26
$\begingroup$

Consider a version of Ricci flow for which short time existence and uniqueness are known, e.g. the Ricci flow on a closed manifold. Does the solution $g_t$ for small $t$ depend continuously on the initial metric?

I thought the answer is "yes" for Ricci flow on a closed manifold but I cannot see why. My immediate interest is the same question for the instantaneously complete Ricci flow on $\mathbb R^2$ studied by Giesen and Topping.

$\endgroup$
4
  • 1
    $\begingroup$ Some time before I needed such statement for Gromov--Hausdorff distance on spaces with positive curvature operator with diameter $\le D$ and volume $\ge v>0$. I asked everyone I know, but no one seems to know. $\endgroup$ Mar 14, 2011 at 23:22
  • $\begingroup$ Anton, do you still need such a statement? $\endgroup$
    – Deane Yang
    Mar 15, 2011 at 4:40
  • 7
    $\begingroup$ @Deane: Yes I need, do you have one for sale? $\endgroup$ Mar 15, 2011 at 13:13
  • 7
    $\begingroup$ @AntonPetrunin it follows from works of Simon and Schultze that you have some weak continuity under your assumptions: it follows from Theorem 4.1 in arxiv.org/pdf/1008.1408.pdf that if a sequence of manifolds satisfy your requirements, then the Ricci flows of those are precompact (on $(0,T)$). I actually proved that in 2D Ricci flow is continuous w.r.t GH perturbations with curvature bounded from below (arxiv.org/abs/1204.5461), sorry for the self promotion. $\endgroup$ Oct 14, 2013 at 4:06

3 Answers 3

21
$\begingroup$

If you use the right topology on the space of metrics, the answer is yes. Basically, this is always true and a consequence of the proof for any theorem on the existence, uniqueness, and regularity of solutions to the initial value problem of a time-dependent PDE. The "right" topology is the one used in the proof.

ADDED: If you can deal with learning the statement of the Nash-Moser implicit function theorem, as say presented in Hamilton's expository article in the Bulletin of the AMS, then his original paper on the 3-d Ricci flow provides a proof for closed manifolds (without boundary). A much simpler proof, relying on standard estimates for the heat equation, was given shortly afterward by DeTurck, and I believe this appears in the same issue as Hamilton's paper.

There is a paper by Shi in JDG that extends this to a complete Riemannian manifold, and I give a different proof of this theorem in papers of mine on $L_p$ convergence of Riemannian manifolds.

I don't know if there is a proof of short-time existence and uniqueness of the Ricci flow in one of Ben Chow's books, but if there is, I'm sure it's a really good and careful presentation.

MORE: I probably overstated the claim that continuous dependence on parameters is proved in these papers. It is more accurate to say that this is a consequence of the arguments in the papers cited. And it is indeed a general principle for PDE's. Almost every proof of existence of solutions to a PDE involves identifying an initial or boundary value problem for which the PDE has a unique solution, and the same techniques used in the proof can be used to show that the solution depends continuously (and, if everything is smooth, smoothly) on the initial or boundary data.

With the Ricci flow, there is a result like the following: Fix a closed manifold and a smooth Riemannian metric $g_0$. Suppose that the Ricci flow $g(t)$ with $g(0) = g_0$ exists for a time interval $[0,T)$ and fix $\tau \in [0,T)$. Let $\|\cdot\|_k$ denote the L_2 Sobolev norm with $k$ derivatives. Given any $\epsilon > 0$, there exists $\delta > 0$ (which depends not only on $\epsilon$ but everything else mentioned so far) such that if $\hat{g}_0$ is a smooth Riemannian metric such that $\|\hat{g}_0-g_0\|_k < \delta$, then the Ricci flow $\hat{g}(t)$ with $\hat{g}(0) = \hat{g}_0$ exists on the interval $[0,\tau]$ and $\|\hat{g}(\tau) - g(\tau)\|_k < \epsilon$.

