4 The limit of the derivative of the K energy

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4 The limit of the derivative of the K energy

In last section, we get a explicit formula of td

dtM(t). Here, we are going to compute the limit lim

t→0td

dtM(t) in Theorem 3.4. In order to do this, we need some combinatoric techniques first.

Let (δ_i, σ_i), i = 0, · · · , p, be a sequence of pair of nonnegative rational numbers.

Let δ0 = 0. We assume that the sequence is “generic” in the sence that

1. All δ_i, i = 0, · · · , p, are distinct numbers. Hence, each δ_i, i = 1, · · · , p, is positive rational number.

2. Define the line ξ_i(x) = δ_i+ σ_ix, i = 0, · · · , p. None of three such lines intersect at the same point.

Now, suppose (δ_i, σ_i), i = 0, · · · , p, are generic, define (i_k, r_k), k = 0, · · · , m, by induction as follows: let i₀ = 0, r₀ = 0. If (i_k, r_k) has been defined, then

1. If for any r > r_k

δ_ik+ σ_ikr < δ_i+ σ_ir, i 6= i_k, then let m = k and stop.

2. If not, then define i_k+1 and r_k+1> r_k such that

δ_ik+ σ_ikr_k+1= δ_ik+1 + σ_ik+1r_k+1 ≤ δ_i+ σ_ir_k+1, (4.1) where i = 1, · · · , p. Note that (i_k, r_k), k = 0, · · · , m, are unique definite since (δ_i, σ_i), i = 0, · · · , p, are generic.

From the process above, we have the obvious.

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Remark. (ik, rk), k = 0, 1, · · · , is a finite sequence. In particular, the sequence stop at (i_m, r_m). Indeed, by the construction of i_k⁰s, we have

σ_i0 > σ_i1 > · · · > σ_ik > · · · .

Hence all i_k⁰s must be distinct. Since 0 ≤ i_k ≤ p, we have at most p + 1 distinct i_k⁰s. The second statement follows from the first item of the construction above.

Let

ξ(x) = min

i≥0(δ_i+ σ_ix). (4.2)

The function ξ(x) is a piecewise linear function, which be non-smooth at r_k, k = 1, · · · , m. And the function ξ(x) is differentiable almost everywhere.

Lemma 4.1 Assume that σ_im = 0, we have

m−1

X

k=0

(−δ_ik + δ_ik+1)(σ_ik + σ_ik+1− 1) = Z ∞

0

ξ⁰(x)(ξ⁰(x) − 1)dx. (4.3)

P roof . Note that ξ ≡ δ_im is a constant function if x large enough.

Z ∞ 0

ξ⁰(x)dx = lim

b→∞

Z b 0

ξ⁰(x)dx = lim

b→∞(ξ(b) − ξ(0)) = δim− δi0. Using the summation by parts, we have

Z ∞ 0

ξ⁰(x)²dx = r₁(σ_i0)²+

m−1

X

k=1

σ_i²

k(r_k+1− r_k) =

m−1

X

k=0

r_k+1(σ²_i

k − σ²_i

k+1) By definition of r_k, k = 0, · · · , m, in (4.1), we have

−δ_ik + δ_ik+1 = (σ_ik − σ_ik+1)r_k+1, for k = 0, · · · , m − 1.

Thus we have

m−1

X

k=0

(−δi_k+ δi_k+1)(σi_k + σi_k+1− 1)

=

m−1

X

k=0

rk+1(σ_i²_k− σ_i²_k+1) − (δim− δi0)

= Z ∞

0

ξ⁰(x)(ξ⁰(x) − 1)dx.

q

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Consider the smooth hypersurface M ⊂ CPⁿ defined by the polynomial F = 0 of degree d. Let X =

n

X

i=0

λ_iZ_i ∂

∂Z_i be the vector field for integers (λ₀, · · · , λ_n) such that

n

X

i=0

λ_i = 0. Let M_t be defined by the equation

F_t(Z₀, · · · , Z_n) = F (t^−λ0Z₀, · · · , t^−λnZ_n). (4.4) We write F_t as

F_t(Z₀, · · · , Z_n) = t^δ

p

X

i=0

a_it^δiZ^α

(i) 0

0 · · · Z_n^α(i)n , (4.5) where δ₀ = 0, and δ_i ≥ 0, i = 1, · · · , p. And

δ = −λ = min

0≤i≤p n

X

k=0

(−λ_k) · α_k⁽ⁱ⁾
.

