THE THEORY OF SUPERSTRING WITH FLUX ON NON-K¨AHLER MANIFOLDS AND THE COMPLEX MONGE-AMP`ERE EQUATION
j.differential geometry
78(2008)369-428
THE THEORY OF SUPERSTRING WITH FLUX ON NON-K¨AHLER MANIFOLDS AND THE COMPLEX
MONGE-AMP`ERE EQUATION
Ji-Xiang Fu&Shing-Tung Yau
Abstract
The purpose of this paper is to solve a problem posed by Stro-minger in constructing smooth models of superstring theory with
?ux.These are given by non-K¨a hler manifolds with torsion.
1.Introduction
The purpose of this paper is twofold.The?rst purpose is to solve an old problem posed by Strominger in constructing smooth models of superstring theory with?ux.These are given by non-K¨a hler mani-folds with torsion.To achieve this,we solve a nonlinear Monge-Amp`e re equation which is more complicated than the equation in the Calabi con-jecture.The estimate of the volume form gives extra complication,for example.The second purpose is to point out the connection of the newly constructed geometry based on Strominger’s equations in realizing the proposal of M.Reid[19]on connecting one Calabi-Yau manifold to an-other one with di?erent topology.In Reid’s proposal,the construction of Clemens-Friedman(see[9])is needed where a Calabi-Yau manifold is deformed to complex manifolds di?eomorphic to connected sums of S3×S3.These are non-K¨a hler manifolds.
There is a rich class of non-K¨a hler complex manifolds for dimensions greater than two.It is therefore important to construct canonical ge-ometry on such manifolds.Since for non-K¨a hler geometry,the complex structure is not quite compatible with the Riemannian metric,it has been di?cult to?nd a reasonable class of Hermitian metric that ex-hibits rich geometry.We believe that metrics motivated by theoretic physics should have good properties.This is especially true for those metrics which admit parallel spinors.The work of Strominger provided such a candidate.In this paper,we provide a smooth solution to the Strominger system.This has been an important open problem through the past twenty years.Our method is based on a priori estimates which
Received05/24/2007.
369
370J.-X.FU &S.-T.YAU
can be generalized to elliptic ?bration over general Calabi-Yau mani-folds.However,in this paper,for the sake of importance in string the-ory,we shall restrict ourselves to complex three-dimensional manifolds.The structure of the equations for higher-dimensional Calabi-Yau man-ifolds is a little bit di?erent.They are also more relevant to algebraic geometry and hence will be treated in a later occasion.
The physical context of the solutions is discussed in a companion paper [3]written jointly with K.Becker,M.Becker,and L.-S.Tseng.Acknowledgement.The authors would like to thank K.Becker,M.Becker,and L.-S.Tseng for useful discussions.J.-X.Fu would also like to thank J.Li and X.-P.Zhu for useful discussions.J.-X.Fu is supported in part by NSFC grant 10471026.S.-T.Yau is supported in part by NSF grants DMS-0244462,DMS-0354737and DMS-0306600.
2.Motivation from string theory
In the original proposal for compacti?cation of superstring [5],Can-delas,Horowitz,Strominger,and Witten constructed the metric prod-uct of a maximal symmetric four-dimensional spacetime M with a six-dimensional Calabi-Yau vacuum X as the ten-dimensional spacetime;they identi?ed the Yang-Mills connection with the SU (3)connection of the Calabi-Yau metric and set the dilaton to be a constant.Adapting the second author’s suggestion of using Uhlenbeck-Yau’s theorem [22]on constructing Hermitian-Yang-Mills connections over stable bundles,Witten [23]and later Horava-Witten [13]proposed to use higher rank bundles for strong coupled heterotic string theory so that the gauge groups can be SU (4)or SU (5).
At around the same time,Strominger [20]analyzed heterotic super-string background with spacetime supersymmetry and non-zero torsion by allowing a scalar “warp factor”for the spacetime metric.He consid-ered a ten-dimensional spacetime that is a warped product of a maximal symmetric four-dimensional spacetime M and an internal space X ;the metric on M ×X takes the form g 0=e 2D (y ) g μν(x )00g ij (y )
,x ∈M,y ∈X ;the connection on an auxiliary bundle is Hermitian-Yang-Mills connec-tion over X :
F ∧ω2=0,F 2,0=F 0,2=0.
Here ωis the Hermitian form ω=√
?12g i ˉj dz i ∧d ˉz j de?ned on the internal space X .In this system,the physical relevant quantities are h =?√?1(ˉ?
??)ω,φ=?12log Ω +φ0,
THE THEORY OF SUPERSTRING 371
and g 0ij
=e 2φ0 Ω ?1g ij ,for a constant φ0.
In order for the ansatz to provide a supersymmetric con?guration,one introduces a Majorana-Weyl spinor so that
δψM = M ?18
h MNP γNP =0,δλ=γM ?M φ ?112h MNP γMNP =0,δχ=γMN F MN =0,
where ψM is the gravitino,λis the dilatino,χis the gluino,φis the dilaton and h is the Kalb-Ramond ?eld strength obeying
dh =α 2
(tr F ∧F ?tr R ∧R ),where α is positive.Strominger [20]showed that in order to achieve spacetime supersymmetry,the internal six manifold X must be a com-plex manifold with a non-vanishing holomorphic three-form Ω;and the anomaly cancellation demands that the Hermitian form ωobey 1√?1?ˉ?ω=α 4
(tr R ∧R ?tr F ∧F )and supersymmetry requires 2d ?ω=√?1(ˉ?
??)log Ω ω.Accordingly,he proposed the system
(2.1)
F H ∧ω2=0;(2.2)
F 2,0H =F 0,2H =0;(2.3)√?1?ˉ?ω=α 4(tr R ∧R ?tr F H ∧F H );(2.4)d ?ω=√?1(ˉ?
