Rotationally symmetric conformal K¨ ahler, Einstein-Maxwell metrics

New York Journal of Mathematics

New York J. Math.26(2020) 334–361.

Rotationally symmetric conformal K¨ ahler, Einstein-Maxwell metrics

Zhiming Feng

Abstract. In this paper, we show that there are non-trivial complete rotationally symmetric conformal K¨ahler, Einstein metrics on B^d and C^d, and there are non-trivial complete rotationally symmetric conformal K¨ahler, Einstein-Maxwell metrics onB^d∗andC^d∗.

Contents

1. Introduction 334

2. Radial conformally K¨ahler, Einstein-Maxwell metrics 337 3. The geodesics of radial conformally K¨ahler metrics 344

4. Proof of Theorem 1.1 350

5. Proof of Theorem 1.2 357

6. Proof of Theorem 1.3 359

References 360

1. Introduction

Let M be a complex m-dimensional K¨ahler manifold endowed with a K¨ahler metric g with respect to an integrable almost complex structure J.

A Hermitian metric eg on (M, J) is called a conformally K¨ahler, Einstein- Maxwell metric (cKEM metric for short) if it satisfies the following three conditions:

(a) There exists a positive smooth functionf on M such thateg= _f¹2g.

(b) The Ricci tensor Ric_eg of the metricgesatisfies Ric_eg(J·, J·) = Ric

eg(·,·).

(c) The scalar curvature s

geof eg is constant.

The Ricci tensors ofeg and g are related by (see e.g. [6], 1.161) Riceg = Ric_g+2(m−1)

f D²f− 1

f²(f∆_gf+ (2m−1)|df|²_g)g,

where D denotes the Levi-Civita connection of g, | · |_g is the tensor norm induced by g, and ∆g :=δd + dδ is the Hodge Laplacian with respect tog.

Received January 29, 2020.

2010Mathematics Subject Classification. 32Q15, 53C55.

Key words and phrases. Conformal K¨ahler, Einstein-Maxwell metrics, Einstein metrics, K¨ahler metrics.

ISSN 1076-9803/2020

334

Lets_g be the Riemann scalar curvature ofg, thus the scalar curvatures

eg of eg is given by

seg = f²s_g−2(2m−1)f∆_gf −2m(2m−1)|df|²_g

= f²s_g+2(2m−1)

m−1 f^m+1∆_gf^−m+1. Note that Ric

eg(J·, J·) = Ric

ge(·,·) if and only ifD²f(J·,·) =√

−1(∂∂f)(·,·), or if and only if

∂(grad^1,0f) =∂(g^j¯qfq¯∂j) = 0.

Setting f to be constant yields constant scalar curvature K¨ahler metric (cscK metric for short), so cKEM metrics with nonconstant f are referred to as non-trivial. For the case of compact, so far, not many examples of non-trivial cKEM metrics are known. Page in [18] and Chen-LeBrun-Weber in [7] have shown that the one-point-blow-up and the two-point-blow-up of CP² admit the conformally K¨ahler Einstein metrics, respectively. The con- formally K¨ahler Einstein metrics also constructed by Apostolov-Calderbank- Gauduchon in [1, 2] on 4-orbifolds and by B´erard Bergery [5] onP¹-bundles over Fano K¨ahler-Einstein manifolds. Non-Einstein cKEM examples are given by LeBrun [16, 17] on CP¹×CP¹ and the one-point-blow-up of CP², by Koca-Tønnesen-Friedman [12] on ruled surfaces of higher genus, and by Futaki-Ono [9] onCP¹×M whereM is a compact constant scalar curvature K¨ahler manifold of arbitrary dimension. For more research on cKEM met- rics, please refer to Apostolov-Maschler [3], Apostolov-Maschler-Tonnesen- Friedman [4], Futaki-Ono [10] and Lahdili [13, 15, 14].

In this paper, we study the existence of cKEM metrics on non-compact manifolds. Specifically, we study the existence of cKEM metrics on rota- tionally invariant domains. Let

B^d:={z= (z₁, . . . , z_d)∈C^d:kzk² =

d

X

j=1

|z_j|² <1}, B^d∗:=B^d\ {0}, C^d∗:=C^d\ {0}.

The following Theorem 1.1, Theorem 1.2 and Theorem 1.3 are the main results of this paper.

Theorem 1.1. Ford≥2 letgF be a K¨ahler metric on a domain Ω =B^d or C^d associated with the K¨ahler form ω_F =√

−1∂∂Φ_F, namely g_F(X, Y) = ω_F(X, J Y) for all X, Y ∈T_zΩ and each z∈Ω, where Φ_F(z, z) = F(t) and t= lnkzk².

For a givena <0, setF(−∞) = 0,x=F⁰(t),ϕ(x) =F⁰⁰(t),f =ax+1>

0, eg= _f¹2g_F and ψ(d, u) =

d−2

X

k=0

(k+ 1) Γ(d−1)Γ(d+ 1)

Γ(d+ 1 +k)Γ(d−1−k)u^k.

(i) egis a complete Ricci flat metric onC^d, that isRic

eg = 0, if and only if ϕ(x) =x(1 +ax)²ψ(d, ax), x∈[0,−1

a).

(ii) For a given λ < 0, eg is a complete Einstein metric on B^d, that is Ric_eg =λg, if and only ife

ϕ(x) =x(1 +ax)²ψ(d, ax)− λ

d+ 1x²ψ(d+ 1, ax), x∈[0,−1 a).

