Bounding scalar curvature and diameter along the Kähler Ricci flow (after Perelman) and some

applications

Natasa Sesum, Gang Tian

Abstract

In this short note we present a result of Perelman with detailed proof. The result states that if g(t) is a Kähler Ricci flow on a compact, Kähler manifold M with c_{1}(M) > 0, the scalar curvature and diameter of (M, g(t)) stay uniformly bounded along the flow, for t ∈ [0, ∞). We learned about this result and its proof from G.Perelman when he was visiting MIT in the spring of 2003. This may be helpful to people studying the Kähler Ricci flow.

1 Introduction

We will consider a Kähler Ricci flow,

\frac{d}{dt}g_{ij̄} = g_{ij̄} − R_{ij̄} = ∂_{i}∂j̄u,  (1)

on a compact, Kähler manifold M, with c_{1}(M) > 0, of an arbitrary complex dimension n. Cao proved in ([1]) that (1) has a solution for all time t. One of the most important questions regarding the Kähler Ricci flow is whether it develops singularities at infinity, that is whether the curvature of g(t) blows up as t → ∞. This question was only answered in the case curvature operator or bisectional curvature is nonnegative (cf. [8], [7], [2]). In 2003, Perelman made a surprising claim that the scalar curvature of g(t) does not blow up as t → ∞. He also showed us a sketched proof.This result of Perelman strengthens the belief that the Kähler Ricci flow converges to a Kähler Ricci soliton as t tends to infinity, at least outside a subvariety of complex codimension 2. Partial progress was already made ([?]).

The goal of this paper is to give a detailed proof of Perelman’s bound on a scalar curvature and a diameter.

Theorem 1 (Perelman). Let g(t) be a Kähler Ricci flow (1) on a compact, Kähler manifold M of complex dimension n, with c_{1}(M) > 0. There exists a uniform constant C so that

• |R(g(t))| ≤ C,

• diam(M, g(t)) ≤ C,

• |u|_{C^{1}} ≤ C.

The outline of the main steps of the proof of Theorem 1 is as follows:

1. Getting a uniform lower bound on Ricci potential u(t).

2. Bounding |∇u(t)| and a scalar curvature R(t) by C^{0} norm of Ricci potential u(t). This can be achieved by considering the evolution equations for \frac{|∇u|^{2}}{u+2B} and \frac{−Δu}{u+2B}, where B is a uniform constant such that u + B > 0, whose existence is guaranteed by step 1.

3. Step 2 tells us that \sqrt{u + 2B} is a Lipshitz function and that it is enough to bound diam(M, g(t)) in order to have uniform bounds on |u(t)|_{C^{1}} and scalar curvature R(t).

4. To show that the diamters are uniformly bounded along the flow, we will argue by contradiction. We will assume that the diameters are unbounded as we approach infinity. Using that, we will show that the integral of the scalar curvature over some large annulus is bounded by C · V, where C is a uniform constant and V is a volume of a slightly larger annulus than the one we started with. We can find such an annulus at every time t in the sequence of times for which diamters go to infinity. By choosing similar cut off functions as in the proof of Perelman’s noncollapsing theorem in [12] we will show that we get a contradiction if the diameters are unbounded as we approach infinity.

The organization of the paper is as follows. In section 2 we will give the proof of Theorem 1. In section 3 we will discuss the convergence of the normalized Kähler Ricci flow, using Perelman’s results.

Acknowledgements: We would like to thank Perelman for his generousity for telling us about his results on the Kähler Ricci flow. The first author would like to thank H.D.Cao for numerous useful discussions and helpful suggestions.

2 Ricci potential u(t)

In this section we will show that there is a uniform lower bound on u(x, t). We will also show that it is enough to bound diameters of (M, g(t)) in order to have Theorem 1.

By taking the trace of (1) we get Δu = n − R. Let ϕ(t) be a metric potential, that is,

g_{ij̄}(t) = g_{ij̄}(0) + ∂_{i}∂j̄ϕ.

Then we can take u(t) = \frac{d}{dt}ϕ(t). Normalize so that

∫_{M} e^{−u} = (2π)^{n}.

Define

μ(g, τ) = inf ; {f| ∫_{M} e^{−f} (4πτ)^{−n}=1}(4πτ)^{−n} ∫_{M} e^{−f}{2τ(R + |∇f|) + f − 2n}dV,

to be Perelman’s functional for g(t) as in [12]. Perelman has proved that μ(g, τ) is achieved. Take f = u and τ = 1/2. Then by monotonicity of μ(g(t)) along the Kähler Ricci flow,

A = μ(g(0), 1/2) ≤ μ(g(t), 1/2)
≤ ∫_{M}(2π)^{−n}e^{−u}(R + |∇u|^{2} + u − n)
= ∫_{M}(2π)^{−n}e^{−u}(−Δu + |∇u|^{2} − n + u)
= ∫_{M}(2π)^{−n}Δe^{−u} − n + (2π)^{−n} ∫_{M} e^{−u}u
= −n + (2π)^{−n} ∫_{M} e^{−u}u.  (2)

We have just proved the following lemma.

Lemma 2. There is a uniform constant C_{1} = C_{1}(A) so that ∫ e^{−u}u ≥ C_{1}.

Define a = −(2π)^{−n} ∫_{M} ue^{−u}dV. In the following claim we will prove a lower bound on a.

Claim 3. Moreover, there is a uniform constant C_{2} > 0 so that a ≥ −C_{2}.