??)ln Ω ω.This system gives a solution of a superstring theory with ?ux that allows a non-trivial dilaton ?eld and a Yang-Mills ?eld.(It turns out D (y )=φand is the dilaton ?eld.)Here ωis the Hermitian form and R is the curvature tensor of the Hermitian metric ω;H is the Hermitian metric and F is its curvature of a vector bundle E ;tr is the trace of the endomorphism bundle of either E or T X .
In [17],Li and Yau observed the following:
1
The curvature F of the vector bundle E in ref.[20]is real,i.e.,c 1(E )=F 2π.But we are used to taking the curvature F such that c 1(E )=√?12πF .So this equation
corrects eq.(2.18)of ref.[20]by a minus sign.2See eq.(56)of ref.[21],which corrects eq.(2.30)of ref.[20]by a minus sign.
372J.-X.FU &S.-T.YAU
Lemma 1.Equation (2.4)is equivalent to
(2.5)d ( Ω ωω2)=0.
In fact,Li and Yau gave the ?rst irreducible non-singular solution of the supersymmetric system of Strominger for U (4)and U (5)prin-ciple bundle.They obtained their solutions by perturbing around the Calabi-Yau vacuum coupled with the sum of tangent bundle and triv-ial line bundles.In this paper,we consider the solution on complex manifolds which do not admit K¨a hler structures.Study of non-K¨a hler manifolds should be useful to understand the speculation of M.Reid that all Calabi-Yau manifolds can be deformed to each other through conifold transition.
An example of non-K¨a hler manifolds X is given by T 2-bundles over Calabi-Yau varieties [2,4,10,12,14].Since we demand that the internal six manifold X is a complex manifold with a non-vanishing holomorphic three form Ω,we consider the T 2?bundle (X,ω,Ω)over a complex surface (S,ωS ,ΩS )with a non-vanishing holomorphic 2-form ΩS .According to the classi?cation of complex surfaces by Enriques and Kodaira,such complex surfaces must be ?nite quotients of K3sur-face,complex torus (K¨a hler),and Kodaira surface (non-K¨a hler).If (X,ω,Ω)satis?es Strominger’s equation (2.4),Lemma 1shows that
d ( Ω ω
ω2)=0.Let ω
= Ω 12
ωω.Then dω 2=0,i.e.,ω is a balanced metric [18].The balanced metric was studied extensively by Michelsohn.She proved that the balanced condition is preserved under proper holomorphic submersions.Note that Alessandrini and Bassanelli [1]proved that this condition is also preserved under mod-i?cations of complex manifolds.Hence if a holomorphic submersion πfrom a balanced manifold X to a complex surface S is proper,S is also balanced (actually π?ω 2is the balanced metric on S ,see Proposition
1.9in [18]).When the dimension of complex manifold is two,the condi-tions of being balanced and K¨a hler coincide.Hence there is no solution to Strominger’s equation (1.4)on T 2bundles over Kodaira surface and we consider T 2-bundles over K 3surface and complex torus only.
On the other hand,duality from M -theory suggests that there is no supersymmetric solution when the base manifold is a complex torus (see
[3]).This class of three manifolds includes the Iwasawa manifold.But the solution to Strominger’s system should exist when the base is K 3surface.In this paper we prove the existence of solutions to Strominger’s system on such torus bundles over K 3surfaces.3.Statement of main result
Let (S,ωS ,ΩS )be a K3surface or a complex torus with a K¨
a hler form ωS and a non-vanishing holomorphic (2,0)-form ΩS .Let ω1and ω2be anti-self-dual (1,1)-forms such that ω12πand ω22πrepresent integral
THE THEORY OF SUPERSTRING 373
cohomology https://www.sodocs.net/doc/b98396925.html,ing these two forms,Goldstein and Prokushkin
[10]constructed a non-K¨a hler manifold X such that π:X →S is a
holomorphic T 2-?bration over S with a Hermitian form ω0=π?ωS +√?12θ∧ˉθand a holomorphic (3,0)-form Ω=ΩS ∧θ(for the de?nition of θ,see section 3).Note that (ω0,Ω)satis?es equation (2.5).(Sethi
pointed out that in papers [6]and [2]similar ansatz was discussed.However the major problem of solving equations was not addressed in the literature.)
Let u be any smooth function on S and let ωu =π?(e u ωS )+√?12
θ∧ˉθ.Then (ωu ,Ω)also satis?es equation (2.5)(see [10]or Lemma 12),i.e.,ωu is conformal balanced.The stability concept can be de?ned on a vector bundle over a complex manifold using the Gauduchon metric [16],and hence for complex manifolds with balanced metrics.Note that the stability concept of the vector bundle depends only on the conformal class of metric.Let V →X be a stable bundle over X with degree zero with respect to the metric ωu .(Such bundles can be obtained by pulling back stable bundles over a K 3surface or a complex torus,see Lemma 16.)According to Li-Yau’s theorem [16],there is a Hermitian-Yang-Mills metric H on V ,which is unique up to positive constants.The curvature F H of the Hermitian metric H satis?es equation (2.1)and (2.2).So (V,F H ,X,ωu )satis?es Strominger’s equations (2.1),(2.2)and (2.4).Therefore we only need to consider equation (2.3).As ω1and ω2are harmonic,ˉ?ω1=ˉ?ω2=0.According to ˉ?-Poincar′e Lemma,we can write ω1and ω2locally as
ω1=ˉ?ξ
=ˉ?(ξ1dz 1+ξ2dz 2)and ω2=ˉ?ζ
=ˉ?(ζ1dz 1+ζ2dz 2),where (z 1,z 2)is a local coordinate on S .Let B = ξ1+√?1ζ1ξ2+√?1ζ2
.We can use B to compute tr R 0∧R 0of the metric ω0(see Proposition
8)and tr R u ∧R u of the metric ωu (see Lemma 14).Then we reduce equation (2.3)to √?1?ˉ?e u ∧ωS ?α 2?ˉ?(e ?u tr(ˉ?B ∧?B ?·g ?1))?α 2
?ˉ?u ∧?ˉ?u (3.1)=α 4tr R S ∧R S ?α 4tr F H ∧F H ?12( ω1 2ωS + ω2 2ωS )ω2S 2!