By (4.4), we have

X(Z^α

(i) 0

0 · · · Z_n^α(i)n ) = −(δ_i+ δ)Z^α

(i) 0

0 · · · Z_n^α(i)n (4.6) for i = 0, · · · , p.

The sequence (δi, α⁽ⁱ⁾_k ), i = 0, · · · , p, k = 0, · · · , n, be the pair of nonnegative rational numbers which satisfies

1. All δ_i, i = 0, · · · , p, are distinct;

2. None of the three lines defined by δ_i+ α⁽ⁱ⁾_k x for i = 0, · · · , p, intersect at the same point, where k = 0, · · · , n.

So we may assume that the choice of (λ0, · · · , λn) is generic. Without loss of generality, we may assume that a₀ = 1, and 0 = δ₀ < δ₁ < · · · < δ_p. We also assume that a1, · · · , ap are all nonzero. Moreover, since M is smooth, we see that for each 0 ≤ k ≤ n, there is an 0 ≤ i ≤ p such that α⁽ⁱ⁾_k = 0.

Let

U_i = {[Z₀, · · · , Z_n] ∈ CPⁿ

|Z_i| > 1

2|Z_j|, j = 0, · · · , n}.

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Then ∪Ui = CPⁿ. Let Pi = {Zi = 0} and Pij = Pi∩ Pj for i 6= j and i, j = 0, · · · , n.

Let σ > 0 be chosen so that σ < 1 dmin

i≥1(δ_i). Let d(· , ·) be the distance induced by the Fubini-Study metric ω on CPⁿ, and define

V_ij^t = {z ∈ CPⁿ

d(z , P_ij) < |t|^σ}, i 6= j, i, j = 0, · · · , n.

By (4.5), we have t^−δF_t → Z^α

(0) 0

0 · · · Z_n^α(0)n as t → 0. Intuitively, M_t goes to the hyperplanes defined by Z^α

(0) 0

0 · · · Z_n^α(0)n = 0.

Lemma 4.2 There is a σ₁ > σ such that for any 0 ≤ k ≤ n and [Z₀, · · · , Z_n] ∈ (M_t− ∪ⁿ_i,j=0V_ij^t) ∩ U_k,

one can find a unique l 6= k such that  

Z_l Z_k 

< |t|^σ1 for t small enough, where [Z0, · · · , Zn] ∈ Mt.

P roof . Since [Z₀, · · · , Z_n] ∈ U_k, we have |Z_j| < 2|Z_k|, j = 0, · · · , n. By (4.5) we have

|Z^α

(0) 0

0 · · · Z_n^α(0)n | ≤ 2^d

p

X

i=1

a_i|t|^{mini≥1(δi)}|Z_k|^d. (4.7) Suppose for any l 6= k, we have

 

Z_l Z_k 

≥ |t|^σ1, then

|Z^α

(0) 0

0 · · · Z_n^α(0)n | ≥ |t|^σ1d|Z_k|^d. But we choose σ₁ by

σ < σ₁ < 1 dmin

i≥1(δ_i).

So we get a contradiction to (4.7). Hence we get the existence part.

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For the uniqueness, suppose there are l, m 6= k such that 

 Z_l Z_k 

< |t|^σ1,  

Z_m Z_k 

< |t|^σ1

for t small enough, then [Z₀, · · · , Z_n] ∈ V_lm^t . This is a contradiction. So we are

done. q

Now, we will prove that for t small enough, the connected component of M_t\ ∪ⁿ_i,j=1V_ij^t are graphs of functions over ˜P_i, where

P˜_i = P_i−[

j6=i

V_ij^t.

