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While reading "Hopper, Andrews - The Ricci Flow in Riemannian Geometry" I came across Shi's global derivative estimates, which posed two problems for me:

  1. For a manifold (M,g) with curvature tensor $Rm$: how exactly are $\left| Rm\right|$ and $\left| \nabla^k Rm\right|$ at a point $p\in M$ defined? Is there some kind of standard tensor norm you can use here?

  2. Since Shi's result [for a ricci flow solution $(M,g(t))_{ t\in [0,T]}$] only gives us bounded curvature derivatives $\left| \nabla ^k Rm (t)\right|$ for times $t>0$ , I want to know under which circumstances the derivatives $\left| \nabla ^k Rm (0)\right|$ are also bounded. (Is e.g. the compactness of $(M,g(0))$ sufficient for this bound?)

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 Regarding 2, there is a variant of Shi's estimates when one assumes in addition bounds for the derivatives of the curvatures of the initial metric (this is written up e.g. in the Morgan-Tian book) – Robert Haslhofer Oct 12 at 22:37

2 Answers

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  1. The norm defined by the Riemannian metric $g$ on the tangent and cotangent bundles naturally induces a norm on each tensor bundle. This follows from the fact that given vector spaces $V$ and $W$ with inner products, there is a naturally induced inner product on the vector space $V\otimes W$.

  2. The metric at $t = 0$ is initially prescribed data and therefore has nothing to do with the Ricci flow itself. So your question is equivalen to: Given a Riemannian metric $g$, when do the covariant derivatives of order $k$ of Riemann curvature have pointwise bounded norm? A sufficient condition is that $g$ can be written in local co-ordinates as $g_ij\,dx^i\,dx^j$, where the function $g_{ij}$ are $C^{k+2}$ functions of the co-ordinates $x^1, \dots x^n$.

EDIT: My answer to #2 is incomplete, since we're working on a noncompact manifold. You also need uniform pointwise upper and lower bounds on the eigenvalues of $g_{ij}$, as well as its derivatives up to order $k+2$ with respect to local co-ordinates.

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Did you perhaps mean $C^{k+1}$ for $g_{ij}$ at the end there? – Glen Wheeler Oct 12 at 19:17
Oy. Actually, I meant $C^{k+2}$. Many thanks for asking. – Deane Yang Oct 12 at 20:09
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. – malik Oct 13 at 12:22
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. – Deane Yang Oct 13 at 15:39
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 sure, on a closed manifold with smooth metric you have bounds for everything. – Robert Haslhofer Oct 14 at 14:58
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For the second question, maybe it suffices to assume that the curvature tensor is $C^k$ (or $C^{k,\alpha}$ to use the elliptic theory). Then one may prove the existence of harmonic coordinates, in which $g_{ij}$ is in $C^{k+2,\alpha}$. The next step is to study the modified Ricci flow (by Deturk's trick), the usual parabolic theory implies that $g_{ij}$ is in parabolic $C^{k+2,\alpha}$ and hence the curvature tensor is bounded in $C^k$.

The existence of harmonic coordinates is only local, however i don't know how to use Deturk' trick locally. So the above argument contains some gap.

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