Bounded curvature (derivatives) and Shi's estimates - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T18:55:47Z http://mathoverflow.net/feeds/question/109474 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/109474/bounded-curvature-derivatives-and-shis-estimates Bounded curvature (derivatives) and Shi's estimates malik 2012-10-12T17:47:19Z 2012-10-13T15:36:52Z <p>While reading "Hopper, Andrews - The Ricci Flow in Riemannian Geometry" I came across Shi's global derivative estimates, which posed two problems for me:</p> <ol> <li><p>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?</p></li> <li><p>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?)</p></li> </ol> http://mathoverflow.net/questions/109474/bounded-curvature-derivatives-and-shis-estimates/109476#109476 Answer by Deane Yang for Bounded curvature (derivatives) and Shi's estimates Deane Yang 2012-10-12T18:10:24Z 2012-10-13T15:36:52Z <ol> <li><p>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 $V\otimes W$.</p></li> <li><p>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 $x^1, \dots x^n$.</p></li> </ol> <p>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.</p> http://mathoverflow.net/questions/109474/bounded-curvature-derivatives-and-shis-estimates/109516#109516 Answer by Hao Yin for Bounded curvature (derivatives) and Shi's estimates Hao Yin 2012-10-13T05:41:35Z 2012-10-13T05:41:35Z <p>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 $C^k$.</p> <p>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. </p>