Geometric Aspects of Quantum Condensed Matter

November 06, 2013 ——————————————————————————– Lecture IV y The Category of Vector Bundles y Operations on Vector Bundles ——————————————————————————–

Giuseppe De Nittis Department Mathematik (room 02.317)

¨ +49 (0)9131 85 67071 @ [email protected] W gdenittis.wordpress.com/courses/ Outline

1 The Category of Vector Bundles Recommended Bibliography Structure of Isomorphisms Triviality and Global Frames Pullback Maps

2 Operations on Vector Bundles Functorial Operations on Vector Spaces Construction of New Vector Bundles Ker and Im Vector Bundles Introduction to Vector Bundles Theory

“Classical” monographies:

[At] Atiyah, M. F.: K -Theory. Benjamin Inc., 1967

[Ha] Hatcher, A.: Vector Bundles and K -Theory. 2009 http://www.math.cornell.edu/~hatcher/#VBKT

[Hu] Husemöller, D.: Fibre Bundles. Springer, 1994

[Ka] Karoubi, M.: K -Theory: An Introduction. Springer, 1978

[LM] Luke, G. & Mishchenko, A. S.: Vector Bundles and Their Applications. Kluwer Ac. Pub., 1998

[MS] Milnor, J. W. & Stasheff, J. D.: Characteristic Classes. Princeton U. Press, 1974

... and many others ... , Outline

1 The Category of Vector Bundles Recommended Bibliography Structure of Isomorphisms Triviality and Global Frames Pullback Maps

2 Operations on Vector Bundles Functorial Operations on Vector Spaces Construction of New Vector Bundles Ker and Im Vector Bundles We recall that:

(i)A vector bundle morphism (ˆf ,f ) is a fibre preserving map that is linear on each fibre.

ˆf E 0 - E

π0 π (π ◦ˆf = f ◦ π0) ? ? Y - X f

ˆ 0 K such that f : Ey → Ef (y) is -linear. y ˆ (ii)A(vector bundle) X-morphism is given by (f ,IdX ) ˆf E 0 - E

0 π (π ◦ˆf = π ) 0 π -  X The category of vector bundles over the field K, denoted VecK, has as its objects vector bundles and as its arrows vector bundle morphisms. Composition is composition of morphisms of vector bundles.

For each space X, let VecK(X) denote the subcategory of vector bundles over X and X-morphisms.

m For each m ∈ N let VecK denote the full subcategory of rank m vector bundles (over K).

m Finally, the subcategory VecK(X) of rank m vector bundles over X is m the intersection VecK(X) ∩ VecK.

We recall that an isomorphism between vector bundles (E 0,π0,X) and (E ,π,X) is a pair of X-morphisms ˆf : E 0 → E and gˆ : E → E 0 such that ˆ ˆ f ◦ gˆ = IdE 0 and gˆ ◦ f = IdE .

m We “abuse” of the symbol VecK(X) to denote equivalence classes (modulo isomorphisms) of rank m vector bundles over X. Moreover, ∞ m VecK(X) = ∪m=1VecK(X)

We anticipate that, given a map f : Y → X then the pullback of this map ∗ f : VecK(X) → VecK(Y ) provides a a (contravariant) functor. THEOREM ˆ Let f be a X-morphism between two rank m vector bundles (E1,π1,X) and ˆ ˆ (E2,π2,X). Then f is an isomorphism if and only if f |x is a vector space −1 −1 isomorphism between the fibers π1 (x) and π2 (x) for eachx ∈ X.

Proof. (⇒) It is immediate because if ˆf is an isomorphism then there exists the ˆ−1 ˆ −1 inverse gˆ = f and by definition gˆ|x = (f |x ) . ˆ −1 (⇐) Let gˆ|x := (f |x ) and define a function gˆ : E2 → E1 by requiring that ˆ ˆ −1 ˆ ˆ g(p) := g|x (p) if p ∈ π2 (x). The function g is clearly an inverse of f but we need to prove the continuity. Let U ⊂ X be an open neighborhood of x ∈ X ˆ Km −1 and hj : U × → πj (U) (j = 1,2) local trivializations. It suffices to prove ˆ −1 −1 g : π2 (U) → π1 (U) is continuous for every such U. The composition ˆ −1 ˆ ˆ m m h2 ◦ f ◦ h1 : U × K → U × K has the form (x,v) 7→ (x,Mx (v)) where m U 3 x 7→ Mx ∈ Lin(K ) is continuous. By construction the inverse relation ˆ −1 ˆ Km Km −1 h1 ◦ gˆ ◦ h2 : U × → U × has the form (x,v) 7→ (x,Mx (v)) and the −1 Km application U 3 x 7→ Mx ∈ Lin( ) is also continuous (continuity of the ˆ −1 ˆ inverse of a bijection). Then h1 ◦ gˆ ◦ h2 and hence gˆ are continuous.  Outline

