Geometric Interpretation of $Q$-curvature - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T19:50:16Z http://mathoverflow.net/feeds/question/69371 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/69371/geometric-interpretation-of-q-curvature Geometric Interpretation of $Q$-curvature Viktor Bundle 2011-07-02T23:06:38Z 2011-07-12T23:03:25Z <p>Let $(M,g)$ be a Riemannian manifold of dimension $n>2$. Thanks to the late T.Branson we have the following definition of the so-called $Q$-curvature:</p> <p>$Q= \Delta R + \frac{n^3-4n^2+16n-16}{4(n-1)(n-2)^2} R^2 - \frac{8(n-1)}{(n-2)^2}|Ric|^2.$</p> <p>Here $\Delta = -div\nabla$, $R$ is the scalar curvature, and $|Ric|$ is the norm of the Ricci tensor. There has been much research on $Q$ curvature since its discovery in the eighties, motivated in large part by the conformal transformation properties that it possesses. A question that appears to be open, though, is whether or not there is a (relatively) concise geometric interpretation of this scalar curvature invariant. For $R$ we have the nice interpretation that it determines the rate at which the growth of a ball around a point differs from the flat case. Similarily the Ricci tensor measures the deviation of a solid angle from the Euclidean case. Can you think of a geometric interpretation of $Q$-curvature that is similarily elegant?</p> http://mathoverflow.net/questions/69371/geometric-interpretation-of-q-curvature/70188#70188 Answer by Paul for Geometric Interpretation of $Q$-curvature Paul 2011-07-12T23:03:25Z 2011-07-12T23:03:25Z <p>I think this is more like a remark than an answer. I gave seminar on Q-curvature (more precisely, Q-curvature flow) twice. In both seminars, I was asked, "What is the geometric meaning of Q-curvture? For example, if Q-curvature is zero, what can we conclude about the manifold $M$?" I was surprised that same question has been raised, and how little we know about Q-curvature. But my answer to their questions is this: "If Q-curvature is zero, then we are solving a 4th order partial differential equation. But we know very little about 4th order PDF, even though it is elliptic. However, if we impose more condition on the manifold $M$, we may be able to get something. For example, if $M$ is locally conformally flat with zero Q-curvature, then the Euler-characteristic of $M$ must be zero, which follows from the Chern-Gauss-Bonnet Theorem in 4 dimension: $$\chi(M)=\frac{1}{8\pi^2}\int_M(|W|^2+Q),$$ where $W$ is the Weyl tensor."</p>