,where g =(g i ˉj )is the Ricci-?at metric on S associated to the K¨a hler form ωS and g ?1is the inverse matrix of g ;R S is the curvature of g .
374J.-X.FU &S.-T.YAU
Taking wedge product with ωu and integrating both sides of the above equation over X ,we obtain
(3.2)α X {tr R S ∧R S ?tr F H ∧F H }∧ωu ?2 X
( ω1 2ωS + ω2 2ωS )ω2S 2!∧ωu =0.When S =T 4,R S =0.We obtain immediately
Proposition 2.There is no solution of Strominger’s system on the torus bundle X over T 4if the metric has the form e u ωS +√?12
θ∧ˉθ.This situation is di?erent if the base is a K 3surface.If E is a stable bundle over S with degree 0with respect to the metric ωS ,then V =π?E is also a stable bundle with degree 0over X with respect to the Hermitian metric ωu .In this case,equation (3.1)on X can be considered as an equation on S .Integrating equation (3.1)over S ,(3.3)α S {tr R S ∧R S ?tr F H ∧F H }=2 S ( ω1 2ωS + ω2 2ωS )ω2S 2!.As S tr R S ∧R S =8π2c 2(V )=8π2×24,and S tr F H ∧F H =8π2×(c 2(E )?12c 21(E ))≥0,we can rewrite equation (3.3)as (3.4)α 24? c 2(E )?12c 21(E ) = S
ω12π 2ωS + ω22π 2ωS ω2S 2!.For a compact,oriented,simply connected four-manifold S ,the Poincar′e duality gives rise to a pairing
Q :H 2(S ;Z )×H 2(S ;Z )→Z
de?ned by
Q (β,γ)=
S β∧γ.
We shall denote Q (β,β)by Q (β).Then for an integral anti-self-dual (1,1)-form ω12π,the intersection number Q (ω12π)can be expressed as ? S
ω12π 2ω2S 2!.
On the other hand,the intersection form on K 3surface is
given by [7]3 0110
⊕2(?E 8),where E 8=????????????20?1020?1?102?1?1?12?1?12?1?12?1?12?1?12????????????
.
THE THEORY OF SUPERSTRING375
Hence Q(ω1
2π)∈{?2,?4,?6,...}.
We shall use the following convention for vector bundles over a com-pact oriented four-manifold:
κ(E)=c2(E)for SU(r)bundle E,
=c2(E)?1
2
c21(E)for U(r)bundle E,
=?1
2
p1(E)for SO(r)bundle E.
Then(3.4)implies
(3.5)α (24?κ(E))+
Q
ω
1
2π
+Q
ω
2
2π
=0,
which means that there is a smooth functionμsuch that
(3.6)α
4
tr R S∧R S?
α
4
tr F H∧F H?
1
2
( ω1 2+ ω2 2)
ω2S
2!
=?μ
ω2S
2!
and
S
μω2
2!
=0.Inserting(3.6)into(3.1),we obtain the following
equation:
(3.7)
√
?1?ˉ?e u∧ωS?α
2?ˉ?(e?u tr(ˉ?B∧?B?·g?1))?
α
2
?ˉ?u∧?ˉ?u+μ
ω2S
2!
=0,
where tr(ˉ?B∧?B?·g?1)is a smooth well-de?ned(1,1)-form on S.In particular,whenω2=nω1,n∈Z,
tr(ˉ?B∧?B?·g?1)=√
?11+n
2
4
ω1 2ω
S
ωS
(see Proposition11).Hence if we set f=1+n2
4 ω1 2ω
S
,we can rewrite
equation(3.7)as the standard complex Monge-Amp`e re equation:
(3.8)Δ(e u?α
2
fe?u)+4α
det u iˉj
det g iˉj
+μ=0,
where u iˉj denotes?2u
?z i?ˉz j and =2g iˉj?2
?z i?ˉz j
.We shall solve equation
(3.7)by the continuity method[24].Our main theorem is
Theorem3.The equation(3.7)has a smooth solution u such that
ω =e uωS?α
2
√
?1e?u tr(ˉ?B∧?B?·g?1)+α
√
?1?ˉ?u
de?nes a Hermitian metric on S.
Our solution u satis?es
S
e?4u
1
4=A 1.Actually we can prove
that inf u≥?ln(C1A)(see Proposition21)where A must be very small (see Proposition22)and our solution u must be very big.
376J.-X.FU&S.-T.YAU
Theorem 4.Let S be a K3surface with a Ricci-?at metricωS.
Letω1andω2be anti-self-dual(1,1)-forms on S such thatω1
2π,ω2
2π
∈
H2(S,Z).Let X be a T2-bundle over S constructed byω1andω2.Let E be a stable bundle over S with degree0.Supposeω1,ω2andκ(E) satisfy condition(3.5).Then there exist a smooth function u on S and a Hermitian-Yang-Mills metric H on E such that(V=π?E,π?F H,X,ωu) is a solution of Strominger’s system.
Since it is easy to?nd(ω1,ω2,κ(E))which satis?es condition(3.5), this theorem provides?rst examples of solutions to Strominger’s system on non-K¨a hler manifolds.
