1. Vector Bundles
Definition 1.1. A family E overX is locally trivial if for any x∈X,there exists an open neigh- borhood U of x so thatE|U is trivial. A locally trivial family E over X is called a (topological) vector bundle overX.A morphism of vector bundles onX is a morphism of the families.
If V is a finite dimensional normed vector space over k, an isomorphism h: E|U → U ×V of continuous family overU is called a local trivialization ofE overU modeled onV.
Definition 1.2. A functionf : X →B from a topological space X to a set B is called a locally constant function if for anyx∈X,there exists an open neighborhoodU ofxsuch thatf(y) =f(x) for anyy∈U.
Lemma 1.1. Let X and Y be topological spaces so that Y is a discrete topological space. Let f :X →Y be a function. Thenf is continuous if and only iff is a locally constant function.
Proof. Suppose that f is a continuous function. Let x ∈X and y = f(x). Since f is continuous and{y} is open,U =f−1({y}) is open inX.For any z∈U, f(z) =y. We find thatf is a locally constant function.
Supposef is a locally constant function. LetV be any open subset ofY.We writeV =S
y∈V{y}.
Eitherf−1({y}) is empty or f−1({y}) is nonempty for y∈V.Whenf−1({y}) is empty, it is open in X. Suppose f−1({y}) is nonempty for some y ∈ X. Let x ∈ f−1({y}). Since f is a locally constant function, there exists an open neighborhood U of xso that f(U) = {y}. Hence U is an open neighborhoodxcontained inf−1({y}) for each x∈f−1({y}),i.e. xis an interior point off− This proves that all points off−1({y}) are interior points of f−1({y}). Hence f−1({y}) is open in X.SetUy=f−1({y}) for ally∈V.Thenf−1(V) =S
y∈V Uy.Since any union of open sets is open,
f−1(V) is open inX.This shows thatf is continuous.
Corollary 1.1. Let f :X →Zbe a locally constant function from a topological space toZ. Here Zis equipped with the subspace topology induced fromR.Thenf is continuous.
Proof. One can easily verify that the subspace topology ofZinduced fromRis the discrete topology
onZ.Hence by the previous lemma,f is continuous.
Lemma 1.2. Let p:E →X be a vector bundle over X.The function rE : X →Z sendingx to dimkExis a locally constant function.
Proof. Sincep:E→X is a vector bundle onX,for anyx∈X,there exists an open neighborhood U ofxso that E|U is isomorphic toU×V for some finite dimensionalk-vector space V (with the Euclidean topology). Supposeh:E|U →U ×V is such an isomorphism. Then hinduces a linear isomorphismh:Ey → {y} ×V for anyy∈U. Therefore, for anyy∈U,
dimkEy= dimk({y} ×V) = dimkV.
Hence rE(x) = dimk(V) for any x∈U. As a consequence, rE :U →Z is a constant function; in
other words,rE is a locally constant function.
Definition 1.3. A vector bundlep:E→X has a rank ifrE:X→Zdefined byx7→dimkEx is a constant function. In this case,r=rE(x) forx∈X is called the rank ofE.
Lemma 1.3. Let f :X →Zbe a locally constant function on a connected spaceX. Thenf is a constant function.
Proof. Let x0 ∈ X and A = {x ∈ X : f(x) = f(x0)}. Since x0 ∈ A, A is nonempty. Since A=f−1({f(x0)}) andf is continuous,Ais closed. Lety0∈Abe any point;f(y0) =f(x0).Sincef is a locally constant function, there exists an open neighborhoodUy0 ofy0inX so thatf :Uy0 →Z is given by f(y) = f(y0) for any y ∈ Uy0. Therefore f(y) = f(y0) = f(x0) for any y ∈ U0. This
1
shows thaty ∈A for anyy∈U0. This implies thatUy0 ⊆A. Hence A=S
y0∈AUy0 is open inX.
We see thatAis a nonempty open and closed set in X.SinceX is connected,A=X.
