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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 '12 at 22:37
    
As additional information: my understanding is that the variant of Shi's local Bernstein-type estimate that Robert Haslhofer is referring to, which appeared in Morgan and Tian's book, is due at least in part to Peng Lu. –  Bennett Chow Oct 16 '13 at 7:49

3 Answers 3

up vote 6 down vote accepted
  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 '12 at 19:17
    
Oy. Actually, I meant $C^{k+2}$. Many thanks for asking. –  Deane Yang Oct 12 '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 '12 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 '12 at 15:39
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sure, on a closed manifold with smooth metric you have bounds for everything. –  Robert Haslhofer Oct 14 '12 at 14:58

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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Following up on (1) in Ben McKay's answer, taking $\operatorname{Rm}$ to be a covariant $4$-tensor, $\nabla^{k}\operatorname{Rm}$ is a (covariant) $(k+4)$-tensor, i.e., a section of $\bigotimes^{k+4}T^{\ast}M$. So in coordinates, $|\nabla^{k}\operatorname{Rm}|^{2}=g^{i_{1}j_{1}}\cdots g^{i_{k+4}j_{k+4}}\nabla_{i_{5}\cdots i_{k+4}}^{k}R_{i_{1}i_{2}i_{3}i_{4} }\nabla_{j_{5}\cdots j_{k+4}}^{k}R_{j_{1}j_{2}j_{3}j_{4}}$ or with respect to an orthonormal frame $\{e_{i}\}$, $$ |\nabla^{k}\operatorname{Rm}|^{2}=(\nabla_{e_{i_{5}}\cdots e_{i_{k+4}}} ^{k}\operatorname{Rm})(e_{i_{1}},e_{i_{2}},e_{i_{3}},e_{i_{4}})(\nabla _{e_{i_{5}}\cdots e_{i_{k+4}}}^{k}\operatorname{Rm})(e_{i_{1}},e_{i_{2} },e_{i_{3}},e_{i_{4}}), $$ summing over repeated indices.

For a $C^{\infty}$ metric $g$ on a closed manifold $M$, we have $C_{m} \doteqdot\sup_{M}\left\vert \nabla^{m}\operatorname{Rm}\right\vert <\infty$; but these constants also depend on $g$. Given $K$, under the restriction $\left\vert \operatorname{Rm}\right\vert \leq K$, it is easy to construct metrics with $\left\vert \nabla\operatorname{Rm}\right\vert $ arbitrarily large. For example, on the noncompact $\mathbb{R}\times S^{1}$ consider $g_{\varepsilon}\doteqdot dr^{2}+\left( 1+\varepsilon^{2}\sin\left( r/\varepsilon\right) \right) ^{2}d\theta^{2}$, where $\varepsilon\in (0,1/2]$. Then $R\left( g_{\varepsilon}\right) =\frac{2\sin\left( r/\varepsilon\right) }{1+\varepsilon^{2}\sin\left( r/\varepsilon\right) }$ satisfies $\left\vert R\left( g_{\varepsilon}\right) \right\vert \leq \frac{8}{3}$. Note that $\left\vert \nabla R\left( g_{\varepsilon}\right) \right\vert =|\frac{\partial}{\partial r}R\left( g_{\varepsilon}\right) |=\frac{2}{\varepsilon}\frac{\left\vert \cos\left( r/\varepsilon\right) \right\vert }{\left( 1+\varepsilon^{2}\sin\left( r/\varepsilon\right) \right) ^{2}}$, so that $\sup_{\mathbb{R}\times S^{1}}\left\vert \nabla R\left( g_{\varepsilon}\right) \right\vert \geq\frac{2}{\varepsilon}$. For $\varepsilon=j^{-1}$ we can take quotients to yield a sequence of metrics on the compact $S^{1}\times S^{1}$ with $|R(g_{j^{-1}})|\leq\frac{8}{3}$ and $\sup_{S^{1}\times S^{1}}\left\vert \nabla R\left( g_{j^{-1}}\right) \right\vert \geq2j$.

The idea behind the derivative of curvature estimate is ubiquitous in geometric analysis and goes back to Bernstein in PDE and Bochner in geometry. Given a tensor $T$, $\frac{1}{2}\Delta\left\vert T\right\vert ^{2}=\left\vert \nabla T\right\vert ^{2}+\left\langle \Delta T,T\right\rangle $; if $T=\nabla U$ then we use $\Delta\nabla U=\nabla\Delta U+\left[ \Delta,\nabla\right] U$, where $\left[ \Delta,\nabla\right] $ involves curvature. For example, the fundamental lemma of geometric analysis is $$\frac{1}{2}\Delta\left\vert \nabla u\right\vert ^{2}=\left\vert \nabla^{2}u\right\vert ^{2}+\left\langle \nabla\left( \Delta u\right) ,\nabla u\right\rangle +\operatorname{Ric} \left( \nabla u,\nabla u\right) . $$ Under Ricci flow for $g$, modulo $\operatorname{Ric}\ast T^{\ast2}$ terms, we have $\frac{1}{2}(\frac{\partial }{\partial t}-\Delta)\left\vert T\right\vert ^{2}=-\left\vert \nabla T\right\vert ^{2}+\langle(\frac{\partial}{\partial t}-\Delta)T,T\rangle$.

