Robbin and Salamon, Spectral Flow

8/2/2019 Robbin and Salamon, Spectral Flow

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The Spectral flow and the Maslov index

Joel Robbin and Dietmar Salamon

Mathematics Institute

University of Warwick

Coventry CV4 7AL

Great Britain

and

Mathematics DepartmentUniversity of Wisconsin

Madison, WI 53706

USA

September 1992

1 Introduction

Atiyah, Patodi and Singer [3] studied operators of form

DA =d

dt A(t)

when A(t) is a first order elliptic operator on a closed odd-dimensional manifoldand the limits

A = limt

A(t)

exist and have no zero eigenvalue. A typical example for A(t) is the div-grad-curloperator on a 3-manifold twisted by a connection which depends on t. Atiyah etal proved that the Fredholm index of such an operator DA is equal to minus thespectral flow of the family {A(t)}tR. This spectral flow represents the netchange in the number of negative eigenvalues of A(t) as t runs from to .

This Fredholm index = spectral flow theorem holds for rather general families{A(t)}tR of self-adjoint operators on Hilbert spaces. This is a folk theorem thathas been used many times in the literature, but no adequate exposition has yetappeared. We give such an exposition here as well as several applications. Moreprecisely, we shall prove the following theorem.

This research has been partially supported by the SERC.

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Theorem A Assume that, for each t, A(t) is an unbeounded, self-adjoint op-erator on a Hilbert space H with time independent domain W = dom(A(t)).Assume, moreover, that W is a Hilbert space in its own right with a compactdense injectionW H and that the norm of W is equivalent to the graph normof A(t) for every t. Assume further that the map R L(W, H) : t A(t)is continuously differentiable with respect to the weak operator topology. As-sume finally that A(t) converges in the norm topology to invertible operatorsA L(W, H) as t tends to . Then the operator

DA : W1,2(R, H) L2(R, W) L2(R, H)

is Fredholm and its index is given by the spectral flow of the operator familyA(t).

The assumptions of Theorem A imply that the spectrum of A(t) is discreteand consists only of eigenvalues. Hence the spectral flow is well defined andwe shall give the precise definition in Section 4. We shall also prove that Theo-rem A remains valid if A(t) is perturbed by a family of compact linear operatorsC(t) : W H which is continuous in t with respect to the norm topology andconverges to zero as t tends to . The operators C(t) in this perturbation arenot required to be self-adjoint when regarded as unbounded operators on H.

A drawback of Theorem A is the assumption that the domain of A(t) beindependent oft. This assumption will in general exclude the case of differentialoperators on manifolds with boundary. Theorem A will remain valid for asuitable class of operator families with time dependent domains but the preciseconditions on how the domain is required to vary with time will be a topic for

future research.The Fredholm index = spectral flow theorem is used in a generalization

of Morse theory known as Floer Homology. Both Morse theory and Floerstheory are used to prove the existence of critical points of a nonlinear functionalf via topological arguments. In either case the operators A appear as theHessian off at critical points x and the operators A(t) represent the covariantHessian of f along a gradient flow line x(t), i.e. a solution of

x = f(x),

which connects x = limt x(t) with x+ = limt+ x(t). Here both the

gradient f(x) and the Hessian A = 2f(x) are taken with respect to a suitable

metric on the underlying (finite or infinite dimensional) manifold. The operatorDA arises from linearizing the gradient flow equation.In the case of Morse Theory, the function f is bounded below and each

operator A(t) has only finitely many negative eigenvalues and hence has a welldefined index, namely the dimension of the negative eigenspace. In this case thespectral flow can be expressed as the index difference so the Fredholm index