Corollary 1.2. Let p: E →X be a vector bundle over a connected space X,the function rE is always a constant function. Hence a vector bundle over a connected space always has a rank.
Definition 1.4. A vector bundle of rank one on a topological space is called a line bundle.
Example 1.1. LetPnk be the projective space overk. LetLbe the subspace of the product bundle Pnk ×kn+1 consisting of points ([x], v) where v = λx for some λ ∈ k and for some x ∈ [x]. Let p:L→Pnk be the restriction of the canonical projectionPnk ×kn+1→Pnk toL.Thenp:L→Pnk is a line bundle onPnk.
Let us recall the definition of projective space. The n-dimensional projective space over kis the spacekn+1\ {0}modulo the relation∼defined below. Two elementsxandy inkn+1\ {0} are said to be equivalent if there existsλ∈k\ {0}so thatx=λy.Ifxandy are equivalent, we writex∼y.
The equivalent class of x= (x0,· · ·, xn) in kn+1\ {0} is denoted by [x] = (x0 :· · ·:xn). We equip Pnk with the quotient topology.
Definition 1.5. Let X be a topological space and R is an equivalence relation onX. The set of all equivalent classes ofX moduloR is denoted by X/R. The quotient map π :X →X/R is the function sendingxto its equivalent class [x].The quotient topology on X/Ris the collection of all subsetsU ofX/Rso thatπ−1(U) is open in X.
You can verify that the quotient topology onX/Ris a topology and the quotient map π:X → X/Ris continuous.
For each 0 ≤ i ≤ n, let Ui be the set of all points (x0 : · · · : xn) in Pnk such that there exists a representative (x0,· · · , xn) of (x0 : · · · : xn) such that xi 6= 0. This set is well-defined. Let (x00,· · ·, x0n) be another representative of (x0 : · · ·: xn). Then there exists a nonzero number λ in k so that x0j = λxi for 0 ≤ j ≤ n. Since xi 6= 0 and λ 6= 0, x0i 6= 0. (k is a field.) Therefore Ui
is well-defined. Let π :kn+1\ {0} →Pnk be the quotient map. For each 0 ≤i≤n, letUi0 be the subset of kn+1 consisting of vectors (x0,· · · , xn) so that xi 6= 0. Then Ui0 =π−1i (k\ {0}) is open by the continuity of the projection mapπi:kn+1→ksending (x0,· · ·, xn) to xi together with the openness of the setk\ {0} ink. For each 0≤i≤n, we have
π−1(Ui) =Ui0∩(kn+1\ {0}).
Thereforeπ−1(Ui) is open inkn+1\ {0}for each 0≤i≤n.By definition (of quotient topology),Ui is open inPn+1k for 0≤i≤n.One can easily see thatSn
i=0Ui=Pnk.Hence {U0,· · ·, Un} forms an open cover forPnk.
Let [x] = (x0 : · · · : xn) be a point of Ui for 0 ≤ i ≤ n. There is a unique representative (x0,· · ·, xn) of [x] such thatxi= 1.This representative of [x] is denoted by [x]i inkn+1\ {0}.The fiberp:L→Pnk over [x] is the set
L[x] ={([x], λ[x]i)∈Pnk×kn+1:λ∈k}.
It is obvious thatL[x]has a structure of an one dimensional vector space. For each 0≤i≤n,we let hi:p−1(Ui)→Ui×k
be the function sending ([x], λ[x]i) to ([x], λ).
Lemma 1.4. For each 0 ≤ i ≤ n, the function hi : p−1(Ui) → Ui×k is an isomorphism in the category of family ofk-vector spaces overX.
Proof. By definition, hi is a bijection. For each [x]∈Ui,it is not difficult to see that the function hi:L[x] → {[x]} ×kis a linear isomorphism. Let us show thathi is a homeomorphism.
To check thathiis continuous, we only need to check thath−1i (U×I) is open for anyU open in Ui and for anyI open ink. The reason is because open sets of the formU×Iforms a base for the product topology onUi×k.Notice that for each 0≤i≤n,
h−1i (U×I) = (U×ki−1×I×kn−i)∩p−1(Ui).