In order to do this, we first let Qi = {[Z0, · · · , Zn]

[Z0, · · · , Zi−1, 0

i, Zi+1, · · · , Zn] ∈ ˜Pi}, for i = 0, · · · , n. By the setting (1.4) and (1.5) in first section, we have

ψ(x₀, · · · , x_n) = min

0≤i≤p(δ + δ_i + α⁽ⁱ⁾₀ x₀+ · · · + α⁽ⁱ⁾_n x_n), (4.8) and

ψ_k(x) = min

0≤i≤p(δ + δ_i+ α⁽ⁱ⁾_k x), (4.9)

for k = 0, · · · , n.

Lemma 4.3 For σ > 0 small enough, there is a constant ε0 > 0 such that the solutions of z₁ of f = 0 satisfies

|z1− ϕ^k_i| ≤ |ϕ^k_i||t|^ε0

for i = 1, · · · , α⁽ⁱ₁^k)− α⁽ⁱ₁^k+1), k = 0, · · · , m − 1. Furthermore, the balls B_i^k = {z ∈ C

|z − ϕ^k_i| ≤ |ϕ^k_i||t|^ε0} for i = 1, · · · , α⁽ⁱ₁^k)− α⁽ⁱ₁^k+1), k = 0, · · · , m − 1, do not intersect each other if t small enough.

P roof . Without loss of generality, we assume that (z1, · · · , zn) = (Z₁

Z₀, · · · ,Z_n Z₀) on U₀. Then F_t= 0 can be written as

f =

p

X

i=0

a_it^δiz^α

(i) 1

1 · · · z_n^α(i)n = 0 (4.10)

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with a₀ = 1 and δ₀ = 0. The sequence (δ_i, α₁⁽ⁱ⁾), i = 0, · · · , p, is assumed to be a generic sequence.

For (z₁, · · · , z_n) ∈ ˜P₁∩ U₀, we have

|z1| ≥ |t|^σ,

for i = 2, · · · , n. The indices i and k are always set by i = 1, · · · , α₁^(ik)− α⁽ⁱ₁^k+1), k = 0, · · · , m − 1, in this proof, unless otherwise stated. We choose ε₁ > 0 such that

ε1 < min

0≤k≤m min

i6=ik,ik+1

(δ + δi+ α⁽ⁱ⁾₁ rk+1− ϕ1(rk+ 1)).

Define f_k and g_k as follows

f_k = a_ikt^δikz₁^α(ik)1 · · · z_n^α(ik)n + a_ik+1t^δik+1z^α

(ik+1) 1

1 · · · z_n^α(ik+1)n , and

g_k = f − f_k. Let ϕ^k_i be the (α⁽ⁱ₁^k)− α⁽ⁱ₁^k+1)) - th roots of

−a_ik+1 ai_k

t^{δik+1−δik}z^α

(ik+1) 2 −α^(ik)₂

2 · · · z_n^{α(ik+1)n −α(ik)n} . Then we have

|t|^rk+1+Cσ ≤ |ϕ^k_i| ≤ |t|^rk+1−Cσ for some constant C independent of t. And we also have

|t|^δ|g_k| ≤ |t|^{ψ1(rk+1)+ε1−dσ} on B_i^k and

|t|^δ|f_k| ≥ |t|^{ψ1(rk+1)+ε0+dσ}

on ∂B^k_i. We choose σ and ε₀ small enough such that ε₁− dσ > 3

4ε₁ and ε₀ ≤ 1 4ε₁. So we have dσ < 1

4ε₁, and then

|f_k| ≥ |t|^{ψ1(rk+1)−δ+ε0+dσ} > |t|^{ψ1(rk+1)−δ+ε0+} 1 4^ε1

≥ |t|^{ψ1(rk+1)−δ+}

2

4^ε1 > |t|^{ψ1(rk+1)−δ+(ε1−dσ)} ≥ |g_k|

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on ∂B_i^k. By the Rouch´e Theorem, fk and f have the same number of solutions in B_i^k. Since f_k has only one solution in B_i^k, there is only one solution z₁ of f = 0 satisfies

|z₁− ϕ^k_i| ≤ |ϕ^k_i||t|^ε0. Suppose there are two balls B_i^k and B_i^k1

1 such that B_i^k ∩ B_i^k1

1 6= φ, then for each z ∈ B_i^k∩ B^k_i1¹, we have

|ϕ^k_i − ϕ^k_i¹

1| ≤ |ϕ^k_i − z| + |z − ϕ^k_i¹

1| ≤ |t|^ε0(|ϕ^k_i| + |ϕ^k_i¹

1|).