1 The Category of Vector Bundles Recommended Bibliography Structure of Isomorphisms Triviality and Global Frames Pullback Maps

2 Operations on Vector Bundles Functorial Operations on Vector Spaces Construction of New Vector Bundles Ker and Im Vector Bundles Definition (k-frame) Let (E ,π,X) be a rank m vector bundle bundle. We say that (E ,π,X) has a k-frame( k 6 m) if there are {ψ1,...,ψk } ⊂ Γ(E ) k such that spanhψj (x) | j = 1,...,ki ' K for all x ∈ X.

T Example (Frame for the trivial bundle): The trivial vector bundle X × Km → X has a m-frame (global frame) defined by the canonical sections sj (x) := (x,ej ) where ej := (0,0,...,1,...,0,0) j-th coordinates.

THEOREM A rankm vector bundle (E ,π,X) is trivial (i.e. isomorphic to ... ) if and only if there exists a (global)m-frame. Proof (sketch of). (⇒) Let ˆf be the isomorphism between (E ,π,X) and the trivial m ˆ−1 vector bundle X × K → X. The compositions ψj := f ◦ sj are continuous maps such that π ◦ ψj = IdX , hence ψj ∈ Γ(E ). ˆ−1 Moreover, f |x is a linear isomorphism for each x ∈ X and so −1 spanhψ1(x),...,ψm(x)i = π (x) for all x ∈ X.

(⇐) Let {ψ1,...,ψm} ⊂ Γ(E ) be a global m-frame for (E ,π,X) and for each x ∈ X set the linear isomorphism m −1 gx : {x} × K → π (x) defined by gx (x,ej ) = ψj (x). The collection of these maps define a map gˆ : X × Km → E between total space given by gˆ(x,∑j cj ej ) = ∑j cj ψj (x) (where cj ∈ K). This map is fiber preserving and continuous (composition of continuous functions), hence a vector bundle X-morphism. Moreover, it is an isomorphism since it restricts to a linear isomorphism to each fiber.  Parallelizability of the Spheres

Let T → Sd be the tangent bundle over the d-dimensional sphere. The span of the sphere σ(Sd ), also called Radon-Hurwitz numbers (1922-23), is the maximum number of linearly independent vector fields (i.e. of sections in Γ(E )). The computation of σ(Sd ) is an “hard problem”. The solution was achieved in [Adams, J. F.: Ann. of Math 75, (1962)] using K -theory:

d = k 2γ+4δ − 1  σ(Sd ) = 2γ + 8 δ − 1 where k odd  0 6 γ < 3 .

The sphere is parallelizable if σ(Sd ) = d.

d = 1 2 3 4 5 6 7 8 9 10 11 ... σ(Sd ) 1 0 3 0 1 0 7 0 1 0 3 ...

The sphere is parallelizable only for d = 1,3,7. Outline

1 The Category of Vector Bundles Recommended Bibliography Structure of Isomorphisms Triviality and Global Frames Pullback Maps

2 Operations on Vector Bundles Functorial Operations on Vector Spaces Construction of New Vector Bundles Ker and Im Vector Bundles The pullback bundle construction applies to the category of vector bundles.

THEOREM (Pullback vector bundle-1) Let (E ,π,X) be a rank-m, K-vector bundle andf : Y → Xa ∗ map. The pullback bundle (f E ,πf ,Y ) has the structure of a a K ˆ rank-m, -vector bundle and the pair (fE ,f ) is a vector bundle morphism. Moreover, this structure is unique, andf ∗E restricts to a linear isomorphism to each fiber.

ˆ ∗ The canonical morphism (fE ,f ):(f E ,πf ,Y ) → (E ,π,X)

ˆf f ∗E E - E

ˆ πf π (π ◦ fE = f ◦ πf ) ? ? Y - X f ˆ is defined by fE (y,p) = p, for all p ∈ E . Proof (sketch of). (Vector space structure) ∗ −1 −1 The fibre of f E over y ∈ Y is πf (y) = {y} × π (f (y)). For λ ∈ K and −1 (y,p1),(y,p2) ∈ πf (y) one defines

(y,p1) + λ (y,p1) := (y,p1 + λp2) ˆ ˆ Since fE (y,pj ) = pj , the restriction fE |y is a linear isomorphism between −1 −1 −1 πf (y) and π (f (y)). This uniquely defines the structure of πf (y).