Since the set U ×ki−1×I×kn−i−1 is open inPnk ×kn+1, h−1i (U ×I) is open in p−1(Ui).Let us show thathi is an open mapping.
LetU be an open subset ofUiandIbe open ink.ThenU×kj−1×I×kn−jis open inPnk×kn+1. Open sets of the formU×kj−1×I×kn−j for 0≤j≤nforms a subbase for the product topology ofPnk ×kn+1.Then U×kj−1×I×kn−j∩p−1(Ui) is open inp−1(Ui) for anyi, j.We assume that U is open inUi.If not, we may takeU∩Ui.Notice that
hi(U×kj−1×I×kn−j∩p−1(Ui)) =
(U× {λ∈k: (λ, xj)∈k2, λxj ∈I} ifj6=i
U×I ifj=i
are both open in Ui×k. Since any finite intersection of open sets is open and any open sets in Pnk×kn+1is a finite intersection of open sets of the formU×kj−1×I×kn−j, hiis an open mapping.
We prove thathi is a homeomorphism. It is obvious thathi is a morphism of families overUi.We
complete the proof of our assertion.
Remark. Let us prove that{λ∈k: (λ, xj)∈k2, λxj ∈I} is an open subset ofk for any I open in k. We consider the function m : k2 → k defined by m(λ, µ) = λµ. Then m is continuous. Let πi :k2 →k be the projection ofk2 onto the i-th component. Since we equip k2 with the product topology,πi are continuous open mapping. One can check that
{λ∈k: (λ, µ)∈k2, λµ∈I}=π1(m−1(I)).
Since m is continuous and I is open in k2, m−1(I) is open in k2. Since π1 : k2 → k is an open mapping,π1(m−1(I)) is open in k.
This lemma implies that p : L → Pnk is a line bundle over Pnk. This line bundle p : L → Pnk is called the tautological line bundle overPnk.
2. Section
Letp:E→X be a vector bundle overX.A section ofE over an open setU ofX is a continuous functions : U →E so that s(x) ∈Ex for allx ∈U or equivalently, p◦s = idU1. The set of all sections ofE overU is denoted by Γ(U, E). WhenU =X,Γ(X, E) is simply denoted by Γ(E).
For everys1, s2∈Γ(E), we define a functions1+s2:X →E by (s1+s2)(x) =s1(x) +s2(x).
For each s∈Γ(E),we define a function −s: X →E by (−s)(x) =−s(x) for x∈E.We define a function 0 :X→E by 0(x) = 0x inEx for eachx∈X.For any continuous functionf :X→k,we define a functionf s:X→E by
(f s)(x) =f(x)s(x), x∈X.
Proposition 2.1. Letp:E→X be a vector bundle. Letf :X →kbe continuous and s, s1, s2∈ Γ(E).
(1) s1+s2∈Γ(E).
(2) −s∈Γ(E).
(3) 0∈Γ(E).
(4) f s∈Γ(E).
Then Γ(E) is aC(X)-module. HereC(X) is the space of allk-valued continuous functions onX.
Corollary 2.1. Letp:E→X be a vector bundle overX andU be an open subset ofX.Then the space of sections Γ(U, E) ofE overU is a C(U)-module.
1The notion of sections can be defined whenp:E→Xis a continuous family of vector spaces overX.
LetE be a vector bundle of rankroverX.For eachx∈X,we choose an open neighborhoodU so thatE|U ∼=U×V,whereV is ak-vector space of dimensionk. Leth:E|U ∼=U×V be a local trivialization ofE over U.Lets:U →E be a section ofE overU. Sinces(p)∈Ep for any p∈U, there exists a continuous vector valued functionf :U →V so thath(s(p)) = (p,f(p)) for all p∈U.
We obtain a map Φ : Γ(U, E) toC(U, V) sendingstof.HereC(U, V) be the space of all continuous functions fromU intoV.