Proof. Let u_{−} = min{u, 0} and u_{+} = max{0, u}. Then we have

a = −(2π)^{−n} ∫_{M} ue^{−u}dV = −(2π)^{−n} ∫_{M} u_{−}e^{−u_{−}}dV − (2π)^{−n} ∫_{M} u_{+}e^{−u_{+}}dV
≥ −(2π)^{−n} ∫_{M} u_{+}e^{−u_{+}}dV ≥ −C_{2},

for some constant C_{2} ≥ 0, since f(x) = xe^{−x} is a bounded function for x ≥ 0.

Remark 4. It is pretty well known that the scalar curvature is uniformly bounded from below along the flow. We may assume R > 0.

Proof. Function u(t) satisfies,

∂_{i}∂j̄u_{t} = gj̄ − R_{ij̄} + \frac{d}{dt}∂_{i}∂j̄ ln det(g_{ij̄} + ∂_{i}∂j̄ϕ)^{n}
= ∂_{i}∂j̄(u + Δu),

which implies

\frac{d}{dt}u = Δu + u + a,  (3)

where we can choose a = − ∫ ue^{−u}(2π)^{n} ≤ C, unifomly bounded from above by the previous lemma.

Lemma 5. Function u(t) is uniformly bounded from below.

If the Ricci potential u is very negative for some time t_{0}, say u(t_{0}) ≤ −2(n + C_{1}), from (3), by Lemma 2 and Remark 4 we have

\frac{d}{dt}u = n − R + u + a ≤ n + C_{1} + u < 0,  (4)

at t = t_{0}, which implies that u(t) stays very negative for t ≥ t_{0}. In other words, if for some y_{0} we have u(y_{0}) << 0 at some time t_{0}, u(y) << 0 for all y in some neighbourhood U of y_{0}, at time t_{0}. By (4), u(y) << 0 continues holding in U, for all t ≥ t_{0}. Then \frac{d}{dt}u ≤ C + u implies

u(t)(z) ≤ e^{t−t_{0}}(C + u(t_{0})) ≤ −C̃e^{t},  (5)

for t ≥ t_{0}, for all z ∈ U. Then ϕ̇ = u yields

ϕ(t)(z) ≤ ϕ(t_{0})(z) − C̃e^{t−t_{0}} ≤ −C_{1}e^{t},  (6)

for big enough t and all z ∈ U. On the other hand, ∫_{M} e^{−u(t)} = (2π)^{n}, which tells us u(t) can not be very negative everywhere on M, that is, there is a uniform constant C_{2} such that u(x_{t}, t) = max_{M} u(t) ≥ −C_{2} and Vol_{g(0)}(U) < Vol_{g(0)}(M). Since ϕ̇(t) = u(t), from (3) we get

u(x_{t}, t) − ϕ(x_{t}, t) ≤ max_{M}(u(·, 0) − ϕ(·, 0)) + C̃t,

which implies,

\frac{d}{dt}(u(t) − ϕ(t)) = n − R + a ≤ C̃,

by Lemma (2) and Remark 4. This implies

max_{M} ϕ(t) ≥ −C_{3} − C̃t,  (7)

for a uniform constant C_{3}.

By taking a trace of g(t) = g(0) + ∂ ¯∂ϕ(t) at time t = 0, we get

−Δ_{0}ϕ(t) = −tr_{g(0)}g(t) + n ≤ n.

Consider a fixed metric g(0). By Green’s formula applied to ϕ(t) we have

ϕ(x_{t}, t) = \frac{1}{Vol_{0}(M)} ∫_{M} ϕ(y, t)dV_{0} − \frac{1}{Vol(M)} ∫_{M} Δ_{0}ϕ(y, t)G_{0}(x_{t}, y)dV_{0}
≤ \frac{Vol_{0}(M\U)}{Vol_{0}(M)} sup_{M} ϕ(·, t) + \frac{Vol_{0}(U)}{Vol_{0}(M)} ∫_{U} ϕ(y, t)dV_{0} + C
≤ \frac{Vol_{0}(M\U)}{Vol_{0}(M)} sup_{M} ϕ(·, t) − C_{4}e^{t} + C,

for t ≥ t_{0}, where ϕ(x_{t}, t) = max_{M} ϕ(y, t) and G_{0} is Green’s function associated with metric g(0) (recall that ∫_{M} G_{0}(x_{t}, y)dV_{y} = const). Since \frac{Vol_{0}(M\U)}{Vol_{0}(M)} < 1, we get

max_{M} ϕ(·, t) ≤ −C_{5}e^{t} + C_{6},

for some uniform constants C_{5}, C_{6}. This together with (7) yields a contradiction for big values of t. Therefore, there exists a uniform lower bound on u(t), that is, there is some constant B such that u(x, t) ≥ B for all (x, t) ∈ M × [0, ∞).

Standard computation gives the following evolution equations for Δu and |∇u|^{2}.

□(Δu) = \frac{d}{dt}Δu − Δ^{2}u = −|∇ ¯∇u|^{2} + Δu,  (8)

□(|∇u|^{2}) = \frac{d}{dt}|∇u|^{2} − Δ|∇u|^{2} = −|∇∇u|^{2} − |∇ ¯∇u|^{2} + |∇u|^{2}. (9)

Proposition 6. There is a uniform constant C so that

|∇u|^{2} ≤ C(u + C),  (10)
R ≤ C(u + C).  (11)

Proof. We will first prove an estimate (10) which we will need in the proof of Lemma 11. By Lemma 5 we may assume u(x, t) > −B. The proof resembles the arguments in [19] and [?]. If H = \frac{|∇u|^{2}}{u+2B}, by (8) and (9) we get,