In particular, if























a=−1, λ=−(4d−2), F(t) =e^t=kzk²,

f(t) = 1−e^t= 1− kzk², theneg is an Einstein metric on B^d.

(iii) The complete rotationally symmetric cKEM metric eg on B^d or C^d must be an Einstein metric.

Theorem 1.2. Ford≥2letgF be a K¨ahler metric on a domainΩ =B^d∗ or C^d∗ associated with the K¨ahler form ω_F =√

−1∂∂Φ_F, namely g_F(X, Y) = ωF(X, J Y) for ∀X, Y ∈ TzΩ and each z ∈ Ω, where ΦF(z, z) = F(t) and t= lnkzk².

For a given a < 0, set x = F⁰(t), ϕ(x) = F⁰⁰(t), f = ax+ 1 > 0 and eg= _f¹2g_F.

For a constant β ≤0 and any x₀ ∈(0,−¹_a), let ϕ(x) = (1 +ax)^2d−1

x^d−1

Z x x0

(x−u)u^d−2 d(d−1)(1 +au)²−βu

(1 +au)^2d+1 du (1) for x∈(x0,−¹_a).

(i) If β = 0and ϕ(x) is defined on (x0,−¹_a) by (1), then eg is a complete cKEM metric on C^d∗ with zero scalar curvature.

(ii) If β < 0 and ϕ(x) is defined (x0,−¹_a) by (1), then eg is a complete cKEM metric on B^d∗ with negative scalar curvature 2β.

Theorem 1.3. For d≥ 2 let gF be a K¨ahler metric on a rotationally in- variant domain Ω ⊂C^d associated with the K¨ahler form ω_F =√

−1∂∂Φ_F, namely g_F(X, Y) = ω_F(X, J Y) for ∀X, Y ∈ T_zΩ and each z ∈ Ω, where ΦF(z, z) =F(t) and t= lnkzk².

Let x=F⁰(t), ϕ(x) =F⁰⁰(t),f =x and eg= _f¹2gF.

For a constant β ≤0 and any x₀ >0, let ϕ(x) = x^d

Z x x0

(x−u)d(d−1)u−β

u^d+2 du (2)

= x− β

d(d+ 1) +β−d²x₀

dx^d₀ x^d+(d²−1)x₀−β

(d+ 1)x^d+1₀ x^d+1. (3) (i) If β = 0 and ϕ(x) is defined on (0, x0) by (3), then eg is a complete cKEM metric on C^d∗ with zero scalar curvature.

(ii) If β < 0 and ϕ(x) is defined on (0, x₀) by (3), then eg is a complete cKEM metric on

Ω :={z∈C^d:kzk² >1}

with negative scalar curvature 2β.

Remark 1.4. The complete cKEM metrics in Theorems 1.2 and 1.3 are not Einstein metrics.

Remark 1.5. The complete cKEM metric eg with zero scalar curvature defined in Theorem 1.3 is not proportional to the complete cKEM metric eg with zero scalar curvature defined in Theorem 1.2.

The paper is organized as follows. In Section 2, by the momentum pro- filesϕ(x) (refer to Hwang-Singer [11]) of rotationally symmetric K¨ahler met- ricsg_F, we derive ordinary differential equations for rotationally symmetric cKEM metrics. In Section 3, in order to discuss the completeness of met- rics, we give the geodesics of radial conformally K¨ahler metrics. In Section 4, Section 5 and Section 6, by using the conclusions of Section 2 and Sec- tion 3, we obtain proofs for Theorem 1.1, Theorem 1.2 and Theorem 1.3, respectively.

2. Radial conformally K¨ahler, Einstein-Maxwell metrics For convenience, we need Lemma 2.1 and Remark 2.2 of [8], namely:

Lemma 2.1. Let

T ≡(T_i¯j) = ∂²ΦF

∂z^t∂¯z,

where ΦF(z, z) =F(t), t= lnr², r=kzk, z ∈C^d. Then T = F⁰

r²Id+F⁰⁰−F⁰

r⁴ z^tz, (4)

detT = (F⁰)^d−1F⁰⁰

r^2d , (5)

and

T⁻¹ = r²

F⁰Id+ ( 1 F⁰⁰ − 1

F⁰)z^tz, (6)

where z= (z1, . . . , z_d), z= (z1, . . . , z_d), andz^t denotes transpose of z.

Remark 2.2. From (4), we obtain that Φ_F(z, z) = F(t) with t = lnkzk² is a K¨ahler potential on the domain Ω\ {0} if and only if F⁰(t) > 0 and F⁰⁰(t)>0 for d >1, orF⁰⁰(t)>0 for d= 1.

Lemma 2.3. Under the situation of Lemma 2.1, if a Hermitian matrix T is positive definite, f is a function in tand

∂ ∂f

∂zT⁻¹

= 0,

then there exist constants a, b such that f =aF⁰(t) +b.

Proof. Let x=F⁰(t) and ϕ(x) =F⁰⁰(t). By Lemma 2.1, we have

∂f

∂z = df dxϕ(x) z

r², and

∂f

∂zT⁻¹ = df dxz.

Thus

∂ ∂f

∂zT⁻¹

=∂ df

dx

z= 0, that is

∂ df

dx

= 0.

Since

∂ df

dx

= d²f

dx²ϕ(x)∂t and ϕ(x)>0, so

d²f dx² = 0.

Then there exist constants a, bsuch that f =ax+b.

By Lemma 2.3, if g_F is the K¨ahler metric associated with the K¨ahler formω_F =√

−1∂∂F(t) witht= logkzk² on a rotationally invariant domain Ω⊂C^d, andeg= _f¹2g_F is a rotationally symmetric cKEM metric on Ω, then f must be formsf =aF⁰(t) +b.