Lemma 2.1. The function Φ : Γ(U, E)→C(U, V) is aC(U)-module isomorphism.
Proof. We leave it to the reader to verify that Φ is a C(U)-module homomorphism. Let us find the inverse function of Φ. For each vector valued continuous function f : U → V, we define a section s:U →E bysf(p) =h−1(p,f(p)) for p∈U. Since f : U →V is continuous, the function (idU,f) : U → U ×V sending p to (p,f(p)) is also continuous. Since h : π−1(U) → U ×V is a homeomorphism, h−1 : U ×V → E|U ⊆ E is continuous. Since the composition of continuous functions is continuous,
sf :U −−−−→(idU,f) U×V h
−1
−−−−→ E
is continuous. We define Ψ :C(U, V)→Γ(U, E) by sendingf tosf.Then it is not difficult to verify
that Ψ is the inverse to Φ.
We conclude that
Theorem 2.1. Letp:E →X be a vector bundle of rankrover X.For eachx∈X,there exists an open neighborhoodU ofxso that Γ(U, E) is isomorphic toC(U, V) as aC(U)-module for some r-dimensionalk-vector spaceV.
Let V be any r-dimensional k-vector space (with any norm topology). Fix an ordered basis {v1,· · ·, vr}.The functionα:V →knsendinga1v1+· · ·+arvrto (a1,· · · , ar) is a linear isomorphism and also a homeomorphism. We can equipV an inner product so thatαis an isomorphism of Hilbert spaces overk. Since any two topologies on a finite dimensionalk-vector space coincides, the topology onV defined by the inner product coincides with the given norm topology. For each vector valued continuous function f : U → V defined on an open set U of a topological space X, we have a continuouskr-valued function (hf, v1i,· · ·,hf, vri) onU.This determines an isomorphism of vector spaces:
C(U, V)→C(U, kr).
Furthermore,αdetermines an isomorphism of trivial bundle X×V →X×kn
sending (p, v) to (p, α(v)) on any topological spaceX.IfEis a vector bundle of rankron a topological space X and h: E|U →U ×V is a local trivialization of E modeled on a r-dimensional normed spaceV overr,then the composition
E|U −−−−→h U×V −−−−−→(idU,α) U×kr
is a local trivialization of E over U modeled on kr. Hence if E is a vector bundle of rank r over X,for each x∈X,we can always find an open neighborhoodU ofxso that E|U is isomorphic to U×kr.
From now on, we will only consider trivialization of vector bundles modeled onkrfor somer∈N. 3. isomorphisms of Vector Bundles
LetV andW be any finite dimensionalkvector space equipped with the Euclidean topology. The space of allk-linear maps fromV to W is denoted by Homk(V, W). On Homk(V, W), we consider the norm topology defined by
kTk= sup
kvk=1
kT(v)k.
Then Homk(V, W) becomes a Banach space overk.
Let us consider the product familiesX×V andX×W.LetpX :X×V →XandqX:X×W →X and qW : X×W →X be the projections respectively. If ϕ: X×V →X×W is a morphism of families overX, thenpX =qX◦ϕwhich implies thatqX(ϕ(x, v)) =xfor any (x, v)∈X×V.For every (x, v)∈X×V,we defineT(x)v=qW(ϕ(x, v)). Letx∈X be fixed. Sinceϕ is a morphism, for everyv1, v2∈V,
T(x)(v1+v2) =qW(ϕ(x, v1+v2)) =qW(ϕ(x, v1) +ϕ(x, v2))
=qW(ϕ(x, v1)) +qW(ϕ(x, v2))
=T(x)v1+T(x)v2. For anya∈kandv∈V,
T(x)(av) =qW(ϕ(x, av)) =qW(aϕ(x, v)) =aqW(ϕ(x, v)) =aT(x)v.
This shows thatT(x) :V →W is a linear map. We obtain a continuous function T :X →Homk(V, W).
