Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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:

$Q= \Delta R + \frac{n^3-4n^2+16n-16}{4(n-1)(n-2)^2} R^2 - \frac{8(n-1)}{(n-2)^2}|Ric|^2.$

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?

share|improve this question
    
In four dimensions the total $Q$ curvature integral makes up the non-conformally invariant part of the geometric side of Chern-Gauss-Bonnett formula. This gives a global interpretation. I still haven't seen a local interpretation. –  Viktor Bundle Jul 2 '11 at 23:42
add comment

1 Answer

up vote 1 down vote accepted

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."

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.