Bergman kernel along the Kähler–Ricci flow and Tian’s conjecture

There exists a constant C>0 such that the scalar curvature R of g⁢(t) satisfies the estimate

for all t≥0 and all x∈M. Here constant C depends only on the lower bound of R⁢(g⁢(0)).

Proof.

By directly computing, we have the evolution of R:

Let Rmin⁢(0) be the minimum of R⁢(x,0) on M. If Rmin≥0, then, by the maximum principle, we have R⁢(x,t)≥0 for all t>0 and all x∈M.

Now suppose Rmin⁢(0)<0. Set F⁢(x,t)=R⁢(x,t)-Rmin⁢(0). Then, F⁢(x,0)≥0 and F satisfies

Hence, it follows again from the maximum principle that F≥0 on M×[0,∞], i.e.,

for all t>0 and all x∈M. ∎

Let g^i⁢j¯⁢(s), 0≤s<1, and gi⁢j¯⁢(t), 0≤t<∞, be solutions to the Kähler–Ricci flow

respectively. Then g^i⁢j¯⁢(s) and gi⁢j¯⁢(t) are related by

Proposition A.3 (Ye [34])

Consider the Kähler–Ricci flow (A.2) on a Fano manifold M. Then there are positive constants A and B depending only on the dimension m, a non-positive lower bound for R⁢(g⁢(0)), a positive lower bound for Vol⁡(M,g⁢(0)) and an upper bound for Sobolev constant Cs⁢(M,g⁢(0)) such that for each t>0 and all f∈W1,2⁢(M) we have

Consequently, let L>0 and assume R≤1r2 on a geodesic ball B⁢(x,r) with 0<r≤L. Then there holds

Proof.

By the monotonicity of the 𝒲-entropy functional and the flow (A.1) in Lemma A.2, one can show that

where constants A and B depend only on the quantities stated in the proposition. By the relation between (A.1) and (A.2), we have

At last, to prove κ-noncollapsing, we can assume r=1, since we can scale the metric with factor 1r2. That is, g¯=1r2⁢g. Thus, we have

and R¯≤1 on the geodesic ball Bg¯⁢(x,1). Then a standard argument implies the lower bound of Volg¯⁡(Bg¯⁢(x,1)). One can find more details in [34]. ∎

Function u⁢(t) is uniformly bounded from below, that is,

where constant C depends only on the constants in Proposition A.3.

Proof.

Since Δ⁢u=m-R, we have

Denote f=e-u. We have

Multiplying fp (p≥1) to both sides of (A.4) and integrating by parts, we get

That is,

Since R has a uniform lower bound, we have

Then, by Proposition A.3 and Moser’s iteration, we deduce

Hence,

This provides a pointwise lower bound of u. ∎

Denote a=-(2⁢π)-m⁢∫Me-u⁢u⁢𝑑μ⁢(t). Then

where the constant C depends only on the volume of g⁢(0), a lower bound of R⁢(g⁢(0)) and an upper bound of the Sobolev constant Cs=Cs⁢(M,g⁢(0)).

There is a uniform constant C so that

where the constant depends only on Vol⁡(M,g⁢(0)), the L2-Sobolev constant Cs of g⁢(0) and upper bounds of |R⁢(g⁢(0))| and |∇⁡u|⁢(0).

Proof.

This is essentially a parabolic version of Yau’s gradient estimate in [25]. By Lemma A.4, we have u⁢(x,t)≥-C. Choosing B=C+1, then u⁢(x,t)+B≥1. Let H=|∇⁡u|2/(u+B). In order to show (A.6), we only need to estimate an upper bound for H. By directly computing, we have

For each T>0, suppose H attains its maximum at (x0,t0) on M×[0,T]. If t0=0, the upper bound of H follows easily by the bound for |∇⁡u|⁢(g⁢(0)). Assume t0>0. Then at (x0,t0) we have

Substituting these into (A.8), we obtain

On the other hand, since ∇⁡H⁢(x0,t0)=0, we have

Thus, at (x0,t0),

Combining with (A.9), we have

Hence, H⁢(x0,t0)≤2⁢(B-a). Let T→∞ and note that a is bounded. We complete the proof of (A.6).

Now we turn to the proof of (A.7). Our goal is to prove that -Δ⁢u is bounded by C⁢(u+C), which yields (A.7), since Δ⁢u=n-R. Let K=-Δ⁢uu+B, where B is a uniform constant as above. Similar computation as before gives that

Combining this with (A.8), we have

For each T>0, suppose 2⁢H+K attains its maximum at (x0,t0) on M×[0,T]. If t0=0, the upper bound of 2⁢H+K follows easily by the bound for |∇⁡u|⁢(0) and |R|⁢(0). Assume t0>0. Then at (x0,t0) we have

Thus,

Here we have used the fact that u+B≥1. Hence, |Δ⁢uu+B|⁢(x0,t0)≤C. Now for each (x,t)∈M×[0,T],

Letting T→∞ finishes the proof. ∎

There exists a constant C depending only on the constant of Lemma A.6 such that each of u⁢(y,t), R⁢(y,t) and |∇⁡u|2⁢(y,t) is no greater than C⁢(distt2⁡(x^,y)+1) for all t>0 and y∈M, where u⁢(x^,t)=minx∈M⁡u⁢(x,t).

Proof.

By Lemma A.6, we only need to estimate u⁢(x,t). Actually, by (A.6), we have

Hence, we have

where u⁢(x^,t)=minx∈M⁡u⁢(x,t). On the other hand, since ∫Me-u⁢𝑑μ⁢(t)=(2⁢π)m, we have

So

The other two inequalities R⁢(y,t)≤C⁢(distt2⁡(x^,y)+1) and |∇⁡u|2⁢(y,t)≤C⁢(distt2⁡(x^,y)+1) follow from this and Lemma A.6. ∎

For each ϵ>0, if diam⁡(M,g⁢(t))≥Cϵ, we can find B⁢(k1,k2) such that

Here we can choose

and C is the constant in (A.10).

Proof.

Denote

and assume diam⁡(M,g)≥2k0⁢4[Vϵ]+1+1. Then we will show that for each k02≤k≤k02⁢4[Vϵ], there exists B⁢(k1,k2) so that 2⁢k≤k1<k2≤6⁢k+1 and

Otherwise, by (A.10)

Thus,

On the other hand, there must be some 0≤l≤[Vϵ] such that

Otherwise,

Getting together all the above arguments implies the lemma. ∎

For each 0<k1<k2<∞, there exist r1,r2 and a uniform constant C such that 2k1≤r1≤2k1+1, 2k2-1≤r2≤2k1 and

where B⁢(r1,r2)={z∈M:r1≤distt⁡(z,x^)≤r2} and the constant C depends only on the constant in Corollary A.7.

Proof.

First of all, since

we have

Here S⁢(r) denotes the geodesic sphere of radius r centered at x^ with respect to g⁢(t).

Hence, we can choose r1∈[2k1,2k1+1] such that

Similarly, there exists r2∈[2k2-1,2k2] such that

Next, by integration by parts and Corollary A.7,

Therefore, since R=-Δ⁢u+m, it follows that

proving Lemma A.9. Here C is the constant in Corollary A.7. ∎

There exists a constant ϵ0>0. If 0<ϵ<ϵ0, there is no B⁢(k1,k2) such that