Timeline for Bound on $Q$ implies bound on $R$?
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10 events
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| Jul 17, 2011 at 21:12 | comment | added | Deane Yang | Thanks! I guess one approach would be to argue by contradiction and use a blow-up argument. But you usually still need some assumptions on integral bounds on curvature. In dimension 4 you get some for free from topological invariants. But in higher dimensions this usually needs to be an explicit assumption, no? | |
| Jul 17, 2011 at 17:16 | comment | added | Viktor Bundle | Deane: an example of this type of sequence would be one where $g_n = u^\frac{4}{n-4}g$, where $u$ is a bubble in the sense of Hebey and Robert. Here we are also assuming that the Yamabe constant of $g$ is negative. I believe there is hope for such a bound, because it might be impossible for $\Delta R$ to blow-up at the same rate as $R^2$ and $|Ric|^2$. | |
| Jul 16, 2011 at 2:09 | comment | added | Deane Yang | By now, I'm rather confused. First, do you know examples of sequences where the scalar curvature does not change sign and becomes unbounded but the $Q$-curvature does not? Second, why do you believe it should be possible to bound the scalar curvature in terms of the $Q$-curvature only (and not the traceless Ricci)? | |
| Jul 15, 2011 at 22:38 | comment | added | Viktor Bundle | Deane: In higher dimensions you could be able to use the maximum principle to get a bound in terms of $Q$ and $E$ (traceless Ricci tensor), but the aim is to get it to depend on $Q$ only. | |
| Jul 15, 2011 at 22:19 | comment | added | Viktor Bundle | Deane: In four dimensions the formula is special because it contains the conformal Laplacian of the scalar curvature, so, yes, higher dimensions are different. By "translate to infinity" I mean to rule out $R$ becoming unbounded like the sequence $\{R+n\}$. The idea is that if the Laplacian of $R$ is controlled it my prevent blow-up. | |
| Jul 15, 2011 at 14:22 | comment | added | Deane Yang | I haven't checked the constants in $n$-dimensional case but I found a formula for $Q$ in dimension 4. In dimension 4 the maximum principle gives an upper bound for $R$ in terms of $Q$ and the traceless part of Ricci. I'm sure that's known to everybody who works with $Q$-curvature. Does this work in higher dimensions? | |
| Jul 15, 2011 at 13:51 | comment | added | Deane Yang | Could you explain in more detail what you mean by the scalar curvature "translates its way to infinity"? | |
| Jul 15, 2011 at 13:38 | comment | added | Viktor Bundle | Deane: The mode situation is where there is a sequence of metrics with constant $Q$ curvature, where the constant is bounded uniformly. The changing signs hypothesis is there to prevent situations where the scalar curvature simply "translates its way to infinity". | |
| Jul 15, 2011 at 2:03 | comment | added | Deane Yang | Why do you believe such a sup norm bound exists? Is there an analogous situation where it does? And how does "all of the scalar curvatures change sign" help? | |
| Jul 15, 2011 at 1:38 | history | asked | Viktor Bundle | CC BY-SA 3.0 |