8
$\begingroup$

According to corollary 8.5.3 in [1], "There are no topological obstructions to negative Ricci or scalar curvature in dimensions at least $3$." More specifically, theorem 4.1 on page 8 in [2] says "Any smooth compact manifold $M^n$, $n\geq 3$ has a metric with negative scalar curvature."

This is surprising to me — so surprising that I find it hard to imagine. To cure my lack of imagination, I'll ask for an example...

For some $n\geq 3$, what is an example of a Riemannian metric defined everywhere on the sphere $S^n$ with everywhere-negative scalar curvature? I'm hoping for something written in the form $$ ds^2 = \sum_{jk} g_{jk}(x) dx^j\,dx^k $$ with the coefficients $g_{jk}(x)$ given by explicit functions of the coordinates $x$, in some set of coordinate-patches that collectively cover the sphere.


References:

[1] Tuschmann and Wraith (2010), Moduli Spaces of Riemannian Metrics, https://link.springer.com/book/10.1007/978-3-0348-0948-1

[2] Notes by Li on Schoen (2017), "Topics in Scalar Curvature," https://geometrysummer.math.uconn.edu/wp-content/uploads/sites/2312/2018/06/Schoen_spring_2017__Topics_in_scalar_curvature.pdf

$\endgroup$
1
  • 1
    $\begingroup$ The result is also mentioned in Wikipedia en.wikipedia.org/wiki/Scalar_curvature, weirdly under positive scalar curvature. According to the linked page on prescribed scalar curvature, by work of Kazdan and Warner, "If the dimension of M is three or greater, then any smooth function ƒ which takes on a negative value somewhere is the scalar curvature of some Riemannian metric" (M is assumed to be a smooth closed manifold). $\endgroup$
    – GFR
    Aug 2, 2020 at 14:54

1 Answer 1

9
$\begingroup$

I used a CAS for the scalar curvature but I am sure it is not too hard to compute it by hand.

On $S^3$ take the Berger metric (also known as a squashed sphere), \begin{equation} g=\eta _1^2+b^2\eta_2^2+c^2\eta_3^2 \end{equation} with $\eta_i$ left-invariant forms on $SU(2)\simeq S^3$.

A possible parametrisation is \begin{equation} \begin{split} \eta _1 &= \sin \psi \, \mathrm{d} \theta - \cos \psi \sin \theta \, \mathrm{d} \phi ,\\ \eta _2 &=\cos \psi \, \mathrm{d} \theta + \sin \psi \sin \theta \, \mathrm{d} \phi,\\ \eta _3 &= \mathrm{d} \psi + \cos \theta \,\mathrm{d} \phi, \end{split} \end{equation}
$b,c$ are constants. The round metric on $S^3$ has $b=c=1$. We have $\theta\in[0,\pi]$, $\phi\in[0,2\pi)$, $\psi\in[0,4\pi)$.

The scalar curvature of this metric is \begin{equation} s= - \frac{1}{2 b^4c^4} [b^8+(c^4-1)^2-2b^4(c^4+1)] \end{equation} which clearly can be made negative by an opportune choice of $b,c$. For example for $b=1$, $s=2-c^4/2$ is negative for $c^4>4$.

$\endgroup$
3
  • 1
    $\begingroup$ Thank you for the answer (+1)! This seems to be exactly what I wanted. For etiquette, I'll wait a while before accepting it, and in the meantime I'll spend some time checking it so I can reap some intuition. By the way, the expression you wrote for $s$ can also be written $$ s = \frac{4B-(1+B-C)^2}{2BC} $$ with $B=b^4$ and $C=c^4$. $\endgroup$ Aug 3, 2020 at 1:16
  • $\begingroup$ Just curious: what does CAS stand for? Computer-Aided-...? $\endgroup$ Aug 3, 2020 at 1:18
  • 1
    $\begingroup$ Glad it was useful. It stands for Computer Algebra System! $\endgroup$
    – GFR
    Aug 3, 2020 at 9:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .