Geometric picture of scalar curvature - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T21:19:07Z http://mathoverflow.net/feeds/question/119641 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/119641/geometric-picture-of-scalar-curvature Geometric picture of scalar curvature Alex_K 2013-01-23T10:26:46Z 2013-01-29T22:15:07Z <p>In first course differential geometry you learn, that Ricci-curvature is something like a mean-value of the curvature endomorphism, because it's a trace, and the scalar curvature is again a mean-value of the Ricci curvature, again because it's a trace. I'm now interested, what examples of manifolds can be given, with big Ricci-curvature but small scalar curvature, i.e. how can one describe in a picture what scalar curvature forgets what Ricci curvature still sees. The same question one can ask for ricci curvature and the curvature endomorphism.</p> http://mathoverflow.net/questions/119641/geometric-picture-of-scalar-curvature/120065#120065 Answer by Renato G Bettiol for Geometric picture of scalar curvature Renato G Bettiol 2013-01-27T22:57:33Z 2013-01-27T22:57:33Z <p>Have you taken a look at <a href="http://en.wikipedia.org/wiki/Scalar_curvature" rel="nofollow">wikipedia page for Scalar curvature</a>? [BTW, always a great resource!] There you can find the standard geometric interpretation of Scalar curvature, as measuring the volume distortion on balls of small radius, compared to Euclidean balls of such radius. Analogously, Ricci curvature in a direction measures volume distortion of small rods'' along that direction (and then these interpretations make sense together in terms of scalar curvature being an average of Ricci in all directions). I think that's possibly the most geometric picture you can get (apart from interpretations from mathematical general relativity).</p> <p>As for what "scalar curvature forgets, but Ricci still sees" you can think in terms of topological obstructions for these curvatures to have a certain sign; e.g., manifolds with positive Ricci curvature (bounded from below) must be compact (you even have an estimate for their diameter) and have finite fundamental group. Instead, manifolds with positive scalar curvature can be much wilder (in particular, they might not have finite fundamental group). Also, you can look at the problem of prescribing Ricci vs. prescribing scalar curvature, one is clearly way more flexible than the other, see <a href="http://mathoverflow.net/questions/30035/some-questions-about-scalar-curvature" rel="nofollow">this post</a> regarding Kazdan-Warner's stuff for scalar curvature and compare with obstructions to positive (and non-negative Ricci curvature). Although endowing a given manifold with a metric with positive/nonnegative ricci may be impossible, it certainly always has tons of metrics with negative ricci curvature (see <a href="http://www.jstor.org/discover/10.2307/2118552?uid=3739256&amp;uid=2129&amp;uid=2&amp;uid=70&amp;uid=4&amp;sid=21101716283427" rel="nofollow">this paper</a>), in that such metrics are actually $C^0$-dense in the space of all metrics.</p> http://mathoverflow.net/questions/119641/geometric-picture-of-scalar-curvature/120066#120066 Answer by Otis Chodosh for Geometric picture of scalar curvature Otis Chodosh 2013-01-28T00:06:51Z 2013-01-29T22:15:07Z <p>I'll add a few things which I've found helpful to get intuition to what Renato's already written. </p> <p>Scalar curvature has the very simple geometric interpretation as the volume defect of small balls (as Wikipedia says) $$Vol(B_r(p)) = Vol^{\mathbb{R}^n}(B_r^{\mathbb{R}^n}(0)) \left( 1- \frac{R_p}{6(n+1)} r^2 + O (r^2) \right)$$ as $r\to 0$. </p> <p>It's an instructive exercise to try to prove this. Hint: see problem 3 &amp; 4 in <a href="http://www.msri.org/c/document_library/get_file?uuid=1ec12439-821c-42f5-865f-130b81a7be7a&amp;groupId=14404" rel="nofollow">this problem set</a> [LINK SEEMS TO BE BROKEN -- if anyone wants a copy they can send me an email] from a <a href="http://www.msri.org/web/msri/scientific/workshops/summer-graduate-workshops/show/-/event/Wm9228" rel="nofollow">MSRI summer school</a>. I think the problems are due to Justin Corvino.</p> <p>As Renato has already discussed a bit, the question of "how flexible" scalar curvature is a great place to try to get some intuition. You might see:</p> <p><a href="http://www.ihes.fr/~gromov/PDF/1%5B77%5D.pdf" rel="nofollow">These notes</a> of Gromov, see p 87. The whole thing is great for all sorts of curvature intuition.</p> <p><a href="http://www.math.uchicago.edu/~shmuel/QuantCourse%20/Guthsystmet.pdf" rel="nofollow">This article</a> by Larry Guth, specifically ch 3. He discusses the "Geroch Conjecture" which asks if the $n$-dimensional torus admits a metric of positive scalar curvature (in two dimensions, it clearly does not, by Gauss-Bonnet). He gives references to the Schoen-Yau proof of the positive mass theorem and the Geroch Conjecture, both of which are great things to look at. </p> <p>As for "difference between Ricci and scalar curvature" one might consider the questions of which 3-manifolds admit metrics positive scalar/Ricci curvature. By <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.jdg/1214436922" rel="nofollow">Hamilton - Three-manifolds of positive Ricci curvature</a>, only compact manifolds which are covered by $S^3$ can admit metrics of $Ric > 0$. On the other hand, the work of Perelman shows that the $3$-manifolds with positive scalar curvature are exactly connect sums of $S^3,S^3/\Gamma, S^2\times S^1$. See <a href="http://arxiv.org/pdf/0907.2444v2.pdf" rel="nofollow">this paper</a> of Fernando Marques. </p> <p>See also <a href="http://projecteuclid.org/DPubS?service=UI&amp;version=1.0&amp;verb=Display&amp;handle=euclid.bams/1183553962" rel="nofollow">the Yamabe problem</a>. The Yamabe problem is whether or not you can conformally change a metric so that it has constant scalar curvature. The answer is yes, by the combined efforts of Yamabe (whose original solution had an error), Trudinger (who found the error and initiated the path towards the eventual resolution), Aubin (who solved the problem for $n\geq6$ and not locally conformally flat) and finally Schoen (who dealt with the other cases with an amazing solution using ideas from general relativity). </p> <hr> <p>Some more:</p> <p>In addition to the Geroch conjecture work by Schoen-Yau (see also the works of Gromov-Lawson <a href="http://www.ams.org/mathscinet-getitem?mr=569070" rel="nofollow">here</a> and <a href="http://www.ams.org/mathscinet-getitem?mr=577131" rel="nofollow">here</a>), there have been a huge number of work on the question of which manifolds admit positive scalar curvature. See <a href="http://users.ictp.it/~pub_off/lectures/lns009/Stolz/Stolz.pdf" rel="nofollow">this survey</a> paper of Stolz. In addition, you might read the part about his conjectures on positive Ricci curvature. In spite of the fact that $Ric>0$ seems like a much stronger condition than $R>0$. However, there are no known examples of simply connected manifolds admitting positive scalar curvature but not positive Ricci curvature! In order to understand Stolz's conjectures you'll probably want to read about the spin-theoretic obstruction to positive scalar curvature. </p> <p>In particular, due to Schoen-Yau/Gromov-Lawson if two manifolds admits metrics of positive scalar curvature, then their connect sum does as well. As far as I know, the corresponding result is not known for positive Ricci.</p> <p>You also asked about positive curvature operator vs positive Ricci. For works on positive curvature operator, you should of course start with Hamilton's <a href="http://intlpress.com/JDG/archive/1986/24-2-153.pdf" rel="nofollow">paper</a> "4-manifolds with positive curvature operator." and then the later works of <a href="http://www.icm2006.org/proceedings/Vol_II/contents/ICM_Vol_2_32.pdf" rel="nofollow">Bohm-Wilking</a> and then <a href="http://math.stanford.edu/~brendle/icm2010.pdf" rel="nofollow">Brendle-Schoen</a> to see that positive curvature operator (and other conditions, e.g. positive isotropic curvature, $1/4$-pinched sectional curvature) is very restrictive. See Stolz's paper (section 4.6) for some examples of $Ric>0$, and you can find some examples admitting $Ric>0$ but not positive curvature operator. </p> <hr> <p>Of course, this answer is quite skewed towards the "global comparison geometry" point of view. One can also ask about analytic/metric consequences of curvature bounds. It turns out here, you probably need at least Ricci lower bounds, scalar curvature bounds are not really enough, as far as I know. See <a href="http://www.math.ucsb.edu/~wei/paper/06survey.pdf" rel="nofollow">this survey</a> (which also discusses some of the things above) or <a href="http://library.msri.org/books/Book30/files/colding.pdf" rel="nofollow">this one</a>.</p> <hr> <p>Finally, you might be interested in the work of <a href="http://math.berkeley.edu/~lott/LottVillani.pdf" rel="nofollow">Lott-Villani</a> and <a href="http://www-wt.iam.uni-bonn.de/~sturm/papers/paper47.pdf" rel="nofollow">Sturm</a>, who (building on work of many people, who I won't try to cite but you can read their papers for this) show that Ricci lower bounds have an amazing interpretation in the "metric-measure" sense. They show that Ricci lower bounds can be detected by "convexity of the entropy functional" as smooth measures "move along geodesics." Furthermore, this property is preserved under measured <a href="http://en.wikipedia.org/wiki/Gromov%E2%80%93Hausdorff_convergence" rel="nofollow">Gromov-Hausdorff convergence</a> (which is a very weak! On the other hand, of course one could try to to show that scalar curvature lower bounds are preserved under Gromov-Hausdorff convergence, but as far as I know this is an open problem. </p> http://mathoverflow.net/questions/119641/geometric-picture-of-scalar-curvature/120070#120070 Answer by Ben Crowell for Geometric picture of scalar curvature Ben Crowell 2013-01-28T01:01:30Z 2013-01-28T01:01:30Z <p>One thing the scalar curvature forgets is all the information about the coordinates you were using. The notion of having a "big Ricci curvature" is one that can only be defined in a particular coordinate system. For instance, in coordinates $(x,y)$, $R_{xx}$ and $R_{yy}$ could be big, but in some other set of coordinates $(u,v)$, $R_{uu}$ and $R_{vv}$ could be small or even zero. Because the scalar curvature is a scalar, its value is coordinate-independent.</p> <p>I assume you had Riemannian spaces in mind, but there are some very well-motivated examples in relativity. For example, when Schwarzschild originally wrote down the metric for the vacuum region surrounding a spherically symmetric body, the metric had a singularity at a certain radius $r>0$, which would be external to the body if the body was very compact. In the coordinates he was using, the singularity was present in the Riemann tensor. However, decades later it was discovered that the singularity could be removed by switching to different coordinates. A hint of this nonphysical character of the singularity was that there was no singularity in any scalar measure of curvature. The scalar curvature $R^{ab}R_{ab}$ was automatically zero because the Einstein field equations require the Ricci tensor to be zero in a vacuum. However, there are other scalar measures of curvature such as Kretschmann invariant $R^{abcd}R_{abcd}$, and these also vanished.</p> <p>The Kretschmann invariant does blow up at $r=0$ in the Schwarzschild solution, and this is now interpreted as the physical singularity at the center of a black hole.</p>