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= spectral flow formula is

index DA = (A+) (A) ()

where denotes the number of negative eigenvalues counted with multiplicity.In the finite dimensional case it turns out that the operator DA is onto (foreach connecting orbit) if and only if the unstable manifold of x intersectsthe stable manifold of x+ transversally (cf. [24]). So in this case the spaceM(x, x+) = Wu(x) Ws(x+) of connecting orbits is a finite dimensionalmanifold whose dimension is the difference of the Morse indices. On the onehand this follows from finite dimensional transversality arguments. On theother hand this can be proved by using an infinite dimensional implicit functiontheorem in a suitable path space where DA appears as the linearized operatorand its kernel as the tangent space to the manifold of connecting orbits. Full

details of this second approach can be found in [26].In Floers theory (see [10] and [11] for example) the operators A can have

both infinite index and infinite coindex so the right hand side of equation () isundefined. The spectral flow can still be defined as the number of eigenvaluesof A(t) which cross zero as t runs from to +. The counting is doneso that each negative eigenvalue which becomes positive contributes +1 andeach positive eigenvalue which becomes negative contributes 1. We make thisprecise in section 4.

In Floers theory the initial value problem for the gradient flow equation isnot wellposed and hence there are no stable or unstable manifolds. However, onecan still prove that the operator DA is onto for suitable generic perturbationsof the function f. It then follows from an infinite dimensional implicit function

theorem that the space of connecting orbits is a manifold. Its dimension is theFredhom index of DA and hence, by Theorem A, agrees with the spectral flowof the 1-parameter family of the Hessians A(t). Even in cases where the indexand coindex of A are both infinite, this spectral flow can still be viewed asan index difference and this leads to Floers relative Morse index. Floer thenproceeds to analyse the properties of these manifolds of connecting orbits toconstruct a chain complex generated by the critical points of f and graded bythe relative Morse index. The boundary operator is defined by counting theconnecting orbits (with appropriate signs) when the index difference is 1. Thehomology groups of this chain complex are called Floer homology. Detailsof this construction can be found in [10], [11], [24], [25], and, for the finitedimensional analogue, in [26].

For example, Floer homology can be used to study closed orbits of Hamil-

tonian systems. These closed orbits are critical points of the symplectic action.Like Morse theory, Floers theory constructs a gradient flow for this action andstudies orbits connecting critical points. A connecting orbit joining two crit-ical points in this infinite dimensional gradient flow is a cylinder joining twoclosed orbits in a finite dimensional symplectic manifold. The cylinder satis-fies a certain nonlinear partial differential equation. Linearization along such a

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connecting orbit gives an operator DA. In this case the spectral flow along theconnecting orbit is the difference of the Maslov indices of the two critical pointsat the ends. See [24] and [25], for example. This linearization is an example ofthe Cauchy-Riemann operators studied in section 7.

The fact that the spectral flow is sometimes the difference of two Maslovindices is not surprising, since the spectral flow can be thought of as an infinitedimensional analogue of the Maslov index for Lagrangian paths. The graph ofa path of symmetric matrices A : [T, T] Rnn is a path of Lagrangian sub-spaces in R2n. Its endpoints are transverse to Rn 0 if and only if the matricesA = A(T) are nonsingular. In this case the Maslov index is the intersec-tion number of the path Gr(A) with the Maslov cycle of those Lagrangiansubspaces which intersect the horizontal Rn 0 in a nonzero subspace. TheMaslov index can be expressed in the form

(Gr(A)) = 12

sign(A+) 12 sign(A)

and agrees with the spectral flow of the matrix family A(t).Our main application is an index theorem for the Cauchy-Riemann operator

S, =

t J0

t+ S

on the infinite cylinder [0, 1] R with general non-local boundary conditions

((0, t), (1, t)) (t)

where S(s, t) = S(s, t)T R2n2n is symmetric and (t) is a Lagrangian pathin (R2n R2n, (0) 0). We prove that S, is a Fredholm operator betweensuitable Sobolev spaces and express the Fredholm index in terms of the relativeMaslov index for a pair of Lagrangian paths.

Theorem B index S, = (Gr(1), ).

Here 1(t) = (1, t) is a path of symplectic matrices determined by S viaJ0s = S with (0, t) = 1l. This generalizes a theorem of Floer [8] for S = 0and local boundary conditions = 0 1 and a theorem in [25] for periodicboundary conditions. Both theorems play an important role in Floer homologyfor Lagrangian intersections [9] and for symplectic fixed points [11].