To exhibit the idea, consider a solution to the heat equation $\frac{\partial f}{\partial t}=\Delta f$ with $\left\vert f\right\vert \leq K$ on a static manifold with $\operatorname{Ric}\geq0$. We have $\frac{1}{2}(\frac{\partial}{\partial t}-\Delta) ( f^{2}) =-\left\vert \nabla f\right\vert ^{2}$ and $\frac{1}{2}(\frac{\partial}{\partial t}-\Delta)\left\vert \nabla f\right\vert ^{2}\leq-\left\vert \nabla^{2}f\right\vert ^{2}$. Assuming the maximum principle holds, $\frac{1}{2}(\frac{\partial}{\partial t}-\Delta)(t\left\vert \nabla f\right\vert ^{2}+\frac{1}{2}f^{2})=-t\left\vert \nabla\nabla f\right\vert ^{2}\leq0$. Hence $\left\vert \nabla f\right\vert ^{2}\leq \frac{K^{2}}{2t}$ for $t>0$.

For Ricci flow, the computations are similar: $\frac{1}{2}(\frac{\partial }{\partial t}-\Delta)\left\vert \operatorname*{Rm}\right\vert ^{2} \leq-\left\vert \nabla\operatorname*{Rm}\right\vert ^{2}+C_{0}\left\vert \operatorname{Rm}\right\vert ^{3}$ and $\frac{1}{2}(\frac{\partial}{\partial t}-\Delta)\left\vert \nabla\operatorname*{Rm}\right\vert ^{2}\leq-\left\vert \nabla^{2}\operatorname*{Rm}\right\vert ^{2}+C_{1}\left\vert \operatorname{Rm} \right\vert \left\vert \nabla\operatorname*{Rm}\right\vert ^{2}$, so that $F=t\left\vert \nabla\operatorname*{Rm}\right\vert ^{2}+\left\vert \operatorname*{Rm}\right\vert ^{2}$ satisfies $$ \frac{1}{2}(\frac{\partial}{\partial t}-\Delta)F\leq(C_{1}t\left\vert \operatorname*{Rm}\right\vert -\frac{1}{2})\left\vert \nabla\operatorname*{Rm} \right\vert ^{2}+C_{0}\left\vert \operatorname*{Rm}\right\vert ^{3}. $$ Assume $M$ is closed and $\left\vert \operatorname*{Rm}\right\vert \leq K$. Applying the maximum principle yields $\left\vert \nabla\operatorname*{Rm}\right\vert ^{2}\leq \frac{C_{2}K^{2}}{t}$ for $t\in(0,(2C_{1}K)^{-1}]$.

Following up on Deane Yang's answer, Shi's first derivative estimate is local and says that if $g\left( x,t\right) $, defined (only locally) in $B_{g\left( 0\right) }(p,r)\times \lbrack0,T]$, satisfies $\left\vert \operatorname*{Rm}\right\vert \leq K$, then $\left\vert \nabla\operatorname*{Rm}\right\vert \leq C_{n}K\left( \frac{1}{r^{2}}+\frac{1}{t}+K\right) ^{1/2}$ in $B_{g\left( 0\right) }(p,\frac{r}{2})\times(0,T]$, where $C_n$ depends only on $n$. E.g., taking $r=cK^{-1/2}$ and $t=cK^{-1}$, we obtain $\left\vert \nabla\operatorname*{Rm}\right\vert (x,cK^{-1})\leq C_{n}K^{3/2}$ in $B_{g\left( 0\right) }(p,\frac{c}{2}K^{-1/2})$.

One way to prove Shi's estimate (localizing in a way following Hamilton's Formation of Singularities paper) is as follows. Let $G=t(16K^{2}+\left\vert \operatorname*{Rm}\right\vert ^{2})\left\vert \nabla\operatorname*{Rm}\right\vert ^{2}$. One computes that $(\frac{\partial }{\partial t}-\Delta)G\leq\frac{1}{t}\left( -c_{3}K^{-4}G^{2}+C_{4} K^{4}\right) $ for $t\in(0,K^{-1}]$. Essentially, because of the good quadratic term involving $G^{2}$ on the right side, this equation is amenable to localization, i.e., multiplication by a cutoff function.

See Bing-Long Chen's paper using Perelman's time-dependent localization to prove that any complete ancient solution to the Ricci flow must have $R\geq0$ ($R>0$ unless $\operatorname{Ric}=0$). In dimension 3, by a localization inspired by the Hamilton-Ivey estimate, Chen proved that any complete ancient solution must have nonnegative sectional curvature.

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