To prove this, you can't just study the PDE satisfied by the difference of the two metrics, because like the Ricci flow itself, this PDE is highly degenerate due to the invariance under the action of diffeomorphisms. You have to use the DeTurck trick or some variant of it to make the PDE an honest nonlinear heat equation. Once you do that, the above follows by applying $L_2$ energy estimates satisfied by the "normalized" difference.

Now that I've written this, I guess I can see why this is a reasonable question. Somebody probably should write up the details (not me, I'm way oversubscribed already).

COMMENT: There are many people who are much more expert in the Ricci flow than me, and I had always taken it for granted that these people understand the existence and uniqueness proof using PDE theory at least as well as me. I'm beginning to realize that all the experts know how to study the long time behavior of the Ricci flow (much better than me) but are not so familiar with the technical details of the short-time argument.

FINAL COMMENT: It appear to me that Terry Tao's remarks below answer the question rather succinctly and better than me. I went a bit astray.

YET ONE MORE: Terry Tao is obviously a counterexample to my statement above about experts on the Ricci flow.

$\endgroup$
9
  • 3
    $\begingroup$ In other words, «if you put the correct topologies on the domain and codomain of your map, then the map is continuous» :) $\endgroup$ Mar 14, 2011 at 22:12
  • $\begingroup$ As far as I can see the topology used in the proofs is that of smooth convergence on compact sets; is it the "right one"? Also what would be the reference for the general principle you formulated: "existence, uniqueness, and regularity of solutions to the initial value problem of a time-dependent PDE implies continuous dependence on the initial condition". $\endgroup$ Mar 14, 2011 at 22:14
  • $\begingroup$ Lang's books on differential geometry are one reference for this material. $\endgroup$
    – Dan Ramras
    Mar 14, 2011 at 23:24
  • $\begingroup$ @Dan: you mean that Lang's "Foundations of Differential Geometry" discusses when a PDE is well-posed? I would be very surprised... would you kindly give a page reference? $\endgroup$ Mar 15, 2011 at 0:01
  • 15
    $\begingroup$ After applying the de Turck trick, Ricci flow becomes a quasilinear parabolic equation that can be solved by a contraction mapping method, which automatically gives continuous dependence on the data. See Remark 2.2.3 of Vol. II. of my book at terrytao.wordpress.com/books/… . There is though an additional step required, namely to verify that the de Turck transformation is bicontinuous, but this should be relatively easy in a smooth enough topology. $\endgroup$
    – Terry Tao
    Mar 15, 2011 at 4:56
17
$\begingroup$

Per Ricardo Andrade's nice suggestion, this is a combination of 4 answers, some of which does not directly address the original question; with the separate answers, as he said, it was getting difficult to parse this thread. I've also reordered some of the paragraphs.

Lack of detailed and self-contained expository proof of short time existence for Ricci flow in the literature. In regards to Deane Yang's answer, in our books we unfortunately do not have a complete, self-contained proof of short time existence of the Ricci flow on closed manifolds with smooth initial data. Our exposition follows DeTurck, and Hamilton's interpretation (in his Formation of Singularities paper), of reducing it to a strictly parabolic system. In his blog, Terence Tao gave a nice description of the proof of short time existence for parabolic systems in relation to Ricci flow. It would be nice to see a really detailed proof in the literature though.

In response to Igor Belegradek's first comment:

My memory wasn't precise on this. In fact, what Luen-Fai Tam asked (in 2009) specifically was about appealing to standard theory for parabolic systems (for existence and uniqueness) after applying DeTurck's trick and whether I had a reference for this standard theory. (It is in this sense that our exposition is not complete.) I didn't know a detailed statement and proof of this theory that applies on manifolds. I assumed that what he had in mind was extending the existence theorem to the unbounded curvature noncompact case or perhaps some other nonstandard application, where knowing the nuts and bolts of the proof would be useful.

