The usual Ricci flow and normalized Ricci flow for surfaces are $$ \partial_t g = -2Kg $$ and $$ \partial_t g = -2Kg + 2sg,$$ where $K$ is the Gaussian curvature and $s$ is its average. The latter equation can be solved on the interval $[0,\infty)$ and converges to a metric of constant Gaussian curvature.

Now consider the reverse flow $$\partial_t = 2Kg - 2sg,$$ which should function as some sort of reverse flow. What can you say about its long-time existence?

I had a discussion about this with someone, who argued that there should be no time-reversal of that flow as the flow would not know "where to go" if you start in, say, a sphere of constant curvature (because of the above convergence result; this flow should move the metric away from a constant curvature metric).

But clearly, at least for round spheres the above equation has a unique solution, as $\partial_t g = 0$. So why can't a round sphere, for instance, be a repelling fix point of the flow... Writing this, I realize that this flow, if it existed, would have one crucial disadvantage: $g(t)$ would not depend continuously on $g$ in the $C^k$-topology.

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    $\begingroup$ One remark, if the metric $g$ is not analytic, then there is no time-reversal, by a result of Bando (the metric becomes instantaneously analytic under Ricci flow). $\endgroup$ – Ian Agol Sep 28 '13 at 18:24
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    $\begingroup$ Even if the metric is analytic, there may still be no time-reversal. Even the classical linear heat equation $u_t = u_{xx}$ on the circle (i.e., periodic in $x$) does not have time reversal for most analytic initial data. Only if the Fourier coefficients decay extremely rapidly, more rapidly than is required for analyticity, will there be even short-time existence for the time-reversed flow. $\endgroup$ – Robert Bryant Sep 28 '13 at 21:33
  • $\begingroup$ However, one can show that if such backward solution exist, it is unique see [arxiv.org/abs/0906.4920] $\endgroup$ – Thomas Richard Oct 6 '13 at 7:46
  • $\begingroup$ See the related MSE question, "Reversing the Ricci flow" $\endgroup$ – Joseph O'Rourke Nov 29 '13 at 3:37

I've combined two answers into one.

Bando and Kotschwar's real analyticity in space and time. As Ian Agol wrote, Shigetoshi Bando, by way of summable derivative of curvature estimates of Bernstein type, proved that for a solution $(M,g(t)),$ $t\in(0,T)$, to the Ricci flow on a closed manifold, at each $t$ the metric $g(t)$ is real analytic (in space). These summable estimates and the consequent real analyticity result can be localized; see arXiv.1111.0355 by Brett Kotschwar. Also, in arXiv.1210.3083, Kotschwar proves that $g(t)$ is real analytic in both space and time. This involves careful estimates for $t^{k+2\ell}|\nabla^{k}\partial_{t}^{\ell}\operatorname{Rm}|^{2}$. As Thomas Richard wrote, backwards uniqueness was obtained by Kotschwar in arXiv.0906.4920.

The King-Rosenau solution is the only nonround ancient solution $S^2$. The following somewhat unrelated facts come to mind regarding long time existence backwards. By this I mean an ancient solution to the Ricci flow on a closed surface. Either it is flat or $R>0$. In the latter case, if it is also Type I, Hamilton proved that the solution must be a round shrinking sphere or its $\mathbb{Z}_2$ quotient. If it is Type II, then Daskalopoulos, Hamilton and Sesum proved that it must be the King-Rosenau solution or its $\mathbb{Z}_2$ quotient. Note that the King-Rosenau solution is rotationally symmetric and invariant under a reflection. Since (loosely speaking) its backwards limits are two opposing cigar steady solitons, one may think of the King-Rosenau solution as a heteroclinic orbit joining two cigars (fixed points modulo conformal diffeomorphisms) to the round sphere (a fixed point modulo scaling).

The cigar is the only nonflat noncompact 2-dimensional ancient solution. This additional comment is on $2$-dimensional noncompact long time solutions to the backward Ricci flow, i.e., ancient solutions to the forward Ricci flow. We indicate a key idea in the proof of Daskalopoulos and Sesum (Intern. Math. Res. Notices 2006) of their result that any complete noncompact nonflat ancient solution to the Ricci flow $(M^{2},g(t))$ with bounded curvature and finite width must be a cigar soliton.

Since $R>0$, there exists $f$ such that $\Delta f=-R$ (e.g., see A. Huber, Comment. Math. Helv. 1957). Because $n=2$ implies $\operatorname{Ric}=\frac {R}{2}g$, we can write a Bochner formula as $$ \Delta\left( R+|\nabla f|^{2}\right) =2|\operatorname{Ric}+\nabla^{2} f|^{2}+4\frac{|\operatorname{div}(\operatorname{Ric}+\nabla^{2}f)|^{2}}{R}+H, $$ where the trace Harnack $H\doteqdot\Delta R+2|\operatorname{Ric}|^{2}-\frac {1}{2}\operatorname{Ric}^{-1}(\nabla R,\nabla R)\geq0$ is nonnegative since the solution is ancient. Remarkably, the right side of the display is the sum of three nonnegative terms, whereas the left side is a divergence. Now take a suitable exhaustion $\Omega_{i}$ of $M$. Integrating the display yields that $\int_{\partial\Omega_{i}}\langle\nabla(R+\left\vert \nabla f\right\vert ^{2}),\nu\rangle ds$ is nonnegative, where $\nu$ is the unit outward normal to $\partial\Omega_{i}$. It can be shown that this boundary integral tends to zero as $i\rightarrow\infty$. This implies $\operatorname{Ric}+\nabla^{2}f=0$ and, by the classification of $2$-dimensional steady Ricci solitons, we must be on a cigar.

See S.-C. Chu (Comm. Anal. Geom. 2007) for the case where the width is not finite.

Added December 13, 2013. Consider $2$-dimensional complete noncompact nonflat ancient solutions with bounded curvature.

(1) By Richard Hamilton, there are no such Type I solutions.

(2) By Sun-Chin Chu, there are no such solutions with infinite width. His work is based on the works of Wan-Xiong Shi and of Lei Ni and Luen-Fai Tam.

It would be interesting to see if there are other proofs of (2). For example, assume that $u$ on $M^{2}\times(-\infty,0)$ satisfies $\frac{\partial u}{\partial t}=-\Delta u$, $u>0$, and $\lim_{t\rightarrow0}u(t)=\delta_{x_{0} }$.

(i) Can one prove that $\frac{d}{dt}\int uRd\mu\leq0$? Note that $\int \frac{\partial}{\partial t}(uRd\mu)=\int(-R\Delta u+u\Delta R)d\mu$ is likely to be zero (can we prove spatial decay of the backward heat kernel?).

(ii) Does there exist $\alpha(t)$ with $\lim_{t\rightarrow-\infty}\alpha(t)=0$ such that $u(x,t)\leq\alpha(t)$? Presumably one needs to use the infinite width assumption here. Note that $\int(-\frac{\partial}{\partial t} )(ud\mu)=\int(\Delta u+uR)d\mu$, while $\int uRd\mu\geq0$ is a bad sign for showing $\int ud\mu$ decays backward in time.

If the answers to (i) and (ii) are yes, then $R(x_{0},0)\leq\alpha(t)\int Rd\mu(t)\rightarrow0$ as $t\rightarrow-\infty$, which would imply that the solution is flat, a contradiction.


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