Theorem 2.4. Let gF be a K¨ahler metric on a rotationally invariant do- main Ω ⊂ C^d associated with the K¨ahler form ω_F = √

−1∂∂Φ_F, namely gF(X, Y) =ωF(X, J Y), where ΦF(z, z) =F(t), t= lnr², r² =kzk².

Set x=F⁰(t), ϕ(x) =F⁰⁰(t), f =ax+b=aF⁰(t) +b >0 and eg= _f¹2g_F. (i) The scalar curvature s_eg is constant for eg if and only if

x^d−1 f^2d−1ϕ

⁰⁰

(x) = x^d−1 f(x)^2d−1

d(d−1)

x − k

eg

f(x)²

, (7)

where k

eg = ¹₂s

eg is a constant.

(ii) eg is an Einstein metric, that is Ric

eg =λeg, if and only if ϕ

f 00

(x) + λ

f(x)³ = 0 (8)

for d= 1 and ϕ⁰(x) +

d−1

x − a

ax−b −(2d−1)a ax+b

ϕ(x) +

d+ 4abd−λ

2a(ax−b) − λ 2a(ax+b)

= 0. (9) for d >1, where λis a constant.

Proof. Let k

eg = ¹₂s

eg andkgF = ¹₂sgF. Then

keg = f²k_gF + 2(2d−1)f4_gFf−d(2d−1)|df|²_g

F

= f²k_gF −2(2d−1)

d−1 f^d+14_gFf^−d+1, where

4_gF = Tr

T⁻¹_∂z^∂t²∂¯z

, k_gF =−Tr

T^−1∂2_∂z^{log det}t∂¯z ^T

, for the expression of T, see Lemma 2.1.

(i) Since

∂f^1−d

∂zi

= (1−d)f^−ddf dx

dx dt

∂t

∂zi

=a(1−d)f^−dϕz_i r² and

∂²f^1−d

∂z_i∂z_j =a(1−d)ϕ

(f^−dϕ)⁰_xzjzi

r⁴ +f^−dδijr²−zjzi

r⁴

, by (6), it follows that

4_gFf^1−d= Tr

T⁻¹∂²f^1−d

∂z^t∂z

=a(1−d)ϕ r⁴Tr

r²

xId+x−ϕ xϕ z^tz

f^−dr²Id+ ((f^−dϕ)⁰_x−f^−d)z^tz

=a(1−d)f^−dϕ r² Tr

r² xI_d+

ϕ⁰ ϕ −ad

f − 1 x

z^tz

=a(1−d)f^−dϕ ϕ⁰

ϕ −ad

f +d−1 x

.

Using (2.11) of [8], we obtain keg = f²kg_F −2(2d−1)

d−1 f^d+14_gFf^−d+1

= f²kgF + 2a(2d−1)f

ϕ⁰+

d−1 x −ad

f

ϕ

= f²

d(d−1)

x −ϕ⁰⁰−2(d−1)

x ϕ⁰−(d−1)(d−2)

x² ϕ

+ 2a(2d−1)f

ϕ⁰+

d−1 x −ad

f

ϕ

, that is,

ϕ⁰⁰+

2(d−1)

x −2a(2d−1) f

ϕ⁰ +

(d−1)(d−2)

x² −2a(2d−1) f

d−1 x −ad

f

ϕ+ k

eg

f² = d(d−1)

x ,

or

x^d−1 f^2d−1ϕ

00

= x^d−1 f^2d−1

d(d−1)

x − k

ge

f²

. (ii) Using

Riceg = Ric_gF +2(d−1)

f D²f − 1

f²(−2f4_gFf + (2d−1)|df|²_g

F)g_F, we get

Ric_gF +2(d−1)

f D²f = 1

f²(λ−2f4_gFf+ (2d−1)|df|²_g

F)g_F, namely,

2(d−1) f

√−1∂∂f

= 1

f²(λ−2f4_gFf+ (2d−1)|df|²_g

F)√

−1∂∂Φ_F +√

−1∂∂log detT or

2(d−1) f

∂²f

∂z^t∂z = 1

f²(λ−2f4_gFf

+ (2d−1)|df|²_g

F)∂²ΦF

∂z^t∂z +∂²log detT

∂z^t∂z . (10) Let

h(t) := log detT(t) = (d−1) logF⁰(t) + logF⁰⁰(t)−dt.

Note that

∂t

∂z^t

∂t

∂z = z^tz r⁴ ,

∂²t

∂z^t∂z = 1

r²I_d−z^tz r⁴ , h⁰(t) = (d−1)F⁰⁰(t)

F⁰(t) +F⁰⁰⁰(t)

F⁰⁰(t) −d= (d−1)ϕ(x)

x +ϕ⁰(x)−d, and

h⁰⁰(t) =

(d−1) ϕ(x)

x 0

+ϕ⁰⁰(x)

