Invariant complex structures on S3× S3 Текст научной статьи по специальности «Математика»

3 3

Invariant complex structures on S x S

Daurtseva N.A. (natali0112@ngs.ru)

KemSU

1. Introduction

Consider 6-dimensional manifold S3 x S3. The construction of complex structures on the product of even-dimensional spheres, is known [1]. Its belong to Calabi and Eckmann and

induced by structure of Hopf bundle S 2"+1 ———> CPn. Recall this construction. The base and

the fiber of the bundle S3 x S3 ———> CP1 x CP1 are complex. CP1 admits two complex structures J1X=iX and

J.iX= - iX, tXe CP1, and fiber T2 =S1 x S1 admits two-parametric family of complex structures:

Ja,c Ux= Vx ,

c c 2 . 2

J v=-a +c Tj-aV

<Ja,c ' x x ' x ,

cc

where a and c are real, c^0; Uand V are vector fields on S3xS3, associated with action S1 on the first and second multipliers S3x S3, respectively. If complex structures on the base and fiber are fixed, one can choose holomorphic transition functions and obtain the complex structure on S3x S3. If the base has complex structure (J1, J1), the fiber has Ja,c , we will denote the corresponding structure as Ja,c too.

One can consider S3x S3 as homogeneous space U(2)/U(1)*U(2)/U(1). It is known that all the complex structures, constructed above are invariant with respect to the action of U(2)*U(2). Moreover, all U(2)xU(2)- invariant almost complex structures are depleted by above. On the other hand, one can consider S3x S3 as Lie group SU(2)*SU(2). There are more SU(2)xSU(2) - invariant almost complex structures then previous ones. Namely, left-invariant almost complex structure corresponds to each linear endomorphism

I : su(2)xsu(2) ^ su(2)xsu(2), I2= -1 Therefore, family of SU(2)x SU(2) - invariant almost complex structures on S3xS3 is 18-parametric, and of U(2)xU(2) - invariant is 2-parametric. So the question emerges, if there exists some complex structure in this 18 - parametric space, which differs with known ones. There exists result with respect to homogeneous compact complex manifolds.

Theorem (Grauert H.,Remmert R. [2]). Each compact homogeneous manifoldMis holomorphic bundle over the homogeneous projective algebraic manifold V wit complex-parallelizable fiber F.

Clear, that construction of bundle described above satisfies to the theorem conditions, but it is not obvious, if there exists another structure of holomorphic bundle for S3x S3 with such properties. In this paper author has proved.

Theorem. Class of the left-invariant complex structures on SU(2)xSU(2) is depleted by complex structures, generated by Hopf bundle.

2. Proof of theorem.

- n 3„

Let J is left-invariant almost complex structure on SU(2)*SU(2). As su(2)xsu(2)= R R23, it can be unique determined by the linear endomorphism

I : R13xR23^ R13XR23,

I z= -1

Let (e1, e2, e3, e4, e5, e6) is standard basis of R1 xR2 . In this basis matrix of I is of view:

I=

ÍA B ^ C D

2 a3 ^ Í ^ ¿2 ¿3 ^ Í c u 1 c2 C ^ í dj d2 d 3 N

where A= a4 a5 a6 , B = ¿4 ¿5 ¿6 , C= C4 c5 C6 , D= d 4 d5 d 6

V a7 a8 a9 j V ¿7 ¿8 ¿9 j V C7 c8 c9 j v d 7 d8 d9 j

Lemma 1 Each left-invariant almost complex structure of SU(2) x SU(2) is integrable if and only if its matrix has a view:

i ( A 0 ^ I=AIqA , A= 1 ,

I 0 A2 J

In the basis (e1, e2, e3, e4, e5, e6), where I0e1= e4, I0e2= e3, I0e5= e6, and A¡e SO(3), i=1,2.

