A complete Riemannian metric for which the Ricci flow has no solution

Question

What is an example of a connected complete Riemannian manifold such that the Ricci flow has no solution with the given Riemannian metric as initial data? As Terry Tao points out, it is easy to construct an example if you don't assume the manifold is connected.

Answer 1

I think this question depends on the precise definition of "initial" metrics.

In the bounded curvature case, thanks to Shi's estimate, an initial metric can be well-defined by smooth convergence, i.e., a metric $g(0)$ is said to be an initial metric of the Ricci flow $g(t),t>0,$ if $g(t)\to g(0)$ in $C_{loc}^\infty$-topology.

However, when we consider the nonexistence problem and manifolds with unbounded curvature, the definition of an initial metric is not uniquely defined.

There are at least two cases one can conceive:

1. $g(0)$ is complete, $Rm(0)$ is pointwise bounded but not uniformly bounded.(See following comments for more precise descriptions.) In this case, we have more general existence results than Shi, proven firstly by Deane Yang in https://eudml.org/doc/82313 (Thm 9.2) and then by Guoyi Xu in http://arxiv.org/pdf/0907.5604v3.pdf (Cor 1.2) They showed that bounded curvature is not a necessary condition for solutions to exist. The convergence is still smooth here.
2. $g(0)$ is incomplete, say $Rm(0)$ is pointwise bounded except at one point. Felix Schulze and Miles Simon constructed expanding solitons coming out from certain "cones". http://arxiv.org/pdf/1008.1408.pdf. They showed the Gromov-Hausdorff distance of $g(t)$ and $g(0)$ goes to zero as $t\to 0$ and took this as the definition of initial metric. For an expanding soliton $g(t)$ of general type, the optimal convergence on regular portions is $C_{loc}^{1,\alpha}$(instead of $C_{loc}^\infty$) even both the curvature and volume behave very well. (This can be derived by using harmonic coordinates).

So, to prove a manifold $(M,g)$ admits no short-time solution, we need to indicate which convergence is involved. For example, $(M,g)$ might not be a $C_{loc}^{1,\alpha}$ initial data but indeed a $C^{0}$ initial data for some solutions.

Let's go back to the original question: how to find a manifold (with unbounded curvature) such that NO Ricci flow can converge to it in ANY sense?

It is obvious that we should replace "ANY" by a specfic term like "Gromov-Hausdorff" before we go further. Even in this case, it is not easy to check, for example, Terry Tao's intuitive example is the one we want. (This is probably due to my poor math ability...)

Actually, Thomas Richard and I constructed an example related to this goal. We found a complete manifold (a rotationally symmetric infinite cusp) which is the last time slice of a Ricci flow $g(t)$. That is, we constructed a Ricci flow $g(t)$ which exists only up to time $T$ and $g(T)$ is a complete smooth metric. In this sense we can say that no solution could flow out from our $(M,g(T))$ as a continuation. But we still don't know can there be a solution, although not a continuation of $g(t)_{t\leq T}$, defining on $t>T$ and converging to $g(T)$ in certain sense when $t\to T$. Our example is similar to Terry's suggestions, however, I haven't conceive a way to show it is "unflowable". In fact, I can imagine a solution flows out from such cusp, hence I guess this question might be harder than people expected. (The difference between "un-continuable" and "unflowable" should be noted.)

Edit: More words about Xu's work are added. Some stupid typos are fixed. The last paragraph is revised.

Edit:(9 June) Add a reference (Deane Yang's article). I apologize for missing this pioneer work at the first time.