Download Differential Harnack Inequalities and the Ricci Flow (EMS by Reto Müller PDF

By Reto Müller

In 2002, Grisha Perelman awarded a brand new type of differential Harnack inequality which consists of either the (adjoint) linear warmth equation and the Ricci movement. This resulted in a totally new method of the Ricci movement that allowed interpretation as a gradient move which maximizes various entropy functionals. The objective of this publication is to give an explanation for this analytic software in complete element for the 2 examples of the linear warmth equation and the Ricci circulation. It starts with the unique Li-Yau outcome, provides Hamilton's Harnack inequalities for the Ricci move, and ends with Perelman's entropy formulation and space-time geodesics. The ebook is a self-contained, smooth creation to the Ricci circulate and the analytic how you can learn it. it truly is basically addressed to scholars who've a easy introductory wisdom of research and of Riemannian geometry and who're interested in extra learn in geometric research. No earlier wisdom of differential Harnack inequalities or the Ricci circulation is needed. A booklet of the ecu Mathematical Society (EMS). dispensed in the Americas by way of the yankee Mathematical Society.

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Extra resources for Differential Harnack Inequalities and the Ricci Flow (EMS Series of Lectures in Mathematics)

Example text

5), gives the contradiction 0≥( − ∂t )F (x0 , t0 ) ≥ 0 + 2 n F (x0 , t0 ) F (x0 , t0 ) − > 0. nt 2 Hence x0 can only be on ∂M. In this case the strong maximum principle applied to F implies ∂F ∂ν (x0 , t0 ) > 0. However in the moving frame described above ∂F (p, t) = t∇n |∇f |2 − ∂t f = 2t ∂ν n ∇n ∇k f ∇k f − ∂t (∇n f ) k=1 n−1 = 2t ∇n ∇k f ∇k f = −2t IIp (∇f, ∇f ), k=1 where we used the Neumann boundary condition ∇n f = 0 on ∂M and the expression for the second fundamental form we derived above.

1. Let u be a positive solution of the heat equation ✷u = 0. Define f = − log u and w = 2 f − |∇f |2 . Then (∂t − )w = −2 |Hess(f )|2 + Ric(∇f, ∇f ) − 2 ∇w, ∇f . Proof. 4 we showed that f . In the |∇f |2 = 2 ∇ f, ∇f + 2 Ric(∇f, ∇f ) + 2 |Hess(f )|2 , thus a direct computation gives (∂t − )w = 2∂t (∂t f ) − 2 (∂t f ) + ∂t |∇f |2 − = 2∂t (∂t f − f ) + ∂t |∇f |2 − = −∂t |∇f |2 − = −2 ∇(∂t + |∇f |2 |∇f |2 |∇f |2 f ), ∇f − 2 Ric(∇f, ∇f ) − 2 |Hess(f )|2 . ✷ This is precisely the assertion of the lemma.

1 The Li–Yau Harnack inequality and putting this together gives |∇f |2 = 2 ∇ f, ∇f + 2 Ric(∇f, ∇f ) + 2 |Hess(f )|2 2 ≥ 2 ∇ f, ∇f − 2K |∇f |2 + ( f )2 . 3) Hence we get F =t |∇f |2 − (∂t f ) 2 ≥ t 2 ∇ f, ∇f − 2K |∇f |2 + ( f )2 − ∂t ( f ) n 2t −F 2 − t∂t n t 1 2 = −2 ∇F, ∇f − 2Kt |∇f |2 − F + F 2 + ∂t F. t nt = 2t ∇ −F t , ∇f − 2Kt |∇f |2 + Subtracting ∂t F on both sides yields the desired inequality. e. an (n − 1)-dimensional submanifold of the n-manifold M. Let ν ∈ C ∞ (N, Rn ), with |ν(p)| ≡ 1 and ν(p) ∈ Tp N ⊥ := {μ ∈ Tp M | μ, v = 0 for all v ∈ Tp N } ⊂ Tp M, be a normal vector field to N in T M.

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