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Research Group ‘Global Analysis’

* * * * * *

SO(3)ir-geometries in dimension five and seven – results and open problems

Thomas Friedrich (Humboldt-Universit¨at zu Berlin) Srn´ı Winter School, January 2008

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Introduction

Fix a subgroup G ⊂ SO(n) and consider a Riemannian manifold (Mn, g, R) equipped with a G-structure R.

Examples:

G = U(n) ⊂ SO(2n) −→ almost hermitian geometry.

G = U(n) ⊂ SO(2n + 1) −→ contact geometry G = G2 ⊂ SO(7) −→ G2-geometry in dimension 7

First Question: Does there exist a metric connection ∇c preserving the structure R such that the torsion

Tc(X, Y, Z) := g(∇cXY − ∇cY X − [X, Y ], Z) is totally skew-symmetric ? → characteristic connection.

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Second Question: Study the curvature and the spin geometry of the new connection ∇c.

In particular, we look for solutions of type II string equations involving a spinor field (super-symmetry) Ψ and a non-trivial ‘B-field’ Tc:

Ricc = λ · g , δ(Tc) = 0 ,

cΨ = 0, Tc · Ψ = µ · Ψ.

Approach: We construct in a systematic way solutions in type II superstring theory starting from non-integrable geometric structures.

Results: In dimension 5 ≤ n ≤ 8 for contact, almost hermitian, G2- and Spin(7)-geometries (Agricola, Friedrich, Ivanov, . . . ).

Reference: the ’Srni lectures’ of Ilka Agricola, 2006.

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In dimension n = 5 one usually considers contact geometries; they are related to the subgroup U(2) ⊂ SO(5) (Fr/Ivanov, Crelle J., 2004).

New Geometries: Consider the subgroup SO(3)ir ⊂ SO(5). Study the geometry of 5-manifolds with such a structure. Apply the described method to them.

First Results:

• M. Bobienski, P. Nurowski: started this program during their stay in Berlin (Crelle J., 2006)

• S. Chiossi, A. Fino: SO(3)ir-structures on Lie groups (J. Lie Th., 2007)

Aim of this lecture: discuss some topological and geometric problems for SO(3)ir-structures in dimensions n = 5,7.

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SO(3)

ir

-structures in dimension five

The group SO(5) contains two subgroups isomorphic to SO(3),

SO(3)st ⊂ SO(5), SO(3)ir ⊂ SO(5) .

The subgroup SO(3)ir is the SO(3)-action on S02(R3) = R5. The generators of the Lie algebra so(3)ir ⊂ so(5) are

X1 = e13 + √

3e23 + e45, X2 = 2e14 + e35 X3 = −e15 + √

3e25 − e34 .

Question: Under which conditions a compact oriented 5-manifold M5 admits an SO(3)st- or an SO(3)ir-structure ?

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First case: SO(3)st-structures

In order to formulate the condition, we need some invariants.

Definition: The semi-characteristics (Kervaire) are defined by

k(M5) :=

2

X

i=0

dimR H2i(M5;R)

mod 2 ,

ˆ

χ2(M5) :=

2

X

i=0

dimZ2 Hi(M5;Z2)

mod 2 .

Theorem:(Lusztig-Milnor-Peterson 1969)

k(M5) − χˆ2(M5) = w2(M5) ∪ w3(M5) . In particular, if M5 is spin, then k(M5) = ˆχ2(M5).

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Theorem: A compact, oriented 5-manifold admits an SO(3)st-structure (i.e. two vector fields) if and only if

w4(M5) = 0 , k(M5) = 0 .

Proof: E. Thomas in 1967 for spin manifolds (w4(M5) = 0 = ˆχ2(M5)), M.F. Atiyah in 1969 for the general case.

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Second case: SO(3)ir-structures

Example 1: M5 = SU(3)/SO(3) has an SO(3)ir-structure.

Some topological properties of this space:

• M5 is simply connected and a rational homology sphere.

• M5 does not admit any Spin- or SpinC-structure.