Proof: Assume, that there exists left-invariant complex structure which is difference with described ones in lemma. Let I is such complex structure, assume that det CjQ. Matrix for I, in the standard basis is:

'A - (A2 + 1)C

I =

vC - CAC- j

Write some needed for proof conditions, emerging from integrability requirement of I:

[IX, IY]- [X, Y]- I[IX, Y]- I[X, IY] = 0 c4c8 - c5c7 - a1c3 + a7c1 - a5c3 + a8c2=0, c4c9 - c6c7 + aic2 - a4ci - a6c3 + a9c2=0, c5c9 - c6c8 + a2c2 + a3c3 - a5c1 - a9c1=0, c2c7 - c1c8 - a1c6 + a7c4 - a5c6 + a8c5=0, c3c7 - c1c9 + a1c5 - a4c4 - a6c6 + a9c5=0, c3c8 - c2c9 + a2c5 - a5c4 + a3c6 - a9c4=0, a4a8 - a5a7 - a1a3 + a1a7 - a3a5 + a2a8=0, a2a7 - a1a8 - a1a6 + a4a7 - a5a6 + a5a8=0,

1. 2.

3.

4.

5.

6. 7.

9.

a1a5 - a2a4 - 1 - a1a9 + a72 - a5a9 + a82=0,

10. a4a9 - a6a7 + a1a2 - a1a4 - a3a6 + a2a9=0,

22

11. a3a7 - a1a9 + 1 + a1a5 - a4 + a5a9 - a6 =0,

12. a1a6 - a3a4 + a1a8 - a4a7 - a6a9 + a8a9=0,

13. a5a9 - a6a8 - 1 - a1a5 + a22 - a1a9 + a32=0,

14. a3a8 - a2a9 + a2a5 - a4a5 - a4a9 + a3a6=0,

15. a2a6 - a3a5 + a2a8 - a5a7 - a7a9 + a3a9=0.

By assumption det C^0, the first string of matrix C has even if one non-vanishing element, let it be c1. Then for vectors

1

Vi=

c c — c c

4 9 6 7

A (

- aA

a1 + a9

- a.

0 0 0

V2=

c5c9 c6c8

a5 a 9

1 ( c 4 c8 c5 c7 + a7 ^

a 2

a3 0

0

0

V3=

c

1

a8

- a1 - a5 0

0

0

the fourth coordinate of vectors Iv,, Iv2, Iv3 is vanish, it follows from conditions 1, 2, 3:

/V1=

a1 (c4c9 c6 c7 )

+ a6 a7 - a4a9

1 ( a1 (c5 c9 c6 c8 )

a4(c4c9 c6c7) 1

I 1 a9 a3 1

a7(c4c9 c6 c7)

c1

0

c4(c4c9 - c6 c7)

c1

c7 (c4c9 c6 c7)

c1

IV

I a3 a4 a1 a6

I c1 c 9 c 3 c 7

+ c3 c 4 c1c6

IV2=

\

I 1 I a6 8 a 5 a 9

a4(c5c9 c6 c8)

c1

a7(c5c9 -c 6 c8)

c1

0

c4 (c5 c9 -c6 c8)

c 1

c7 (c5 c9 -c 6 c8)

I a2a9 a3a8

I a3 a5 a2 a6

I c2c9 c3c8

+ c3 c5 c 2 c6

a1 (c4c8 c5 c7)

a4 (c4c8 c5c7 )

I a5 a7 a4a8

I a1 a8 a2 a7

a7(c4c8 c5 c7)

+1 + a2 a4 - a1 a5

0

c4 (c4c8 - c5c7 )

c7(c4c8 c5 c7)

I c1c8 c2c7

+ c 2 c 4 c1c5

Let n2 : R1 x R2 ^ R2 , then vectors n2(Iv1), n2(Iv2), n2(Iv3) are linearly dependent. Obviously, that from det C ^ 0 it follows, that n2(/v;)^0, so there exist X1, k2, X3 not equal to zero simultaneously, such that X1n2(Iv1)+ l2n2(Iv2)+ X3n2(Iv3)=0. Let X2j0 and

a=

C4(c4C9 C6 C7)

+ c1c9 - c3c7, b=

c4(c54C9 C6C8)

+ c2c9 - c3c8, c=

c4(c4c8 C5C7)

+

„„ „„ J_C7(C4C9 C6____u7 V^5^9

c^- c2c7, d--

c7(c5c9 c6 cg)

+ c3c4 - c1c6, e=-+ c3c5 - c2c6,

on(o4oq O5on )

-+ c2c4 - c1c5. Therefore, system

c,

(*)

has to have non-zero solution.

[â1 a + Ä2 b + A3 c = 0, IA1 d + A2 e + A3 f = 0

c

c

c

c

c

c

c

c

3

c

c

c

C

C

C

1

c

C

det

a c

d f.

