IfE and F are Banach bundles over the spaceX, we denote by Fk(E,F), k∈ Z, the space of morphisms betweenE andF that are bounded Fredholm operators of indexk in each fibre and byF(E,F) the union of all these spaces. As in the first chapter we will use the convention to denote the model spaces ofE andF byE andF, respectively.
5.1.1 Definition. Thesupport of the morphismL∈ L(E,F)is defined by
suppL={λ∈X :Lλ∈/GL(Eλ,Fλ)}.
As next step we prove thatsuppL actually is a closed subset ofX.
5.1.2 Lemma. Let L ∈ L(E,F) be a morphism between the Banach bundles p : E → X and q:F →X. Then the set
{λ∈X:Lλ∈GL(Eλ,Fλ)} ⊂X is open.
Proof. At first we note that if GL(E, F) is not empty, we have a Banach space isomorphism L(F)→ L(E, F)which mapsGL(F)bijectively ontoGL(E, F). Hence we obtain thatGL(E, F) is open from the well known fact that GL(F) is open for any Banach space F. Moreover, in order to prove the lemma we can assume without loss of generality that suppL 6=X which in particular implies thatGL(E, F)6=∅.
Now, givenλ0∈/suppL, we can choose an open neighbourhoodU ofλ0 and trivialisationsϕ, ψ as in (1.1) such that
U 7→ L(E, F), λ7→ψλ◦Lλ◦ϕ−1λ is continuous. Since ψλ0 ◦Lλ0 ◦ϕ−1λ
0 ∈ GL(E, F) and GL(E, F) is open, we can find a neighbourhoodV ⊂U ofλ0 such that ψλ◦Lλ◦ϕ−1λ ∈GL(E, F)and hence Lλ ∈GL(Eλ,Fλ) for allλ∈V.
Now we begin the construction of the index bundle and assume in the following that X is a paracompact space unless otherwise stated. Let L ∈ F(E,F) be a Fredholm morphism and V ⊂ F be a direct subbundle that is transversal toim(L)in the sense that
im(Lλ) +Vλ=Fλ for allλ∈X. (5.1)
Using proposition 1.2.6 there exists a fibrewise projectionP ∈ L(F)such thatim(P) =Vand im(I−P)is a direct subbundle ofF as well. By the property (5.1) ofV we obtain a surjective bundle morphism
E−→ FL −−−→I−P im(I−P).
If moreover the kernels of this morphism are complemented subspaces of the corresponding fibres, then they build a direct subbundleE(L,V)of E by corollary 1.2.10 whose total space is given by
a
λ∈X
{u∈ Eλ:Lλu∈ Vλ}.
Note that this is in particular the case if we can choose a finite dimensional transversal bundleV. The next theorem ensures that such bundles really exist already under quite moderate assumptions. It is the first of two technical results which we need in order to construct the index bundle.
5.1.3 Theorem. Let the base spaceX be compact andL∈F(E,F)a Fredholm morphism. Then there exists a finite dimensional trivial subbundle V ⊂ F such that (5.1)holds overX.
Proof. As a first step we show that the assertion holds ifF is trivial.
Letψbe a global trivialisation ofF. Letλ0∈X andUλ0⊂X be an open neighbourhood ofλ0 such thatE is trivial on Uλ0 by means of a trivialisationϕ. Let
L˜=pr2◦ψ◦L◦ϕ−1:Uλ0×E→F
denote the corresponding family of bounded Fredholm operators with respect to these trivi-alisations. SinceL˜λ0 is Fredholm, there existsVλ0 ⊂F,dimVλ0 <∞, andWλ0⊂E closed such that
im( ˜Lλ0)⊕Vλ0 =F, ker( ˜Lλ0)⊕Wλ0 =E.
Now consider
Aλ:Wλ0×Vλ0 →F, Aλ(w, v) = ˜Lλw+v.
Because ofAλ0∈GL(Wλ0×Vλ0, F)and the continuity ofA:Uλ0→ L(Wλ0×Vλ0, F), there exists a neighbourhoodU˜λ0 ⊂Uλ0 ofλ0 such thatAλ∈GL(Wλ0×Vλ0, F)and hence
im( ˜Lλ) +Vλ0 =F for all λ∈U˜λ0.
By compactness we can now coverX by a finite number of neighbourhoodsU˜λi,i= 1, . . . , n, such that for eachithere exists a finite dimensional subspaceVλi such that
im(pr2◦ψλ◦Lλ) +Vλi =F for allλ∈U˜λi, i= 1, . . . , n.