4.Geometric model
In this section,we take the geometric model of Goldstein and Pro-kushkin for complex non-K¨a hler manifolds with an SU(3)structure[10]. We summarize their results as follows:
Theorem5.[10]Let(S,ωS,ΩS)be a Calabi-Yau2-fold with a non-vanishing holomorphic(2,0)?formΩS.Letω1andω2be anti-self-dual
(1,1)-forms on S such thatω1
2π∈H2(S,Z)andω2
2π
∈H2(S,Z).Then
there is a Hermitian3-fold X such thatπ:X→S is a holomorphic T2-?bration over S and the following holds:
1.For any real1-formsα1andα2de?ned on some open subset of S
that satisfy dα1=ω1and dα2=ω2,there are local coordinates x and y on X such that dx+idy is a holomorphic form on T2-?bers and a metric on X has the following form:
(4.1)g0=π?g+(dx+π?α1)2+(dy+π?α2)2,
where g is a Calabi-Yau metric on S corresponding to the K¨a hler formωS.
2.X admits a nowhere vanishing holomorphic(3,0)-form with unit
length:
Ω=((dx+π?α1)+i(dy+π?α2))∧π?ΩS.
3.If eitherω1orω2represents a non-trivial cohomological class then
X admits no K¨a hler metric.
4.X is a balanced manifold.The Hermitian form
(4.2)ω0=π?ωS+(dx+π?α1)∧(dy+π?α2)
corresponding to the metric(4.1)is balanced,i.e.,dω20=0.
5.Furthermore,for any smooth function u on S,the Hermitian met-
ric
ωu=π?(e uωS)+(dx+π?α1)∧(dy+π?α2)
is conformal balanced.Actually(ωu,Ω)satis?es equation(2.5).
THE THEORY OF SUPERSTRING 377
Goldstein and Prokushkin also studied the cohomology of this non-K¨a hler manifold X :
h 1,0(X )=h 1,0(S ),
h 0,1(X )=h 0,1(S )+1;
In particular,
h 0,1(X )=h 1,0(X )+1.
Moreover,
b 1(X )=b 1(S )+1,when ω2=nω1,
b 1(X )=b 1(S ),when ω2=nω1;
b 2(X )=b 2(S )?1,when ω2=nω1,
b 2(X )=b 2(S )?2,when ω2=nω1
and
χ(X )=0.
The above topological results can be explained as follows.Let L 1be a holomorphic line bundle over S with the ?rst Chern class c 1(L 1)=
[?ω12π].Then we can choose a Hermitian metric h 1on L 1such that its curvature is √?1ω1.Let S 1={v ∈L 1|h 1(v,v )=1}which is a circle bundle over S .Locally we write ω1=dα1U for some real 1-form α1U on some open subset U on S .Such α1U de?ne a connection on S 1,i.e.,there is a section ξU on S 1such that ξU =√?1α1U ?ξU .
The section ξU de?nes a local coordinate x U on ?bers of S 1|U ,i.e.,we can describe the circle S 1by e √?1x U ξU .If we write ω1=dα1V on
another open set V of S ,then there is another section ξV such that (4.3) ξV =√?1α1V ?ξV
and this section ξV de?nes another coordinate x V on ?ber of S 1|V .On U ∩V ,d (α1U ?α1V )=0and there is a function f UV such that (4.4)d f UV =α1U ?α1V .
On the other hand,on U ∩V ,there is also a function g UV on U ∩V such that ξV =e √?1g UV ξU .We compute
ξV = (e √
?1g UV ξU )=(√?1dg UV +√?1α1U )?(e √?1g UV ξU )=(√?1dg UV +√?1α1U )?ξV .
Comparing the above equality with (4.3),we get
(4.5)?dg UV =α1U ?α1V .
378J.-X.FU &S.-T.YAU
So combining (4.4),we ?nd
(4.6)g UV =f UV +c UV ,
where c UV is some constant on U ∩V .On U ∩V ,from e
ix U ξU =e ix V ξV =e √?1x V e √?1g UV ξU ,we obtain
(4.7)x U =x V +g UV +2kπ=x V +f UV +c UV +2kπ.
(4.4)and (4.7)imply (4.8)
dx U ?dx V =d f UV =?α1U +α1V .So dx U +α1U is a globally de?ned 1-form on X .We denote it by dx +α1.
We construct another line bundle L 2with the ?rst Chern class [?ω22π].
Similarly,we write locally ω2=dα2,and de?ne a coordinate y on ?bers such that dy +α2is a well-de?ned 1-form on the circle bundle S 1of L 2.On X ,ω1=d (dx +α1)and ω2=d (dy +α2),and so [ω1]=[ω2]=0∈H 2(X,R ).When ω2=nω1,d (n (dx +α1)?(dy +α2))=0.So
[n (dx +α1)?(dy +α2)]∈H 1(X,R ).Finally we de?ne θ=dx +α1+√?1(dy +α2).
Then θis a (1,0)-form on X ,see [10]or the next section.Because d ˉθ
=ω1?√?1ω2is a (1,1)-form on X ,its (0,2)-component ˉ?ˉθ=0.So [ˉθ]∈H 0,1ˉ?
(X )～=H 1(X,O ).5.The calculation of tr R ∧R
In order to calculate the curvature R and tr R ∧R ,we express the Hermitian metric (4.1)in terms of a basis of holomorphic (1,0)vector ?elds.Hence we need to write down the complex structure of X .Let {U,z j =x j +√?1y j ,j =1,2}be a local coordinate in S .The horizontal
lifts of vector ?elds ??x j and ??y j ,which are in the kernel of dx +π?α1and dy +π?α2,are X j =??x j ?α1 ??x j ??x ?α2 ??x j ??y
for j =1,2,Y j =??y j ?α1 ??y j ??x ?α2 ??y j ??y
for j =1,2.The complex structure ?I
on X is de?ned as ?IX
j =Y j ,?IY j =?X j ,for j =1,2,?I ??x =??y ,?I ??y =???x
.
