Maxwell-Einstein metrics on Certain C P 1 Bundles


 In this paper, we prove that for certain fiber bundle there is a Maxwell-Einstein metric conformally related to any given Kähler class.


Introduction
In every Kähler class of a compact almost homogeneous manifold with two ends we found an unique Calabi extremal metric in [Gu1,2], Moreover, we found an unique extremal metric in a given Kähler class on certain CP 1 bundle if the function Φ there is positive. We realized in [Gu5] that this is equivalent to the geodesic stability of the Kähler class.
In this paper, we shall prove that for these kind of manifolds, there is another natural Kähler metric in the given Kähler class.
Definition For any given Kähler class, there is a Maxwell-Einstein metric conformally related to the Kähler class if h = u −2 g is an Hermitian metric with a constant scalar curvature such that u is the Hamiltonian function of a holomorphic vector field related to a Kähler metric g in the given Kähler class.
THEOREM For any Kähler class on a compact almost homogeneous manifold with two ends, there is at least one Maxwell-Einstein metrics in the given Kähler class.
We notice that when the complex dimension of the manifold M is 2, some extremal metrics, e. g., on CP 2 blow up a point, are actually the same as the Maxwell-Einstein metrics. Moreover, in that case, the corresponding Hermitian metrics are actually Hermitian-Einstein as Riemannian manifolds.
Let p : L → M be a holomorphic line bundle over a compact complex Kähler manifold M and h a hermitian metric of L. Denote by L 0 the open subset L − {0-section} of L and let s ∈ C ∞ (L 0 ) R be defined by s(l) = log |l| h (l ∈ L 0 ), where | | h is the norm defined by h. Now we consider a function τ = τ (s) ∈ C ∞ (L 0 ) R which is only depending on s and is monotoneincreasing with respect to s.
LetJ be the complex structure of L and J be the complex structure of M . Now we consider a Riemannian metric on L 0 of the form where g(l) = p * g τ (s(l)) (m) with m = p(l) ∈ M , g τ is a one parameter family of Riemannian metrics on M . This form of the metrics is general by using the function τ as the length of the geodesics perpendicular to the generic orbits. Define a function u on L 0 depending only on τ by u(τ ) 2 =g(H, H), where H be the real vector field on L 0 corresponding to the R * action on L 0 .
Lemma 1.(Cf [KS1,2], [Gu2 p.2257]) Suppose that the range of τ contains 0. Theng is Kähler if and only if g 0 is Kähler and g t = g 0 − U B, where B is the curvature of L with respect to h, U = τ 0 u(τ )dτ Throughout this paper, we assume that (1)L is a compactification of L 0 andg is the restriction of a Kähler metric ofL to L 0 .
(2) the range of τ contains 0 and (3) the eigenvalues of B with respect to g τ are constants on M .
(4) the traces of the Ricci curvature r of g on each eigenvector space of B are constant.
The condition (4) here is much more general than that in [Gu1,2] in which we have: (4)' the eigenvalues of r are constants. Our results cover some results which appeared in recent years. For example, if g has a constant scalar curvature and B has only one eigenvalue.
By abusing the language, we call the constants in (4) the trace eigenvalues.
Let (z 1 , ..., z n ) be a system of holomorphic local coordinates on M . n = dim C M . Using a trivialization of L 0 , we take a system of holomorphic local coordinates (z 0 , ..., z n ) on L 0 such that ∂/∂z 0 = H − √ −1JH. Here we notice that z 0 is correspond to w 1 in [Gu4 p.552]. s can be regarded as Rez 0 near the considered point. So s is the x 1 in [Gu4 p.552].
As in [Gu2] we let ϕ = u 2 as a function of U and we let F be the Kähler potential as in [Gu4 p.552], then by comparing [Gu2 Lemma 2] (or the Lemma 4 below) with [Gu4 p.552] we immediately have: [Gu4 p.552], i.e., Lemma 3. U is the Legendre transformation of s.
Here we use the Legendre transformation in [Gu4] instead of the moment map in [Gu2] since we need the new insight in the last section.
The Ricci curvature at this point is In particular, we have the scalar curvaturẽ where ϕ = u 2 as a function of U and ∆(U ) = Q i,j r 0 ij g ij τ (U ) . We also have ϕ (min U ) = 2, ϕ (max U ) = −2.
Lemma 5.(Cf. [FMS], [Mb], [Gu2 Lemma 3]) We can also regard U as a moment map corresponding to (g,JH) and g τ just be the symplectic reduction ofg at U (τ ).g is extremal if and only ifR = aU + b for some a, b ∈ R.
Let M 0 = U −1 (min U ) and M ∞ = U −1 (max U ), they are complex submanifolds, since they are components of the fixed point set of Lemma 6.(Cf. [Gu2 Lemma 4]) Suppose that there is another Kähler metricsg ∨ onL in the same Kähler class which is of form (1) on L 0 . Let be the corresponding metric and the corresponding functions of s. Then there is a unique corresponding τ ∨ such that where Proof: Letg −g ∨ =∂∂φ, theñ for 1 ≤ i, j ≤ n, so at min U (or max U )g ij =g ∨ ij , therefore there is a τ 0 such that g ∨ τ ∨ (τ 0 ) = g 0 . By choosing τ ∨ such that τ ∨ (τ 0 ) = 0, one sees that min U ∨ = min U, max U ∨ = max U as desired.
The last statement follows from the fact that the scalar curvature R is finite on both M 0 and M ∞ .
Q. E. D. We need normalization in this paper. By rescaling we can choose U ∨ = a 2 1 U + a 2 for any a 1 > 0 and a 2 ∈ R, allowing us to assume that max U − min U = 2 and min U = −1, then max U = 1.