Finally,V :=V1+. . .+Vn defines a finite dimensional subspace of F such that
im(pr2◦ψλ◦Lλ) +V =F for allλ∈X.
Thenψ−1(X×V)is a finite dimensional trivial subbundle of F such that (5.1) holds on all ofX and the assertion is proved in the special case thatF is trivial.
We now turn to the general case. LetUk0,k= 1, . . . , N, be a finite open covering ofX such that F is trivial over each Uk0. Moreover, we choose open sets Uki, i= 1,2, k= 1, . . . , N, such that Uki+1⊂Uki,i= 0,1, and{Uk2} is still an open covering ofX (cf. [Br93, I.12.9]).
ConsiderU10. SinceF is trivial on the compact subspaceU11, we obtain from the special case in which we proved the assertion already that there is a finite dimensional trivial subbundleV0 of F overU11such that (5.1) holds. We choose a complementW0 toV0in F |U1
Note that V1 is a trivial bundle over all of X satisfying (5.1) on U12. Moreover, since F is trivial onUk1, k= 1, . . . N, we obtain by corollary 1.3.13 that the bundle W1is trivial on allUk1 as well.
Next we considerU20and letP ∈ L(F)denote a projection ontoW1. ConsiderP◦L:E → W1
which is again a Fredholm morphism. SinceW1is trivial onU21we can argue as onU10above and obtain a decompositionW1=V2⊕ W2 such that (5.1) holds forP◦LandV2onU22. Moreover, V2 is trivial on X and W2 is trivial on each Uk1, k = 1, . . . , N. Finally, note that V1⊕ V2 is a subbundle ofF which is transversal toimLoverU12∪U22.
Continuing this process we eventually arrive at a finite dimensional trivial subbundleV =LN i=1Vi
ofF overX such that (5.1) holds over all ofX =SN k=1Uk2.
5.1.4 Corollary. Let X be a general topological space and L ∈F(E,F) a Fredholm morphism having a compact support where we assume the Banach bundleF to be trivial. Then there exists a finite dimensional trivial subbundleV ⊂ F which is transversal to imLin the sense of (5.1).
Proof. Arguing as in the first part of the proof of theorem 5.1.3, we can find a finite dimensional trivial subbundleV of F which is transversal toimL over the compact subspacesuppL. But, sinceLis an isomorphism outsidesuppL, it is clear thatV is actually transversal toimLon all ofX.
By using the same argument as in the foregoing proof of corollary 5.1.4 in the special case that F is a product bundle, we obtain the following important observation.
5.1.5 Corollary. If under the assumptions of corollary 5.1.4 the bundleF is a productX×F for some Banach spaceF, then there exists a finite dimensional subspaceV ⊂F such thatX×V is transversal to imL.
Before going on in the definition of the index bundle, we note the following result on the dimension of the bundle E(L,V).
5.1.6 Lemma. LetL∈F(E,F)be a Fredholm morphism between the Banach bundlesE andF andV be a finite dimensional subbundle ofF which is transversal to the image ofL. Then
dimE(L,V) = ind(L) + dimV.
Proof. Consider Lλ for some λ ∈ X as a map between the finite dimensional vector spaces E(L,V)λ=L−1λ (Vλ)andVλ. We obtain
dimE(L,V)λ= dim ker(Lλ|E(L,V)λ) + dim im(Lλ|E(L,V)λ)
= dim kerLλ+ dim(imLλ∩Vλ)
= indLλ+ dim cokerLλ+ dim(imLλ∩ Vλ)
= indLλ+ dim cokerLλ+ dimVλ−dim cokerLλ
= indLλ+ dimVλ.
IfL∈F(E,F)is a Fredholm morphism and V a finite dimensional subbundle ofF which is transversal to imLas in (5.1), then E(L,V)is by construction a subbundle of E and hence the restriction ofLdefines a bundle morphism
L|E(L,V):E(L,V)→ V
which is actually an isomorphism outside the subspace suppL ⊂ X. Hence, if we assume thatX is in addition locally compact,Lhas a compact support andA⊂X is a closed subspace such thatA∩suppL=∅ , we obtain aK-theory class
[E(L,V),V, L|E(L,V)]∈K(X, A).
Our next aim is to show that this element is independent of the choice of the bundleV. Before we need the following lemma.
5.1.7 Lemma. Let E,F be Banach bundles over X, L ∈ F(E,F) a Fredholm morphism and V ⊂ F a finite dimensional subbundle such that (5.1) holds. Moreover, let Y be a further paracompact space andf :Y →X a continuous map. Thenf∗V is transversal to the image of the pullback morphismf∗L:f∗E →f∗F and we have
E(f∗L, f∗V) =f∗E(L,V).