Noncompact case. I think this would be helpful because the (complete) noncompact case for Ricci flow is not well understood, although there is the seminal work of W.-X. Shi on short time existence assuming bounded curvature. B.L. Chen and X.P. Zhu prove uniqueness of the Ricci flow in the noncompact case also assuming bounded curvature. A few years ago, Luen-Fai Tam asked me about a detailed proof of existence, for which I had no answer. So I think there must be some very interesting related problems, in particular, in the noncompact case. Of course, there are the works of Giesen and Topping as mentioned by Igor Belegradek as well as works by Cabazes-Rivas--Wilking and others.

Question about effective estimates. About the continuous dependence issue, to add to what Terence Tao said, I think there may be a way to get some effective estimates, perhaps related to the calculations exposited in Section 5 of Chapter 7 in my book with Peng Lu and Lei Ni.

In response to Igor Belegradek's second comment:

Thanks for the information about applications. I started to try to write up something about continuous dependence along the lines I discussed above. I have found a pdf file (5 pages) in an email attachment I sent Deane Yang on 03-15-2011. Soon thereafter, I decided to cut this section out of the book RF Part IV (still in preparation) and haven't been able to locate the tex file yet, so this unedited pdf file has a few paragraphs of unrelated material. I've just put it at the URL: http://www.math.ucsd.edu/~benchow/ContinuousDependence.pdf (Wayback Machine) It is only the start of an idea and may contain serious errors since it has not been proofread, so usor emptor. Also, other methods, such as the one mentioned by Terence Tao, may be much better.

Continuous dependence is part of broader question of how solutions depend on their initial data as well as the study of families of solutions, especially 1-parameter families. The following is an introduction to some possible tools.

Linearized Ricci flow and families of solutions. Tangential to this (digressing a bit), but perhaps related, is the question of how to apply some apriori estimates for the linearized Ricci flow. In particular, I don't know of really any useful applications of the linear trace Harnack estimate (assumes bounded $\operatorname{Rm}\geq0$ that Hamilton and I obtained (except for the Kaehler analogue of this estimate, which was applied by Lei Ni); also see Ni and Tam for a proof of the linear trace Harnack estimate in the noncompact case which was not proved originally by Hamilton and me. Another possibly related estimate is the pinching estimate by Greg Anderson and me for solutions to the Ricci flow in dimension 3; this is an extension of a pinching estimate of Hamilton for the 3-dimensional Ricci flow. Brendle used the elliptic version of this estimate in his proof of the uniqueness of the Bryant steady soliton. In view of Brendle's work, I actually think/hope that perhaps by combining the linear trace Harnack estimate and the parabolic version of the pinching estimate may be a starting point for an approach to study aspects of 3-dimensional ancient solutions. Of course, there is a whole (in some sense, almost complete) theory on this by Perelman based on Hamilton's earlier works and conjectures.

Possible applications of the Linearized Ricci flow.

December 12, 2013: This is an addition to my previous post. The continuous dependence question brings to mind some questions about what can be proved for ($1$-parameter) families of the Ricci flow. Also, one may ask if there are lower semicontinuity type results for the singular (maximal) time as a function of the initial metric.

Existence of Type II solutions. A question related to the work of Gu and Zhu (arXiv:0707.0033) is: Given two initial metrics $g_{0,0}$ and $g_{1,0}$ on say $S^{3}$, one of which shrinks to a round point and one of which forms an $S^{2}\times\mathbb{R}$ singularity (neck). Is it true that for any smooth $1$-parameter family of metrics $g_{s,0}$, $s\in\lbrack0,1]$, joining the two, there exists $s^{\prime}$ such that the solution $g_{s^{\prime}}(t)$ with $g_{s^{\prime}}(0)=g_{s^{\prime} ,0}$ forms a Type II singularity? For example, if we start with $g_{0,0}$ a round sphere and rotationally and reflectionally symmetric $g_{1,0}$ forming a neckpinch (Angenent and Knopf), with the $g_{s,0}$ having the same symmetries, must we get for some $s^{\prime}$ a peanut forming two opposing Bryant solitons?