In section 2 we discuss the finite dimensional case as a warm-up. In section 3

we prove that DA is a Fredholm operator. In section 4 we characterize thespectral flow axiomatically and prove that the Fredholm index satisfies theseaxioms. In section 5 we review the properties of the Maslov index. In section 6we use the spectral flow and the Maslov index to give a proof of the Morse indextheorem. A special case of this is Sturm oscillation. Finally, in section 7 wediscuss the aforementionned Cauchy-Riemann operators.

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2 The finite dimensional case

Linearization along a connecting orbit of a gradient flow of a Morse functionleads to a differential operator

(DA)(t) = (t) A(t)(t). (1)

The index of this operator is the difference of the Morse indices of the criticalpoints at the two ends. In this example the matrices A(t) may be chosen to besymmetric. In this section we shall prove a more general fact. The hypothesisthat the vector field is a gradient field is dropped. We linearize along an orbitconnecting two hyperbolic critical points. As a result the matrices A(t) will nolonger be symmetric but the limit matrices

A = limt

A(t)

exist and are hyperbolic (no eigenvalues on the imaginary axis). For any matrixB Rnn we define

Es(B) =

v Rn : limt

eBtv = 0

,

Eu(B) =

v Rn : lim

teBtv = 0

.

Then Es(B) is the direct sum of the generalized eigenspaces corresponding toeigenvalues with negative real parts and similarly for Eu(B) with positive realparts. Hence the matrix B is hyperbolic if and only if

Rn = Es(B) Eu(B).

Theorem 2.1 Assume that A : R Rnn is continuous and that the limitmatrices A exist and are hyperbolic. Then formula (1) defines a Fredholmoperator

DA : W1,2(R,Rn) L2(R,Rn)

with indexindex DA = dim E

u(A) dim Eu(A+).

Proof: That the operator is Fredholm follows from the inequality

W1,2(R) c L2(I) + DAL2(R) (2)for a sufficiently large interval I = [T, T]. This estimate is proved in threesteps. Firstly, since = DA + A, the estimate is obvious for I = R:

W1,2(R) c

L2(R) + DAL2(R)

. (3)

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Secondly, if A(t) A0

is a constant hyperbolic matrix, then the associatedoperator DA0 satisfies

W1,2(R) cDA0L2(R). (4)

To see this decompose Rn as a direct sum where each of the summands has allits eigenvalues in one of the two halfplanes. Hence it suffices to treat the specialcase where all the eigenvalues of A0 have negative real part. For L

2(R,Rn)the unique solution of A0 = with L2(R,R

n) is given by

(t) =

t

eA0(ts)(s) ds = (t)

where (t) = e

A0t

for t 0 and (t) = 0 for t < 0. By Youngs inequality,L2 L1L2 .

Since = A0 + we also have

L2 A0L2 + L2 (A0 L1 + 1) L2.

Note in fact that the operator DA0 is bijective since any function in its kernel isan exponential and can lie in L2 only if it vanishes identically. This proves (4).Alternatively, (4) can be proved with Fourier transforms as in Lemma 3.7 below.

Finally, the estimate is proved by a patching argument. It follows from (4)that there exist constants T > 0 and c > 0 such that for every W1,2(R,Rn)

(t) = 0 for |t| T 1 = W1,2(R) cDAL2(R). (5)

Now choose a smooth cutoff function : R [0, 1] such that (t) = 0 for|t| T and (t) = 1 for |t| T 1. Using the estimate (3) for and (5) for(1 ) we obtain

W1,2 W1,2 + (1 )W1,2

c1 (L2 + DA()L2 + DA((1 ))L2)

c2

L2[T,T] + DAL2

.

This proves (2). Since the restriction operator W1,2(R,Rn) L2([T, T],Rn)is compact it follows from Lemma 3.5 below that DA has a finite dimensional

kernel and a closed range.We examine the kernel ofDA. It consists of those solutions of the differentialequation = A which converge to zero as t tends to + and . Considerthe fundamental solution (t, s) Rnn defined by

t(t, s) = A(t)(t, s), (s, s) = 1l,

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and note that (t, s)(s, r) = (t, r). Define the stable and unstable subspaces

Es(t0) =

0 R

n : limt+

(t, t0)0 = 0

,

Eu(t0) =

0 R

n : limt

(t, t0)0 = 0

.