Proof. By the definition of the pullback it is clear that
im((f∗L)λ) + (f∗V)λ= (f∗F)λ for all λ∈X.
Moreover,
{u∈(f∗E)λ: (f∗L)λu∈(f∗V)λ}={u∈ Ef(λ):Lf(λ)u∈ Vf(λ)}, λ∈Y,
and hence the total spaces ofE(f∗L, f∗V)andf∗E(L,V)coincide as sets. Since both bundles are subbundles off∗E this implies that they actually coincide as bundles.
From the foregoing lemma, we obtain immediately the following corollary.
5.1.8 Corollary. Let A⊂X be a closed subspace of the paracompact and locally compact space X andL∈F(E,F)a Fredholm morphism with compact support such that A∩suppL=∅. IfV is a finite dimensional subbundle ofF which is transversal to imL andf : (Y, B)→(X, A) is proper, then
[E(f∗L, f∗V), f∗V, f∗L|E(f∗L,f∗V)] =f∗[E(L,V),V, L|E(L,V)]∈K(X, A).
Now we can prove that the element[E(L,V),V, L|E(L,V)]∈K(X, A)does not depend on the particular choice of a finite dimensional subbundle ofF which is transversal toimL.
5.1.9 Theorem. LetL∈F(E,F)be a Fredholm morphism with compact support acting between the Banach bundlesE andF andA⊂X a closed subspace such thatA∩suppL=∅. We assume that
• eitherX is compact, or
• X is locally compact and paracompact and F a trivial bundle.
If V,W ⊂ F are two finite dimensional subbundles of finite type that are transversal to the image ofL in the sense of (5.1), then
[E(L,V),V, L|E(L,V)] = [E(L,W),W, L|E(L,W)]∈K(X, A).
Proof. Step 1
In a first step we consider the case thatV ⊂ W is a subbundle.
Then E(L,V)⊂E(L,W)is a subbundle as well and hence, using the paracompactness of the base space, we can find a complement bundle E(L,V)⊥, such that
E(L,W) =E(L,V)⊕E(L,V)⊥.
SinceL|E(L,V)⊥ is injective by the definition ofE(L,V), we conclude thatL(E(L,V)⊥)⊂ W is a subbundle and the restriction ofL defines a bundle isomorphism
L|E(L,V)⊥:E(L,V)⊥→L(E(L,V)⊥).
Moreover, we haveVλ⊕L(E(L,V)⊥)λ=Wλ for allλ∈X. Indeed, ifv∈ Vλ∩L(E(L,V)⊥λ), then there exists u ∈ E(L,V)⊥λ ⊂ E(L,W)λ such that Lλu = v. Since v ∈ Vλ we infer u ∈ E(L,V)λ∩E(L,V)⊥λ ={0}and hencev= 0. Moreover, we obtain from lemma 5.1.6 that
dimW −dimV= dimE(L,W)−dimE(L,V) = dimE(L,V)⊥
= dimL(E(L,V)⊥).
Now we have a commutative diagram
E(L,W) L|E(L,W) //W
E(L,V)⊕E(L,V)⊥
ι
OO
L|E(L,V)⊕L|E(L,V)⊥
//V ⊕L(E(L,V)⊥)
ι
OO
whereι denotes the canonical isomorphism. Hence we obtain
[E(L,W),W, L|E(L,W)] = [E(L,V)⊕E(L,V)⊥,V ⊕L(E(L,V)⊥), L|E(L,V)⊕E(L,V)⊥]
= [E(L,V),V, L|E(L,V)],
where we use thatL|E(L,V)⊥:E(L,V)⊥ →L(E(L,V)⊥)is an isomorphism in the last equality.
Step 2
In the second step of the proof we now turn to the general case and consider two finite dimensional bundlesV, W of finite type as in the assertion. Our aim is to use the special case we already proved above in order to show that we can assume without loss of generality thatV andW are trivial and of the same dimension. In order to do so, note at first thatV andW are contained in trivial finite dimensional subbundles ofE by corollary 1.3.5. Hence by the special case proved above, we can assume without loss of generality thatVandWare trivial. If nowdimV= dimW, we are done. If, however, they are not of the same dimension, saydimV <dimW, we can choose a subbundleMofE which is complementary toVinE by corollary 1.2.8. According to corollary 1.3.5 we now can find a(dimW −dimV)-dimensional trivial subbundle ofMand the direct sum of this bundle andV yields a trivial subbundle ofE of the same dimension thanW. By using the first step of our proof once again, we finally obtain that it suffices to prove the assertion of the theorem under the additional assumption thatV andW are trivial and of the same dimension.