Since t small enough, we have

|ϕ^k_i − ϕ^k_i¹

1| < 1

2max{|ϕ^k_i|, |ϕ^k_i¹

1|}.

Say, |ϕ^k_i| < |ϕ^k_i1¹|. This means B_i^k ⊂ B_i^k₁¹, we get a contradiction. So if t is small

enough, B_i^k’s do not intersect each other. q

Proposition 4.4 Using the notation as above, we have Z

Mt∩Q_i n

X

j=0

XFt

|∇F_t|²

j

(F_t)_jωⁿ⁻¹

→ −δα⁽⁰⁾_i − Z ∞

0

ψ⁰_i(x)(ψ⁰_i(x) − 1)dx,

(4.11)

for i = 0, · · · , n, as t → 0.

P roof . In this proof, we omit the constants in an inequality for convenience. So A ≤ B means there is a constant C independent of t such that A ≤ CB. It suffices to prove the case i = 1 because the proof of other cases are similarly. If α⁽⁰⁾₁ = 0, then ϕ⁰₁ ≡ 0, so the proposition holds automatically. Now we assume that α₁⁽⁰⁾≥ 1, and we only prove this property on M_t∩ Q₁∩ U₀.

For the sake of simplicity, let F = F_t. As the setting in Lemma 4.3, the indices i, k are always running in i = 1, · · · , α₁^(ik)−α⁽ⁱ₁^k+1), k = 0, · · · , m−1, unless otherwise

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stated. For fixed i, k, attaching the B_i^k in the above lemma for each p ∈ ˜P1 ∩ U0, we get a bundle ˜B_i^k. On each bundle ˜B_i^k, since |z_i| > |t|^σ, we have

n

X

j=0

( XF

|∇F |²)

j

(F )_j = (XF )₁F₁− (XF )F₁₁

F₁² + o(1)

= −(δ + δ_ik)α⁽ⁱ₁^k)+ (δ + δ_ik+1)α⁽ⁱ₁^k+1) α₁^(ik)− α₁^(ik+1)

−(−δ_ik + δ_ik+1) α⁽ⁱ₁^k)(α⁽ⁱ₁^k)− 1) − α₁^(ik+1)(α⁽ⁱ₁^k+1)− 1)
(α⁽ⁱ₁^k)− α⁽ⁱ₁^k+1))²

+ o(1)

= −δ + −δ_ikα⁽ⁱ₁^k)+ δ_ik+1α⁽ⁱ₁^k+1)+ (δ_ik− δ_ik+1)(α⁽ⁱ₁^k)+ α⁽ⁱ₁^k+1)− 1)

α⁽ⁱ₁^k)− α⁽ⁱ₁^k+1) + o(1) (4.12) as t → 0 for k = 0, · · · , m − 1, where o(1) → 0 as t → 0. The equation (4.12) is also true for p ∈ ˜P₁∩ U_l for l 6= 0 by the same process. Hence the equation holds for p ∈ ˜P₁. If π : Q₁ → ˜P₁ is the projection, and ∂z₁

∂zk

= −F_k F1

→ 0 as t → 0, by (4.10), we have

det π = o(1) (4.13)

as t → 0. Hence by (4.12), (4.13) and the main result in [3] for hypersurfaces, we have