Both subspaces define invariant vector bundles over R. This means that Es(t) =(t, s)Es(s) and Eu(t) = (t, s)Eu(s). Moreover, limt+ E

s(t) = Es(A+)and limt E

u(t) = Eu(A). Hence

dim Es(t) = n dim Eu(A+), dim Eu(t) = dim Eu(A).

Now let (t) be any solution of the differential equation = A. Then (t) =

(t, s)(s) for all t and s. Moreover, |(t)| converges to 0 exponentially fort + whenever (t) Es(t) and |(t)| converges to exponentially fort + whenever (t) / Es(t). Similarly for t . Hence

ker DA (t) = (t, s)(s) and (t) Es(t) Eu(t).

We examine the cokernel of DA. Assume that L2(R,Rn) is orthogonalto the range of DA. Then

(t), (t) dt

(t), A(t)(t) dt = 0

for every W1,2(R,Rn). If(t) = 0 for |t| T then this implies

TT

(t) Tt

A(s)T(s) ds, (s) ds = 0

and hence(t) + A(t)T(t) = 0.

The fundamental solution of this equation is (t, s) = (s, t)T and it is easy tosee that the associated stable and unstable bundles are given by Es(t) = Es(t)and Eu(t) = Eu(t). Hence

range DA (t) = (s, t)T(s) and (t) Es(t) + Eu(t).

In particular the cokernel of DA is finite dimensional. Moreover,

index DA = dim(Es Eu) dim(Es + Eu)

= dim(Es Eu) + dim(Es + Eu) n

= dim Es + dim Eu n

= dim Eu(A) dim Eu(A+).

This proves the theorem. 2

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Remark 2.2 Assume that the matrices A are symmetric. Then the indexformula of Theorem 2.1 can be expressed in terms of the signature as

index DA = 12 sign A 12 sign A

+.

3 Fredholm theory

Assume that W and H are separable real Hilbert spaces with

W H = H W.

Here the inclusion W H is compact with a dense range. Throughout weidentify H with its dual space. We shall not use the inner product on W but

only the norm. Hence we distinguish W from its dual space W

. The notation, will denote the inner product in H when , H and the pairing of Wwith W when W and W.

Fix a family of bounded linear operators

A(t) : W H

indexed by t R. Given a differentiable curve : R W, define DA : R Hby

(DA)(t) = (t) A(t)(t) (6)

for t R. In the intended application W = W1,2 and H = L2 and A(t) is a firstorder linear elliptic differential operator whose coefficients depend smoothly ont. We impose the following conditions:

(A-1) The map A : R L(W, H) is BC1. This means that it is continuouslydifferentiable in the weak operator topology and there exists a constantc0 > 0 such that

A(t)H + A(t)H c0W.

for every t R and every W.

(A-2) The operators A(t) are uniformly self-adjoint. This means that for eacht the operator A(t) when considered as an unbounded operator on H withdom A(t) = W is self-adjoint and that there is a constant c1 such that

2W

c1 A(t)2H + 2H . (7)for every t R and every W.

(A-3) There are invertible operators A L(W, H) such that

limt

A(t) AL(W,H) = 0.

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Remark 3.1 Condition (A-1) implies that the map t A(t) is continuous inthe norm topology but t A(t) is only weakly continuous. We shall use thefact that L2(R, W) implies A L2(R, H).

Remark 3.2 Let A be a self-adjoint operator on H with dense domain W =dom A. Then W is a Hilbert space in its own right with respect to the graphnorm of A and the estimate (7) holds trivially. The inclusion W H iscompact if and only if the resolvent operator (1l A)1 : H H is compactfor every / (A). In this case the spectrum of A is discrete and consists ofreal eigenvalues of finite multiplicity.