• k(M5) − χˆ2(M5) = w2(M5) ∪ w3(M5) = 1

• k(M5) = 1 and χˆ2(M5) = 0

In particular, M5 = SU(3)/SO(3) does not admit any SO(3)st-structure!

Example 2: M5 = S5 has no SO(3)st- or SO(3)ir-structure.

• M5 admits a Spin-structure

• k(M5) − χˆ2(M5) = w2(M5) ∪ w3(M5) = 0

• k(M5) = 1 and χˆ2(M5) = 1

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Example 3: Consider the subgroup H = {(A, A2), A ∈ SO(2)} ⊂ SO(3) × SO(3) as well as the homogeneous space M5 = (SO(3) × SO(3))/H. Then M5 has an SO(3)ir-structure and the following topological data:

• H1(M5;Z) = Z2 , H2(M5;Z) = Z.

• k(M5) = 0 and χˆ2(M5) = 0.

More examples: Bobienski/Nurowski (Crelle Journal 2006): there is a 2-parameter family G6(s, t) of 6-dimensional Lie groups containing SO(2) such that the isotropy representation of M5 = G6(s, t)/SO(2) is the maximal torus Tmax = {(A, A2,1), A ∈ SO(2)} ⊂ SO(3)ir ⊂ SO(5).

The groups are, for example,

G6(s, t) = SO(3) × SO(3), SO(3) × SO(1,2),

R1 × (SO(2) on R4), (SO(2) on R2) × SO(3) .

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The obstructions for SO(3)ir-structures:

The relevant space is X7 := SO(5)/SO(3)ir. Let us list some of its homotopy groups:

π1(X7) = 0 , π2(X7) = 0 , π3(X7) = Z10 , π4(X7) = Z2 .

Consequence: The obstructions for the existence of an SO(3)ir- structure on a compact 5-manifold M5 are in H4(M5;Z10) = H4(M5;Z5) ⊕ H4(M5;Z2) and in H5(M5;Z2) .

Problem: Compute the topological conditions for the existence in general.

The criterion given in Bobenski, math.dg/0601066 is wrong (w4 = 0, k = 0, p1/5 ∈ Z). The space M5 = SU(3)/SO(3) does not satisfy it

!

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Proposition 1: M5 admits an SO(3)ir-structure if and only if there exists a 3-dimensional bundle E3 such that T(M5) = S02(E3).

Proposition 2: Suppose that T(M5) = S02(E3). Then

• p1(M5) = 5 · p1(E3).

• w1(M5) = 0 and w4(M5) = 0.

• w2(M5) = w2(E3) and w3(M5) = w3(E3).

Corollary: If M5 admits an SO(3)ir-structure, then

• w4(M5) = 0 in H4(M5;Z2) ;

• p1(M5)/5 ∈ H4(M5;Z) is integral.

Conjecture: M5 admits an SO(3)ir-structure if and only if p (M5)

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SO(3)

ir

- and G

2

-structures in dimension seven

The real, irreducible 7-dimensional representation S03(R3) = R7 of SO(3) yield an embedding SO(3)ir ⊂ G2 ⊂ SO(7).

The sub-algebra so(3)ir ⊂ g2 ⊂ so(7) is given by

X1 =

r1 5

e12 + 2e34 − 3e56

X2 = − r6

5 e27

r1

2 e14 +

r1

2 e23 +

r 3

10 e35

r 3

10 e46

X3 = − r6

5 e17

r1

2 e13

r1

2 e24 +

r 3

10 e36 +

r 3

10 e45 .

The group G2 preserves the 3-form

ω3 = e567 + e347 + e127 − e146 + e135 − e245 − e236 .

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The following formula proves the inclusion SO(3)ir ⊂ G2 :

∗ X1 ∧ X1 + X2 ∧ X2 + X3 ∧ X3

= − 6

5 ω3 .