THE THEORY OF SUPERSTRING 379
Let
U j
=X j ?√
?1?IX j =X j ?√?1Y j ,U 0=??x ?√?1?I ??x =??x ?√?1??y .Then {U j ,U 0}is the basis of the (1,0)vector ?elds on X .The metric (4.1)takes the following Hermitian form:(5.1) (g i ˉj )001
as U 1and U 2are in the kernel of dx +π?α1and dy +π?α2.Let (5.2)θ=dx +√?1dy +π?(α1+√?1α2).
It’s easy to check that {π?dz j ,θ}annihilates the {U j ,U 0}and is the basis of (0,1)-forms on X .So {π?dz j ,θ}are (1,0)-forms on X .Certainly π?dz j are holomorphic (1,0)-forms and θis not.We need to construct another holomorphic (1,0)-form on X .Because ω1and ω2are harmonic forms on S ,?ω1=?ω2=0.By ˉ?
-Poincar′e Lemma,locally we can ?nd (1,0)-forms ξ=ξ1dz 1+ξ2dz 2and ζ=ζ1dz 1+ζ2dz 2on S ,where ξi and ζj are smooth complex functions on some open set of S ,such that ω1=?ξand ω2=?ζ.Let θ0=θ?π?(ξ+√?1ζ)=(dx +√?1dy )+π?(α1+√?1α2)?π?(ξ+√?1ζ).
We claim that θ0is a holomorphic (1,0)-form.By our construction,θ0is the (1,0)-form.But dθ=d (dx +√?1dy +π?(α1+√?1α2))=π?(ω1+√?1ω2)is a (1,1)-form on X .So
(5.3)
?θ=0and ?θ=dθ=π?(ω1+iω2).Thus,
?θ0=?θ??π?(ξ+√
?1ζ)=π?(ω1+√?1ω2)?π?(ω1+√?1ω2)=0.So θ0is a holomorphic (1,0)-form and {π?dz j ,θ0}forms a basis of holo-morphic (1,0)-forms on X .Let
?j =ξj +√?1ζj for j =1,2
and
?U j =U j +?j U 0for j =1,2.
Then {?U
j ,U 0}is dual to {π?dz j ,θ0}because U j is in the kernel of θ.It’s the basis of holomorphic (1,0)-vector ?elds.The metric g 0then
380J.-X.FU&S.-T.YAU becomes the following Hermitian matrix:
(5.4)
H X=?
?
g1ˉ1+|?1|2g1ˉ2+?1?2?1
g2ˉ1+?2?1g2ˉ2+|?2|2?2
?1?21
?
?=
g+B·B?B
B?1
,
where g is the Calabi-Yau metric on S and B=(?1,?2)t.
According to Strominger’s explanation in[20],when the manifold is not K¨a hler,we should take the curvature of Hermitian connection on the holomorphic tangent bundle T https://www.sodocs.net/doc/b98396925.html,ing the metric(5.4),we compute
the curvature to be
R=?(?H X·H?1X)=
R1ˉ1R1ˉ2
R2ˉ1R2ˉ2
,
where
R1ˉ1=R S+?B∧(?B?·g?1)+B·?(?B?·g?1),
R1ˉ2=?R S B+(?g·g?1)∧?B??B∧(?B?·g?1)B
?B?(?B?·g?1)B+B(?B?·g?1)∧?B+??B, R2ˉ1=?(?B?·g?1),
R2ˉ2=??(?B?·g?1)B+(?B?·g?1)∧?B,
and R S is the curvature of Calabi-Yau metric g on S.It is easy to check that tr(?B∧(?B?·g?1)+B·?(?B?·g?1))??(?B?·g?1)B+(?B?·g?1)∧?B=0.So tr R=π?tr R S.
Proposition6([11]).The Ricci forms of the Hermitian connections on X and S have the relation tr R=π?tr R S.
Remark7.In the above calculation,we don’t use the condition that the metric g on S is Calabi-Yau.
Proposition8.
(5.5)tr R∧R=π?(tr R S∧R S+2tr??(?B∧?B?·g?1)).
Proof.Fix any point p∈S,we pick B such that B(p)=0.Otherwise, B(p)=0and we simply replace B by B?B(p).Hence in the calculation of tr R∧R at p,all terms containing the factor B will vanish.Thus tr R∧R
=tr R S∧R S+2tr R S∧?B∧(?B?·g?1)
+2tr?g·g?1∧?B∧ˉ?(?B?·g?1)+2tr??B∧?(?B?·g?1). Proposition8follows from the next two lemmas.q.e.d.
THE THEORY OF SUPERSTRING381 Lemma9.
tr??(?B∧?B?·g?1)=tr R S∧?B∧(?B?·g?1)
+tr?g·g?1∧?B∧ˉ?(?B?·g?1)
+tr??B∧?(?B?·g?1).
Proof.
tr??(?B∧?B?·g?1)
=?tr?(?B∧?(?B?·g?1))
=tr??B∧?(?B?·g?1)+tr?B∧?(?B?∧?g?1)
=tr??B∧?(?B?·g?1)?tr?B∧?(?B?·g?1∧?g·g?1)
=tr??B∧?(?B?·g?1)?tr?B∧?(?B?·g?1)∧?g·g?1
+tr?B∧(?B?·g?1)∧ˉ?(?g·g?1)
=tr??B∧?(?B?·g?1)?tr?B∧?(?B?·g?1)∧?g·g?1
+tr?B∧(?B?·g?1)∧R S
=tr(??B∧?(?B?·g?1))+tr(R S∧?B∧?B?·g?1)
+tr(?g·g?1∧?B∧?(?B?·g?1).
q.e.d. Lemma10.tr(ˉ?B∧?B?·g?1)is a well-de?ned(1,1)-form on S.
