ESI
The Erwin Schrödinger International Boltzmanngasse 9
Institute for Mathematical Physics A-1090 Wien, Austria

On Hermitian Surfaces
with J-Invariant Ricci Tensor

Oleg Mu<=skarov

Vienna, Preprint ESI 795 (1999) November 26, 1999

Supported by Federal Ministry of Science and Transport, Austria
Available via anonymous ftp or gopher from FTP.ESI.AC.AT
or via WWW, URL: http://www.esi.ac.at

On Hermitian Surfaces with J-Invariant Ricci
Tensor

Oleg Mu<=skarov

Abstract
We prove that a compact Hermitian surface with J-invariant Ricci tensor is Kähler provided that the difference of its scalar and conformal scalar curvature is constant. In particular, there are no locally homogeneous examples of such surfaces with odd first Betti number.

1 Introduction

The Riemannian version of the Goldberg{Sachs Theorem in General Relativity [8, 10] says that the self-dual Weyl tensor W+ of an Einstein Hermitian metric is degenerate, i. e., at least two of its three eigenvalues coincide. This, together with results of Derdzinski [4] and Boyer [3], implies that in the compact case any such metric is conformally Kähler. A further analysis of Le Brun [7] showed that in fact only a few compact complex surfaces (M; J) may admit Einstein metrics which are Hermitian (i. e., compatible with the complex structure J) but non-

Kähler. The only known example is the Hirzebruch surface F1 ' CP 2 # CP 2

with the Page metric [9] and the other possible candidates are CP 2 # 2CP 2

and CP 2 # 3CP2.

Recently Apostolov and Gauduchon [1] proved that some weak versions of the Einstein condition still imply that W+ is degenerate; in particular, the latter holds when the Ricci tensor of a Hermitian metric is J-invariant and this allowed them to describe all compact complex surfaces with even first Betti number, possibly admitting such non-Kähler metrics (cf. [1, Theorem 2]). It is still unknown, however, whether there are compact Hermitian surfaces with J-invariant Ricci tensor and odd first Betti number. This note aims to show that such surfaces (if any) could not be locally homogeneous.
More precisely, we prove the following:

Supported in part by NSF grant INT-9903302 and by the Erwin Schrödinger International Institute for Mathematical Physics.
1991 Mathematics Subject Classification. Primary 53C55.

Key words and phrases. Hermitian surfaces, J-invariant Ricci tensor.

Theorem 1. Let (M; J; g) be a compact Hermitian surface with scalar curvature s and conformal scalar curvature k. If the Ricci tensor of g is J-invariant and s ? k is constant, then (M; J; g) is a Kähler surface.

Recall that the conformal scalar curvature of a Hermitian surface (M; J; g) is the scalar curvature with respect to g of the canonical Weyl structure associated with the Hermitian structure (g; J) (cf. [5, 11]). Note that in general the function s?k is not constant and it measures the defect for a compact Hermitian surface to be Kähler.

The main point of the proof of Theorem 1 is to show that when s ? k is constant, there is a Kähler metric on (M; J). This, together with [1, Theorem 2], forces the metric g to be Kähler.

2 Preliminaries

Let (M; J; g) be a Hermitian surface, i. e., a Hermitian manifold of real dimension four with complex structure J and Riemannian metric g such that g(JX; JY ) = g(X; Y ) for any vector fields X, Y . Denote by ? the Kähler form of (M; J; g) defined by ?(X; Y ) = g(JX; Y ). We shall always consider M with the orientation determined by J so that the volume form of M is dV = 12? ^ ?. It is well known (cf. [12]) that d? = ! ^?, where ! = ?ffi?ffiJ is the Lee form of (M; J; g). Hence (M; J; g) is a Kähler surface iff ! = 0; it is locally conformally Kähler iff d! = 0 and conformally Kähler iff ! = df for some smooth function f on M (in this case e?fg is a Kähler metric).

Let r, R, æ and s be the Levi{Civita connection, the curvature tensor, the Ricci tensor, and the scalar curvature of the Riemannian metric g, respectively. Here

R(X;Y )Z = [rX ; rY ]Z ? r[X;Y ]Z
æ(X; Y ) = Trace(Z ! R(Z; X)Y ); s = Trace æ:

Recall that the ?-Ricci tensor æ? and the ?-scalar curvature s? of (M; J; g) are defined by

æ?(X; Y ) = TracefZ ! R(X; JZ)JY g; s? = Trace æ?:

Using the first Bianchi identity, we get

æ?(X; Y ) = TracefZ ! 1
2R(X; JY )JZg: (1)