Z

Mt∩Q1

n

X

j=0

XF_t

|∇Ft|²

j

(F_t)_jωⁿ⁻¹

= − δα⁽⁰⁾₁ +

m−1

X

k=0

(δ_ik − δ_ik+1)(α⁽ⁱ₁^k)+ α⁽ⁱ₁^k+1)− 1)
V ol(CPⁿ⁻¹) + o(1)

as t → 0. We know that V ol(CPⁿ⁻¹) = 1. By Lemma 4.1, we get

Z

Mt∩Q1

n

X

j=0

XF_t

|∇F_t|²

j

(F_t)_jωⁿ⁻¹

→ −δα₁⁽⁰⁾− Z ∞

0

ψ₁⁰(x)(ψ₁⁰(x) − 1)dx

as t → 0. By the same argument, we get (4.11) holds for i = 0, · · · , n. q

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Lemma 4.5 Let p be a fixed point in Mt and let d(x, p) be the distance from x ∈ CPⁿ to p defined by the Fubini-Study metric. Let B_p(ε) = {x ∈ CPⁿ

d(x, p) < ε}.

Then there are constants C, σ independent of p and t such that Z

Mt∩Bp(ε)

ωⁿ⁻¹ ≤ Cε²ⁿ⁻²log ε⁻¹ (4.14) for t small enough, where ε = |t|^σ.

P roof . Consider the function ρ : R → R which is defined by

ρ(x) =







1 , if x ∈ [0, 1];

0 , if x ∈ R \ [0, 1].

Then we have

Z

Mt∩Bp(ε)

ωⁿ⁻¹ ≤ Z

Mt

ρ(d(x, p) ε )ωⁿ⁻¹.

Since F_t be the defining function of M_t. Then in the sence of distribution, we have i

2π∂∂ log |Ft|² (Pn

i=0|Z_i|²)^d = [Mt] − dω.

Then we have Z

Mt

ρ(d(x, p) ε )ωⁿ⁻¹

= d Z

CPⁿ

ρ(d(x, p) ε )ωⁿ+

Z

CPⁿ

ρ(d(x, p)

ε )∂∂ log |F_t|² (Pn

i=0|Z_i|²)^dωⁿ⁻¹.

(4.15)

Now we have to estimate the right hand side of (4.15). For the first term, we have Z

CPⁿ

ρ(d(x, p)

ε )ωⁿ≤ Cε²ⁿ. (4.16)

Assume that p ∈ U₀ = {[Z₀, · · · , Z_n]

|Z₀| > 1

2|Z_j|, j = 1, · · · , n}. Then by (4.5), we have

F_t= t^δZ₀^df_t, where f_t → f₀ = z^α

(0) 1

1 · · · z_n^α(0)n 6≡ 0. Note that f_t is defined in Lemma 4.3. If we define dV₀ = ( i

2π)ⁿdz₁ ∧ dz₁ ∧ · · · ∧ dz_n ∧ dz_n is the Euclidean measure and

‧

For the second term of the right hand side of (4.17), we have C for t small enough. By (4.15), (4.16), (4.17), (4.18) and (4.19), we have

Z

Lemma 4.6 There exists a constant C > 0 such that for t small X

ε²ⁿ⁻⁴], satisfying

∪^k=1_m B_pk(ε) ⊃ P_ij.

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By the definition of V_ij^t, we have

∪^k=1_m B_pk(2ε) ⊃ V_ij^t. Hence we have

Z

V_ij^t∩M_t

ωⁿ⁻¹ ≤

m

X

k=1

Z

Mt∩B_pk(2ε)

ωⁿ⁻¹. By Lemma 4.5, we have

Z

V_ij^t∩Mt

ωⁿ⁻¹ ≤

m

X

k=1

(Cε²ⁿ⁻²log ε⁻¹)

≤ C

ε²ⁿ⁻⁴ε²ⁿ⁻²log ε⁻¹

= C|t|^2σlog ε⁻¹. Then

X

i6=j

Z

V_ij^t∩M_t

ωⁿ⁻¹ ≤X

i6=j

C|t|^2σlog ε⁻¹

= C|t|^2σlog |t|⁻¹.

q

Lemma 4.7 There exists a constant C independent of t such that for any measur-able subset E of M_t

 

Z

E

∂ϕ ∧ ∂θ ∧ ωⁿ⁻²

≤ Cp

log |t|⁻¹·p

meas(E) where

ϕ = log |∇F |² (Pn

i=0|Z_i|²)^(d−1), and

θ = − Pn

i=0λ_i|Z_i|² Pn

i=0|Z_i|² .