Conversely, ifT :X→Homk(V, W) is a continuous function, we define ϕT :X×V →X×W
by sending (x, v) to (x, T(x)v).ThenϕT is a morphism of families overX.In other words, we obtain a bijection
Hom(X×V, X×W)→C(X,Homk(V, W)).
Lemma 3.1. Suppose dimkV = dimkW.Let Isok(V, W) be the set of all linear isomorphisms from V toW.Then
(1) Isok(V, W) forms an open subset of Homk(V, W).
(2) The function Isok(V, W)→Iso(W, V) sendingT toT−1is continuous.
When ϕ : X ×V → X ×W is a morphism of families so that ϕ is bijective. If we denote ϕ(x, v) = (x, T(x)v) for any (x, v) ∈X ×V, then T(x) :V →W is a linear isomorphism for any x∈ X and ϕ−1(x, w) = (x, T(x)−1w) for any (x, w) ∈ X×W. The previous lemma tells us that x7→T(x)−1 is continuous. Thereforeϕ−1 is also continuous. This observation allows us to prove the following Theorem.
Theorem 3.1. Letϕ: E →F be a morphism of vector bundles over X.Then ϕ: E → F is an isomorphism if and only ifϕ:E→F is a bijection.
Proof. One direction is obvious. Let us assume thatϕ:E →F is a bijection. Then ϕ−1:F →E exists. Let us show thatϕ−1is continuous. For everyx∈X,we choose an open neighborhoodUxof xso thatE|Ux andF|Ux are trivial families. Sinceϕ:E→F is a bijection,ϕx:Ex→Fxis a linear isomorphism for anyx∈X. We assume that dimkEy = dimkFy =n fory ∈ Ux. We may choose isomorphismshE:E|Ux →Ux×kn andhF :F|Ux →U×kn.We obtain a bijective continuous map
hF◦ϕ◦h−1E :Ux×kn→Ux×kn.
We know thathF◦ϕ◦h−1E is a homeomorphism by the above observation. ThereforehE◦ϕ−1◦h−1F is also continuous. SincehE, hF are both homeomorphisms,ϕ−1:F|Ux→E|Uxis continuous. Since {Ux:x∈X} forms an open cover forX andϕ−1 :F|Ux →E|Ux is continuous for any x∈X, ϕ−1
is continuous.
Proposition 3.1. Letϕ:E→F be a morphism of vector bundles overX and U ={x∈X :ϕx∈Iso(Ex, Fx)}.
ThenU forms an open subset ofX.
Definition 3.1. Let p : E → X be a continuous family of k-vector spaces over X. A set of section{s1,· · · , sr} ofE over an open setU is nowhere dependent if {s1(p),· · ·, sr(p)} is linearly independent inEp for allp∈U.If {s1(p),· · ·, sr(p)} forms a basis forEp for any p∈U, the set of sections{s1,· · ·, sr} ofE overU is called a frame overU.
Theorem 3.2. Letp:E→X be a vector bundle of rankroverX.ThenE is trivial if and only if it admits a frame overX.
Proof. SupposeEis trivial. Leth:E→X×krbe a trivialization. Let{e1,· · · , er}be the standard basis for kr. We define si(x) = h−1(x, ei) for each x∈X. Thensi :X →E is a section of E over X.Sinceh:Ex→ {x} ×kr is a linear isomorphism,{s1(x),· · · , sr(x)} forms a basis forExfor all x∈X.Therefore{s1,· · · , sr}is a frame over X.
Suppose thatE admits a frame{s1,· · · , sr}overX.We define a functiong:X×kr→E by g(x,a) =a1s1(x) +· · ·+arsr(x)
for anya= (a1,· · · , ar)∈kr. Theng(x,·) :kr→Ex is a linear isomorphism for allx∈X by the fact that{s1(x),· · · , sr(x)} forms a basis forEx.Thereforeg:X×kr→E is a bijection. Sinceg
is continuous, by Theorem 3.1,g is an isomorphism (check).