Proof.We take local coordinates(U,z i)and(W,w j)on S such that U∩W=?.Let J=(?w i
?z j
)and
(ω1+√
?1ω2)|U=?(?1dz1+?2dz2)=??1∧dz1+ˉ??2∧dz2,
(ω1+√
?1ω2)|W=?(γ1dw1+γ2dw2)=?γ1∧dw1+ˉ?γ2∧dw2.
Then on U∩W,
ˉ
?γ1ˉ?γ2
∧
dw1
dw2
=
ˉ
??1ˉ??2
∧
dz1
dz2
.
So
ˉ
??1ˉ??2
=
ˉ
?γ1ˉ?γ2
J.
(5.6)
On the other hand,we have
(5.7)g(z)=J t g(w)J,
382J.-X.FU &S.-T.YAU
where g (z )=(g i ˉj (z ))and g (w )=(g i ˉj (w )).Then on U ∩W ,using (5.6),(5.7),we have
tr ˉ?γ1ˉ?γ2
∧ ?ˉγ1?ˉγ
2 ·g ?1(w )=tr(J t )?1 ˉ??1ˉ??2 ∧ ??1??2 ˉJ ?1·ˉJ ·g ?1(z )·J t =tr ˉ??1ˉ??2 ∧ ?ˉ?1?ˉ?2 ·g ?1(z ),which proves that tr(ˉ?B
∧?B ?·g ?1)is a well-de?ned (1,1)-form on S .q.e.d.
Although tr(ˉ?B
∧?B ?·g ?1)is a well-de?ned (1,1)-form on S ,we can not express it by ω1and ω2.But in some particular cases,we can.Proposition 11.When ω2=nω1,n ∈Z ,
(5.8)tr(ˉ?B ∧?B ?·g ?1)=√?14
(1+n 2) ω1 2ωS ωS ,where g is the given Calabi-Yau metric on S and ωS is the corresponding K¨a hler form.
Proof.We recall that locally,
ω1=ˉ?ξ,
ξ=ξ1dz 1+ξ2dz 2,ω2=ˉ?ζ,ζ=ζ1dz 1+ζ2dz 2,?j =ξj +√?1ζj ,for j =1,2,B = ?1?2
,B ?= ˉ?1ˉ?2 .When ω2=nω1,we take ζ=nξ.Then ˉ?ζ
j =n ˉ?ξj ,ˉ?B
= ˉ??1ˉ??2 =(1+n √?1) ˉ?ξ1ˉ?ξ2 and
?B ?= ?ˉ?1?ˉ?2 =(1?n √?1) ?ˉξ1?ˉξ2 .
THE THEORY OF SUPERSTRING 383
Using above equalities,we ?nd
(5.9)tr(ˉ?B ∧?B ?·g ?1)=(1+n 2)tr ˉ?ξ1ˉ?ξ2 ∧ ?ˉξ1?ˉξ2 ·g ?1=1+n 2det g tr ?ξ1?ˉz i d ˉz i ?ξ2?ˉz i d ˉz i ∧ ?ξ1?ˉz j dz j ?ξ2?ˉz j dz j · g 2ˉ2?g 1ˉ2?g 2ˉ1g 1ˉ1 =1+n 2det g tr ?ξ1?ˉz i ?ξ2?ˉz i
∧ ?ξ1?ˉz j ?ξ2?ˉz j · g 2ˉ2?g 1ˉ2?g 2ˉ1g 1ˉ1 d ˉz i ∧dz j .In order to get the global formula,we need to calculate ω1.As ω1is real,(5.10)?ξi ?ˉz j =??ξj ?ˉz i for i,j =1,2.
Since ω1is anti-self-dual,i.e.,ω1∧ωS =0,we have
(5.11)
g 1ˉ1?ξ2?ˉz 2+g 2ˉ2?ξ1?ˉz 1?g 1ˉ2?ξ2?ˉz 1?g 2ˉ1?ξ1?ˉz 2=0.Because
(5.12)ω1∧ω1=?ω1∧?ω1=?ω1?ˉω
1=? ω1 2ωS ω2S 2!,locally we also have (5.13)1
det(g ) ?ξ1?ˉz 1?ξ2?ˉz 2??ξ1?ˉz 2?ξ2?ˉz 1
=18
ω1 2ωS .Now using above (5.10),(5.11)and (5.13),we calculate the component of d ˉz 1∧dz 1in (5.9)to be
(5.14)1+n 2det(g ) g 2ˉ2?ξ1?ˉz 1?ξ1?ˉz 1?g 2ˉ1?ξ1?ˉz 1?ξ2?ˉz 1?g 1ˉ2?ξ2?ˉz 1?ξ1?ˉz 1?g 1ˉ1?ξ2?ˉz 1?ξ2?ˉz 1 =1+n 2det(g ) ?ξ1?ˉz 1 g 1ˉ1?ξ2?ˉz 2+g 2ˉ2?ξ1?ˉz 1 ?g 2ˉ2 ?ξ1?ˉz 1 2?g 1ˉ1?ξ2?ˉz 1?ξ1?ˉz 2 =1+n 2det(g )g 1ˉ1 ?ξ1?ˉz 1?ξ2?ˉz 2??ξ2?ˉz 1?ξ1?ˉz 2
=1+n 28
ω1 2ωS g 1ˉ1.Similarly,the components of d ˉz 2∧dz 1,d ˉz 1∧dz 2and d ˉz 2∧dz 2in (5.9)are 1+n 28 ω1 2ωS g 1ˉ2,1+n 28 ω1 2ωS g 2ˉ1and 1+n 28
ω1 2ωS g 2ˉ2respectively.