ϕ(x), so

∂²log detT

∂z^t∂z = h⁰

r²I_d+h⁰⁰−h⁰ r⁴ z^tz,

∂f

∂z^t = f⁰ϕz^t r²,

∂f

∂z = f⁰ϕz r²,

∂²f

∂z^t∂z = (f⁰ϕ)⁰ϕz^tz r⁴ +f⁰ϕ

I_d r² −z^tz

r⁴

= f⁰ϕ

r² I_d+((f⁰ϕ)⁰−f⁰)ϕ r⁴ z^tz

= aϕ

r²I_d+a(ϕ⁰−1)ϕ r⁴ z^tz,

|df|²_g

F = 2Tr

T⁻¹∂f

∂z^t

∂f

∂z

= 2(f⁰ϕ)² r⁴ Tr

r²

F⁰I_d+ ( 1 F⁰⁰ − 1

F⁰)z^tz

z^tz

= 2(f⁰)²ϕ= 2a²ϕ and

4_gFf = Tr

T⁻¹ ∂²f

∂z^t∂z

= Tr r²

xI_d+ (1 ϕ− 1

x)z^tz aϕ

r²I_d+a(ϕ⁰−1)ϕ r⁴ z^tz

= aϕ r²Tr

r² xI_d+

1 ϕ− 1

x

z^tz

+a(ϕ⁰−1)ϕ r⁴ Tr

r² xz^tz+

1 ϕ− 1

x

z^tzz^tz

= a

ϕ⁰+ (d−1)ϕ x

.

From (10), we get

A

r²I_d+ B

r⁴z^tz= 0, (11)

where

A = xλ−2f4_gFf + (2d−1)|df|²_g

F

f² +h⁰−2a(d−1)ϕ f

= x λ

f² +−2a ϕ⁰+ (d−1)^ϕ_x

f +2(2d−1)a²ϕ f²

!

+(d−1)ϕ

x +ϕ⁰−d−2a(d−1)ϕ f

= (1−2ax f )ϕ⁰+

d−1

x −4a(d−1)

f + 2(2d−1)a²x f²

ϕ +λx

f² −d

= x 1

x −2a f

ϕ⁰+

d−1

x² − 4a(d−1)

xf +2(2d−1)a² f²

ϕ +λ

f² −d x

, (12)

and

B = (ϕ−x)λ−2f4_gFf+ (2d−1)|df|²_g

F

f²

+h⁰⁰−h⁰−2a(d−1)(ϕ⁰−1)ϕ f . (13) Then,

A+B = ϕλ−2f4_gFf+ (2d−1)|df|²_g

F

f² +h⁰⁰−2a(d−1)ϕ⁰ϕ f

= ϕ

( λ

f² +−2a ϕ⁰+ (d−1)^ϕ_x

f +2(2d−1)a²ϕ f² +(d−1)

ϕ x

0

+ϕ⁰⁰−2a(d−1)ϕ⁰ f

= ϕ

ϕ⁰⁰+

d−1 x −2ad

f

ϕ⁰ +

2(2d−1)a²

f² −2a(d−1)

xf −d−1 x²

ϕ+ λ

f²

. Using

det

µI_d− A

r²I_d− B r⁴z^tz

=

µ− A r²

d−1

µ− A+B r²

,

we have that (11) is equivalent toA+B= 0 ford= 1 andA= 0,A+B = 0 ford >1.

Note that if 1

x −2a f

ϕ⁰+

d−1

x² − 4a(d−1)

xf +2(2d−1)a² f²

ϕ+ λ

f² −d

x = 0 (14) and

ϕ⁰⁰(x) +

d−1 x −2ad

f

ϕ⁰ +

2(2d−1)a²

f² −2a(d−1)

xf −d−1 x²

ϕ+ λ

f² = 0, (15) then (14)×(d−1) + (15) gives

ϕ⁰⁰(x) + 2

d−1

x −(2d−1)a f

ϕ⁰+

2d(2d−1)a²

f² − 2a(d−1)(2d−1) xf

+ (d−1)(d−2) x²

ϕ+λd

f² −d(d−1) x = 0.

In fact, this is equation (7), where k

eg =dλ.

Now we prove that if equation (14) holds, then equation (15) must be true. Let

Q₁: =

d−1

x² −^4a(d−1)_xf + ^2(2d−1)a_f2 ²

1

x −^2a_f = d−1

x − a

ax−b −(2d−1)a ax+b , Q₂: =

λ f² −^d_x

1

x −^2a_f =d+ 4abd−λ

2a(ax−b) − λ 2a(ax+b), and

P1 := d−1 x − 2ad

f , P2 := 2(2d−1)a²

f² −2a(d−1)

xf −d−1

x² , P3 := λ f², and

C:=ϕ⁰+Q1ϕ+Q2, D:=ϕ⁰⁰+P1ϕ⁰+P2ϕ+P3. Then,

∂C

∂x =ϕ⁰⁰+Q₁ϕ⁰+ ∂Q₁

∂x ϕ+∂Q₂

∂x , where

∂Q₁

∂x = (2d−1)a²

(ax+b)² + a²

(ax−b)² − d−1 x² ,

∂Q2

∂x = − 4abd−λ

2(ax−b)² + λ 2(ax+b)².

Thus,

∂C

∂x −D= (Q1−P1) ϕ⁰+

∂Q1

∂x −P2

Q1−P1

ϕ+

∂Q2

∂x −P3

Q1−P1

! , where

Q1−P1 =− 2ab a²x²−b²,

∂Q1

∂x −P2

Q1−P1

=Q1,

∂Q2

∂x −P3

Q1−P1

=Q2. This shows that

D= ∂C

∂x −(Q1−P1)C.

Therefore, if C= 0, then D= 0.

Finally, by (15) we have (8), and by (14) we get (9).

3. The geodesics of radial conformally K¨ahler metrics

In this section, we give the geodesics of radial conformally K¨ahler metrics.

To this end, we first give the following Lemma 3.1.

