Download Differential Harnack inequalities and the Ricci flow by Muller R. PDF

By Muller R.

In 2002, Grisha Perelman awarded a brand new form of differential Harnack inequality which contains either the (adjoint) linear warmth equation and the Ricci move. This ended in a totally new method of the Ricci movement that allowed interpretation as a gradient movement which maximizes various entropy functionals. The objective of this e-book is to give an explanation for this analytic device in complete aspect for the 2 examples of the linear warmth equation and the Ricci stream. It starts off with the unique Li-Yau consequence, offers Hamilton's Harnack inequalities for the Ricci move, and ends with Perelman's entropy formulation and space-time geodesics. The e-book is a self-contained, sleek advent to the Ricci circulate and the analytic how to learn it. it truly is basically addressed to scholars who've a simple introductory wisdom of research and of Riemannian geometry and who're interested in additional research in geometric research. No earlier wisdom of differential Harnack inequalities or the Ricci movement is needed. A booklet of the eu Mathematical Society (EMS). dispensed in the Americas by means of the yankee Mathematical Society.

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Example text

Note that since the difference of two connections ∇ (1) and ∇ (2) on M is a tensor, ∇˙ is also a tensor – in contrary to the connection ∇ itself. We now write h = ∂t g and differentiate the equation from the lemma above. This yields 2 ∇˙ X Y, Z + 2h(∇X Y, Z) = X · h(Y, Z) − Z · h(X, Y ) + Y · h(X, Z) − h([Y, Z], X) + h([X, Y ], Z) − h([X, Z], Y ). Hence, we get 2 ∇˙ X Y, Z = (∇X h)(Y, Z) − h(∇X Y, Z) + h(∇X Z, Y ) − (∇Z h)(X, Y ) − h(∇Z X, Y ) − h(∇Z Y, X) + (∇Y h)(X, Z) + h(∇Y X, Z) + h(∇Y Z, X) − h([Y, Z], X) + h([X, Y ], Z) − h([X, Z], Y ) = (∇X h)(Y, Z) − (∇Z h)(X, Y ) + (∇Y h)(X, Z), where the last equality follows directly from the fact that the Lie-brackets satisfy [X, Y ] = ∇X Y − ∇Y X for the Levi-Cività connection.

15) proves that if Lij ≥ 0 at time t0 then Lij ≥ 0 for all t ≥ t0 . However Lij could also start negative and stay negative for all time. As in the trace case, the matrix Harnack inequality becomes an equality for the heat 2 kernel u = (4π t)−n/2 e−|x| /4t on Euclidean space Rn . 12 (quadratic version). 11. Then for any vector field V on M we have Hij (V ) := ∇i ∇j u + u gij + ∇i uVj + ∇j uVi + uVi Vj ≥ 0 on M × (0, T ]. 2t Proof. 6, the minimizing V 0 must be V 0 = − ∇u u and we get Hij (V ) ≥ Hij V 0 = ∇i ∇j u − ∇i u∇j u u + gij ≥ 0.

17) We get the derivative d F˜ (u, t) := N˜ (u, t) = dt u · log u dV + M n = 2t (log u) + M n u dV .

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