# How can we formulate maximal time $T$ in Hyperbolic Kahler Ricci ﬂow

In general, the exact maximal time $T$ of a Riemannian Ricci flow may not be easy to find. However, fortunately, for Kähler-Ricci flows, the maximal time of existence $T$ is explicitly determined by the initial Kähler class $[ω_0]$ and the first Chern class.

Theorem (Tian-Zhang ). Let $(M,ω(t))$ be an Kahler-Ricci flow $∂ω_t/∂t=-Ric(ω_t)$ on a compact Kahler manifold $M$, with $dim_C=n$, then the maximal existence time $T$ is given by

$T=sup \{t:[ω_0 ]-tc_1 (M)>0\}$.

we have Hyperbolic Kahler Ricci ﬂow

$\frac{∂^2}{∂t}g_{i\bar{j}}(t)=−Ri\bar{j}$ So how can we formulate the maximal time $T$ in Hyperbolic Kahler Ricci ﬂow ?

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The Kahler-Ricci flow behaves particularly well as a PDE, making it possible but still not easy to characterize the existence time in terms of global geometric invariants. It's possible that the hyperbolic Kahler-Ricci flow is similar, but since it has not been studied that much yet, I suspect your question is a good challenging research question beyond the scope of MathOverflow. A good way to start is to see what properties, if any, of the Kahler-Ricci flow also hold for the hyperbolic Kahler-Ricci flow. –  Deane Yang Sep 2 '12 at 0:16
Indeed it sounds like a highly nontrivial problem. The key to the Tian-Zhang theorem are $L^\infty$ estimates on the Kahler potential and its time derivative which are obtained by maximum principle. First of all, can the hyperbolic Kahler Ricci flow be reduced to a scalar hyperbolic equation? If not, then most likely there is no such neat result for this flow. If yes, you would still need to find ways to replace the maximum principle arguments for the hyperbolic scalar equation that you'd get. –  YangMills Sep 2 '12 at 3:03
YangMills, I think there is at least a chance that the reduction to a scalar equation still works for the hyperbolic flow. But you're right that there would be no analogue to the maximum principle for a hyperbolic flow (which is usually oscillatory), and this makes it unlikely that a result as good as Tian-Zhang holds. –  Deane Yang Sep 2 '12 at 3:15