Lemma 3.1. LetD be the Chern connection for them-dimensional K¨ahler manifold (M, g), f be a positive real smooth function on M, and De be the Levi-Civita connection for a Riemannian metric eg= _f¹2g on M. For locally holomorphic coordinates z= (z₁, . . . , z_m) of M, letg_i¯j =g(_∂z^∂

i,_∂¯^∂_z

j) and D

_∂

∂z^t

∂

∂z^t

= Ω_C⊗ _∂

∂z^t

∂

∂z^t

, De

_∂

∂z^t

∂

∂z^t

=Ωe_C⊗ _∂

∂z^t

∂

∂z^t

, where

T = (g_ij), Ω_C=

(∂ T)T⁻¹ 0 0 (∂ T)T⁻¹

. Then

Ωe_C =

(∂ T)T⁻¹ 0 0 (∂ T)T⁻¹

− df

f I_2m− 1 f





∂f

∂z^t

∂f

∂z^t



 dz dz +1

f

0 T T 0

dz^t dz^t

∂f

∂z

∂f

∂z

0 T⁻¹ T⁻¹ 0

, where I2m is the identity matrix of order2m.

Proof. Let

z_i=x_i+√

−1y_i, e_i ∈ ∂

∂x1

, . . . , ∂

∂xm

, ∂

∂y1

, . . . , ∂

∂ym

, e^∗_i ∈ {dx₁, . . . ,dxm,dy1, . . . ,dym},

gij =g(ei, ej),egij = 1

f²gij, g_R= (gij),(g^ij) = (gij)⁻¹,(eg^ij) = (egij)⁻¹

and

Γ^k_ij = 1

2g^kl(eigjl+ejgil−elgij), Γe^k_ij = 1

2ge^kl(eigejl+ejegil−elegij), whereΓ^k_ij andΓe^k_ij are the Christoffel symbols of the Riemann metricsg and eg, respectively. Then

Γe^k_ij =Γ^k_ij −f_i

fδ_j^k−f_j

f δ_i^k+f_l fg^klg_ij, where

f_i =e_if, δ^j_i =

(1, i=j, 0, i6=j.

This implies that

De





∂

∂x^t

∂

∂y^t



=Ωe_R⊗





∂

∂x^t

∂

∂y^t



, where

Ωe_R = Ω_R− df

f I2m− 1 f





∂f

∂x^t

∂f

∂y^t



 dx dy

+1 fg_R



 dx^t dy^t



 _∂f

∂x

∂f

∂y

g_R⁻¹,

Ω_R= (Ω^j_i), Ω^j_i = Γ^j_ike^∗_k, dx= (dx1, . . . ,dxm), dx^t= (dx)^t and

∂f

∂x = ∂f

∂x₁, . . . , ∂f

∂x^m

, ∂f

∂x^t = ∂f

∂x t

. Since





∂

∂x^t

∂

∂y^t



=P





∂

∂z^t

∂

∂z^t



, dx dy

= dz dz P⁻¹,

where

P =





Im Im

√−1I_m −√

−1I_m



.

It follows that

Ωe_R= Ω_R−df

f I2m− 1 fP





∂f

∂z^t

∂f

∂z^t



 dz dz P⁻¹

+ 1

fg_R(P^t)⁻¹ dz^t

dz^t

∂f

∂z

∂f

∂z

P^tg⁻¹

R

and Ωe_C=P⁻¹Ωe_RP. By

g_R=

A B

−B A

, T = 1

2(A+√

−1B), and

P⁻¹g_R(P⁻¹)^t

= ¹₄

I_m −√

−1I_m I_m √

−1I_m

A B

−B A

I_m I_m

−√

−1I_m √

−1I_m

=

0 ¹₂(A+√

−1B)

1

2(A−√

−1B) 0

=

0 T T 0

, we get

Ωe_C = P⁻¹Ωe_RP

= P⁻¹Ω_RP−^df_f I_2m−¹_f





∂f

∂z^t

∂f

∂z^t



 dz dz

+_f¹P⁻¹g_R(P^t)⁻¹ dz^t

dz^t

∂f

∂z

∂f

∂z

P^tg⁻¹

R P

= Ω_C− ^df_f I2m−_f¹





∂f

∂z^t

∂f

∂z^t



 dz dz

+_f¹

0 T T 0

dz^t dz^t

∂f

∂z

∂f

∂z

0 T⁻¹ T⁻¹ 0

.

Lemma 3.2. Let gF be a K¨ahler metric on a rotationally invariant do- main Ω ⊂ C^d associated with the K¨ahler form ω_F = √

−1∂∂Φ_F, namely

g_F(X, Y) = ω_F(X, J Y), where Φ_F(z, z) = F(t), t = lnr², r² = kzk². Let f =aF⁰(t) +b >0, eg= _f¹2gF, and

γ(τ) =τ z0, kz₀k² = 1.

Then γ is a geodesic of (Ω,eg).

Proof. Let D be the Chern connection for the K¨ahler manifold (Ω, g), De be the Levi-Civita connection for the Riemannian manifold (Ω,eg).

Set D





∂

∂z^t

∂

∂z^t



= Ω_C⊗





∂

∂z^t

∂

∂z^t



, De





∂

∂z^t

∂

∂z^t



=Ωe_C⊗





∂

∂z^t

∂

∂z^t



, where

Ω_C=

(∂ T)T⁻¹ 0 0 (∂ T)T⁻¹

, ∂:= dz ∂

∂z^t, ∂:= dz ∂

∂z^t, and

Ωe_C =

(∂ T)T⁻¹ 0 0 (∂ T)T⁻¹

(16)

−df

f I_2d− 1 f

∂f

∂z^t

∂f

∂z^t

!

dz dz +1

f

0 T T 0

dz^t dz^t

∂f

∂z

∂f

∂z

0 T⁻¹ T⁻¹ 0

, for the definition of T, refer to Lemma 2.1.

