On a certain change of K¨ ahler metrics Tsuyoshi Yamazaki and Yoshiyuki Watanabe
1 . Introduction
It is well known that every K¨ahler metric ds
2= 2Σg
αβ¯dz
αd¯ z
βis locally expressible in the form
g
αβ¯= ∂
2ρ
∂z
α∂ z ¯
β,
with respect to local complex coordinates (z
α), α = 1, · · · , n, where ρ = ρ(z, z) is a real valued function, called (K¨ahler) potential. ¯
The purpose of this paper is to initiate a study on the K¨ahler metric ˜ g
αβ¯, changing a K¨ahler metric g
αβ¯as follows:
˜
g
αβ¯= f(ρ)g
αβ¯+ f
0(ρ)ρ
αρ
β¯, (1.1)
2000Mathematics Subject Classification. 53B35, 53C55.
Key words and phrases. K¨ahler metrics, a change of K¨ahler metrics, Ricci tensors, constant scalar curvature.
where ρ
α= ∂ρ
∂z
α, f = f (ρ) and f
0(ρ) = df
dρ satisfy f > 0, f + lf
0> 0, (1.2)
denoting l = g
αβ¯ρ
αρ
β¯= ρ
αρ
αand ρ
α= g
αβ¯ρ
β¯. In fact, it can be easily checked that ˜ g
αβ¯satisfy the K¨ahler condition
∂ g ˜
αβ¯∂¯ z
γ= ∂˜ g
α¯γ∂¯ z
β. (1.3)
In particular, if the g
αβ¯is a (locally) flat K¨ahler metric, that is, g
αβ¯=
∂
2t
∂z
α∂ z ¯
β= δ
αβ, where t = Σz
αz ¯
α, then we have
˜
g
αβ¯= ∂
2f(t)
∂z
α∂¯ z
β= ˙ f (t)δ
αβ+ ¨ f (t)¯ z
αz
β, (1.4)
denoting ˙ f(t) = df
dt and ¨ f(t) = d
2f
dt
2. ( Tachibana and Liu [2], Watanabe [3], [4] and Watanabe and Mori [5]). Now, let g
αβ¯be a Fubini metric (cf.
Tachibana [1]) and we study the K¨ahler metric ˜ g
αβ¯, changed by (1.1). In particular, we shall generalize some results in Watanabe [3]. The main theorem is the following
Theorem 1. Let D
nbe C
nor the ball B
n= {(z
α) ∈ C
n| Σz
αz ¯
α< r
2} around the origin of C
n, whose radius is given by r = 2
p
|k| (k 6= 0). Let (D
n, g
αβ¯) be a Fubini space. If the scalar curvature R ˜ of the K¨ahler metric
˜
g
αβ¯, changed by (1.1), satisfies the condition R ˜ = constant, (1.5)
then (D
n, g ˜
αβ¯) is of constant holomorphic curvature or flat.
Preliminary facts will be given in Section 2. In Section 3, we shall com-
pute the curvature tensors of the K¨ahler metric, changed by (1.1). In
Sections 4, 5, and 6, we shall study the K¨ahler metric ˜ g
αβ¯, changing a Fu-
bini metric in the form (1.1). In Section 5, we shall consider such a K¨ahler
manifold, satisfying the condition:
∇ ˜
k∇ ˜
lR ˜
ij− ∇ ˜
l∇ ˜
kR ˜
ij= 0, (1.6)
where ˜ R
ijis the Ricci tensor of ˜ g
αβ¯. The last section is devoted to prove the main theorem.
2.Preliminaries
We agree to adopt the summation convention and the following range of indices throughout the paper:
1 ≤ i, j, k, · · · ≤ 2n, 1 ≤ α, β, γ, · · · ≤ n.