(-c2 - c42 - c2)det C

t0, if det C t 0;

r.1 ] us j

¿2 f bf -

af - cd

ce

\

ae - bd,

bf - ce= - det C

C i C 2 I C 4 C 5 I C 7 C 8

_ C1C 2 + C 4 C5 + C7 C8 ]

---2 , 2 , 2-

c1 + c 4 + c7

c

C

ae

bd= det C ClCs + C 4 C6 + C7 C9

_ C1C3 + C4 C6 + C 7 C9

/3 2-2-2—

Ci + C 4 + C7

¿2-

Hence Tr2(I(X1v1+ l2v2+ a3v3))=0, this possible in only two cases: det C=0 and n1(l1v1+ l2v2+ l3v3) is proper vector for C, corresponding to zero proper value, or A1v1+ l2v2+ a3v3=0. As det C ^ 0, to A1v1+ l2v2+ a3v3=0. This case has a view:

(a4 - a2 )(c1c2 + c4 c5 + c7 c8) + (a7 - a3 )(c1c3 + c4 c6 + c7 c9 ) = 0,

(a4 - a2 )(-c2 - c4 - c2 ) + (a8 - a6 )(c1c3 + c4c6 + c7c9 ) = 0,

(a7 - a3 )(-c2 - c4 - c2) + (a8 - a6 )(-c1c2 - c4c5 - c7c8 ) = 0.

Therefore we obtain

(a8 -a6)(c,c3 + c4c6 + c7c9)

,-y ,-Y — v 8_6/V 13 46 7 9 / _

( ) a4-a2--------, a7-a3---

c2 + c4 + c72

Next, as det C ^ 0, then the second string of matrix has to have a non-vanishing element, let it be c5 (we can choose c4 or c6, it will led to the same results). As analogous one can define three vectors u1, u2, u3:

(a8 - a6)(cic2 + c4c5 + c7c8)

222 c12 + c42 + c72

u1=

-a

a1 + a9 +

c3c 7 - c1c9

- a 6 0

0

0

f a 5 9

u2=

a 2 +

c3 c8 c 2 c9

a 3 0

0

0

u3=

a8 +

a7

c2 c7 - c1c8

- a1 - a5

0 0

Vectors n2(Iu1), n2(Iu2), n2(Iu3) have zero second coordinate, it follows from conditions 4, 5, 6. It means, that these vectors are linearly dependent. By the same reasons as above one can obtain:

_ (a7 - a3)(cjc2 + c4c5 + c7c8) _ (a7 - a3)(c2c3 + c5c6 + c8c9) a8 - a6=--------, a4 - a-=--

222 c 2 + c5 + c8

222 c 2 + c5 + c8

Combine this equalities with (**) we obtain:

a8 -a6=

(a8 - a6)(c1c2 + c4c5 + c7c8)

(c2 + c 42 + c 72)(c 22 + c52 + c82) 22

This is possible only if a8 =a6 or (c1 +c4 +c7 )( c2 +c5 +c8 )=(c1c2+ c4c5+ c7c8) .

Let a8 =a6, then the equalities a2 =a4, a3 =a7 are follows. Conditions 7-15 became the following:

7. a2a6=a3a5,

8. a2a3=a1a6,

9. a1a5 - a22-1-a1a9+a32-a5a9+a62=0,

10. a2a9=a3a6,

11. -a1a9+a32+1+a1a5-a22-a62+a5a9=0,

12. a1a6=a2a3,

C

c

c

c

5

5

0

13. a5aa9-a(^-1-a1a5+ a22+a32-a1a9=0,

14. a2a9=a3a6,

15. a2a6=a3a5.

Express a22, a32, a62 from 9, 11 and 13 we obtain: a22=a1a5+1, a32=a1a9+1, a62=a5a9+1. Square the both parts of the equations 7, 8 and 10, then subtract 8-th from 7-th, 10-th from 8-th and 10-th from 7-th, we obtain the equation system:

(a5 - a9 )(a5 + a9 - a1) = 0, (a5 - a1 )(a5 + a1 - a9) = 0, (a9 - a1 )(a9 + a1 - a5) = 0,

The general solutions of 7-15 have to satisfy to this system. System (***) has following solutions:

1. a1=a5=a9, then a22= a32= a62= a12+1, for such values the equation 12 has no solutions;

2. a5=a9, a1=0, then a62= a52+1, a22= a32=1, for such values condition number 8 is not

hold;

3. a1=a5, a9=0, then a22= a12+1, a32= a62=1, condition number 10 is not hold;

4. a1=a9, a5=0, then a32= a12+1, a22= a62=1, condition number 9 is not hold.

