Timeline for Bounded curvature (derivatives) and Shi's estimates
Current License: CC BY-SA 3.0
14 events
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| Nov 26, 2013 at 7:34 | history | edited | Ben McKay | CC BY-SA 3.0 |
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| Oct 15, 2012 at 1:53 | comment | added | Deane Yang | If you assume the manifold is closed, not only do you have bounds on everything as Robert says, you have easier proofs. In particular, you can refer back to Hamilton's original papers instead of trying to read Shi's paper, which addresses the difficulties that arise when working with a noncompact complete Riemannian manifold | |
| Oct 14, 2012 at 14:58 | comment | added | Robert Haslhofer | sure, on a closed manifold with smooth metric you have bounds for everything. | |
| Oct 14, 2012 at 11:45 | comment | added | malik | This variant of Shi's estimates is indeed very interesting - thank you for the hint. But if I had read this result first, my question would have been the same: What do I need for bounded derivatives of the initial curvature? According to the answers above I now think a closed manifold with smooth metric will do. | |
| Oct 14, 2012 at 4:45 | comment | added | Deane Yang | Robert, thanks for the clarification. | |
| Oct 14, 2012 at 2:28 | comment | added | Robert Haslhofer | I have the feeling this needs some more clarification: In Shi's estimate $|D^k Rm|<C/t^{k/2}$ the bound blows up as $t$ goes to zero. If one however assumes in addition bounds for the derivatives of the initial curvature, then one can in fact prove a uniform estimate $|D^k Rm|<C$ that doesn't deteriorate at $t$ goes to zero (see also my comment above). | |
| Oct 13, 2012 at 15:39 | comment | added | Deane Yang | One does usually assume that the metric $g(0)$ is smooth. In that case, the $k$-th order covariant derivative of curvature is automatically locally continuous and therefore bounded, so having it bounded uniformly on the whole manifold is an extra assumption. The point about Shi's theorem is that even if you don't assume a uniform bound but do assume a uniform bound on the curvature itself, then all the higher order covariant derivatives of curvature become uniformly bounded in positive time. | |
| Oct 13, 2012 at 15:36 | history | edited | Deane Yang | CC BY-SA 3.0 |
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| Oct 13, 2012 at 12:22 | comment | added | malik | First of all thank you for the answers. I was aware of the equivalent formulation to my second question and just wanted to give my motivation for it. But now I seem to misunderstand some (presumably trivial) point here. Aren't these functions $g_{ij}$ always smooth? I thought, this was implied by the definition of a riemannian metric. | |
| Oct 12, 2012 at 21:07 | vote | accept | malik | ||
| Oct 12, 2012 at 20:10 | history | edited | Deane Yang | CC BY-SA 3.0 |
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| Oct 12, 2012 at 20:09 | comment | added | Deane Yang | Oy. Actually, I meant $C^{k+2}$. Many thanks for asking. | |
| Oct 12, 2012 at 19:17 | comment | added | Glen Wheeler | Did you perhaps mean $C^{k+1}$ for $g_{ij}$ at the end there? | |
| Oct 12, 2012 at 18:10 | history | answered | Deane Yang | CC BY-SA 3.0 |