Remark 3.3 Let A be a closed symmetric operator on H with dense domainW = dom A. When A is regarded as a bounded operator from W to H its adjointis a bounded operator from H to W. Since A is symmetric the restriction of

this adjoint to W agrees with A. Thus the adjoint is an extension of A whichwe still denote by A. With this notation we have

A L(W, H) L(H, W).

The condition that A be self-adjoint now means that A H implies W.

Remark 3.4 A symmetric operator A : W H which satisfies (7) is neces-sarily closed but need not be self-adjoint.

We define Hilbert spaces H and W by

H = L2(R, H),

W = L2(R, W) W1,2(R, H)

with norms

2H =

(t)2Hdt,

2W =

(t)2W + (t)

2H

dt.

The inclusion W H is a bounded linear injection with a dense range sinceC0 (R, W) is dense in both spaces. The uniform bound on A(t) from condi-tion (A-1) means that DA defines a a bounded linear operator

DA : W H.

Our first aim is to show that it is Fredholm. The proof relies on the following

Lemma 3.5 (Abstract Closed Range Lemma) Suppose that X, Y, and Zare Banach spaces, that D : X Y is a bounded linear operator, and thatK : X Z is a compact linear operator. Assume that

xX c (DxY + KxZ)

for x X. Then D has a closed range and a finite dimesional kernel.

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For each T > 0 define Hilbert spaces W(T) and H(T) by

H(T) = L2([T, T], H),

W(T) = L2([T, T], W) W1,2([T, T], H)

with norms as above.

Lemma 3.6 For every T > 0 the inclusion W(T) H(T) is a compact oper-ator.

Proof: Choose an orthonormal basis for H and denote by n : H Rn theorthogonal projection determined by the first n elements. Since nn L(H)converges strongly to the identity of H and the inclusion W H is compact,the operator

n

n|W : W H

converges to the inclusion W H in the norm topology. The induced operatorW(T) H(T) : nn can be decomposed as

W(T) W1,2([T, T],Rn) C([T, T],Rn) H(T).

Here the first operator is induced by n, the second is compact by the Arzela-Ascoli theorem, and the last is induced by n. Now

nnH(T) 1l nnL(W,H)W(T).

Hence the inclusion W(T) H(T) is a uniform limit of compact operators andis therefore compact. 2

Lemma 3.7 There exist constants c > 0 and T > 0 such that

W c

H(T) + DAH

for every W.

Proof: The proof is analogous to the proof of (2). The first step is to provethe estimate with T = . For every C0 (R, W)

DA2H =

2H + A

2H 2

,A dt

= 2H + A2H +

, A dt

2H + A2H c0H L2(R,W)

2H +1c1

2L2(R,W) 2H c0H L2(R,W)

2H +1

2c12L2(R,W)

1 +

c0c12

2H

1

2c12W c

2H.

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The second step is to prove the estimate with A replaced by a constantbijective operator A(t) A0. The associated operator DA0 satisfies

W c DA0H (8)

In terms of the Fourier transform the equation DA0 = can be rewritten as

i(i) A0(i) = (i).

Since the operator A0 is symmetric we have

|| 2H |i A0, | i A0HH

and hence

|| H i A0H.With c = A0

1L(H,W) we obtain

W cA0H

ci A0H + c|| H

2ci A0H.

Hence it follows from the Fourier-Plancherel theorem that for every W

2W =

(i)2W +

2(i)2H

d

(1 + 4c2)

i(i) A0(i)2

H

d

= (1 + 4c2)DA02H.

This proves (8).The final step uses a patching argument and is analogous to the patching

argument in the proof of (2). 2

Assume A : W H is self-adjoint and , H such that

A, = ,

for every W. Then W (see Remark 3.3). In other words every weaksolution H of A = with H is a strong solution. The following

theorem says that a similar result holds for the operator DA. Since DA is notnot self-adjoint we must use the formal adjoint operator of DA to define thenotion of a weak solution. The formula

+ A,H + , DAH = 0

for , W shows that DA is the formal adjoint operator of DA.

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Theorem 3.8 (Elliptic regularity) Assume , H satisfy

+ A,H + , H = 0

for every C0 (R, W). Then W and DA = .