□H = \frac{−|∇ ¯∇u|^{2} − |∇∇u|^{2}}{u + 2B} + \frac{|∇u|^{2}(2B − a)}{(u + 2B)^{2}} +
+ \frac{2 ¯∇u · ∇|∇u|^{2}}{(u + 2B)^{2}} − \frac{2|∇u|^{4}}{(u + 2B)^{3}} .  (12)

We can write

\frac{2 ¯∇u · ∇|∇u|^{2}}{(u + 2B)^{2}} − \frac{2|∇u|^{4}}{(u + 2B)^{3}} = (2−ϵ)\frac{¯∇u · ∇H}{u + 2B} +ϵ\frac{¯∇u · ∇|∇u|^{2}}{(u + 2B)^{2}} −ϵ\frac{|∇u|^{4}}{(u + 2B)^{3}} ,

(13)

for some small ϵ > 0. Since

∇īu∇_{i}(∇_{j}u∇j̄u) = ∇īu∇_{i}∇_{j}u∇j̄u + ∇īu∇_{j}u∇_{i}∇j̄u
≤ |∇u|^{2}(|∇∇u| + |∇ ¯∇u|),

by Cauchy Schwartz inequality,

ϵ\frac{|∇u · ∇|∇u|^{2}|}{(u + 2B)^{2}} ≤ Cϵ\frac{|∇u|^{2}(|∇∇u| + |∇ ¯∇u|)|}{(u + 2B)^{3/2}(u + 2B)^{1/2}}
≤ \frac{ϵ}{2} \frac{|∇u|^{4}}{(u + 2B)^{3}} + \frac{2C^{2}ϵ(|∇∇u|^{2} + |∇ ¯∇u|^{2})}{u + 2B}. (14)

Choose ϵ small so that 2\sqrt{C}ϵ < 1/2. Combining (12), (13) and (14) yields

□H ≤ \frac{|∇u|^{2}(2B − a)}{(u + 2B)^{2}} + (2 − ϵ)\frac{¯∇u · ∇H}{u + 2B} − \frac{ϵ}{2} \frac{|∇u|^{4}}{(u + 2B)^{3}} .  (15)

At a point at which H achieves its maximum we have that ∇H vanishes and therefore by maximum principle, an estimate (15) reduces to

0 ≤ \frac{d}{dt}H_{max} ≤ \frac{|∇u|^{2}}{(u + 2B)^{2}} (2B − a − \frac{ϵ}{2} \frac{|∇u|^{2}}{u + 2B}).  (16)

If we assume that

|∇u|^{2} >> u + 2B,  (17)

then a term on the right hand side of (16) becomes negative for large t and we have that H_{max} is decreasing for big values of t, which gives that |∇u|^{2} ≤ C(u+2B) for some constant C. This contradicts (17) and therefore we have (10).

Our next goal is to prove that −Δu is bounded by C(u+C) which yields (11), since Δu = n − R. Let K = \frac{−Δu}{u+2B}, where B is a uniform constant as above. Similar computation as before gives that

□(−\frac{Δu}{u + 2B}) = \frac{|∇ ¯∇u|^{2}}{u + 2B} + \frac{(−Δu)(2B − a)}{(u + 2B)^{2}} + 2\frac{¯∇u · ∇K}{u + 2B} .

Take b > 1. Then

□(\frac{−Δu + b|∇u|^{2}}{u + 2B}) = \frac{−b|∇∇u|^{2} − (b − 1)|∇ ¯∇u|^{2}}{u + 2B} + \frac{(−Δu + b|∇u|^{2})(2B − a)}{(u + 2B)^{2}}

+ \frac{2 ¯∇u∇(\frac{−Δu+b|∇u|^{2}}{u+2B})}{u + 2B}.

Let G = \frac{−Δu+b|∇u|^{2}}{u+2B} and by maximum principle,

\frac{d}{dt}G_{max} ≤ −(b − 1)\frac{|∇ ¯∇u|^{2}}{u + 2B} + \frac{(−Δu + b|∇u|^{2})(2B − a)}{(u + 2B)^{2}}.

In local coordinates,

(Δu)^{2} = (∑_{i} u_{iī})^{2} ≤ n ∑_{i} u_{iī}^{2} = n|∇ ¯∇u|^{2},

and therefore,

\frac{d}{dt}G_{max} ≤ −(b − 1)\frac{(Δu)^{2}}{n(u + 2B)} + \frac{(−Δu + b|∇u|^{2})(2B − a)}{(u + 2B)^{2}}
≤ \frac{(−Δu)}{u + 2B} \{\frac{2B − a}{u + 2B} − \frac{1}{n} \frac{(−Δu)}{u + 2B} \} + \frac{b|∇u|^{2}(2B − a)}{(u + 2B)^{2}}. (18)

By Lemmas 2 and 5 we may assume that \frac{2B−a}{u+2B} is bounded from above by a uniform constant. We have also proved the estimate (10) on |∇u|. If

−Δu >> u + 2B,  (19)

by (18) we would have that \frac{d}{dt}G_{max} < 0 for big values of t. This would imply −Δu(t) ≤ C(u + 2B), for some uniform constant C and all big values of t, which contradicts (19). Therefore, there exists a uniform constant C such that (11) holds.

Proposition 7. There is a C = C(A), such that Vol(B(x, 1)) ≥ C, for any metric g satisfying |R| ≤ 1 on B(x, 1), where ∂B(x, 1) ≠ ∅.