(Local triviality) ˆ m −1 Let hU : U × K → π (U) be a trivializing domain for (E ,π,X). Then we can set ˆ0 −1 Km −1 −1 hU : f (U) × → πf (f (U)) ˆ0 ˆ via hU (y,v) := (y,hU (f (y),v)). This construction provides trivializing domains for the pullback bundle.  THEOREM (Pullback vector bundle-2) Let (E ,π,X) and (E 0,π0,Y ) be two vector bundles. For a map f : Y → X, the vector bundles E 0 and f ∗E areY-isomorphic if and only if there exists a vector bundle morphism (ˆf ,f ) between (E 0,π0,Y ) and (E ,π,X) such that ˆf is an isomorphism on each fibre of E 0.

ˆf f ∗E E - E 6 π f π - ? gˆ Y - X f - 6

0 π π

E 0 - E ˆf Proof (sketch of). We know that if (ˆf ,f ):(E 0,π0,Y ) → (E ,π,X) is a vector bundle morphism then ˆ ˆ 0 ∗ 0 f factors as a composition fE ◦ gˆ where gˆ : E → f E is defined for p ∈ E by gˆ(p) = (π0(p),ˆf (p)).

As before, one can prove that gˆ is a vector bundle morphism over Y .

We know that gˆ is an Y -isomorphism of vector bundles if and only if gˆ|y is a linear isomorphism on each fibre π0−1(y). This is equivalent to ˆf being a fibrewise isomorphism since ˆ ˆ ˆ f |y = (fE ◦ gˆ)|y = fE |y ◦ gˆ|y is a composition of linear isomorphisms. 

ˆ  Remark: If h : E1 → E2 is an X-morphism of vector bundles and if ∗ ˆ ∗ ∗ f : Y → X is a map, then the pullback map f (h): f E1 → f E2 defined by f ∗(hˆ)(y,v) = (y,hˆ(v)) is a Y -morphism of vector bundles. In fact, the linearity of f ∗(hˆ) on each fiber follows from the linearity of hˆ Therefore, ∗ f : VecK(X) → VecK(Y ) is a (contravariant) functor. Outline

1 The Category of Vector Bundles Recommended Bibliography Structure of Isomorphisms Triviality and Global Frames Pullback Maps

2 Operations on Vector Bundles Functorial Operations on Vector Spaces Construction of New Vector Bundles Ker and Im Vector Bundles We wish to prove that every (continuous) operation on vector spaces defines a corresponding operation on vector bundles. Let VSK be the category of vector space over K and VecK(X) be the category of all K-vector bundles over a space X. Obviously VecK(∗) ' VSK with ∗ a point.

DEFINITION (Continuous functor) ∗ Let VSK be the category of dual vector spaces. A functor

∗ ∗ F : VSK × ... × VSK ×VSK × ... × VSK → VSK | {z } | {z } p−times q−times is called continuous provided for each family of (linear) mapsf j : Vj → Wj (1 6 j 6 p + q) the composite map p ! p+q ! p ! p+q ! F(f1,...,fp+q): ∏ Vj × ∏ Wj → ∏ Wj × ∏ Vj j=1 j=p+1 j=1 j=p+1 is continuous (with respect to the usual topology of finite dimensional vector spaces). T Example (Continuous functors): - Direct sum -

n m n+m VSK × VSK 3 (V1,V2) 7−→ V1 ⊕ V2 ∈ VSK .

- Tensor product -

n m nm VSK × VSK 3 (V1,V2) 7−→ V1 ⊗ V2 ∈ VSK .

- Wedge (or exterior) product -

n n n(n−1) VSK × VSK 3 (V ,V ) 7−→ V ∧ V ∈ VSK 2 .

- Duality - n ∗ n ∗ VSK 3 V 7−→ V ∈ VSK .