The space of 3-dimensional ancient $\kappa$-solutions. Certain apriori estimates which may be useful for their study. Perelman's conjecture is that assuming noncompact and positive sectional curvature, the Bryant soliton is the only possibility (Brendle's result works toward this.)

To also expand on the third paragraph of my previous post, one has the following heuristic calculations for the linearized Ricci flow; the applications of the maximum principle need to be justified and may require further assumptions. Given a Riemannian metric invariant $T$ of $g$, let $D_{v}T$ denote the variation of $T$ under a variation $v$ of $g$. We have $D_{v}(-2\operatorname{Ric})=\Delta_{L}v-\mathcal{L}_{\operatorname{G}(v)}g$, where $\Delta_{L}$ is the Lichnerowicz Laplacian and $\operatorname{G}(v)=$ $\operatorname{div}v-\frac{1}{2}\nabla\operatorname{tr}v$ (e.g., $\operatorname{G}(\operatorname{Ric})=0$ gives $\frac{\partial}{\partial t}\operatorname{Ric}=\Delta_{L}\operatorname{Ric}$ under the Ricci flow).

Let $W$ be a time-dependent vector field. Then $\operatorname{G} (\mathcal{L}_{W}g)=\Delta W+\operatorname{Ric}\left( W\right) $ and, by diffeomorphism invariance, $D_{\mathcal{L}_{W}g}(-2\operatorname{Ric} )=-2\mathcal{L}_{W}\operatorname{Ric}=\frac{\partial}{\partial t} (\mathcal{L}_{W}g)-\mathcal{L}_{\frac{\partial W}{\partial t}}g$. Combining the above yields $(\frac{\partial}{\partial t}-\Delta_{L})(\mathcal{L} _{W}g)=\mathcal{L}_{(\frac{\partial}{\partial t}-\Delta-\operatorname{Ric} )\left( W\right) }g$. So, by $\frac{\partial}{\partial t} g=-2\operatorname{Ric}$ and the Bochner formula $\Delta=\Delta_{d} +\operatorname{Ric}$ acting on $1$-forms, if the dual $1$-form $W^{\flat }=g(W)$ satisfies the Hodge Laplacian heat equation $(\frac{\partial}{\partial t}-\Delta_{d})W^{\flat}=0$, where $\Delta_{d}=-(d\circ\delta+\delta\circ d)$, then $(\frac{\partial}{\partial t}-\Delta)|W|^{2}=-2|\nabla W|^{2}\leq0$ and $(\frac{\partial}{\partial t}-\Delta_{L})(\mathcal{L}_{W}g)=0$.

We may try to bound $|\mathcal{L}_{W}g|$. Now, if $g(t)$ is complete, ancient, and not Ricci flat, then $R>0$ (B.-L. Chen). And, if $n=3$ and $R>0$, then $(\frac{\partial}{\partial t}-\Delta_{L})v=0$ implies $(\frac{\partial }{\partial t}-\Delta-\frac{2\nabla R}{R}\cdot\nabla)\frac{\left\vert v\right\vert ^{2}}{R^{2}}\leq0$ (the pinching estimate of Greg Anderson). So, if $|v|\leq CR$ at $t=0$, then $|v|\leq CR$ for $t>0$ (e.g., a bound for $\frac{|\mathcal{L}_{W}g|}{R}$ will propagate forward in time).