Proof: We first prove (in four steps) that the theorem holds under the as-sumption that and are supported in an interval I such that A(t) : W His bijective with

A(t)1L(H,W) c

for t I.

Step 1: W1,2(R, W) and

(t) = A(t)(t) + (t) (9)

where A(t) L(H, W) as in Remark 3.3.

For C0 (R, W)

(t), (t)

H

dt =

(s), A(s)(s) + (s)W,W ds

=

(t),

t

A(s)(s) + (s)

ds

W,W

dt.

Since the derivatives of test functions are dense in L2(R, W) this impliesStep 1.

Now choose a smooth cutoff function : R R such that (t) = 0 for|t| 1, (t) 0 and

= 1. For > 0 define (t) =

1(1t).

Step 2: For > 0 sufficiently small we have = W.

Multiply equation (9) by A1 to obtain = A1 A1 and convolve with:

=

A1

A1

=

A1

+

A1AA1

A1

=

A1

+

A1

where = AA1 H. We have used (uv) = (uv) (uv).

Step 3: There exists a constant c > 0 such that

DA( )H c

for every sufficiently small .

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Use step 2 and the identity = to obtain

DA = A

= A (A1) + A

A1

= A

A1 (A

1)

+ A

A1

.

The second term on the right is bounded in H, uniformly in . For the otherterm we have A1 (t) (A1)(t)W

=

t+t

1

t s

A1(t) A1(s)

(s) ds

W

c

1 t s (s)H ds.Here c is a uniform bound for the derivative ofA1 on I. By Youngs inequalityA1 (A1)L2(R,W) cL1(R)H.This proves Step 3.

Step 4: W and DA = .

It follows from Step 2 and Lemma 3.7 that W c for some constant cindependent of . Choose a sequence 0 such that converges weakly

in W. Let 0 W denote the weak limit. Then converges weakly to 0 inH. On the other hand = converges strongly to in H and hence = 0 W. Now it follows from (9) that DA = A = .

This proves the theorem under the assumption that and are supportedin an interval on which A is bijective. Cover the real axis by finitely many openintervals Ij such that j1l A(t) : W H is bijective with

(j1l A(t))1L(H,W) cj

for t Ij . Now choose a partition of unity j subordinate to the cover. Thenthe function j = j is a weak solution of

j Ajj = j

whereAj = A j1l, j = j + (j + jj).

By the special case j W and hence =

j j W. 2

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Remark 3.9 The previous theorem only requires the estimate of Lemma 3.7with I = R. Hence it continues to hold if the limits A do not exist. Moreover,local regularity does not require bounds on the function A : R Lsym(W, H).However, we cannot dispense with the assumption that A(t) be self-adjoint.

Theorem 3.10 The operator DA is Fredholm.

Proof: By Lemma 3.7 and Lemma 3.6 the operator DA has a finite dimensionalkernel and a closed range. By Theorem 3.8 the cokernel of DA is the kernel ofthe operator DA : W H. Hence the cokernel of DA is finite dimensional. 2

Theorem 3.11 Assume A : R L(W, H) is of class Ck1 with dkA/dtk

weakly continuous and djA/dtj uniformly bounded for 0 j k. If Wand

DA = Wk,2(R, H)

then Wk,2(R, W) Wk+1,2(R, H).

Proof: The proof is by induction on k. Assume k = 1. Then 1 = is a weaksolution of DA1 = 1 with 1 = A + L

2(R, H). To see this note that

1 = A + W1,2(R, W)

with

1 =

A + A

+ = A1 + 1.By Theorem 3.8 this implies that 1 W and hence hence W1

,2(R, W) W2,2(R, H).

Suppose by induction that the statement has been proved for k 1. Let Wk+1,2(R, H) and W with DA = . By what we just proved Wand

DA = A + Wk,2(R, H).

Hence it follows from the induction hypothesis that

Wk,2(R, W) Wk+1,2(R, H).

This proves the theorem. 2

Proposition 3.12 Suppose A(t) is bijective for all t and

A(t)1L(H,W)A(t)L(W,H)A(t)1L(H,H) < 2. (10)

Then DA is bijective.