Proof. Let g(t) be as before, a solution to a normalized Kähler Ricci flow equation, and let g̃(s) be a solution to the equation \frac{d}{ds} g̃(s) = −2Ric(g̃(s)). Reparametrization between these two flows is given by g̃(s) = (1−2s)g(t(s)), where t(s) = − ln(1 − 2s). The first flow has a solution for t ∈ [0, ∞) and the second one has a maximal solution for s ∈ [0, 1/2). The scalar curvature rescales as R(g̃(s)) = \frac{R(g(t(s)))}{1−2s} ≤ \frac{1}{1−2s}. The following improvement of Perelman’s noncollapsing result (noticed by Perelman himself) that requires only a scalar curvature bound can be found in [9]. The result was communicated to Kleiner and Lott by Tian. It says that there is a universal constant κ = κ(g̃(0)), so that for an unnormalized Ricci flow \frac{d}{ds} g̃(s) = −2Ric(g̃(s)), if R(g̃(s) ≥ −\frac{1}{r^{2}} in a ball B_{g̃(s)}(p, r), then Vol_{g̃(s)}B_{g̃(s)}(p, r) ≥ κr^{2n}. The detailed arguments of the proof can be found in [9] and [14], but for the convenience of a reader we will include it here as well. We argue by contradiction, that is, assume there are sequences p_{k} ∈ M and t_{k} → ∞ so that |R| ≤ \frac{C}{r_{k}^{2}} , but Vol(B_{k})r_{k}^{−2n} → 0 as k → ∞, where B_{k} = B_{t_{k}}(p_{k}, r_{k}). Let τ = r_{k}^{2}. Define

u_{k} = e^{C_{k}}ϕ(r_{k}^{−1}dist(x, p_{k}))  (20)

at t_{k}, where ϕ is a smooth function on R, equal 1 on [0, 1/2], decreasing on [1/2, 1] and equal 0 on [1, ∞). C_{k} is a constant to make u satisfy the constraint

(4π)^{n} = e^{2C_{k}}r_{k}^{−2n} ∫_{B(p_{k},r_{k})} ϕ(r_{k}^{−1}dist(x, p_{k}))^{2}dV
≤ e^{2C_{k}}r_{k}^{−2n}Vol(B_{k}).

Since r_{k}^{−n}VolB_{k} → 0, this shows that C_{k} → +∞. We compute

W(u_{k}) = (4π)^{−n}r_{k}^{−2n}e^{2C_{k}} ∫_{B(p_{k},r_{k})}(4|ϕ′(r_{k}^{−1}dist(x, p_{k}))|^{2} − 2ϕ^{2} ln ϕ)dV
+ r_{k}^{2} ∫_{B(p_{k},r_{k})} Ru^{2}(4π)^{−n}r_{k}^{−n}dV − n − 2C_{k}
≤ (4π)^{−n}r_{k}^{−2n}e^{2C_{k}} ∫_{B(p_{k},r_{k})}(4|ϕ′|^{2} − 2ϕ^{2} ln ϕ)dV
+ r_{k}^{2} max_{B_{k}} R − n − 2C_{k}.

Let V (r) = Vol(B(p_{k}, r)). The necessary ingredients of the argument are that

(a) r_{k}^{−2n}Vol(B(p_{k}, r_{k})) → 0.

(b) r_{k}^{2}R is uniformly bounded above.

(c) \frac{Vol(B(p_{k},r_{k}))}{Vol(B(p_{k},r_{k}/2))} is uniformly bounded above.

Suppose that r_{k}^{−2n}Vol(B(p_{k}, r_{k})) → 0 and r_{k}^{2}R ≤ n(n − 1) on B_{k}. If \frac{Vol(B(p_{k},r_{k}))}{Vol(B(p_{k},r_{k}/2))} < 3^{n} for all k, we are done. If not, for a given k we have

that \frac{Vol(B(p_{k},r_{k}))}{Vol(B(p_{k},r_{k}/2))} ≥ 3^{n}. Let r′_{k} = r_{k}/2. We have (r′_{k})^{−2n}Vol(B(p_{k}, r′_{k})) ≤ r_{k}^{−2n}Vol(B(p_{k}, r_{k})) and (r′_{k})^{2}R ≤ n(n − 1) on B(p_{k}, r′_{k}). Replace r_{k} by r′_{k}. If \frac{Vol(B(p_{k},r_{k}))}{Vol(B(p_{k},r_{k}/2))} < 3^{n} we stop. If not, then we repeat the process and replace

r_{k} by r_{k}/2. At some point we will achieve that \frac{Vol(B(p_{k},r_{k}))}{Vol(B(p_{k},r_{k}/2))} < 3^{n}. We then take this new subsequence {p_{k}, r_{k})}_{k=1}^{∞} into further discussion. Hence V (r_{k}) − V (r_{k}/2) ≤ C′V (r_{k}/2). Therefore

∫_{B(p_{k},r_{k})}(4|ϕ′|^{2} − 2ϕ^{2} ln ϕ)dV ≤ C (V (r_{k}) − V (r_{k}/2))
≤ CV (r_{k}/2)
≤ C ∫_{B_{k}} ϕ^{2}dV.

Plugging this into the previous estimate for W and using the constraint (4πτ_{k})^{−n} ∫_{M} e^{−u_{k}}dV_{t_{k}} = 1, we get

W(u_{k}) ≤ C′′ − 2C_{k}.  (21)

Since C_{k} → +∞ and μ(g(t_{k}), r_{k}^{2}) ≤ W(g(t_{k}), u_{k}, r_{k}^{2}), we conclude that μ(g(t_{k}), r_{k}^{2}) → −∞. By the condition (a) we have A ≤ μ(g(t_{k}), r_{k}^{2}) → −∞ which is impossible.