For convenience, let Ωe₁:= df

f I_2d, Ωe₂ := 1 f





∂f

∂z^t

∂f

∂z^t



 dz dz and

Ωe₃ := 1 f

0 T T 0

dz^t dz^t

∂f

∂z

∂f

∂z

0 T⁻¹ T⁻¹ 0

. Then

Ωe_C= Ω_C−Ωe1−Ωe2+Ωe3. Using

γ⁰(τ) =z0

∂

∂z^t +z0

∂

∂z^t, we have

kγ⁰(τ)k²

eg =

z₀ ∂

∂z^t +z₀ ∂

∂z^t, z₀ ∂

∂z^t +z₀ ∂

∂z^t

ge

= 2

f²z₀T z₀^t z=τ z0

= 2

τ²f²zT z^t z=τ z0

= 2F⁰⁰(t) τ²f²(t).

Let

e(τ) : = γ⁰(τ) kγ⁰(τ)k

eg

= τ f(t) p2F⁰⁰(t)

z₀ ∂

∂z^t +z₀ ∂

∂z^t

= τ f(t)

p2F⁰⁰(t) z0 z0





∂

∂z^t

∂

∂z^t



. Then

De_γ⁰e z=τ z0

= τ f(t) p2F⁰⁰(t)

d

dτ log( τ f(t)

p2F⁰⁰(t)) z₀ z₀ + z0 z0

hγ⁰,Ωe_Ci z=τ z0





∂

∂z^t

∂

∂z^t



, where t = logkzk²

z=τ z0 = 2 logτ, and the symbol h·,·i indicates the dual pairing between differential forms and tangent vectors.

By t= 2 logτ, we get d

dτ log( τ f(t)

p2F⁰⁰(t)) = 1 τ

1 +2f⁰(t)

f(t) −F⁰⁰⁰(t) F⁰⁰(t)

. Since

T⁻¹

z=τ z0 =τ² 1

F⁰(t)I_d+ ( 1

F⁰⁰(t) − 1

F⁰(t))z0tz0

and

∂T =

F⁰⁰⁰(t)−F⁰⁰(t)

r⁴ ∂t−2(F⁰⁰(t)−F⁰(t)) r⁶ dzz^t

z^tz +F⁰⁰(t)−F⁰(t)

r⁴ z^tdz+

F⁰⁰(t)

r² ∂t−F⁰(t) r⁴ dzz^t

I_d

= F⁰⁰⁰(t)−3F⁰⁰(t) + 2F⁰(t)

r⁶ dzz^tz^tz+F⁰⁰(t)−F⁰(t) r⁴ z^tdz +F⁰⁰(t)−F⁰(t)

r⁴ dzz^tI_d, it follows that

hγ⁰, ∂Ti|_{z=τ z0} = F⁰⁰⁰(t)−3F⁰⁰(t) + 2F⁰(t)

τ⁶ z₀τ z₀^tτ z₀^tτ z₀ +F⁰⁰(t)−F⁰(t)

τ⁴ τ z₀^tz₀+F⁰⁰(t)−F⁰(t)

τ⁴ z₀τ z₀^tI_d

= F⁰⁰(t)−F⁰(t)

τ³ Id+F⁰⁰⁰(t)−2F⁰⁰(t) +F⁰(t) τ³ z0tz0

and

hγ⁰,(∂T)T⁻¹i|_{z=τ z0} =

F⁰⁰(t)−F⁰(t)

τ I_d+F⁰⁰⁰(t)−2F⁰⁰(t) +F⁰(t) τ z₀^tz₀

× 1

F⁰(t)I_d+ ( 1

F⁰⁰(t) − 1

F⁰(t))z0tz0

= 1 τ

F⁰⁰(t)−F⁰(t)

F⁰(t) I_d+F⁰(t)F⁰⁰⁰(t)−F⁰⁰(t)F⁰⁰(t) F⁰(t)F⁰⁰(t) z₀^tz₀

. Similarly,

hγ⁰,(∂ T)T⁻¹i|_{z=τ z0} = 1 τ

F⁰⁰(t)−F⁰(t)

F⁰(t) I_d+F⁰(t)F⁰⁰⁰(t)−F⁰⁰(t)F⁰⁰(t) F⁰(t)F⁰⁰(t) z₀^tz0

. So

z0 z0

hγ⁰,Ω_Ci

z=τ z0 = z0 z0

A 0 0 B

= F⁰⁰⁰(t)−F⁰⁰(t)

τ F⁰⁰(t) z₀ z₀ , where

A = 1

τ

F⁰⁰−F⁰

F⁰ (t)I_d+F⁰F⁰⁰⁰−F⁰⁰F⁰⁰

F⁰F⁰⁰ (t)z₀^tz₀

,

B = 1

τ

F⁰⁰−F⁰

F⁰ (t)I_d+F⁰F⁰⁰⁰−F⁰⁰F⁰⁰ F⁰F⁰⁰ (t)z₀^tz0

. By direct calculation, we obtain

γ⁰,df

f

z=τ z0

= 1

f(t) df dτ z=τ z0

= 2f⁰(t) τ f(t), z₀ z₀

hγ⁰,Ωe₁i z=τ z0

= 2f⁰(t)

τ f(t) z₀ z₀ , and

z0 z0

hγ⁰,Ωe₂i z=τ z0

= 1

f(t) z0 z0

∂f

∂z^t

∂f

∂z^t

!

