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Ricci blowup at first singular time

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The major analytic task in geometric flows is to understand the structure of singularities.   The first step toward understanding singularities and their development under the flow is a theorem like:

Characterisation of Ricci Flow Singularities.  Suppose \{g(t)\} is a Ricci flow, defined up to time T<\infty.  Assume T is maximal.  Then we have |\max Rm(t)|\rightarrow\infty, where Rm is the Riemann tensor.

That is to say, what goes wrong at a Ricci flow singularity is the curvature tensor.  We could restate the theorem as its contrapositive:

Characterisation of When the Flow Can Be Extended. Suppose that up to time t_0<\infty, we have a bound |\max Rm|\leq C<\infty.  Then the Ricci flow can be extended past t_0.

But of course the curvature tensor is a big nasty gadget, which is none too easy to understand.  We want something easier to check than “Are my curvatures all bounded?”.  Šešum proved that, in fact, one need only look at the trace of the Riemann tensor, which is the Ricci tensor:

Better Characterisation of Ricci Flow Singularities. As above, let T<\infty be the first singular time of a Ricci flow.  Then |\max Ric(t)|\rightarrow\infty.

This is somewhat remarkable, because the Ricci tensor interacts weirdly with Ricci flow.  For example, the condition Ric\geq K is not preserved under the flow.  Yet somehow the Ricci tensor carries the information about when singularities occur.  Proof under the fold.

We start with a ball-collapse estimate.  Ultimately, the estimate derives from Perel’man’s density ideas.

Glickenstein’s Lemma.  Let r>0.  Then we have B_{g(0)}(x, \frac{r}{1+\sqrt{e^{2Ct}-1}})\subset B_{g(t)}(x,r), where C is a bound for the Ricci tensor.

That is to say, balls are collapsing at worst like e^{Ct}.

Now suppose that we have a flow with |Ric|\leq C, but which has a singularity at T<\infty.  We will obtain a contradiction by a blow-up argument.

Since T<\infty, Rm is blowing up, i.e. there is a sequence p_k\in M_{t_k} with |Rm(p_k,t_k)|\rightarrow\infty, t_k\rightarrow T.  Set Q_k=|Rm(p_k,t_k)|.  Consider the rescaled metrics

g_k(t)=Q_kg(t_k+\frac{t}{Q_k})

each of which is also a solution to the Ricci flow.  The pointed manifolds \{(M,g_k(t),p_k)\} converge to a limit (\overline{M},\overline{g}(t),p) for each t.  By the work of Hamilton, (\overline{M},\overline{g}(t)) is an ancient solution to the Ricci flow.

We want to come up with more properties of \overline{g}.  Since |Ric(g(t))|\leq C, we have R(g(t))\leq C as well.  The rescaled scalar curvature is R(g_k(t))=\frac{R(g(t))}{Q_k}, and since Q_k\rightarrow \infty, we see that R(\overline{g}(t))=0.

Under Ricci flow, the scalar curvature evolves as \partial_t R=\Delta R + 2|Ric|^2, so the vanishing of R(\overline{g}(t)) implies the vanishing of Ric(\overline{g}(t)).  Thus (\overline{M},\overline{g}(t)) is a stationary solution to Ricci flow, i.e. \overline{g}(t)\equiv \overline{g}.

Consider the volume of a ball of radius r>0 in the limit metric:

\frac{vol_{\overline{g}}(B(p,r))}{r^n}=\lim \frac{vol_{g_k(0}(B(p_k,r))}{r^n}=\lim \frac{vol_{g(t_k)}B(p_k, \frac{r}{\sqrt{Q_k}})}{(\frac{r}{\sqrt{Q_k}})^n}

Now let \epsilon>0 be arbitrary.  Since t_k\rightarrow T, we can choose k_0 large enough that

  • e^{\frac{C}{\sqrt{n}}|t_k-t_{k_0}|}>1-\frac{\epsilon}{2}
  • (\frac{1}{1+\sqrt{e^{2C(t_{k}-t_{k_0})}-1}})^n>1-\frac{\epsilon}{2}
  • \frac{vol(B(p,r))}{r^n}\geq \frac{vol_{g(t_k)}(B(p_k,\frac{r}{\sqrt{Q_k}}))}{(\frac{r}{\sqrt{Q_k}})^n}-\frac{\epsilon}{2}

for any k>k_0.

By Glickenstein’s lemma, we have B_{g(t_{k_0})}(p_k, \frac{r}{\sqrt{Q_k}(1+\sqrt{e^{2C(t_k-t_{k_0})}-1}}\subset B_{g(t_k)}(p_k,\frac{r}{\sqrt{Q_k}}).    Applying the evolution equation for the volume form, we have the estimate

vol_{g(t_k)}(B_{g(t_{k_0})}(p_k, \frac{r}{\sqrt{Q_k}(1+\sqrt{e^{2C(t_k-t_{k_0})}-1}})\geq (1-\frac{\epsilon}{2})vol_{g(t_{k_0})}(B_{g(t_{k_0})}(p_k,\frac{r}{\sqrt{Q_k}(1+\sqrt{e^{2C(t_k-t_{k_0})}-1}})

Thus we have fixed a single metric in the volume comparison limit:

\frac{vol_{\overline{g}}(B_{\overline{g}}(p,r))}{r^n}>(1-\frac{\epsilon}{2})\frac{vol_{g(t_{k_0})}B_{g(t_{k_0})}(p_k, \frac{r}{\sqrt{Q_k}(1+\sqrt{e^{2C(t_k-t_{k_0})}-1}})}{(\frac{r}{\sqrt{Q_k}})^n}

Now recall that the scalar curvature is the quadratic term in the Taylor expansion for the volume of a ball:

vol_{g(k_0)}(B(p_k,\frac{r}{\sqrt{Q_k}(1+\sqrt{e^{2C(t_k-t_{k_0})}-1}})=\omega_n[\frac{r}{\sqrt{Q_k}(1+\sqrt{e^{2C(t_k-t_{k_0})}-1})}]^n(1-\frac{R(g(t_{k_0}))}{6(n+2)}\frac{r^2}{Q_k(1+\sqrt{e^{2C(t_k-t_{k_0})}-1})^2}+O([\frac{r}{\sqrt{Q_k}(1+\sqrt{e^{2C(t_k-t_{k_0})}-1})}]^4)

(Here \omega_n is the volume of a euclidean ball.) Letting k\rightarrow \infty, we get

\frac{vol(B(p,r))}{r^n}>(1-\frac{\epsilon}{2})^2\omega_n-\epsilon

So that in fact \frac{vol(B(p,r))}{r^n}\geq \omega_n since \epsilon>0 was arbitrary.

On the other hand, the Bishop-Gromov comparison says that Ric\geq 0 means \frac{vol(B(p,r))}{r^n}\leq \omega_n.  So (N,\overline{g}) has the same ball volume as euclidean space.

This in turn implies that (N,\overline{g}) is flat.  So Rm(\overline{g})\equiv 0.  But each |Rm_{g_k}(p_k,0)|=1, so |Rm(p,0)|=1.  Thus we have the desired contradiction.

Written by thecooper

31 January 2009 at 4:58 pm

Posted in moderate, prerequisites required, ricci flow

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