The previous argument, since R(g̃(s)) ≤ \frac{1}{1−2s} implies Vol_{g̃(s)}B_{g̃(s)}(x, \sqrt{1 − 2s}) ≥ κ(1 − 2s)^{n}, which by rescaling implies VolB(x, 1) ≥ κ at metric g(t), where κ is a constant depending only on an initial metric g(0).

Claim 8. There is a uniform constant C such that

u(y, t) ≤ Cdist_{t}^{2}(x, y) + C,
R(y, t) ≤ Cdist_{t}^{2}(x, y) + C,
|∇u| ≤ Cdist_{t}(x, y) + C,

where u(x, t) = min_{y∈M} u(y, t).

Proof. By Lemma 5 we can assume u ≥ δ > 0, since otherwise we can consider u + 2B + δ instead of u. From (10) it follows that \sqrt{u} is a Lipshitz function since |∇(\sqrt{u})| ≤ C = C(δ) and therefore,

|\sqrt{u}(y, t) − \sqrt{u}(z, t)| ≤ \frac{|∇u|(p, t)}{2\sqrt{u}}dist_{t}(y, z)
≤ C̄dist_{t}(y, z),

and therefore,

u(y, t) ≤ (C̃dist_{t}(y, z) + \sqrt{u}(x, t))^{2}
≤ C_{1}dist_{t}^{2}(x, y) + C_{1}u(x, t).

Assume u(x, t) ≥ K(t). Then u(y, t) ≥ K(t) for all y ∈ M and we would have,

(2π)^{n} = ∫_{M} e^{−u}dV_{t} ≤ e^{−K(t)}Vol(M) → 0,

if K(t) → ∞, which is not possible. Therefore, u(x, t) ≤ K, for a constant that does not depend on t and finally

u(y, t) ≤ Cdist_{t}^{2}(y, x) + C̃,  (22)

for some uniform constants C and C̃. Other two estimates in the claim follow from (22) and Proposition 6.

By Claim 8 it follows that if we manage to estimate the diameter, we will get uniform bounds on the scalar curvature and the C^{1} norm of u.

3 A uniform upper bound on diameters

In this section we want to prove the following proposition which will finish the proof of Theorem 1.

Proposition 9. There is a uniform constant C such that diam(M, g(t)) ≤ C.

The proof goes by contradiction argument. Assume that the diameters are unbounded in time. Denote by d_{t}(z) = dist_{t}(x, z) where u(x, t) = min_{y∈M} u(y, t).

Let B(k_{1}, k_{2}) = {z : 2^{k_{1}} ≤ d_{t}(z) ≤ 2^{k_{2}}}. Consider an annulus B(k, k+1). By Claim 8 we have that R ≤ C2^{2k} on B(k, k + 1). Interval [2^{k}, 2^{k+1}] fits 2^{2k} balls of radii \frac{1}{2^{k}}. By Claim 8 and Proposition 7 we have that at time t

Vol(B(k, k + 1)) ≥ ∑_{i} Vol(B(x_{i}, 2^{−k})) ≥ 2^{2k}2^{−2kn}C.  (23)

Claim 10. For every ϵ > 0 we can find B(k_{1}, k_{2}), such that if diam(M, g(t)) is large enough, then

(a) Vol(B(k_{1}, k_{2})) < ϵ and

(b) Vol(B(k_{1}, k_{2})) ≤ 2^{10n}Vol(B(k_{1} + 2, k_{2} − 2)).

Proof. Since Vol_{t}(M) is constant along the flow and therefore uniformly bounded, if diameter is sufficiently big, there is k_{0} such that for all k_{2} ≥ k_{1} ≥ k_{0}, we have that Vol(B(k_{1}, k_{2})) < ϵ. If our estimate (b) did not hold, that is, if

Vol(B(k_{1}, k_{2})) ≥ 2^{10n}Vol(B(k_{1} + 2, k_{2} − 2)),

we would consider B(k_{1} + 2, k_{2} − 2) instead and ask whether (b) holds for that ball. Assume that for every p, at the p-th step we are still not able to find our radii so that (a) and (b) are satisfied. In that case, at the p-th step we would have

Vol(B(k_{1}, k_{2})) ≥ 2^{10np}Vol(B(k_{1} + 2p, k_{2} − 2p)).

In particular, assume we have the above estimate at the p-th step so that k_{1} + 2p + 1 ∼ k_{2} − 2p, which is for 2p ∼ \frac{k_{2}−k_{2}−1}{2} (*). Take k_{1} = k/2 and k_{2} = 3k/2 for k >> 1. In that case (*) becomes p ∼ k/4, k_{1} + 2p ∼ k and k_{2} − 2p ∼ k + 1. Combining this with (23) yields

ϵ > Vol(B(k_{1}, k_{2})) ≥ 2^{10nk/4}Vol(B(k, k + 1)) ≥ 2^{10nk/4}C2^{2k}2^{−2nk}.

This leads to contradiction if we let k → ∞. This finishes the proof of our claim.

For every t for which diameter of (M, g(t)) becomes very big, find k_{1} and k_{2} as in Claim 10. Then we have the following lemma.

Lemma 11. There exist r_{1}, r_{2} and a uniform constant C such that 2^{k_{1}} ≤ r_{1} ≤ 2^{k_{1}+1}, 2^{k_{2}−1} ≤ r_{2} ≤ 2^{k_{2}} and

∫_{B(r_{1},r_{2})} R ≤ CV,

where ϵ > 0 is the same as in Claim 10, B(r_{1}, r_{2}) = {z ∈ M : r_{1} ≤ dist_{t}(z) ≤ r_{2}} and V = Vol(B(k_{1}, k_{2})).