Existence of Maxwell-Einstein Metrics
We recall our definition of the Maxwell-Einstein metrics: For any given Kähler class, there is a Maxwell-Einstein metric conformally related to the Kähler class if h = u −2 g is an Hermitian metric with a constant scalar curvature such that u is the Hamiltonian function of a holomorphic vector field related to a Kähler metric g in the given Kähler class. From Lemma 5, it can be seen that ifg is a Maxwell-Einstein metric, then u = aU + b for some a, b ∈ R.
By Lemma 4 we have that We then have: ∆u = a 2Q (ϕQ) .
We have also: We denote the right side by Φ(U ). We also have: Now, we have aU + b > 0. In particular, b > 0. We might assume that b = 1, then −1 < a < 1.
(12) and (13) imply that Lemma 7. When a is near 1, the right side of (14) is bigger than the left side. When a is near −1, the right side is smaller than the left side.
Proof: When a → 1, the major part of the left side is if Q(−1) = 0. And in this case the major part of the right side is We need more work for the case in which Q(−1) = 0. On the other hand, when a → −1, the major part of the left side is if Q(1) = 0. And in this case the major part of the right side is Similarly, we need more work for the case in which Q(1) = 0. This implies that there is actually at least one a ∈ (−1, 1) such that this identity holds. Therefore, there is also a number c such that both (12) and (13) hold.
which is equivalent to 1!(k−1−2)! (k−1)!(1−a) k−1−1 when a turns to 1. Therefore, by our induction formula we can prove that L k,l is equivalent to when a turns to 1.
The major part of the difference of the two sides of equation (14) comes from By the formula of ∆ near −1 in (7) of Lemma 6, we only need to check that has a negative major part. By (15) it is proportional to: It is determined by the sign of and that is the sign of D 0 − n, which is negative if n > D 0 .
This takes care of the cases with D 0 = n. When D 0 = n, then Q = (1 + x) n−1 and ∆ is a proportion of (1 + x) n−2 . We only need to take care of the major part of L 2n+1,n−1 [L 2n−2,n−2 + A] − L 2n,n−1 [L 2n−1,n−2 + B], here, A and B are the constants from the Q(1) terms.
As above, we know that the power of 1 − a which comes from the major part of L k,l 's cancels out. Therefore, we need also take care of the second major part of those L k,l .
From the induction formula, we see that Each L k−l+s,0 term contributes an (1 − a) s term other than the major term. Therefore, the second major term only comes from the constant term in the first term L k−l,0 . Therefore, the second major term in L k,l is the second term of Now, we see that the second major part of the difference of the two sides of (14) actually come from the major part of L 2n+1,n−1 , and the sign only depends on the sign of the constant term in But (1 + a) −2n+2 Q(1) = (1 + a) −2n+2 2 n−1 . Therefore, the sign of the major part is the sign of −n2 −n+1 + 2 −n+1 , which is the sign of −n + 1 < 0 since n > 1, otherwise, the C * bundle is trivial and M is homogeneous. The same argument also works for a → −1 by the symmetric argument. Q. E. D. We then have: Lemma 8. There is a solution for equation (14).
Q. E. D. This, in particular, concludes our THEOREM.