Consider a complex n dimensional K¨ahler manifold with metric ds
2= Σg
jkdz
jdz
k,
(2.1)
where (z
α) are local complex coordinates and ¯ z
α= z
α¯(=conjugate of z
α).
g
jksatisfy the conditions:
g
αβ= g
α¯β¯= 0, (2.2)
and (1.3). Then (2.1) becomes
ds
2= 2Σg
αβ¯dz
αd¯ z
β. (2.3)
g
jksatisfy the corresponding equations to (2.2). The Christoffel symbols Γ
jkivanish except
Γ
βγα= g
α¯²∂g
β¯²∂z
γ, (2.4)
and their conjugates. As to the curvature tensor R
ijkl, only the components of the form R
αβγ¯δand their conjugates are different from zero, and it holds that
R
αβγδ¯= ∂Γ
βγα∂¯ z
δ,
(2.5)
from which the Ricci tensor R
ijis given by R
β¯γ= R
αβ¯γα= − ∂Γ
βαα∂ z ¯
γ, (2.6)
R
βγ= R
β¯¯γ= 0.
(2.7)
The scalar curvature R = g
jkR
jkis 2g
αβ¯R
αβ¯.
A K¨ahler manifold is called a space of constant holomorphic curvature k if its curvature tensor satisfies
R
αβγ¯ δ¯= k(g
αβ¯g
γ¯δ+ g
αδ¯g
γβ¯), denoting k = R
2n(n + 1) .
3.Curvature tensors
For a K¨ahler metric g
αβ¯= ∂
2ρ
∂z
α∂ z ¯
β, we denote the curvature tensor, the Ricci tensor and the scalar curvature by R
αβγ¯δ, R
αβ¯and R, respectively.
We are now going to compute the curvature tensor ˜ R
αβγ¯δ, the Ricci tensor R ˜
αβ¯and the scalar curvature ˜ R of the K¨ahler metric ˜ g
αβ¯, given by
˜
g
αβ¯= f (ρ)g
αβ¯+ f
0ρ
αρ
β¯, (3.1)
where the dash means differentiation with respect to ρ. As the metric is positive definite, the function should satisfy (1.2). Then ˜ g
αβ¯are given by
˜ g
αβ¯= 1
f
³
g
αβ¯− f
0f + lf
0ρ
αρ
β¯´. (3.2)
From (2.4), we have
Γ ˜
βγα= ˜ g
α¯λ∂˜ g
β¯λ∂z
γ= Γ
βγα+ f
0f (ρ
βδ
γα+ ρ
γδ
βα) + f f
00− 2f
02f (f + lf
0) ρ
αρ
βρ
γ+ f
0f + lf
0ρ
α∇
βρ
γ,
denoting ∇
γρ
β= ∂ρ
β∂z
γ− Γ
γβαρ
α. After some computations, we have
(3.3) ˜ R
αβγδ¯= ∂ Γ ˜
βγα∂ z ¯
δ= R
αβγδ¯+ f f
00− f
02f
2ρ
δ¯(ρ
βδ
γα+ ρ
γδ
βα) + f
0f (g
βδ¯δ
γα+ g
γ¯δδ
βα) + f (f + lf
0)(f f
00− 2f
02)
0− (f f
00− 2f
02)(3f f
0+ lf f
00+ lf
02)
f
2(f + lf
0)
2ρ
αρ
βρ
γρ
δ¯+ f f
00− 2f
02f (f + lf
0) ρ
α(ρ
βg
γδ¯+ ρ
γg
βδ¯) − f
0(f f
00− 2f
02)
f (f + lf
0)
2ρ
αρ
βρ
γρ
¯²∇
¯²ρ
¯δ+ f f
00− 2f
02f (f + lf
0) ρ
βρ
γg
α¯²∇
¯²ρ
δ¯+ f f
00− 2f
02(f + lf
0)
2ρ
αρ
δ¯∇
βρ
γ− f
02(f + lf
0)
2(ρ