- Homomorphism -

n m nm VSK × VSK 3 (V1,V2) 7−→ Hom(V1,V2) ∈ VSK , moreover, Hom(V ,K) = V ∗. - Conjugation - n n VSC 3 V 7−→ V ∈ VSC . Outline

1 The Category of Vector Bundles Recommended Bibliography Structure of Isomorphisms Triviality and Global Frames Pullback Maps

2 Operations on Vector Bundles Functorial Operations on Vector Spaces Construction of New Vector Bundles Ker and Im Vector Bundles THEOREM For each continuous functor

F : VSK × ... × VSK −→ VSK | {z } p−times there exists a family of functors

FX : VecK(X) × ... × VecK(X) −→ VecK(X) | {z } p−times

∗ ∗ one for each spaceX, such thatf ◦ FX ' FY ◦ f (as an Y -isomorphism) for each mapf : Y → X. Moreover, it is required that F ' F∗.

 Remark: This theorem can be proved using the local description of vector bundles in terms of transition functions. T Example (Whitney sum): Given (E1,π1,X) and (E2,π1,X), the Whitney sum (E1 ⊕ E2,π,X) is the prolongation to vector bundles of the direct sum functor. This follows from the fact that the fiber E1 ⊕ E2|x over a point x ∈ X is the direct sum of the fibers E1|x ⊕ E2|x . The usual properties of direct sums of vector spaces prolong to Whitney sums of vector bundles. There are the following isomorphisms:

E1 ⊕ E2 ' E2 ⊕ E1 , E1 ⊕ (E2 ⊕ E3) ' (E1 ⊕ E2) ⊕ E3 .

T Example (Tensor product): Given (E1,π1,X) and (E2,π1,X) the product (E1 ⊗ E2,π,X) is the prolongation to vector bundles of the tensor product functor. The fiber E1 ⊗ E2|x over a point x ∈ X is the tensor product of the fibers E1|x ⊗ E2|x . The usual properties of direct sums of vector spaces prolong to tensor product of vector bundles. The following isomorphisms hold true:

E1 ⊗ E2 ' E2 ⊗ E1 , E1 ⊗ (E2 ⊗ E2) ' (E1 ⊗ E2) ⊗ E3 1 E ⊗ Θ ' E , E1 ⊗ (E2 ⊕ E3) ' (E1 ⊗ E2) ⊕ (E1 ⊗ E3) . here Θ1 = X × K is the trivial line bundle. T Example (Homomorphism bundle): The homomorphism functor is continuous, and so, given (E1,π1,X) and (E2,π1,X) we may define the homomorphism vector bundle (Hom(E1,E2),π,X). The fiber Hom(E1,E2)|x over x ∈ X is the vector space of homomorphisms E1|x → E2|x . A cross ˆ section ψ of Hom(E1,E2) is just a vector bundle X-morphism fψ : E1 → E2. ˆ The continuity of ψ and the continuity of fψ are equivalent to each other.

T Example (Dual and conjugate vector bundles): Given a vector bundle (E ,π,X) we denote with (E ∗,π,X) the dual vector bundle obtained as the prolongation to vector bundles of the continuous dual functor. The fiber ∗ E |x over a point x ∈ X is the dual vector space of the fiber E |x . One has the identities

∗ 1 ∗ E ' Hom(E ,Θ ) , Hom(E1,E2) ' E1 ⊗ E2 .

The vector bundle (E ,π,X) called conjugate vector bundle is obtained as the prolongation to vector bundles of the continuous complex-conjugation functor

(K = C). The fiber E |x over a point x ∈ X has opposite complex structure with respect the fiber E |x . Outline

1 The Category of Vector Bundles Recommended Bibliography Structure of Isomorphisms Triviality and Global Frames Pullback Maps

2 Operations on Vector Bundles Functorial Operations on Vector Spaces Construction of New Vector Bundles Ker and Im Vector Bundles ˆ Let f : E1 → E2 be a morphism of vector bundles over X. We define two bundles: ˆ Ker(f ): this is a subbundle of π : E1 → X with total space ˆ ˆ Ker(f ) := {p ∈ E1 | f (p) = 0} .

ˆ Im(f ): this is a subbundle of π : E2 → X with total space

ˆ ˆ 0 0 Im(f ) := {p ∈ E2 | p = f (p ), p ∈ E1} .

 Remark: In general, Ker(ˆf ) and Im(ˆf ) are not vector bundles because they do not satisfy the property of local triviality.

 Remark: Recall that the rank of a linear transformation f : V → W is dim(V ) − dim(Ker(f )) which equals dim(Im(f )).

ˆ  Remark: A X-morphism f : E1 → E2 has constant rank r provided that ˆ −1 −1 f |x : π1 (x) → π2 (x) is of rank r for each x ∈ X. THEOREM ˆ Let f : E1 → E2 be a X-morphism between K-vector bundles. Assume that ˆf has constant rankr. Then Ker(ˆf ) and Im(ˆf ) are vector bundles over X.