P roof . Since M_t is a submanifold, the Ricci curvature has an upper bound. So by (3.1), there exists a constant C such that

− i

2π∂∂ϕ ≤ Cω. (4.20)

‧

Since M is smooth, we have an estimate

−C log |t|⁻¹ ≤ |ϕ| ≤ C log |t|⁻¹

for some constant C. By (4.20), using integration by parts, we have Z

Since E is a measurable subset of Mt, by Cauchy inequality, we have  P roof of T heorem1.8. By Proposition 4.4, we have

Z

‧

By Lemma 4.7, we have Z By Lemma 4.6, we have

Z

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as t → 0. If M₀ is defined as the zero set of Z^α

(0) 0

0 · · · Z_n^α(0)n = 0 counting the multiplicity, since θ is a bounded function, we have

Z

Mt

θωⁿ⁻¹ = Z

M0

δωⁿ⁻¹+ o(1) as t → 0. Zhiqin Lu [10, Theorem 5.1] showed that

Z

M0

θωⁿ⁻¹ = −δ n. By (3.31) in Theorem 3.4, we have

td

dtM(t) = 2 d

δ(n + 1)(d − 1)

n +

n

X

i=0

Z ∞ 0

ψ⁰_i(x)(ψ⁰_i(x) − 1)dx
+ o(1)

as t → 0. Hence

limt→0td dtM(t)

= 2

d − λ(n + 1)(d − 1)

n +

n

X

i=0

Z ∞ 0

ψ_i⁰(x)(ψ_i⁰(x) − 1)dx

for generic (λ₀, · · · , λ_n). q

Since for a K¨ahler–Einstein manifold, the K energy has a lower bound, Lu[11]

give a general result of theorem 1.8.

Theorem 4.8 (Lu) If M is a K¨ahler–Einstein hypersurface with positive first Chern class, then we have

−λ(d − 1)(n + 1)

n +

n

X

i=0

Z ∞ 0

ψ⁰_i(x)(ψ⁰_i(x) − 1)dx ≤ 0

for any λ0, · · · , λn ∈ R with

n

X

i=0

λi = 0.

Here, we give two effective ways to verify the K stability for hypersurface.

‧

Theorem 4.9 Let M be a compact Fano hypersurface defined by the zeros of the polynomial F of degree d ≥ 1. If one can find a sequence (λ₀, · · · , λ_n) with

n

X

k=0

λ_k= 0 such that λ < 0, then there is no K¨ahler–Einstein metric on M .

P roof . By the equation (1.2), if λ = max

Hence M is not K stable. By theorem 4.8, we know that there is no K¨ahler–Einstein

metric on M . q

Theorem 4.10 Let M be a compact Fano hypersurface on CPⁿ defined by the zeros of polynomial F (Z₀, · · · , Z_n) of degree d ≥ 1. Suppose that for some k = 0, · · · , n, we have α⁽ⁱ⁾_k = 0 for all i = 0, · · · , p. Then there is no K¨ahler–Einstein metric on M .

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P roof . Without losing generality, we assume that F miss the term Z0. Write

F (Z0, · · · , Zn) =

p

X

i=0

aiZ^α

(i) 1

1 · · · Z_n^α(i)n .

Take λ_i = −i, i = 1, · · · , n, and λ₀ = n(n + 1)

2 . From the equation (1.2), we have λ = max

0≤i≤p(

n

X

k=0

λ_kα⁽ⁱ⁾_k ) = max

0≤i≤p(

n

X

k=1

λ_kα⁽ⁱ⁾_k ) < 0.

By theorem 4.9, there is no K¨ahler–Einstein metric on M . q

‧ 國

立 政 治 大 學

‧

N a tio na

l C h engchi U ni ve rs it y

在文檔中 K 穩定性與熱帶幾何之研究 - 政大學術集成 (頁 33-49)