¯²∇
¯²ρ
δ¯)ρ
α∇
βρ
γ+ f
0f + lf
0(∇
βρ
γ)g
α¯²∇
¯²ρ
δ¯− f
0f + lf
0ρ
αρ
λR
λβγδ¯. Next, from (2.6) we have
(3.4) ˜ R
β¯δ= − R ˜
αβα¯δ= R
βδ¯− (n + 1)f
0f g
β¯δ− f f
00− 2f
02f (f + lf
0) lg
βδ¯− (n + 1)(f f
00− f
02) f
2ρ
βρ
¯δ− f (f + lf
0)(f f
00− 2f
02)
0− (f f
00− 2f
02)(3f f
0+ lf f
00+ lf
02) f
2(f + lf
0)
2lρ
βρ
¯δ− f f
00− 2f
02f (f + lf
0) ρ
βρ
¯δ− f f
00− 2f
02(f + lf
0)
2ρ
βρ
¯²∇
¯²ρ
¯δ− f f
00− 2f
02(f + lf
0)
2ρ
αρ
δ¯∇
βρ
α+ f
02(f + lf
0)
2(ρ
¯²∇
¯²ρ
δ¯)ρ
α∇
βρ
α− f
0f + lf
0(∇
βρ
α)g
α¯²∇
¯²ρ
δ¯+ f
0f + lf
0ρ
αρ
λR
λβαδ¯. Finally, the scalar curvature ˜ R is given by
(3.5) ˜ R = 2˜ g
βδ¯R ˜
βδ¯= 1
f R − 2(n + 1)(f f
00− f
02)
f
3l
− 2f (f + lf
0)(f f
00− 2f
02)
0− 2(f f
00− 2f
02)(3f f
0+ lf f
00+ lf
02)
f
3(f + lf
0)
2l
2− 2(f f
00− 2f
02)
f
2(f + lf
0) l − 2n(n + 1)f
0f
2− 2n(f f
00− 2f
02)
f
2(f + lf
0) l − 2(f f
00− 2f
02)
f(f + lf
0)
2(ρ
¯δρ
¯²∇
¯²ρ
δ¯+ ρ
αρ
β∇
βρ
α)
− 2f
0f (f + lf
0) g
βδ¯g
α¯²(∇
βρ
α)∇
¯²ρ
δ¯+ 2f
02f (f + lf
0)
2g
β¯δ(ρ
¯²∇
¯²ρ
¯δ)ρ
α∇
βρ
α− f
0f(f + lf
0) lR + 2(n + 1)f
0(f f
00− f
02)
f
3(f + lf
0) l
2+ 2(n + 1)f
02f
2(f + lf
0) l
+ 2f f
0(f + lf
0)(f f
00− 2f
02)
0− 2f
0(f f
00− 2f
02)(3f f
0+ lf f
00+ lf
02)
f
3(f + lf
0)
3l
3+ 2f
0(f f
00− 2f
02)
f (f + lf
0)
3lρ
δ¯ρ
²¯∇
¯²ρ
δ¯+ 4f
0(f f
00− 2f
02) f
2(f + lf
0)
2l
2+ 2f
0(f f
00− 2f
02)
f (f + lf
0)
3lρ
αρ
β∇
βρ
α− 2f
03f(f + lf
0)
3(ρ
¯δρ
¯²∇
¯²ρ
δ¯)ρ
αρ
β∇
βρ
α+ 2f
02f (f + lf
0)
2g
α¯²(ρ
β∇
βρ
α)ρ
δ¯∇
¯²ρ
¯δ− 2f
02f (f + lf
0)
2ρ
λρ
αρ
βρ
δ¯R
λβαδ¯.
4 . A K¨ ahler manifold of constant holomorphic curvature A K¨ahler metric with non zero constant holomorphic curvature k defined in the complex number n-space C
nis called a Fubini metric. The potential function is given by
ρ = 2
k log
³1 + k 4 t
´, denoting t = Σz
αz ¯
α. Then putting Θ = 1 + k
4 t, we have g
αβ¯= 1
2Θ
2³
Θδ
αβ− k 4 z ¯
αz
β´. (4.1)
First from (4.1), we have
g
αβ¯= 1
2e
k2ρδ
αβ− k 2 ρ
αρ
β¯, and
ρ
α= ∂ρ
∂z
α= z ¯
α2Θ , (4.2)
from which we obtain
g
αβ¯= 2Θ
³δ
αβ+ k 4 z
αz ¯
β´, (4.3)
ρ
α= g
α¯²ρ
¯²= z
αΘ, (4.4)
e
k2ρ= l
0= 1 + k 2 l > 0, (4.5)
and
l = ρ
αρ
α= t 2 . Taking account of (2.4), we have
Γ
βγα= − k
2 (δ
γαρ
β+ δ
βαρ
γ).