Theorem: SO(3)ir ⊂ SO(7) is the stabilizer of two symmetric tensors in S4(R7). The first polynomial is (x21 + . . . + x27)2 and the second polynomial is given by the formula

4

15x41 + 4

15x42 + 120x32x5 120x31x6 + 60x1`

10x3x4x5 + (6x22 + 5x23 5x24)x6 (4x2x4 + 2

15x4x5 + 2

15x3x6)x7´

60x2`

5x23x5 5x24x5 10x3x4x6 + 2

15x3x5x7 2

15x4x6x7´ +

15`

25x43 + 25x44 30x24x27 + 10x23(5x24 3x27) + 9x27(10x25 + 10x26 + x27)´ + 2x21`

4

15x22 + 10

15x23 180x2x5 60x3x7 +

15(10x24 + 30x25 + 30x26 + 9x27)´ +2x22`

10

15x23 + 60x3x7 +

15(10x24 + 30x25 + 30x26 + 9x27)´

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Some consequences:

• An SO(3)ir-structure on a Riemannian 7-manifolds is defined by a special symmetric tensor (polynomial) of degree 4.

• Any SO(3)ir-structure on a Riemannian 7-manifold (M7, g) induces a unique G2-structure ω3.

• There are 4 basic classes W1,W2,W3,W4 of G2-structures on a 7- dimensional Riemannian manifold (Fernandez/Gray 1982). They are given by the G2-components of the representation R7 ⊗ (so(7)/g2).

• A G2-manifold admits a characteristic connection if and only if it is of type W1 ⊕ W3 ⊕ W4 (d ∗ ω3 = θ ∧ ∗ω3 – Fr/Ivanov 2002).

• Denote by Vk the real, (2k + 1)-dimensional, irreducible representation of SO(3). The SO(3)ir-representations decompose into

W1 = V0 , W2 = V2 ⊕ V4 , W3 = V2 ⊕ V4 ⊕ V6 , W4 = V3 .

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Theorem: There is a bijection between

• SO(3)ir-structures admitting a characteristic connection;

• G2-structures of type W1⊕W3⊕W4 such that the holonomy hol(∇c) ⊂ so(3)ir of their characteristic connection is contained in so(3)ir.

Problem: Construct 7-dimensional Riemannian manifolds with SO(3)ir- structure such that the underlying G2-structure is of type

W1 ⊕ W3 ⊕ W4 = V0 ⊕ V2 ⊕ V3 ⊕ V4 ⊕ V6 .

Can any Vα-type be realized ?

An equivalent formulation: Construct G2-structures on Riemannian 7- manifolds with characteristic connection such that hol(∇c) ⊂ so(3)ir and of a fixed Vα-type.

Remark: A parallel G -manifold (i.e. ∇c = ∇g,Tc = 0) cannot have a

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Theorem:

A compact, 7-dimensional SO(3)ir-manifold of type W4 = V3 and hol(∇c) ⊂ so(3)ir , Tc 6= 0 does not exist. Equivalently, a compact G2-manifold of type W4 and hol(∇c) ⊂ so(3)ir , Tc 6= 0 does not exist.

Sketch of the proof:

Up to a conformal change of the metric, the universal covering splits into Y 6 × R1, where Y 6 is a nearly K¨ahler manifold (see Agricola/Friedrich, J. Geom. Phys. 2006). Then we conclude that

hol(∇c) ⊂ su(3) ∩ so(3)ir = so(2) .

In particular, Y 6 is nearly K¨ahler with a reduced, 1-dimensional characteristic holonomy, a contradiction (Belgun/Moroianu, Ann. Glob.

Anal. Geom. 2002).

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The basic example: X7 = SO(5)/SO(3)ir . Geometric properties of the basic example:

• X7 admits an SO(3)ir-structure. It is not symmetric.

• The underlying G2-structure is of type W1 (nearly parallel). In particular, it realizes the type V0. Moreover, X7 admits one real Killing spinor. This spinor field is ∇c-parallel.

Theorem: (TF 2006)

X7 is the unique G2-manifold of type W1 ⊕ W3 ⊕ W4 such that

hol(∇c) ⊂ so(3)ir , ∇cTc = 0 , Tc 6= 0 .