384J.-X.FU &S.-T.YAU
So we obtain
tr(ˉ?A ∧?A ?·g ?1)=√?11+n 24
ω1 2ωS ωS .q.e.d.
6.Reduction of Strominger’s system
Consider a 3-dimensional Hermitian manifold (X,ω0,Ω)as described in the section 2.Let ωS be the Calabi-Yau metric on S .Let θ=dx +α1+√?1(dy +α2),
then the Hermitian form ω0in (4.2)is ω0=π?ωS +√?12
θ∧ˉθ.Because Ω ω=1,and ω1and ω2are anti-self-dual,we use (5.3)to compute
d ( Ω ω0ω20)(6.1)=d (π?ω2S +√?1π?ωS ∧θ∧ˉθ)=√?1π?ωS ∧dθ∧ˉθ?√?1π?ωS ∧θ∧d ˉθ=√?1π?ωS ∧((ω1+√?1ω2)∧ˉθ?(ω1?√?1ω2)∧θ)
=0.
According to Lemma 1,(ω0,Ω)is the solution of equation (2.4).Let u be any smooth function on S and let (6.2)ωu =π?(e u ωS )+√?12
θ∧ˉθ.Then Ω 2ωu =ω30ω3u =1e
2u and
Ω ωu ω2u =ω20+(e u ?1)ω2S .
Using (6.1),we obtain
d ( Ω ωu ω2u
)=dω20+d (e u ?1)∧ω2S =0because e u is a function on S .Hence we have proven the following
Lemma 12([10]).The metric (6.2)de?ned on X satis?es equation (2.5)and so satis?es equation (2.4).
Let V be a stable vector bundle over X with degree 0with respec-tive to the metric ωu .According to Li-Yau’s theorem [16],there is a Hermitian-Yang-Mills metric H on V ,which is unique up to constant.
THE THEORY OF SUPERSTRING 385
Then (V,H,X,ωu )satis?es equation (2.1),(2.2)and (2.4)of the Stro-minge’s system.Hence to look for a solution to Strominger’s system,we need only to consider equation (2.3):
(6.3)√?1?ˉ?ωu =α 4
(tr R u ∧R u ?tr F H ∧F H ),where R u is the curvature of Hermitian connection of metric ωu on the holomorphic tangent bundle T X .De?ne the Laplacian operator with respective to the metric ωS as
ψω2S 2!
=√?1?ˉ?ψ∧ωS .Lemma 13.√?1?ˉ?ωu = e u ·ω2S 2!+12( ω1 2ωS + ω2 2ωS )ω2S 2!
https://www.sodocs.net/doc/b98396925.html,ing (5.3)and (5.12),we compute √?1??ωu =√?1??(e u ωS +√?12θ∧θ)=√?1?ˉ?e u ∧ωS ?12
ˉ?θ∧?ˉθ= e u ·ω2S 2!+12( ω1 2ωS + ω2 2ωS )ω2S 2!
.q.e.d.
Lemma 14.tr R u ∧R u =π?tr R S ∧R S +2√?1π?(?ˉ?u ∧?ˉ?u )+2π?(?ˉ?(e ?u ρ)),where ρ=?√?1tr(ˉ?B
∧?B ?·g ?1).Proof.In the proof of Proposition 8we don’t use the condition that
ωS is K¨a hler.So if we replace metric g by e u g ,we can still obtain:(6.4)tr R u ∧R u =π?(tr R u S
∧R u S +2√?1?ˉ?(e ?u ρ)),here R u S denotes the curvature of Hermitian connection of the metric e u g on holomorphic tangent bundle T S .So
R u S
=ˉ??u ·I +R S and
(6.5)tr R u S
∧R u S =tr R S ∧R S +2?ˉ?u ∧?ˉ?u,here we use the fact that tr R S =0because the Hermitian metric g is the Calabi-Yau metric on S .Inserting (6.5)into (6.4),we have proven the lemma.q.e.d.From Lemma 13and 14,we can rewrite equation (6.3)as
(6.6)√?1?ˉ?e u ∧ωS ?√?1α 2?ˉ?(e ?u ρ)?α 2?ˉ?u ∧?ˉ?u =α 4tr R S ∧R S ?α 4tr F H ∧F H ?12( ω1 2+ ω2 2ωS )ω2S 2!
.
386J.-X.FU &S.-T.YAU
Proposition 15.There is no solution of Strominger’s system on the torus bundle X over T 4if the metric is e u ωS +√?12θ∧ˉθ
.Proof.Wedging the left-hand side of equation (6.6)by ωu and inte-grating over X ,we get (6.7) X √?1?ˉ?e u ∧ωS ?α 2?ˉ?(e ?u ρ)?α 2?ˉ?u ∧?ˉ?u ∧ωu =0because ?ωu =?(e u )∧ωS +2θ∧(ω1?√?1ω2).When S =T 4,R T 4=0.Integrating both sides of (6.6)and applying (6.7),we get (6.8)α X tr F H ∧F H ∧ωu +12 X
( ω1 2ωS + ω2 2ωS )ω22!∧ωu =0.Certainly
(6.9)2
X ( ω1 2ωS + ω2 2ωS )ω2S 2!∧ωu >0.
On the other hand,it is well-known that
tr F 2H 8π2
=12c 21(V )?c 2(V )=12r c 21(V )?12r (2rc 2(V )?(r ?1)c 21(V )),where r is a rank of the bundle V and that
(2r (c 2(V )?(r ?1)c 21(V ))∧ωu =r 4π|F 0|2ω3u 3!
,where F 0=F H ?1r tr F H ·i d V .So
tr F 2H ∧ωu =8π22r c 21(V )?|F 0|2ω3u 3!