(4.6)
From (4.1), (4.2) and (4.6), we have
∇
γρ
β= k
2 ρ
βρ
γ.
(4.7)
Taking account of (2.5), we have R
αβγδ¯= − k
2 (δ
αγg
β¯δ+ δ
αβg
γδ¯).
(4.8)
Moreover taking account of (2.6), we have R
β¯δ= (n + 1)k
2 g
βδ¯. (4.9)
Finally, taking account of (4.9), we have
R = n(n + 1)k.
In the sequel, we consider the K¨ahler metric ˜ g
αβ¯, deforming g
αβ¯in (1.1).
Then we compute the following
˜
g
βδ¯= f
2e
k2ρδ
αβ+
³f
0− k
2 f
´ρ
αρ
β¯, (4.10)
Γ ˜
βγα=
³f
0f − k
2
´
(ρ
γδ
βα+ ρ
βδ
γα) + 2(f f
00− 2f
02+
k2f f
0)
f(f + lf
0) e
kρρ
α¯ρ
βρ
γ, (4.11)
(4.12) ˜ R
αβγδ¯=
³f
0f − k
2
´
(g
β¯δδ
γα+ g
γδ¯δ
αβ) + f f
00− f
02f
2(δ
γαρ
βρ
δ¯+ δ
βαρ
γρ
δ¯)
+ F
f (f + lf
0) (ρ
αρ
γg
βδ¯+ ρ
αρ
βg
γδ¯) +
·
1
f
2(f + lf
0)
2 nF
0f (f +lf
0)−F (3f f
0+lf
02+lf f
00)
o+
k2
F (f + lf
0)
2¸
ρ
αρ
βρ
γρ
¯δ, where we denote F by
F = f f
00− 2f
02+ k 2 f f
0. (4.13)
Moreover, we obtain
R ˜
βδ¯=
½
(n + 1)
³k 2 − f
0f
´
− F l f(f + lf
0)
¾
g
β¯δ−
·
(n + 1)(f f
00− f
02)
f
2+ F
f(f + lf
0) + l f
2(f + lf
02)
n
F
0f (f + lf
0)
−F (3f f
0+ lf
02+ lf f
00)
o+
k2
lF (f + lf
0)
2¸
ρ
βρ
δ¯, and
R ˜
βδ¯= µg
βδ¯+ µ
0ρ
βρ
¯δ, (4.14)
where µ is a function given by µ = (n + 1)
³k
2 − f
0f
´
− F l f (f + lf
0) . (4.15)
Finally, the scalar curvature ˜ R is given by R ˜ = 2nµ
f + 2l(f µ
0− f
0µ) f (f + lf
0) . (4.16)
5.A K¨ ahler metric satisfying a certain condition
Let D
nbe C
nor the ball B
n= {(z
α) ∈ C
n| Σz
αz ¯
α< r
2} around the origin O of C
n, whose radius is given by r = 2
p
|k| (k 6= 0). Let (D
n, g
αβ¯) be a Fubini space. In this section, let us consider the K¨ahler metric ˜ g, satifying the condition (1.6). Then by the Ricci’s formula, we have
R ˜
hjR ˜
hikl+ ˜ R
ihR ˜
hjkl= 0, (5.1)
from which
R ˜
αjR ˜
αikl+ ˜ R
αj¯R ˜
α¯ikl+ ˜ R
iαR ˜
αjkl+ ˜ R
iα¯R ˜
α¯jkl= 0.
