Download e-book for kindle: Canonical Metrics in Kaehler Geometry by Gang Tian, M. Akveld

By Gang Tian, M. Akveld

There has been basic development in complicated differential geometry within the final twenty years. For one, The uniformization idea of canonical Kähler metrics has been verified in larger dimensions, and lots of purposes were discovered, together with using Calabi-Yau areas in superstring conception. This monograph offers an advent to the speculation of canonical Kähler metrics on complicated manifolds. It additionally provides a few complex themes now not simply discovered elsewhere.

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Then f satisfies the equations i) Ric + Hess(f ) − ii) R + f− n 2τ iii) |∇f |2 + R − g 2τ = 0, = 0, f τ ≡ const, 26 1 Foundational material iv) ∂t f = |∇f |2 = −∂τ f . Proof. 14) implies gij gij + (LX g)ij = − + ∇i Xj + ∇j Xi , τ τ g hence −2Rij = 2∇i ∇j f − τij , which proves i). The second equation is again the trace of the first one. Because ∇g = 0, we can continue by taking ∇ of i) as in the steady case to get the same equation, namely ∇j Rij − ∇i R + Rip ∇p f = 0, where now Rip = −∇i ∇p f + gip 2τ .

10) Define the distance function s(f ) = dist(f, X) = sup{ (f − k) | k ∈ ∂X, 0 ∈ Sk X} if f ∈ X, if f ∈ X. To estimate the derivative of s(f ) we need the following lemma. 9. Let g be a smooth function of t ∈ R and y ∈ Rm , and let f (t) = sup{g(t, y) | y ∈ Y } where Y ⊂ Rm is compact. Then f (t) is Lipschitz (and hence differentiable almost everywhere) and with Y (t) = {y ∈ Y | g(t, y) = f (t)} we have ∂t f (t) ≤ sup{∂t g(t, y) | y ∈ Y (t)}. Proof. Choose a sequence of times tj decreasing to t0 such that lim tj t0 f (tj ) − f (t0 ) f (t) − f (t0 ) = lim sup .

Because F (x, 0) = 0 for all x ∈ M, t0 must be strictly positive. e. (x0 , t0 ) ∈ UT , then obviously ∇F (x0 , t0 ) = 0, F (x0 , t0 ) ≤ 0, ∂t F (x0 , t0 ) ≥ 0. 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.

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