Moreover, for $n\geq2$ and $\operatorname{Rm}\geq0$ and $v\geq0$ are bounded, then $\operatorname{div}^{2}v+\left\langle \operatorname{Ric},v\right\rangle +\frac{\operatorname{tr}v}{2t}\geq0$ (Hamilton's linear trace Harnack). We have $(\frac{\partial}{\partial t}-\Delta)\operatorname{div}W=\left\langle \operatorname{Ric},\mathcal{L}_{W}g\right\rangle $ (since $\operatorname{tr} (\Delta_{L}v)=\Delta(\operatorname{tr}v)$) and $\operatorname{div} ^{2}(\mathcal{L}_{W}g)=2\Delta\operatorname{div}W+\left\langle \nabla R,W\right\rangle +\left\langle \operatorname{Ric},\mathcal{L}_{W} g\right\rangle $. Assume that $v=\mathcal{L}_{W}g+A\operatorname{Ric}\geq0$ for some $A\geq0$ at $t=0$. Then $$ 2\frac{\partial\operatorname{div}W}{\partial t}+\left\langle \nabla R,W\right\rangle +\frac{\operatorname{div}W}{t}+\frac{A}{2}(\frac{\partial R}{\partial t}+\frac{R}{t})\geq0. $$ In view of the assumption $\mathcal{L}_{W}g\geq-A\operatorname{Ric}$, the fact that $-\frac{\partial\operatorname{div}W}{\partial t}$ has some upper bound and that $\operatorname{div}W=\frac{1}{2}\operatorname{tr}(\mathcal{L}_{W}g)$ gives some bound for $|\mathcal{L}_{W}g|$ backward in time. If the derivative estimates $\left\vert \nabla R\right\vert \leq CR^{3/2}$ and $|\frac{\partial R}{\partial t}|\leq CR^{2}$ hold, then they may be applied.

Naturality of the linear trace Harnack from the space-time point of view. It is the variation of the space-time metric and satisfies the space-time Lichnerowicz Laplacian heat equation.

December 15, 2013: This second follow-up is about how the linear trace Harnack arises naturally when looking at $1$-parameter families of solutions to the Ricci flow from the space-time point of view. It may not be related to continuous dependence problems, but rather it may be related to studying spaces of solutions.

Let $g(t,s)$ be a $2$-parameter family of metrics satisfying $\frac{\partial }{\partial t}g=-2\operatorname{Ric}+\mathcal{L}_{W}g$. In Section 6 of arXiv:0211350 (Sun-Chin Chu) a function $f(t,s)$ and vector field $W(t,s)$ are defined by $e^{-f}d\mu_{g(t,s)}=d\mu_{g(t,0)}$, which is independent of $s$, and by $W=\operatorname{tr}_{1,2}^{g(t,s)}(\nabla_{g(t,s)}-\nabla_{g(t,0)})$, with corresponding space-time metric: $$ \hat{g}(X,Y)=g(X,Y),\quad\;\hat{g}(T)=g(W)+df,\quad\;\hat{g}(T,T)=R+|W|^{2} +2\frac{\partial f}{\partial t}+N $$ for space vectors $X,Y$.

The variation of $\hat{g}$ is the linear trace Harnack: Let $T=\frac{\partial }{\partial t}$. The following are at $s=0$. If $\frac{\partial}{\partial s}g=v$, then $$ \frac{\partial}{\partial s}\hat{g}(X,Y)=v(X,Y),\quad\frac{\partial}{\partial s}\hat{g}(T)=\operatorname{div}v,\quad\frac{\partial}{\partial s}\hat {g}(T,T)=\operatorname{div}^{2}v+\left\langle \operatorname{Ric}% ,v\right\rangle . $$ The second inequality follows from $\frac{\partial f}{\partial s} =\frac{\operatorname{tr}v}{2}$ and $\frac{\partial W}{\partial s} =g^{-1}(\operatorname{div}v-\frac{1}{2}d\operatorname{tr}v)$. From $\frac{\partial R}{\partial s}=\operatorname{div}^{2}v-\Delta\operatorname{tr} v-\left\langle \operatorname{Ric},v\right\rangle $ and $\frac{\partial }{\partial s}(\frac{\partial f}{\partial t})=\frac{\partial}{\partial t} (\frac{\operatorname{tr}v}{2})=\langle\operatorname{Ric},v\rangle+\frac {\Delta\operatorname{tr}v}{2}$ (since $\operatorname{tr}(\Delta_{L} v)=\Delta\operatorname{tr}v$), we obtain the third inequality.