Proof. We will first prove the existence of r_{1}, such that 2^{k_{1}} ≤ r_{1} ≤ 2^{k_{1}+1} and

VolS(r_{1}) ≤ 2\frac{V}{2^{k_{1}}} ,  (24)

where S(r) is a metric sphere of radius r. We have that,

\frac{d}{dr} Vol(B(r)) = VolS(r).  (25)

Assume that for all r ∈ [2^{k_{1}}, 2^{k_{1}+1}] we have Vol(S(r)) ≥ 2\frac{V}{2^{k_{1}}} . Integrate (25) in r. Then,

Vol(B(k_{1}, k_{1} + 1)) = ∫_{2^{k_{1}}}^{2^{k_{2}}} Vol(S(r))dr
> 2\frac{V}{2^{k_{1}}} 2^{k_{1}} = 2V = 2Vol(B(k_{1}, k_{2})),

which is not possible, since k_{2} >> k_{1} by the proof of Claim 10. If for all r ∈ [2^{k_{2}−1}, 2^{k_{2}}] we have that Vol(S(r)) ≥ 2\frac{V}{2^{k_{2}}} , similarly as above we would get Vol(B(k_{2} − 1, k_{2})) > V = B(k_{1}, k_{2}), which is not possible. Therefore, there exists r_{2} ∈ [2^{k_{2}−1}, 2^{k_{2}}] such that

VolS(r_{2}) ≤ 2\frac{V}{2^{k_{2}}} .  (26)

Estimates (24), (26), together with bounds on ∇u obtained in Claim 8 imply

∫_{B(r_{1},r_{2})} R = ∫_{B(r_{1},r_{2})}(R − n) + nVol(B(r_{1}, r_{2}))
= − ∫_{B(r_{1},r_{2})} Δu + nVol(B(r_{1}, r_{2}))
≤ ∫_{S(r_{1})} |∇u| + ∫_{S(r_{2})} |∇u| ≤ \frac{V}{2^{k_{1}}} C2^{k_{1}+1} + \frac{V}{2^{k_{2}}} C2^{k_{2}+1}
= C̃V < C̃ϵ.

We can now finish the proof of Proposition 9.

Proof of Proposition 9. The proof of the proposition is similar to the proof of Perelman’s noncollapsing theorem from [12]. Assume diam(M, g(t)) is not uniformly bounded in t, that is, there exists a sequence t_{i} → ∞ such that diam(M, g(t_{i})) → ∞. Let ϵ_{i} → 0 be a sequence of positive numbers. By Claim 10 we can find sequences k_{1}^{i} and k_{2}^{i} such that

Vol_{t_{i}}B_{t_{i}}(k_{1}^{i}, k_{2}^{i}) < ϵ_{i},  (27)
Vol(B_{t_{i}}(k_{1}^{i}, k_{2}^{i})) ≤ 2^{10n}Vol(B(k_{1}^{i} + 2, k_{2}^{i} − 2)).  (28)

For each i, find r_{1}^{i} and r_{2}^{i} as in Lemma 11. Let ϕ_{i} be a sequence of cut off functions such that ϕ(z) = 1 for z ∈ [2^{k_{1}^{i}+2}, 2^{k_{2}^{i}−2}] and equal zero for z ∈ (−∞, r_{1}^{i}] ∪ [r_{2}^{i}, ∞). Let u_{i}(x) = e^{C_{i}}ϕ_{i}(dist_{t_{i}}(x, p_{i})) such that (2π)^{−n} ∫_{M} u_{i}^{2} = 1. This implies

(2π)^{n} = e^{2C_{i}} ∫_{M} ϕ_{i}^{2}
≤ e^{2C_{i}}Vol_{t_{i}}B_{t_{i}}(k_{1}^{i}, k_{2}^{i} + 1)
≤ e^{2C_{i}}ϵ_{i}.

Since ϵ_{i} → 0, this is possible only if lim_{i→∞} C_{i} = −∞. By Perelman’s monotonicity formula,

A ≤ W(g(t_{i}), u_{i}, 1/2)
= (2π)^{−n}e^{2C_{i}} ∫_{B_{t_{i}}(r_{1}^{i},r_{2}^{i})}(4|ϕ′_{i}(dist_{t_{i}}(y))|^{2} − 2ϕ_{i}^{2} ln ϕ_{i})dV_{t_{i}}
+ (2π)^{−n} ∫_{B_{t_{i}}(r_{1}^{i},r_{2}^{i})} Ru_{i}^{2}dV_{t_{i}} − 2n − 2C_{i}.  (29)

First of all by Lemma 11 and (28) we have

∫_{B_{t_{i}}(r_{1}^{i},r_{2}^{i})} Ru_{i}^{2} ≤ e^{2C_{i}} ∫_{B_{t_{i}}(r_{1}^{i},r_{2}^{i})} R
≤ C̃e^{2C_{i}}Vol_{t_{i}}B_{t_{i}}(k_{1}^{i}, k_{2}^{i})
≤ C̃e^{2C_{i}}2^{10n}Vol_{t_{i}}B_{t_{i}}(k_{1}^{i} + 2, k_{2}^{i} − 2)
≤ C̃2^{10n} ∫_{M} u_{i}^{2}dV_{t_{i}} = C̃2^{10n}(2π)^{n}.