For any Hermitian surface the Ricci and the ?-Ricci tensor are related by

æ(X; Y ) ? æ?(X; Y ) = 1
2(L(JX; JY ) ? L(X; Y )) + s ? s?
4 g(X; Y ); (2)

where L = r! + 12! ? !: This follows by the well-known expression of the commutator [RX;Y ; J ] by means of L (cf. [12]). In particular (cf. [11]),

æ(X; Y ) + æ(JX; JY ) ? æ?(X; Y ) ? æ?(JX; JY ) = s ? s?
2 g(X; Y ): (3)

We also have (cf. [12])
s ? s? = 2ffi! + jj!jj2: (4)

The Riemannian metric g induces a metric on the bundle ?2M of 2-vectors on M by g(X1^X2; X3^X4) = det(g(Xi; Xj)). The curvature operator R is the symmetric endomorphism of ?2M defined by g(R(X; Y ); Z^W ) = ?R(X; Y; Z; W ). The Hodge star operator defines an endomorphism ? of ?2M with ?2 = Id. Hence ?2M = ?+M ? ??M , where ?+M (resp. ??M ) is the subbundle of ?2M corresponding to the eigenvalue +1 (resp. ?1) of ?. Then we have the following SO(4)-splitting of the curvature operator:

R = s
12Id + B + W+ + W?

where B is the Kulkarni{Nomizu extension of the traceless Ricci tensor of g and W? = 12 (W ? ?W) are respectively the self-dual and anti-self-dual parts of the Weyl tensor. Next we shall consider W+ as a symmetric, traceless endomorphism of ?+M and we shall say that W+ is degenerate if at each point at least two of its three eigenvalues coincide. As is well known [1,3] this is equivalent to the vanishing of the self-dual part of d!, i. e., g being locally conformally Kähler metric in the compact case.

The conformal scalar curvature k of a Hermitian surface (M; J; g) is defined by k = 3g(W+(?); ?). It is well known that k is the scalar curvature with respect to g of the canonical Weyl structure associated with the Hermitian structure (g; J) (cf. [5,11]). The conformal scalar curvature is conformally covariant of weight ?2 and it is related to the scalar and ?-scalar curvature of (M; J; g) by k = 3s??s
2 , since s? = 2g(R(?); ?) (cf. (1)). Hence the equality (4) can be rewritten as

s ? k = 3
2(2ffi! + jj!jj2): (5)

Integrating (5) shows that a compact Hermitian surface is Kähler if and only if s ? k = 0.

3 Proof of Theorem 1

We divide the proof of Theorem 1 into two lemmas.

Lemma 1. The Ricci tensor of a Hermitian surface (M; J; g) is J-invariant if and only if

æ(X; Y ) ? æ?(X; Y ) = s ? k
6 g(X; Y ) (6) for any vector fields X and Y .

Proof. Suppose that the Ricci tensor of g is J-invariant. Then by [1,Theorem 2] it follows that the self-dual Weyl tensor W+ is degenerate. Hence the 2-form d! is anti-self-dual (cf. [1]) and the identity (2) shows that the ?-Ricci

tensor æ? is symmetric, i. e., it is J-invariant (cf. (1)). Now (3) implies (6) since æ and æ? are J-invariant and s ? k = 32 (s ? s?):

Conversely, if (6) holds, then the ?-Ricci tensor æ? is symmetric, i. e., it is J-invariant. Hence (6) shows that the Ricci tensor æ is J-invariant too.

Denote by æA and æ?A the 1-forms defined by æA(X) = æ(A; X) and æ?A(X) = æ?(A; X), where A is the dual vector field of the Lee form !. Set

?(X; Y ) = 2!((rXJ)(Y )) (7)

for any vector fields X and Y . Since rX? = 12(X ^ JA + JX ^ A), cf. [6], it follows that ? is a (1,1)-form with respect to J . In the next lemma we compute the differential d? of ? by means of æA, æ?A and the Kähler form ? of (M; J; g). Lemma 2. For any Hermitian surface (M; J; g) we have
d? = 2(æ?A ? æA) ^ ?: (8)

Proof. It is easy to check that the (1,1)-form ? is given by

? = jj!jj2? ? ! ^ ! ffi J (9)

This shows that if !x = 0 at a point x 2 M , then (d?)x = 0 and (8) is fulfilled. Assume now that !x 6= 0, i. e., A 6= 0 in a neighbourhood of x. Let B be a unit vector field around x which is perpendicular to A and JA. To prove the lemma, we have to check (8) at the following triples of vector fields: (A; JA; C) and (D; B; JB), where C 2 fB; JBg and D 2 fA; JAg. To do this, we first note that
(rAJ)(X) = (rJAJ)(X) = 0 (10)

for any vector field X. Hence ?(A; X) = ?(JA; X) = 0 and from (9) it follows that
d?(A; JA; C) = ??([A; JA]; C) = jj!jj2g([A; JA]; JC): (11)