Space-time proof of the linear trace Harnack.

Moreover, Clairaut's theorem implies the linear trace Harnack formula: Let $\hat{v}=\frac{\partial}{\partial s}\hat{g}$. Then $$ \frac{\partial}{\partial t}\hat{v}=\frac{\partial^{2}}{\partial t\partial s}\hat{g}=\frac{\partial^{2}}{\partial s\partial t}\hat{g}=\tilde{\Delta}% _{L}\hat{v}, $$ where $\tilde{\Delta}_{L}$ is a space-time Lichnerowicz Laplacian. From a modification of this one can prove the linear trace Harnack estimate assuming bounded nonnegative $\operatorname{Rm}$ and $v$: Under the Ricci flow and $(\frac{\partial}{\partial t}-\Delta_{L})v=0$, we have $$ \operatorname{div}^{2}v+\langle\operatorname{Ric},v\rangle+2\langle \operatorname{div}v,X\rangle+v(X,X)+\frac{\operatorname{tr}v}{2t}\geq0. $$

The linear trace Harnack related to $\mathcal{L}$-geometry and Perelman's entropy functional.

December 16, 2013: This third follow-up discusses relations between the linear trace Harnack and Perelman's $\mathcal{L}$-length and his energy integrand. This may not be surprising since Perelman's geometry presumably should naturally occur when studying spaces (e.g., $1$-parameter families) of solutions of the Ricci flow.

Let $g(\tau)$ be a solution to the backward Ricci flow $\frac{\partial }{\partial\tau}g=2\operatorname{Ric}$. Given a path $\gamma:\left[ 0,\bar{\tau}\right] \rightarrow\mathcal{M}$, we have Perelman's $\mathcal{L} $-length $\mathcal{L}_{g}(\gamma)=\int_{0}^{\bar{\tau}}\sqrt{\tau} (R_{g}(\gamma(\tau),\tau)+|\gamma^{\prime}(\tau)|_{g(\tau)}^{2})d\tau$. Let $v(\tau)$ be a solution to the linearized backward Ricci flow $\frac{\partial }{\partial\tau}v=-\Delta_{L}v$, where $\Delta_{L}$ is the Lichnerowicz Laplacian and consider the variation $\frac{\partial}{\partial s}g=v$.

Using $\frac{\partial R}{\partial s}=\operatorname{div}^{2}v-\Delta \operatorname{tr}v-\langle\operatorname{Ric},v\rangle$, we obtain $$ \frac{\partial}{\partial s}\mathcal{L}_{g}(\gamma)=\int_{0}^{\bar{\tau}} \sqrt{\tau}(\operatorname{div}^{2}v-\Delta\operatorname{tr}v-\langle \operatorname{Ric},v\rangle+v(\gamma^{\prime},\gamma^{\prime}))d\tau. $$ Since $\frac{\partial}{\partial\tau}\operatorname{tr}v=-\Delta \operatorname{tr}v-2\langle\operatorname{Ric},v\rangle$ and $\frac{d}{d\tau }(\operatorname{tr}v(\gamma(\tau),\tau))=\frac{\partial}{\partial\tau }\operatorname{tr}v+\langle\nabla\operatorname{tr}v,\gamma^{\prime}\rangle$, integrating by parts yields $$ \frac{\partial}{\partial s}\mathcal{L}_{g}(\gamma)=\int_{0}^{\bar{\tau}} \sqrt{\tau}(L(v,\gamma^{\prime})-\frac{\operatorname{tr}v}{2\tau})d\tau +\sqrt{\bar{\tau}}\operatorname{tr}v\left( \gamma\left( \bar{\tau}\right) ,\bar{\tau}\right) +\int_{0}^{\bar{\tau}}2\sqrt{\tau}\langle W,\gamma ^{\prime}\rangle d\tau, $$ where $L(v,X)=\operatorname{div}^{2}v+\langle\operatorname{Ric},v\rangle -2\langle\operatorname{div}v,X\rangle+v(X,X)$ is the (steady version of the) linear trace Harnack and where $W=\operatorname{div}v-\frac{1}{2}% \nabla\operatorname{tr}v=\frac{\partial}{\partial s}\operatorname{tr} _{1,2}^{g}(\nabla_{g(s,\tau)}-\nabla_{g(0,\tau)})$ is related to DeTurck's trick.