By (28) we also have

e^{2C_{i}} ∫_{B_{t_{i}}(r_{1}^{i},r_{2}^{i})}(4|ϕ′_{i}(dist_{t_{i}}(y))|^{2} − 2ϕ_{i}^{2} ln ϕ_{i})dV_{t_{i}} ≤
≤ Ce^{2C_{i}}Vol_{t_{i}}B_{t_{i}}(k_{1}^{i}, k_{2}^{i}) ≤ e^{2C_{i}}C2^{10n}Vol_{t_{i}}B_{t_{i}}(k_{1}^{i} + 2, k_{2}^{i} − 2)
≤ C2^{10n} ∫_{M} u_{i}^{2} = C2^{10n}(2π)^{n}.

By (29) we get

A ≤ C̄ − 2C_{i} → −∞,

as i → ∞ and we get a contradiction. Therefore, there is a uniform bound on (M, g(t)), which gives us uniform bounds on scalar curvatures and |u(y, t)|_{C^{1}}.

4 Convergence of the Kähler Ricci flow

In this section we will apply Perelman’s boundness results on the scalar curvature and the diameter of the Kähler Ricci flow to show its sequential convergence towards singular metrics that are smooth and satisfy the Kähler Ricci soliton equation outside a singular set. We would like to understand more closely a singular metric that we get as a limit in Theorem 12. Our final goal as a long term project is to prove that a singular set S is a variety, that is, that it has some structure than only being a closed set.

Theorem 12. Let g(t) be a Kähler Ricci flow on a compact, Kähler manifold M, given by

\frac{d}{dt}g_{ij̄} = g_{ij̄} − R_{ij̄} = ∂_{i}∂j̄u.

Assume the Ricci curvature is uniformly bounded along the flow. For every sequence t_{i} → ∞ there is a subsequence so that (M, g(t_{i} + t)) converges to (M_{∞}, g_{∞}(t)) in the following sense:

(i) M_{∞} is smooth outside a singular set S which is of codimension at least 4 and the convergence is smooth off S.

(ii) A singular metric g_{∞}(t) satisfies a Kähler Ricci soliton equation g_{∞} − Ric(g_{∞}) = ∂ ¯∂f_{∞}, with (f_{∞})_{ij} = (f_{∞})ī ; j̄ = 0 outside S.

(iii) A potential function f_{∞} is smooth off S and there is a uniform constant C so that |f_{∞}(t)|_{C^{1}(M_{∞}\S)} ≤ C.

In [15] we have proved that if the Ricci curvatures are uniformly bounded, i.e. |Ric| ≤ C for all times t, then for every sequence t_{i} → ∞ there exists a subsequence such that (M, g(t_{i} + t)) → (M_{∞}, g_{∞}(t)) and the convergence is smooth outside a singular set S, which is at least of codimension four and M_{∞} is a smooth manifold off S. We also showed that a singular set S does not depend on time t. Moreover, g_{∞}(t) solves the Kähler-Ricci flow equation on M_{∞}\S. We want to show that g_{∞}(t) is actually a Kähler Ricci soliton off the singular set.

Due to Perelman we have the following uniform estimates for the Kähler Ricci flow: there are uniform constants C and κ such that for all t,

1. |u(t)|_{C^{1}} ≤ C,

2. diam(M, g(t)) ≤ C,

3. |R(g(t))| ≤ C,

4. (M, g(t)) is κ-noncollapsed.

This together with the uniform lower bound on the Ricci curvatures along the flow gives a uniform Sobolev constant, that is, there is a uniform constant C_{S} so that for any v ∈ C_{0}^{1}(M) we have that

(∫_{M} v^{\frac{4n}{2n−2}} dV_{g(t)})^{\frac{2n}{2n−2}} ≤ C_{S} ∫_{M} |∇v|^{2}dV_{g(t)},  (30)

for all times t ≥ 0. This enables us to work with integral estimates. The proof of Theorem 12 will be completed after we prove Proposition 13 and Proposition 14.

Proposition 13. A singular metric g_{∞} that arises in Theorem 12 satisfies the Kähler Ricci soliton equation on M_{∞}\S.

Proof. Notice that μ(g(t), 1/2) ≤ W(g(t), u(t), 1/2) ≤ C, for a uniform constant C, due to Perelman’s estimates mentioned above (recall that u(t) is the Ricci potential of metric g(t)). This yields a finite lim_{t→∞} μ(g(t), 1/2). Let f_{t} be a minimizer of W with respect to metric g(t) and let f_{t}(s) be a solution of the equation

\frac{d}{ds}f_{t}(s) = −Δf_{t}(s) + |∇f_{t}(s)|^{2} − R(s) + 2n,  (31)

for s ∈ [0, t]. Fix A > 0. By monotonicity of Perelman’s functional W we also have

0 ≤ (2π)^{−n} ∫_{0}^{A} ∫_{M} |R_{jk̄} + ∇_{j} ¯∇_{k}f_{t_{i}+A}(t_{i} + s) − g_{jk̄}|^{2}e^{−f_{t_{i}+A}(t_{i}+s)}dV_{t_{i}+s}ds +
+ (2π)^{n} ∫_{0}^{A} ∫_{M}(|∇_{j}∇_{k}f_{t_{i}+A}(t_{i} + s)|^{2} + | ¯∇_{j} ¯∇_{k}f_{t_{i}+A}(t_{i} + s)|^{2})e^{−f_{t_{i}+A}(t_{i}+s)}dV_{t_{i}+s}ds
= W(g(t_{i} + A), f_{t_{i}+A}, 1/2) − W(g(t_{i}), f_{t_{i}+A}(t_{i}), 1/2)
≤ μ(g(t_{i} + A), 1/2) − μ(g(t_{i}), 1/2),

where the right hand side of the previous estimate tends to zero as i → ∞, since there is a finite lim_{t→∞} μ(g(t), 1/2). Moreover, for any compact set K ⊂ M_{∞}\S we have

lim_{i→∞} ∫_{0}^{A} ∫_{K} |R_{jk̄}(t_{i} + s) + ∇_{j} ¯∇_{k}f_{t_{i}+A}(t_{i} + s) − g_{jk̄}(t_{i} + s)|^{2}e^{−f_{t_{i}+A}(t_{i}+s)}dV_{t_{i}+s}ds +