Thus (5.1) is equivalent to
R ˜
αλ¯R ˜
αβγ¯δ+ ˜ R
βα¯R ˜
α¯λγ¯ δ¯= 0 (conj.), (5.2)
by virtue of (4.12) and (4.14).
Now substituting (4.8) and (4.10) into the left hand side of (5.2), we can
see that it reduces to the following.
(5.2)
0½
µ
0³f
0f − k 2
´
+ lµ
0F
f (f + lf
0) + µF
f (f + lf
0) − µ(f f
00− f
02) f
2¾
(ρ
¯λρ
γg
β¯δ−ρ
βρ
δ¯g
γ¯λ) = 0.
Now we assume that n ≥ 2. Then we have
½
µ
0³f
0f − k 2
´
+ lµ
0F
f (f + lf
0) + µF
f (f + lf
0) − µ(f f
00− f
02) f
2¾
= 0, (5.3)
taking account of f (ρ) ∈ C
∞(R). (5.3) gives
µ
0f
³f
0− k
2 f
´+ lµ
0f F
f + lf
0− lµf
0F
f + lf
0− µf
0³f
0− k
2 f
´= 0.
Thus we obtain
½
f
0− k
2 f + lF f + lf
0¾
(µ
0f − µf
0) = 0.
(5.4)
In the followings, we consider two cases, that is, Case where f
0− k 2 f + lF
f + lf
0= 0 and Case where µ
0f − µf
0= 0.
Case : Assume that f
0− k
2 f + lF
f + lf
0= 0 in an open subdomain ∆
n1of D
n. Taking account of (4.13),
we have
lf
00f + f
0f − lf
02f
2− k
2 = 0.
(5.5)
Putting g = f
0f (= (logf )
0) in (5.5), we have lg
0+ g − k
2 = 0.
(5.6)
Putting h = lg in (5.6), we have h
0− k
2 h − k 2 = 0.
(5.7)
The general solution of (5.7) is given by
h = ce
k2ρ− 1,
where c is an integral constant. Therefore the general solution of (5.6) is given by
lg = ce
k2ρ− 1,
where c is an integral constant. Therefore the general solution of (5.5) is given by
f = ae
bk2ρµ
1 − e
−k2ρk
¶b−1
, (5.8)
where a, b are integral constants.
If b 6= 1, then f and f + lf
0satisfy
f (0) = 0, (f + lf
0)(0) = 0,
because of ρ(0) = 0. Thus in this case the solution does not satisfy the condition (1.2) where ∆
n13 O.
Next, suppose that b = 1. Then from (1.2), we see that a > 0 where
∆
n13 O, that is,
(5.8)
0f = ae
k2ρ(a > 0), where ∆
n13 O.
Case : Suppose that
µ
0f − µf
0= 0 (5.9)
holds in a subdomain ∆
n2of D
n. First by (4.10) and (4.14) we see that the potential function satisfying (5.9) gives an Einstein metric. By (4.15), we have
(5.10) F
f (f + lf
0)
½³
n + 2 + k
2 l
´f
2+ nlf f
0¾
+ l
(f + lf
0)
2½
F
0f (f + lf
0) − F
³3f f
0+ lf
02+ lf f
00+ k
2 lf f
0´¾= 0.
Setting
σ = F
f (f + lf
0) , (5.11)
we have
lf σ
0+
½³n + 2 + k
2 l
´f + nlf
0¾
σ = 0.
(5.12)
If ∆
n2contains the origin O, putting l(0) = 0 in (5.12) we have σ(0) = 0,
(5.13)
because of f(0) > 0.
If l > 0, then multiplying (5.12) by l
n+1f
n−1, we have
l
n+2f
nσ
0+
³n + 2 + k
2 l
´l
n+1f
nσ + nl
n+2f
n−1f
0σ = 0, which implies
(l
n+2f
nσ)
0− (n + 1)k
2 (l
n+2f
nσ) = 0.