Following Perelman and introducing the dilaton $f$, we have that if $\frac{\partial f}{\partial s}=h$, then the variation of his energy integrand is $$ \frac{\partial}{\partial s}(R+2\Delta f-\left\vert \nabla f\right\vert ^{2})=L(v,\nabla f)+2\left( \Delta-\nabla f\cdot\nabla\right) (h-\frac {\operatorname{tr}v}{2})-2\langle v,\operatorname{Ric}+\nabla^{2}f\rangle. $$ When the variation is measure preserving, i.e., $\frac{\partial}{\partial s}(e^{-f}d\mu_{g})=0$, then the second term on the right side drops out since then $h=\frac{\operatorname{tr}v}{2}$. The third term vanishes on a steady Ricci soliton structure.

$\endgroup$
3
  • $\begingroup$ When you say "Luen-Fai Tam asked me about a detailed proof of existence, for which I had no answer", did you mean short time existence when the initial metric is complete and has bounded curvature? Doesn't Shi's paper contain enough detail? $\endgroup$ Oct 14, 2013 at 0:48
  • $\begingroup$ Continuous dependence is needed in studying spaces of metrics satisfying various curvature conditions, e.g. it provides a quick proof that the space of nonnegatively curved metrics on $S^2$ is weakly contractible because any spheroid flows into a contractible neighborhood of the round metric. Thus it would be good to have continuous dependence recorded in the literature. $\endgroup$ Oct 14, 2013 at 14:19
  • $\begingroup$ Thank you for making the computation available. If I may suggest, it would be helpful to include the $C^\infty$ case because this is the topology many people like to use on the spaces of metric. This should be immediate once $C^k$ case is established. $\endgroup$ Oct 15, 2013 at 13:58
6
$\begingroup$

A complete proof of continuous dependence of Ricci flow was published recently by Eric Bahuaud, Christine Guenther & James Isenberg. Let me record their main theorem:

Theorem A (Continuous Dependence of the Ricci Flow) Let $(M, g_0)$ be a smooth compact Riemannian manifold. Let $g_0(t)$ be the maximal solution of the Ricci flow (1) for $t\in [0,\tau (g_0))$, $\tau (g_0) \le \infty$. Choose $0 < \tau < \tau(g_0)$. Let $k \ge 2$. There exist positive constants $r$ and $C$ depending only on $g_0$ and $\tau$ such that if $$\|g_1 − g_0\|_{h^{k+2,\alpha}} \le r,$$ then for the unique solution $g_1(t)$ of the Ricci flow starting at $g_1$, the maximal existence time satisfies $\tau (g_1)\ge \tau$, and $$\|g_1(t) − g_0(t)\|_{h^{k,\alpha}} \le C\|g_1 − g_0\|_{h^{k+2,\alpha}}$$ for all $t \in [0,\tau ]$.

Their proof involves the De-Turck trick and standard semi-group theory in parabolic PDE. They also cite this MO post in their paper.

$\endgroup$
1
  • 1
    $\begingroup$ Yes, I am aware of their paper, but thank you anyway for taking the time to link it to this question. $\endgroup$ Sep 6, 2020 at 17:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.