+ ∫_{0}^{A} ∫_{K}(|∇_{j}∇_{k}f_{t_{i}+A}(t_{i} + s)|^{2} + | ¯∇_{j} ¯∇_{k}f_{t_{i}+A}(t_{i} + s)|^{2})e^{−f_{t_{i}+A}(t_{i}+s)}dV_{t_{i}+s}ds = 0,

which implies for almost all s ∈ [0, A] and almost all x ∈ K,

• lim_{i→∞} |R_{jk̄}(t_{i} + s) + ∇_{j} ¯∇_{k}f_{t_{i}+A}(t_{i} + s) − g_{jk̄}(t_{i} + s)|^{2}e^{−f_{t_{i}+A}(t_{i}+s)} = 0.

• lim_{i→∞} |∇_{j}∇_{k}f_{t_{i}+A}(t_{i} + s)|^{2}e^{−f_{t_{i}+A}(t_{i}+s)} = 0.

Let D_{l} ⊂ M_{∞} be a sequence of open sets where S ⊂ D_{l} and D̄_{l} → S as l → ∞. We know that g(t_{i} + t) → g_{∞}(t) smoothly on M_{∞}\D_{l}. Function f_{t_{i}+A}(t_{i} + s) satisfies an evolution equation (31) and we have the geometries g(t_{i} + s) are uniformly bounded on M_{∞}\D_{l} for all i and all s ∈ [0, A] (those bounds depend on l, that is, on the closeness to the singular set S). Standard parabolic estimates as in [15] and [16] give there are uniform constants C(p, q, l) so that |\frac{∂^{p}}{∂s^{p}} ∇^{q}f_{t_{i}+A}(t_{i} + s)| ≤ C(p, q, l). Using the uniform derivative bounds from above, by Arzela Ascoli theorem and diagonalizing the sequence (by letting l bigger and bigger) we can find a subsequence so

that f_{t_{i}+A}(t_{i} + s) →^{C^{k}(M_{∞}\S)} f_{∞}(s). Moreover, this limit f_{∞}(s) satisfies

R_{jk̄}(g_{∞}(s)) + ∇_{j} ¯∇_{k}f_{∞}(s) − (g_{∞})_{jk̄}(s) = 0.  (32)
∇_{i}∇_{j}f_{∞} = ¯∇_{i} ¯∇_{j}f_{∞} = 0.  (33)

Relations (32) and (33) give,

sup_{M_{∞}\S} |D^{2}f_{∞}| ≤ C < ∞.  (34)

The previous estimate helps us prove the following proposition.

Proposition 14. There is a constant C so that |f_{∞}|_{C^{1}(M_{∞}\S)} ≤ C.

Proof. We will first mention few results and some notation from [3] and [4], which our proof will rely upon.

A point y ∈ M_{∞} is called regular, if for some k, every tangent cone at y is siometric to R^{2n}. Let R_{k} denote the set of k-regular points and put R = ∪_{k}R_{k}, the regular set. A point y ∈ M_{∞} is acalled singular, if it is not regular. Denote by S the set of singular points. In [3] it has been shown that under the assumption |Ric(g(t))| ≤ C, we have R = R_{n}. Moreover, one of the results in [5] is that dim S ≤ n − 4. The ϵ-regular set, R_{ϵ}, consists of those points y, such that every tangent cone, (Y_{y}, y_{∞}), satisfies d_{GH}(B_{1}(y_{∞}, B_{1}(0)) < ϵ. In [3] it was shown that for the uniform bound on the Ricci curvatures, there is an ϵ_{0}, so that for every ϵ < ϵ_{0}, R_{ϵ} = R and R_{ϵ} ∩ S = ∅. That means we can write

M_{∞} = R_{ϵ} ∪ S,

for ϵ < ϵ_{0}. Fix x_{0} ∈ R. By Theorem 3.9 in [4], there exists C(x_{0}) ⊂ R, with ν(M_{∞}\C(x_{0})) = 0 (ν is the unique limit measure, which in our noncollapsed case is exactly Hausdorff measure), such that for all y ∈ C(x_{0}) and ϵ > 0, there exists a minimal geodesic from x_{0} to y, which is contained in Ṙ_{ϵ}. If we choose ϵ small enough, R_{ϵ} = R and R is an open set. This means for almost all y ∈ R there is a minimal geodesic, call it γ, connecting x_{0} and y, all contained in R. For such y, we have

|Df_{∞}(x_{0}) − Df_{∞}(y)| ≤ C sup_{M_{∞}\S} |D^{2}f_{∞}|length(γ) ≤ C̃,

since we have an estimate (34) and since length(γ) ≤ D, where D is a uniform bound on the diameters of (M, g(t)). Since |Df_{∞}(x_{0})| is a finite number, we get |Df_{∞}|(x) ≤ C̃_{1} for almost all x ∈ R. On the other hand, since f_{∞} is a smooth function on R = M_{∞}\S, we get

sup_{M_{∞}\S} |Df_{∞}| ≤ C̃_{1}.

By similar arguments, we also have

sup_{M_{∞}\S} |f_{∞}| ≤ C̃_{2}.

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