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Topological properties of the basic example:

• X7 is simply connected and a rational homology sphere,

H1(M7;Z) = 0 , H2(M7;Z) = 0 , H3(M7;Z) = Z10 , H4(M7;Z) = 0 , H5(M7;Z) = 0 , H6(M7;Z) = 0 .

• k(X7) = 1 and χˆ2(X7) = 0 .

• All Stiefel-Whitney classes wi(X7) and the Pontrjagin class p1(X7) are trivial.

• X7 is not parallelizable, but admits 4 vector fields (Goette/Kitchloo/Shankar 2002). In particular, there is no further reduction of the frame bundle to a subgroup of SO(3)ir.

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Existence of SO(3)

ir

-structures in dimension seven

Problem: Study the conditions for the existence of a topological SO(3)ir- structure on a compact 7-manifold. The relevant space is

Z18 = SO(7)/SO(3)ir .

The homotopy groups of Z18:

π1(Z18) = Z2 , π2(Z18) = Z2 , π3(Z18) = Z28 , π4(Z18) = 0 , π5(Z18) = Z2 , π6(Z18) = Z2 .

Consequence: Let M7 be a compact, oriented 7-manifold. Then the obstructions for the existence of an SO(3)ir-structure are in

H2(M7; ) , H3(M7; ) , H4(M7; ) , H6(M7; ) , H7(M7; ) .

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Proposition: M7 admits an SO(3)ir-structure if and only if there exists a 3-dimensional bundle E3 such that T(M7) = S03(E3).

Now we compute again characteristic classes.

Theorem: Let F7 be a real, oriented, 7-dimensional vector bundle over some space Y and suppose that there exists a 3-dimensional bundle E3 such that F7 = S03(E3). Then we have:

• w2(F7) = w3(F7) = w5(F7) = w7(F7) = 0 ;

• w4(F7) = w22(E3) and w6(F7) = w32(E3) ;

• p1(F7)/14 = p1(E3) is an integral cohomology class and p1(F7)/14 = w4(F7) mod 2.

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If F7 = T(M7) is the tangent bundle of some 7-manifold, then w1(M7) = w2(M7) = w3(M7) = 0 implies the vanishing of w4(M7) and w6(M7) (use the Wu formulas !).

Corollary: Let M7 be an oriented, compact 7-manifold. If it admits an SO(3)ir-structure, then

• all Stiefel-Whitney classes are trivial.

• The Pontrjagin class p1(M7) is divisible by 28.

Remark: The criterion is only necessary, but not sufficient. Again the highest obstruction in H7(M7;Z2) is missing.

Remark: This obstruction cannot be χˆ2(M7). Indeed, the sphere S7 is parallelizable and χˆ2(S7) = 1. On the other hand, SO(5)/SO(3)ir admits an SO(3)ir-structure, but χˆ2(SO(5)/SO(3)ir) = 0. However, both manifolds have k = 1.

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Bundles over twistor spaces

Let X4 be a compact, oriented 4-manifold such that

w3(X4) = 0, 6σ(X4) − 2χ(X4) ≡ 0 mod 28

holds. Denote by Z6 its twistor space and consider a principal S1-bundle M7 → Z6. Then all Stiefel-Whitney classes of M7 vanish and the Pontrjagin class p1(M7) is divisible by 28.

Remark: The condition 0 = w3(X4) ∈ H3(X4;Z2) = H1(X4;Z2) is satisfied for example if

• X4 is simply connected.

• X4 is a spin manifold (apply Wu’s formula !).

Example: Consider X4 = A · (S1 × S3) #B ·(S2 × S2). X4 is spin and σ(X4) = 0, χ(X4) = 2(B + 1 − A).

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Example: Consider X4 = CP2 #k · (−CP2) = blow up of CP2 in k points. X4 is simply connected and

σ(X4) = 1 − k, χ(X4) = 3 + k, 6σ(X4) − 2χ(X4) = −8k .

In particular, del Pezzo surfaces (k = 7) satisfy the necessary conditions.

These surfaces admit K¨ahler-Einstein metrics with positive scalar curvature.

More example: X4 = A · (S2 × S2) # B · (K3) , X4 = A · (S1 × S3) # B · (K3) , . . .