Step 3
By corollary 1.3.9V andW are homotopic and hence there exists a finite dimensional subbundle Mofπ∗F such thatM |X×{0}=V andM |X×{1}=W. Moreover,Mis trivial by [Hu94, 3.4.4]
since its restriction toX× {0}is trivial. Consider the bundle morphismπ∗L:π∗E →π∗F. Our next aim is to extend M to a larger bundle over X×I which is transversal to im(π∗L) over X×I. Since we have to argue differently for both kinds of base spaces we consider, we split the proof at this point.
Step 3 a): X compact
Consider P π∗L : π∗E → M0, where P : π∗F → M0 denotes a projection onto a complement M0 ofMin π∗F which exists according to corollary 1.2.8. P π∗Lis a Fredholm morphism and by theorem 5.1.3, we can find a finite dimensional subbundleM00of M0 which is transversal to im(P π∗L: π∗E → M0) overX ×I. Taking the direct sum ofM00 and Mwe finally obtain a finite dimensional subbundle ofπ∗F that is transversal to im(π∗L: π∗E → π∗F)and contains Mas a subbundle.
Step 3 b): X non compact butF trivial
We consider again P π∗L:π∗E → M0, whereP :π∗F → M0 is as in step 3 a). NowP π∗Lis a Fredholm morphism but its support is non compact unlessX is compact. But sinceπ∗F andM are trivial, M0 is trivial as well according to corollary 1.3.13. Now we can argue as in the first part of the proof of theorem 5.1.3 and obtain a finite dimensional trivial subbundle M00 ofM0 that is transversal to imP π∗L over the compact set (suppL)×I⊂X×I. But since P π∗L is surjective outside(suppL)×I, we infer thatM ⊕ M00 is transversal toimπ∗Lover all ofX×I.
Step 4
We denote by ι0 :X ,→ X× {0} ⊂X ×I and ι1 : X ,→X× {1} ⊂X×I the inclusions and note that
ι∗0=ι∗1:K(X×I, A×I)→K(X, A).
Using corollary 5.1.8 and once again the first step of our proof, we finally obtain
[E(L,V),V, L|E(L,V)] = [E(L, ι∗0(M ⊕ M00)), ι∗0(M ⊕ M00), L|E(L,ι∗
0(M⊕M00))]
= [E(ι∗0π∗L, ι∗0(M ⊕ M00)), ι∗0(M ⊕ M00), ι∗0π∗L|E(ι∗
0π∗L,ι∗0(M⊕M00))]
=ι∗0[E(π∗L,M ⊕ M00),M ⊕ M00, π∗L|E(π∗L,M⊕M00)]
=ι∗1[E(π∗L,M ⊕ M00),M ⊕ M00, π∗L|E(π∗L,M⊕M00)]
= [E(ι∗1π∗L, ι∗1(M ⊕ M00)), ι∗1(M ⊕ M00), ι∗1π∗L|E(ι∗1π∗L,ι∗1(M⊕M00))]
= [E(L, ι∗1(M ⊕ M00)), ι∗1(M ⊕ M00), L|E(L,ι∗1(M⊕M00))]
= [E(L,W),W, L|E(L,W)].
Because of theorem 5.1.3 we now can define the index bundle over compact base spaces as follows:
5.1.10 Definition. Let L ∈ F(E,F) be a Fredholm morphism between the Banach bundles E andF over the compact baseX andA⊂X a closed subspace such that A∩suppL=∅. We call the element
ind(L) = [E(L,V),V, L|E(L,V)]∈K(X, A)
the index bundleof Lwith respect to A, where V ⊂ F is any finite dimensional subbundle such that (5.1)holds.
5.1.11 Remark. Note that suppL=X if L∈Fk(E,F), k6= 0. Hence the case A6=∅ is only of interest ifL∈F0(E,F).
Finally, in the special case that F is a trivial bundle, we can extend the definition of the index bundle to more general base spaces as follows.
5.1.12 Definition. Let X be locally compact and paracompact, A ⊂ X closed, F a trivial bundle and L∈ F0(E,F) a Fredholm morphism of index 0 having a compact support such that A∩suppL=∅. Then the index bundleof Lwith respect to Ais defined by
ind(L) = [E(L,V),V, L|E(L,V)]∈K(X, A),
where V ⊂ F is any finite dimensional subbundle of finite type which is transversal toimL in the sense of (5.1).
5.1.13 Remark. Since suppL = X for any L ∈ Fk(E,F), k 6= 0, we restrict to F0(E,F) in definition 5.1.12.