By Huai-Dong Cao, Xi-Ping Zhu.

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**Example text**

We have the following useful estimate on its derivative. 3. d ϕ(t) ≤ sup dt ∂ψ (y, t) | y ∈ Y satisfies ψ(y, t) = ϕ(t) . ∂t Proof. Choose a sequence of times {tj } decreasing to t for which lim tj →t ϕ(tj ) − ϕ(t) dϕ(t) = . tj − t dt Since Y is compact, we can choose yj ∈ Y with ϕ(tj ) = ψ(yj , tj ). By passing to a subsequence, we can assume yj → y for some y ∈ Y . By continuity, we have ϕ(t) = ψ(y, t). It follows that ψ(yj , t) ≤ ψ(y, t), and then ϕ(tj ) − ϕ(t) ≤ ψ(yj , tj ) − ψ(yj , t) = ∂ ψ(yj , t˜j ) · (tj − t) ∂t for some t˜j ∈ [t, tj ] by the mean value theorem.

To see the converse, first note that we may assume Z is compact. This is because we can modify the vector field N (ϕ, t) by multiplying a cutoff function which is everywhere nonnegative, equals one on a large ball and equals zero on the complement of a larger ball. The paths of solutions of the ODE are unchanged inside the first large ball, so we can intersect Z with the second ball to make Z convex and compact. If there were a counterexample before the modification there would still be one after as we chose the first ball large enough.

D. -P. ZHU Since Mαβ ≥ 0 and Nαβ ≥ 0, we must have v ∈ null (Nαβ ) and ∇i v ∈ null (Mαβ ), for all i. The first inclusion shows that null (Mαβ ) ⊂ null (Nαβ ), and the second inclusion shows that null (Mαβ ) is invariant under parallel translation. To see null (Mαβ ) is also invariant in time, we first note that ∆v = ∇i (∇i v) ∈ null (Mαβ ) and then g kl ∇k Mαβ · ∇l v α = g kl ∇k (Mαβ ∇l v α ) − Mαβ ∆v α = 0. Thus we have 0 = ∆(Mαβ v α ) = (∆Mαβ )v α + 2g kl ∇k Mαβ · ∇l v α + Mαβ ∆v α = (∆Mαβ )v α , and hence 0= ∂ (Mαβ v α ) ∂t = (∆Mαβ + Nαβ )v α + Mαβ = Mαβ ∂v α .

### A complete proof of the Poincare and geometrization conjectures - application of the Hamilton-Perelman theory of the Ricci flow by Huai-Dong Cao, Xi-Ping Zhu.

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