dz dτ

dz dτ

z=τ z0

= f⁰(t)

τ f(t) z0 z0 z₀^t

z₀^t

z0 z0

= 2f⁰(t)

τ f(t) z₀ z₀

and

z₀ z₀

hγ⁰,Ωe3i z=τ z0

= 1

f(t) z₀ z₀

0 T T 0

z₀^t z₀^t

∂f

∂z

∂f

∂z

0 T⁻¹ T⁻¹ 0

z=τ z0

= 2f⁰(t)F⁰⁰(t)

τ³f(t) z₀ z₀

0 T⁻¹ T⁻¹ 0

z=τ z0

= 2f⁰(t)

τ f(t) z0 z0 . Thus,

z0 z0

hγ⁰,Ωe_Ci z=τ z0

= z0 z0

hγ⁰,Ω_C−Ωe1−Ωe2+Ωe3i z=τ z0

= 1

τ

F⁰⁰⁰(t)−F⁰⁰(t)

F⁰⁰(t) −2f⁰(t) f(t)

z0 z0

. Finally, we have

d

dτ log( τ f(t) p2F⁰⁰(t))

z=τ z0

z0 z0

+ z0 z0

hγ⁰,Ωe_Ci z=τ z0

= 0, namely,

De_γ⁰ γ⁰ kγ⁰k

eg

z=τ z0

= 0,

which implies that γ is a geodesic of (Ω,eg).

4. Proof of Theorem 1.1

4.1. Proof of part (i) and part (ii) of Theorem 1.1.

Proof of part (i) and part (ii) of Theorem 1.1 . Since F⁰(t) > 0 and F⁰⁰(t)>0 for t >−∞, using F(−∞) = 0 it follows that

t→−∞lim x(t) = lim

t→−∞F⁰(t) = 0.

Because the metricgF is smooth atz= 0, soϕ(0) = 0 andϕ⁰(0) = 1. By (9), we get



















ϕ⁰(x) +

d−1

x −_ax−1^a − ^(2d−1)a_ax+1 ϕ(x) +

d+ _2a(ax−1)^4ad−λ −_2a(ax+1)^λ

= 0, ϕ(0) = 0.

(17)

(i) Forλ= 0, letϕ(x) =x(ax+ 1)²ψ(d, u) withu=ax, then ψ(d,0) = 1.

According to (17), we get















(d−2)u²−2(d−2)u+d ψ(d, u)−d +u(1−u²)^dψ_du(d, u) = 0,

ψ(d,0) = 1,

(18)

which implies

ψ(d,−1) = d

4d−6, ψ(d,1) = d

2. (19)

Solving the equation (18), we have ψ(d, u) =

d−2

X

k=0

(k+ 1) Γ(d−1)Γ(d+ 1)

Γ(d+ 1 +k)Γ(d−1−k)u^k (20)

= d(1−u)(1 +u)^2d−3 u^d

Z u 0

v^d−1

(1−v)²(1 +v)^2d−2dv. (21) Thus,

ψ(d,−u) = d(1 +u)(1−u)^2d−3 u^d

Z u 0

v^d−1

(1 +v)²(1−v)^2d−2dv. (22) Hence,

ϕ(x) =x(1 +ax)²ψ(d, ax)>0, x∈(0,−1 a).

For a given x₀ ∈(0,−_a¹), let t=

Z x(t) x0

dv

ϕ(v), F(t) = Z x(t)

0

vdv ϕ(v). Then

x→0lim⁺t=−∞, lim

x→(−_a¹)⁻

t= +∞, which implies that theg_F is defined onC^d.

For anyz₀∈C^dwith kz₀k²= 1, from Lemma 3.2, ray C:z(τ) =τ z0, τ ∈[0,+∞) is a geodesic with respect to the metric eg. Using

z⁰(τ) = (z0

∂

∂z^t+z0

∂

∂z^t)|_{z=τ z0}

and (4), the square norm of the tangent vector of C atτ z_o with respect to the metriceg is

|z⁰|²

eg(τ) =

z₀ ∂

∂z^t+z₀ ∂

∂z^t, z₀ ∂

∂z^t +z₀ ∂

∂z^t

eg

z=τ z0

= 2

z₀ ∂

∂z^t, z₀ ∂

∂z^t

eg

_{z=τ z}

0

= 2

(1 +aF⁰(t))² ×z₀

F⁰(t)

r² I_d+F⁰⁰(t)−F⁰(t)

r⁴ τ z₀^tτ z_o

z₀^t

= 2

(1 +aF⁰(t))²

F⁰(t)

r² +F⁰⁰(t)−F⁰(t) r⁴ τ²

= 2F⁰⁰(t)

(1 +aF⁰(t))²τ² = 2e^−tF⁰⁰(t) (1 +aF⁰(t))², wherer² =kτ z₀k² =τ² =e^t. So the length ofC is

l =

Z +∞

0

|z⁰|

eg(τ)dτ =

√ 2 2

Z +∞

−∞

pF⁰⁰(t) 1 +aF⁰(t)dt

=

√2 2

Z −¹

a

0

1 (1 +ax)p

ϕ(x)dx= +∞.

This shows that the metriceg is complete onC^d.

(ii) For λ <0, let ϕ(x) =x+µ(λ)x²h(u) with u=axand µ(λ) = (4d−2)a−λ

d+ 1 .

Ifλ= (4d−2)a, thenϕ(x) =x is a solution of the equation (17).