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Construction of T

max

⊂ SO(3)

ir

-structures

The algebraic fact

Tmax = diag(A, A2, A3,1) = SO(3)ir ∩ SU(3) ⊂ G2 , A ∈ SO(2)

yields the following

Theorem: Let M5 be a 5-manifold with an Tmax = {(A, A2,1), A ∈ SO(2)} ⊂ SO(3)ir ⊂ SO(5)-structure R and denote by ρ the 2- dimensional representation of Tmax given by ρ(A) = A3 , A ∈ SO(2).

Then M7 := R ×ρ R2 admits an Tmax ⊂ SO(3)ir ⊂ SO(7)-structure.

M7 is a complex vector bundle over M5.

Remark: In a similar way we can consider 5 manifolds with an {(A, A3,1), A ∈ SO(2)} ⊂ SO(5)- or an {(A2, A3,1), A ∈ SO(2)} ⊂ SO(5)-structure, Then a vector bundle M7 over M5 again admits an Tmax ⊂ SO(3)ir ⊂ SO(7)-structure.

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Construction of SO(3)

ir

-structures via twistor theory

The algebraic fact

Tmax = diag(z, z2, z3) = SO(3)ir ∩ SU(3) ⊂ G2

yields the following

Proposition: Let Y 6 be a 6-manifold such that its tangent bundle splits T Y 6 = (E ⊕ E2 ⊕ E3)R, where E is a complex, 1-dimensional bundle.

Then any S1-bundle M7 → Y 6 admits a topological Tmax ⊂ SO(3)ir- structure.

Consider a compact spin 4-manifold X4. Then the twistor space Z6 = P(S) is the projective spin bundle and there exists the tautological bundle H → Z6. The tangent bundle is given by

T(Z6) = Tv ⊕ Th, Tv = H2 .

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Ansatz: Th = H1 ⊕ H3.

Then any S1-bundle over Z6 admits a Tmax ⊂ SO(3)ir-structure.

Theorem: If Th = H1 ⊕ H3, then 9σ(X4) = 10χ(X4).

Conversely, if 9 σ(X4) = 10χ(X4), then

• c1(Th) = c1(H1 ⊕ H3), c2(Th) = c2(H1 ⊕ H3).

Moreover, the following conditions are equivalent:

• Th splits into H1 ⊕ H3.

• Th splits into the sum of two line bundles.

Remark: If 9 σ(X4) = 10χ(X4) = 0, then Z6 is parallelizable.

Consequently, the interesting case is 9 σ(X4) = 10χ(X4) 6= 0

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Problem to handle: Suppose that 9 σ(X4) = 10χ(X4) 6= 0. Then we have to decide whether or not the complex 2-dimensional bundle Th over Z6 splits. These bundles are basically given by the homotopy classes

[Z6 , P(H)] = [Z6 , S4] Theorem: (Steenrod classification theorem)

Let Z6 be a compact 6-dimensional manifold and consider two SU(2)- principal fiber bundles with the same Chern class, c2(P1) = c2(P2) ∈ H4(Z6 ; Z). Then there exist a cohomology class

δ(P1, P2) ∈ H5(Z6 ; Z2)/Sq2 H3(Z6 ; Z2)

such that δ(P1, P2) = 0 if and only if P1 and P2 are isomorphic over Z6 − {point}. In this case, the last obstruction to P1 = P2 over Z6 is in H6(Z6 ; π5(SU(2)) = H6(Z6 ; Z2) = Z2.

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Corollary: Let X4 be a compact, oriented 4-manifold such that

• X4 is spin.

• 9 σ(X4) = 10 χ(X4) 6= 0.

Then the twistor space Z6 of X4 is not parallelizable. The tangent bundle splits into

T(Z6) = H1 ⊕ H2 ⊕ H3 . over the 4-skeleton of Z6

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Examples:

The spaces

X4 = 20 · (S1 × S3) # 5 · (K3) and

X4 = (1 + A) · (S1 × S3) # A · (S2 × S2) satisfy the condition 9 σ(X4) = 10 χ(X4).

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