103596/singular-scalar-curvature/103597#103597 Answer by Anton Petrunin for Singular scalar curvature Anton Petrunin 2012-07-31T11:32:48Z 2012-08-01T10:42:46Z <p>Take the graph $t=\sqrt{x^2+y^2+z^2}$ with induced intrinsic metric. In $(x,y,z)$-coordinates, the scalar curvature is $$\frac C{x^2+y^2+z^2}.$$</p> <p>Are you happy?</p> http://mathoverflow.net/questions/103596/singular-scalar-curvature/103616#103616 Answer by Cristi Stoica for Singular scalar curvature Cristi Stoica 2012-07-31T14:37:18Z 2012-07-31T14:37:18Z <p>You can use the warped product of two Riemannian manifolds, with a warping function which vanishes on a compact region.</p> <p>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 $B\times_f F$.</p> <p>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 $$s = s_B + \frac {s_F}{f^2} + 2\dim F\frac{\Delta f}{f} + \dim F(\dim F - 1)\frac{\langle grad f, grad f\rangle_B}{f^2}.$$</p> <p>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 $$\Delta f+\frac{s_B}{2\dim F}f=0.$$</p> <p>More about warped products in O'Neill's "Semi-Riemannian geometry: with applications to relativity", and in <a href="http://arxiv.org/abs/1105.3404" rel="nofollow">this paper</a>.</p> http://mathoverflow.net/questions/103596/singular-scalar-curvature/103628#103628 Answer by Ben Crowell for Singular scalar curvature Ben Crowell 2012-07-31T19:35:20Z 2012-07-31T19:48:57Z <p>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.)</p> <p>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.</p> <p>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 $s\rightarrow 0$.</p>