Ifλ6= (4d−2)a, from (17), we get















u(1−u²)h⁰(u) +

(d−1)u²−2(d−1)u+d+ 1 h(u)

−(d+ 1) = 0, h(0) = 1.

(23)

Comparing (23) with (18), we obtain h(u) =ψ(d+ 1, u), so ϕ(x) = x+µ(0)x²h(u)− λ

d+ 1x²h(u)

= x(1 +ax)²ψ(d, ax)− λ

d+ 1x²ψ(d+ 1, ax)>0.

forx∈(0,−¹_a].

Let

t= Z x(t)

−¹_a

dv

ϕ(v), F(t) = Z x(t)

0

vdv ϕ(v).

Then

x→0lim⁺t=−∞, lim

x→(−¹

a)⁻

t= 0, which implies that thegF is defined onB^d.

For anyz₀∈C^dwith kz₀k²= 1, let

C:z(τ) =τ z₀, τ ∈[0,1). Then the length of the geodesic C:

l = Z ₁

0

|z⁰|

eg(τ)dτ =

√2 2

Z ₀

−∞

pF⁰⁰(t) 1 +aF⁰(t)dt

=

√ 2 2

Z −¹_a 0

1 (1 +ax)p

ϕ(x)dx= +∞.

This proves that the metric egis complete.

Finally, if a=−1 andλ= (4d−2)a=−(4d−2), then ϕ(x) =x, thus F(t) =e^t=kzk² andf(t) = 1 +aF⁰(t) = 1−e^t= 1− kzk².

The proof of (20) is given below.

Proof of (20). Forα >0, let







u(1−u²)^dψ_du(u) +

(α−2)u²−2(α−2)u+α ψ(u)−α= 0, ψ(0) = 1.

(24) and

ψ(u) =

+∞

X

k=0

c_ku^k. Then

(u−u³)dψ

du =c₁u+ 2c₂u²+

+∞

X

k=3

(kc_k−(k−2)ck−2)u^k, and

(α−2)u²−2(α−2)u+α ψ(u)

=αc₀+{αc₁−2(α−2)c₀}u+{(α−2)c₀−2(α−2)c₁+αc₂}u² +

+∞

X

k=3

{(α−2)ck−2−2(α−2)ck−1+αc_k}u^k,

and

+∞

X

k=3

{(α+k)c_k−2(α−2)ck−1+ (α−k)ck−2}u^k+α(c₀−1) +{(α+ 1)c1−2(α−2)c0}u

+{(α−2)c₀−2(α−2)c₁+ (α+ 2)c₂}u² = 0.

Thus,























α(c₀−1) = 0,

(α+ 1)c₁−2(α−2)c₀= 0,

(α+ 2)c₂−2(α−2)c₁+ (α−2)c₀ = 0,

(α+k)c_k−2(α−2)ck−1+ (α−k)ck−2 = 0, k≥3.

(25)

Solving (25), we have























c₀= 1, c1= 2^α−2_α+1, c₂= 3^{(α−2)(α−3)}_(α+1)(α+2),

c_k= (k+ 1)(α−2)···(α−k−1)

(α+1)···(α+k) , k≥3,

(26)

that is,

ψ(u) =

+∞

X

k=0

(k+ 1) Γ(α−1)Γ(α+ 1)

Γ(α+ 1 +k)Γ(α−1−k)u^k. (27) Remark 4.1. From (24), we have

ψ⁰(u) = (α−2)u²−2(α−2)u+α

u³−u ψ(u)− α

u³−u, ψ⁰(0) = 2(α−2)

α+ 1 ,

and

ψ⁰⁰(u) = ∂

∂u

(α−2)u²−2(α−2)u+α u³−u

ψ(u)

− ∂

∂u α

u³−u

+ (α−2)u²−2(α−2)u+α u³−u ψ⁰(u)

= (α−2)(α−3)u³−3(α−2)(α−3)u²+ (3α²−9α)u−α(α+ 1)

u²(u−1)(u+ 1)² ψ(u)

+(−α²+ 5α)u+α(α+ 1) u²(u−1)(u+ 1)² . Thus,

ψ⁰⁰(u)−2((α−3)u−α)

u(1 +u) ψ⁰(u) =

−(α−2)(α−3)u²−2α(α−2)u+α²−α

u²(1 +u)² ψ(u)− α−α² u²(1 +u)². Therefore, the equation (24) is equivalent to























u²(1 +u)^{2 d}_du^2ψ2(u)−2u(1 +u){(α−3)u−α}^dψ_du(u) +

(α−2)(α−3)u²−2α(α−2)u+α²−α ψ(u) +(α−α²) = 0,

ψ(0) = 1, ψ⁰(0) = ^2(α−2)_α+1 .

(28)

4.2. Proof of part (iii) of Theorem 1.1.

Proof of part (iii) of Theorem 1.1. Let the scalar curvature for eg be equal to 2β.

From the metric g_F is smooth at z= 0, we get ϕ(0) = 0 and ϕ⁰(0) = 1.

By (7), we have

ϕ(x) = (1 +ax)^2d−1 x^d−1

Z x 0

(x−u)χ(u)du, (29) where

χ(u) := u^d−2 d(d−1)(1 +au)²−βu

(1 +au)^2d+1 . (30)

From (29) it follows thatϕis a polynomial in x,ϕ(x)>0 for x∈(0,−_a¹),

x→0lim⁺ ϕ(x)

x = lim

x→0⁺

Rx

0 χ(u)du dx^d−1 = 1 and

lim

x→(−_a¹)⁻

ϕ(x) = −β

2d(2d−1)a² ≥0.