Proof (sketch of). ˆ The statement refers to a local question, then f , E1 and E2 can be restricted to a trivialization domain. Then, there is a X-morphism ˆf : X × Km → X × Kn ˆ m n such that f (x,v) = (x,f x (v)) where x 7→ f x ∈ Hom(K ,K ) is a map. For each x ∈ X, the rank of f x is r.

At x0 ∈ X, f x0 : V1 ⊕ V2 −→ W1 ⊕ W2 | {z } | {z } Km Kn where V2 = Ker(f x0 ) and W1 = Im(f x0 ). Moreover, dim(V1) = dim(W1) = r and dim(V2) = m − r, dim(W2) = n − r. For each x ∈ X we define ωx V := V1 ⊕ V2 ⊕ W2 −→ W1 ⊕ W2 ⊕ V2 =: W | {z } | {z } Km Kn

by ωx |V 1 := f x |V 1 ⊕ 0 ⊕ 0, ωx |V 2 := f x |V 2 ⊕ 0 ⊕ IdV 2 , ωx |W 2 := 0 ⊕ IdW 2 ⊕ 0.

Observe that ωx0 is an isomorphism. ...  ...

Isomorphisms form an open subset of Hom(V ,W ) and the function x 7→ ωx is continuous then ωx is a linear isomorphism for each x ∈ U ⊂ X, where U is some open neighborhood of x0. Let µx : W → V be the inverse of ωx for each x ∈ U . Then x 7→ µx is continuous. ˆ We prove the triviality of Ker(f ) (restricted to U ). We observe that (v1,v2), with vj ∈ Vj , is a point in ∈ Ker(f x ) if and only if ωx (v1,v2,0) = (0,0,v2). This means that (v1,v2) = µx (0,0,v2), namely Ker(f x ) = µx (V 2). Therefore, the ˆ map (x,v2) 7→ (x, µx (0,0,v2)) is a U -isomorphism U × V2 ' Ker(f )|U with inverse (x,v) 7→ (x,ωx (v,0)). ˆ We prove the triviality of Im(f ) (restricted to U ). One has f x (v1,0) = 0 if and only if ωx (v1,0,0) = 0. Therefore, f x |V 1 : V 1 → Im(f x ) is an isomorphism for each x ∈ U since ωx is a monomorphism. Therefore, (x,v1) 7→ (x,f x (v1,0)) ˆ is a U -isomorphism U × V1 ' Im(f )|U with inverse (x,w) 7→ (x, µx (w,0)) since ωx |V 1 = f x |V 1 ⊕ 0 ⊕ 0 for each x ∈ U .   Remark: The usual terminology of exact sequences carries over to vector bundles and morphisms of constant rank. A sequence

ˆf gˆ E1 −→ E2 −→ E3 is exact provided Im(ˆf )= Ker(gˆ). E.g. given a short exact sequence

ˆf gˆ 0 −→ E1 −→ E2 −→ E3 −→ 0 one has E1 ' Ker(gˆ).

COROLLARY ˆ Let f : E1 → E2 be a X-morphism between K-vector bundles. (i) If ˆf is injective (fiber monomorphism) then Im(ˆf ) is a vector bundle; (ii) If ˆf is surjective (fiber epimorphism) then Ker(ˆf ) is a vector bundle;

Proof (sketch of).

In both cases the X-morphism ˆf has evidently constant rank.  THEOREM Let E → X be a rankm vector bundle and ˆf : E → E be a X-endomorphism such that ˆf 2 = ˆf(idempotent). Then Then, ˆ the set of x ∈ X with Rk(f x ) = r is open and closed in X. If X is connected Ker(ˆf ) and Im(ˆf ) are vector bundles (of constant rank) and E ' Ker(ˆf ) ⊕ Im(ˆf ).

Proof (sketch of). ˆ ˆ ˆ ˆ On each fibre Idx = f x + (Idx − f x ) implies Rk(f x ) + Rk(Idx − f x ) = m. The set ˆ of all x ∈ X with Rk(f x ) = r is at the same time the set of all x ∈ X with ˆ ˆ Rk(f x ) 6 r and Rk(Idx − f x ) 6 m − r, which is a closed set, and the set of all ˆ ˆ x ∈ X with Rk(f x ) > r − 1 and Rk(Idx − f x ) > m − (r − 1), which is an open set.