Thus the general solution is given by
l
n+2f
nσ = ce
(n+1)k2 ρ, (5.14)
where c is an integral constant. Since the function f satisfies (5.12), by the argument of continuity of the left hand side of (5.14), we can conclude that c = 0 if ∆
n2contains the origin O. Thus we have
σ(ρ) = 0, (5.15)
together with (5.14) if ∆
n23 O. Since
µ
f
0−
k2f f
2¶0
= F
f
3, we have f
0− k
2 f − Cf
2= 0 (C = constant) (5.16)
together with (5.11). (5.16) has the general solution f
2= 1
ce
−k2ρ+ d ,
(5.17)
where constants c(> 0) and d satisfy the condition c + d > 0, that is, when O ∈ ∆
n2,
f
2= 1
ce
−k2ρ+ d (c > 0, c + d > 0).
(5.18)
Note that in the case of (5.18) the corresponding K¨ahler manifold (∆
n2, g) ˜ is of constant holomorphic curvature dk
2 .
Finally, the function f given by (5.8)
0does not satisfy (5.17) with d 6=
0 and can be smoothly connected only the solution of (5.18) for d = 0.
Conversely the solution of (5.18) for d 6= 0 or a 6= 1
c does not satisfy (5.8)
0. Thus we obtain the following
Theorem 2. Let D
nbe C
nor the ball B
n= {(z
α) ∈ C
n| Σz
αz ¯
α< r
2} around the origin of C
n, whose radius is given by r = 2
p
|k| (k 6= 0). Let (D
n, g
αβ¯) be a Fubini space. If the K¨ahler metric ˜ g
αβ¯, changed by (1.1) satisfies the condition (1.6), then it is of constant holomorphic curvature or flat.
6.Proof of the main theorem
By assumption ˜ R is constant in (4.16). Multiplying (4.16) by f (f +lf
0), we have
R ˜
2 f (f + lf
0) = lµ
0f + nf µ + (n − 1)lf
0µ.
(6.1)
Putting l = 0 in (6.1), we have R ˜
2 f (0) − nµ(0) = 0, (6.2)
because of f (0) > 0. For l > 0, we multiply (6.1) by l
n−1f
n−2. Then we have
R ˜
2 (f
nl
n−1+ l
nf
n−1f
0) = l
nf
n−1µ
0+ nl
n−1f
n−1µ + (n − 1)l
nf
n−2f
0µ.
By
³
R ˜
2n l
nf
n´0= R ˜
2 (f
nl
n−1+ l
nf
n−1f
0) + nk 2
³
R ˜ 2n l
nf
n´and
(l
nf
n−1µ)
0= l
nf
n−1µ
0+ nl
n−1f
n−1µ + (n − 1)l
nf
n−2f
0µ + nk
2 (l
nf
n−1µ), we obtain
³
R ˜
2n l
nf
n− l
nf
n−1µ
´0= nk 2
³
R ˜
2n l
nf
n− l
nf
n−1µ
´, from which
R ˜
2n l
nf
n= l
nf
n−1µ + Ce
nk2 ρ, where C is an integral constant. Then we have
l
nf
n−1³R ˜
2n f − µ
´= Ce
nk2 ρ. (6.3)
But we can see that C = 0, taking limit of the left hand side of (6.3) of l as l tends to 0. Therefore we have
R ˜
2n f
2(f + lf
0) + (n + 1)
³f
0f − k
2
´
f (f + lf
0) (6.4)
+l
³f f
00− 2f
02+ k
2 f f
0´= 0.
Putting l = 0 in (6.4), we have f
0(0) − k
2 f (0) + R ˜
2n(n + 1) f (0)
2= 0.
(6.5)
Multiplying (6.4) by f
n−2l
n, we have R ˜
2n (f
n+1l
n+ f
nl
n+1f
0) + (n + 1)f
n−1l
nf
0+ l
n+1f
n−1f
00+ (n − 1)f
n−2l
n+1f
02− (n + 1)k
2 f
nl
n− nk
2 f
n−1l
n+1f
0= 0,
from which
½
l
n+1f
n−1³f
0− k
2 f + R ˜
2n(n + 1) f
2´¾0