.Now according to equation (2.2),F H ∧ω2u =0and c 1(V )∧ω2u
=0.Thus c 21(V
)∧ωu =?|c 1(V )|2ω3u 3!and
(6.10) X tr F 2H ∧ωu =?4π2r X |c 1(V )|2ω3u 3!? X |F 0|2
ω3u 3!≤0.Inserting (6.9)and (6.10)into (6.8),we get a contradiction.q.e.d.This situation is di?erent if the base is a K 3surface.At ?rst we observe
Lemma 16.Let E be a stable vector bundle over S with degree 0with respective to the Calabi-Yau metric ωS .Then V =π?E is also a stable vector bundle over X with degree 0with respective to Hermitian metric ωu for any smooth function u on S .
THE THEORY OF SUPERSTRING 387
Proof.According to the Donaldson-Uhlenbeck-Yau theorem,there is a unique Hermitian-Yang-Mills metric H on E up to constant.Since we assume that the degree of E is zero,the curvature F H of H satis?es the equation
F H ∧ωS =0.
For the metric π?H on V =π?E ,the curvature π?(F H )satis?es
π?F H ∧ω2u
=π?(F H ∧ωS )∧(π?(e 2u ωS )+π?(e u )θ∧ˉθ)=0.So π?H is also the Hermitian-Yang-Mills metric on V =π?E with degree 0.Thus V is a stable vector bundle over X with respective to the
Hermitian metric ωu for any smooth function u .
q.e.d.We also have the following observation:
Proposition 17([3]).Let (V,H )be a Hermitain-Yang-Mills vector bundle over (X,ωu )with gauge group SU (r ).If (X,ωu ,V,H )is the solution to Strominger’s system,then there is a Hermitian-Yang-Mills vector bundle (E,H )over S and a ?at line bundle L over X such that
V =π?E ?L.
When we restrict ourselves to consider such a vector bundle (V =π?E,π?F H )over X ,we see that equation (6.6)on X can be considered as an equation on S .Integrating equation (6.6)over S ,we get (6.11)α S {tr R S ∧R S ?tr F H ∧F H }=2 S ( ω1 2ωS + ω2 2ωS )ω2S 2!.As S tr R S ∧R S =8π2c 2(V )=8π2×24,and S tr F H ∧F H =8π2×(c 2(E )?12c 21(E ))≥0,we can rewrite equation (6.11)as (6.12)α (24?(c 2(E )?12c 21(E )))= S
( ω12π 2ωS + ω22π 2ωS )ω2S 2!.Using notations of section 1,above equation implies:(6.13)α (24?κ(E ))+ Q ω12π +Q ω22π
=0.This equation implies that there is a smooth function μsuch that
(6.14)α 4tr R S ∧R S ?α tr F H ∧F H ?12( ω1 2+ ω2 2ωS )ω2S 2!=?μω2S 2!
and S μω2S 2!=0.Inserting (6.14)into (6.6),we obtain the following
equation:
(6.15)√?1?ˉ?e u ∧ωS ?√?1α 2?ˉ?(e ?u ρ)?α 2?ˉ?u ∧?ˉ?u +μω2S 2!=0where μis a smooth function satisfying the integrable condition S μ=0and ρ=√?1tr(ˉ?B
∧?B ?·g ?1)is a smooth well-de?ned real (1,1)-form on S .In the next section we will use the continuity method to
388J.-X.FU &S.-T.YAU
solve equation (6.15).We will prove that equation (6.15)has a smooth solution u .
Theorem 18.Let S be a K 3surface with a Calabi-Yau metric ωS .
Let ω1and ω2be anti-self-dual (1,1)-forms on S such that ω12π∈H 2(S,Z )
and ω22π∈H 2(S,Z ).Let X be a T 2-bundle over S constructed by ω1and ω2.Let E be a stable bundle over S with degree 0.Suppose that ω1,ω2and κ(E )satisfy the condition (6.13).Then there exists a smooth function u on S and a Hermitian-Yang-Mills metric H on E such that (V =π?E,π?F H ,X,ωu )is a solution of Strominger’s system.
Proof.Because we assume that E is a stable bundle over S with degree 0with respect to the Calabi-Yau metric ωS ,according to the Donaldson-Uhlenbeck-Yau theorem,there is a unique Hermitian-Yang-Mills metric H on E up to constant such that the curvature F H of metric H satis?es
F 2,0H =F 0,2H =0,F H ∧ωS =0.
So we have π?F 2,0H =π?F 0,2H =0and according to Lemma 16,we also have π?F H ∧ω2u =0.Now according to our assumption,(ω1,ω2,E )satis-?es the condition (6.13),and hence there is a function μsatisfying equa-
tion (6.14).Then we solve equation (6.15).According to Theorem 19in the next section,there exists a smooth solution u of equation (6.15).Combining equation (6.15)with (6.14),we know that u is the solution
of equation (6.6).So (π?F H ,ωu )satis?es equation (2.3).On the other hand,according to Lemma 12,the metric ωu =e u ωS +√?12θ∧ˉθon X sat-is?es equation (2.4).Thus we have proven that (V =π?E,π?F H ,X,ωu )
satisfy all equations of Strominger’s system.q.e.d.
7.Solving the equation
As above section,we let
ρ=?√?1tr(ˉ?B
∧?B ?·g ?1).In this section,we want to prove
Theorem 19.The equation
(7.1)√?1?ˉ?e u ∧ωS ?√?1α 2?ˉ?(e ?u ρ)?α 2?ˉ?u ∧?ˉ?u +μω2S 2!=0has a smooth solution u such that
ω =e u
ωS +α 2e ?u ρ+α √?1?ˉ?u de?nes a Hermitian metric on S .