So, when a6=a8 the corresponding almost complex structure is not integrable.

2 2 2 2 2 2 2

Let (C1C2+ C4C5+ C7C8) =(C1 + C4 + C7 )( C2 + C5 + C8 ). If «1=(C1, C4, C7), n2=(C2, C5, C8),

then we write this condition as: (n1, n2)2=( n1, n1)( n2 , n2), this is possible if and only if cos2 Z (n1; n2)=1 or n1=0 or n2=0. In all this cases det C=0.

Therefore condition det C=0 is necessary for integrability of I.

Now, as det C=0, there exists v1gR13x{0}, such that C^1(v1)=0, then n2(I v1)=0.

As I2= -1 then Iv1 and v1 are not linear dependent, hence rankC<1. If C=0, then from I2= -1 it follows that A2= -1, det2 A= -1. So C^0. It follows that rankC=1.

As rankC=1, there exists non-zero element of matrix C. Without lost of generality we can think that it is c1. Then v2=(-c2, c1, 0, 0, 0, 0), v3=(-c3, 0, c1, 0, 0, 0) form basis of 2-dimensional subspace in R13x{0}, invariant with respect to I. Let try to understand how integrable almost complex structure acts on the vector v1=(c1, c2, c3, 0, 0, 0), which is orthogonal to < v2, v3>, with respect to standard metric in R13xR23.

(Iv1, v2) = —2 (a1c1+ a2c2+ a3c3)+ a4c1+ a5c2+ a6c3=[by conditions 3 and

2]= — (c2(c5c9-c6c8)+ c1(c4c9-c6c7))=0, as rankC=1.

By analogous (Iv1, v3) =0.

Discuss as above, one can show that det B=0, rankB=1 and there exists 2-dimensional plane < v5, v6>e {0}xR23 , which is invariant with respect to I. Also, one can show that for vector v4e {0}x R23, orthogonal to this plane the Iv4 is orthogonal to < v5, v6>. Therefore, Iv1=a1v1+^1v4+y1v5+¿1v6, Iv4=a2v1+p2v4+y2v1+S2v2, for some a¡,y¡, ¿¡gR, /'=1,2. Using condition I2=-1 we can show that y1=S1= y2=S2=0. So vectors v1, v4 form I-invariant plane.

We don't know, how integrable almost complex structure acts on the invariant planes <v2, v3>, < v5, v6>. Let M1=v1/||v1||, u4= v4/||v4||, u2, u3 is basis of <v2, v3>, u5, u6 is basis of <v5, v6>, such that (u1,

u2, u3, u4, u5,

u6) form orthonormal basis and [u2,u3]=u1, [u1,u2]=u3, [u1,u3]=-u2, [u5,u6]=u4, [u4,u5]=u6, [u4,u6]=-u5. The matrix of almost complex structure, satisfying to the above that such almost complex structure integrable. Therefore, if A1 is transition matrix from ^1(^1), ^1(^2), ^1(^3) to TC1(u1), TC1(u2), n1(u3), and A2 from ^2(^4), ^2(^5), ^2(^6) to n^u),

n2(u5), n2(u6), A1, A2eSO(3) then matrix of integrable almost complex structure in the

standard basis is I=AIa CA'\ A=

( Aj 0 ^

v 0 A2 j

All almost complex structures from lemma 1 one can obtain from the construction described in the introduction. So the theorem follows from lemma 1.

Corollary. Integrable almost complex structures form 6-parametric family.

( 1 0 ^

Proof: Fix parameters a and c. Operator A doesn't change Ia,c if A¿ is: ,

V0 J

O¡eSO(2). So, choice of vectors u1, u2, u3, u4, u5, u6 is defined by four parameters.

References:

[1] Kobayashi S., Nomizu K. Foundations of differential geometry. Vol.2, Interscience Publishers. New York, London, 1969

[2] Grauert H., Remmert R. Über kompakte homogene komplexe Mannigfaltigkeiten.-Arch.Math., 1962, 13, S. 498-507