On the other hand, g([A; JA]; JC) = L(A; C)?L(JA; JC), since (rAJ)(A) = 0, and from (11) and (2) we conclude that

d?(A; JA; C) = 2?(æ?A ? æA) ^ ??(A; JA; C)

Now consider a triple of the form (D; B; JB), where D 2 fA; JAg. We shall prove that the identity (8) holds when D = A; the case D = JA is similar. To do this we first note that ?(X; B) = ?g(X; JB)jj!jj2 for any vector field X. Hence

d?(A; B; JB) = A(?(B; JB)) ? ?([A; B]; JB) + ?([A; JA]; B)
= A(g(A;A)) ? g([A; B]; B)jj!jj2 ? g([A; JA]; JB)jj!jj2
= 2g(rAA;A) + g(rBA; B)jj!jj2 + g(rJBA; JB)jj!jj2
= ?jj!jj2ffi! + L(A; A) ? L(JA; JA):

>From (2) it follows that

d?(A; B; JB) = 2?(æ?A ? æA) ^ ??(A; B; JB)

and the lemma is proved.

Now we are ready to prove Theorem 1.
Let (M; J; g) be a compact Hermitian surface with J-invariant Ricci tensor such that s ? k is constant. If s ? k <= 0, then integrating the identity (5) we obtain Z

M jj!jj2dV j <= 0;

which shows that ! = 0. Hence (M; J; g) is a Kähler surface.
Assume now that s?k > 0. Then from (9) it follows that ? = s?k
3 ?+? is a

positive (1,1)-form. By Lemma 1 we have æ?A ? æA = k?s
6 ! and using Lemma 2, we get d? = 0. Hence ? is a closed positive (1,1)-form, i. e., ? determines a Kähler metric on (M; J). In particular the first Betti number of M is even and by [1,Theorem 2] we conclude that the metric g is conformally Kähler. Let g = f?2g0, where g0 is a Kähler metric on (M; J) and f is a positive function. Denote by s0 and k0 the scalar and the conformal scalar curvature of (M; J; g0), respectively. Then k0 = f?2k (see, e. g., [5]) and s0 = f?2(s + 6?ff ), where

? is the Laplacian of g (cf. [2, 1.161]). Since g0 is a Kähler metric, we have s0 = k0 (cf. (5)) and therefore 6?f + (s?k)f = 0. From the maximum principle it follows that s ? k = 0, which contradicts the assumption s ? k > 0. This completes the proof of Theorem 1.

Acknowledgements: The author would like to thank Vestislav Apostolov for many helpful discussions related to this work.

References

[1] V. Apostolov and P. Gauduchon, The Riemannian Goldberg{Sachs Theorem, Int. J. Math., 8 (1997), 421{439.

[2] A. Besse, Einstein Manifolds, Ergeb. Math. Grenzgeb., Vol. 10, Springer, Berlin, Heidelberg, New York, 1987.

[3] C. P. Boyer, Self-dual and anti-self-dual Hermitian Einstein metrics on compact complex surfaces, Contemp. Math. 71 (1988), 105{114.

[4] A. Derdzinski, Self-dual Kähler manifolds and Einstein manifolds of dimension four, Comp. Math. 49 (1983), 405{433.

[5] P. Gauduchon, La 1-forme de torsion d'une variété hermitienne compacte, Math. Ann. 267 (1984), 495{518.

[6] S. Kobayashi and K. Nomizu, Foundations of Differential Geometry I, II, Interscience Publishers (1963).

[7] C. Le Brun, Einstein Metrics on Complex Surfaces, in Geometry and Physics (Aarhus 1995), eds. J. Andersen, J. Dupont, H. Pedersen and A. Swann, Lect. Notes in Pure Appl. Math., Marcel Dekker, 1996.

[8] P. Nurowski, Einstein equations and Cauchy{Riemann geometry, Ph. D. Thesis, SISA/ISAS, Trieste, 1993.

[9] D. Page, A compact rotating gravitational instanton, Phys. Lett. 79B (1979), 235{238.

[10] M. Przanowski and B. Broda, Locally Kähler gravitational instantons, Acta Phys. Pol. B14 (1983), 637{661.

[11] F. Tricerri, L. Vanhecke, Curvature tensors on almost-Hermitian manifolds, Trans. Amer. Math. Soc., 267 (1981), 365{398.

[12] I. Vaisman, Some curvature properties of complex surfaces, Ann. Mat. Pura Appl. 32 (1982), 1{18.

Institute of Mathematics, Bulgarian Academy of Sciences,
Acad. G. Bonchev Str. bl. 8, 1113 Sofia, Bulgaria
E-mail address: muskarov@math.bas.bg