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Proof: Let ker DA. Then it follows from Theorem 3.11 that

W1,2(R, W) W2,2(R, H).

Henced2

dt22H = 4A

2H + 2, A. (11)

The right hand side is continuous and hence the function t (t)2H is C2.

By (10) this function is convex. Since it is integrable on R it must vanish. HenceDA is injective. The same argument with A replaced by A shows that DA isonto. 2

Corollary 3.13 If A(t) is bijective for all t then DA has Fredholm index 0.

Proof: The operator family A(t) satisfies the hypotheses of Proposition 3.12for small . 2

Example 3.14 If we drop the hypothesis that each A(t) is self-adjoint thenDA need not be Fredholm. For example, let H = W

1,2(S1) L2(S1), W =W2,2(S1) W1,2(S1), and let A0 : W H be defined by

A0(u, v) = (v, u).

This operator A0 : W H has a compact resolvent and is the infinitesimalgenerator of a strongly continuous one parameter group U(t) L(H) of unitaryoperators on H. Define

A(t) = A0 b(t)1lwhere b(t) = 1 for t 1 and b(t) = 1 for t 1. Then the Cauchy problem

(t) = A(t)(t), (0) = 0 W,

is well-posed and all solutions converge to zero exponentially as t tends to .Hence the kernel of DA is infinite dimensional. Hence the operator DA is notFredholm.

In contrast to the previous example lower order perturbations of A alwaysproduce Fredholm operators. The perturbation is a multiplication operatorinduced by C(t) : W H. We assume that the function C : R L(W, H) iscontinuous in the norm topology such that C(t) : W H is a compact operator

for every t andlim|t|

C(t)L(W,H) = 0.

Remark 3.15 If B : R L(H, H) is strongly continuous and converges to 0in the norm topology as t tends to then the operator family C(t) = B(t)|Wsatisfies the above requirements.

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Lemma 3.16 The operator

W H : C

is compact.

Proof: First assume that C is compactly supported in an interval I. Chooseprojection operators n : H R

n as in Lemma 3.6. Then the operator

Cn(t) = nnC(t) : W H

converges in the norm topology to the operator C(t) L(W, H) and the conver-gence is uniform in t since C : R L(W, H) is continuous in the norm topology.The multiplication operator induced by Cn can be decomposed as

W W1,2(I,Rn) L2(I,Rn) H.

Here the first operator is induced by n C, the second is compact, and the lastis induced by n. Since the operator Cn : W H converges to C : W Hin the norm topology it follows that the operator C is compact. In the generalcase use a cutoff function to approximate C in the norm topology by operatorswith compact support. 2

Corollary 3.17 The operator DA+C : W H defined by

DA+C = A C

is Fredholm. It has the same index as DA:

index DA+C = index DA.

Remark 3.18 Assume that the curve A : R L(W, H) is continuous in thenorm topology and satisfies (A-2) and (A-3). We were not able to prove underthese assumptions that DA is Fredholm.

4 The spectral flow

We continue the notation of the previous section. For A : R L(W, H) andt R we define the crossing operator by

(A, t) = PA(t)P|kerA

where P : H H denotes the orthogonal projection onto the kernel of A. Acrossing for A is a number t R for which A(t) is not injective. The setof crossings is compact. A crossing t is called regular if the crossing opera-tor (A, t) is nonsingular. It is called simple if it is regular and in addition

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dim ker A = 1. If t0

is a simple crossing then there is a unique real valuedC1 function = (t) defined near t0 such that (t) is an eigenvalue of A(t)and (t0) = 0. This function is called the crossing eigenvalue. For a simplecrossing the crossing operator (A, t0) is given by multiplication with (t0) andhence (t0) = 0.

Theorem 4.1 Assume A satisfies (A-1), (A-2), (A-3) and has only regularcrossings. LetDA be defined by (6). Then the set of crossings is finite and

index DA = t

sign (A, t) (12)

where the sum is over all crossings t andsign denotes the signature (the numberof positive eigenvalues minus the number of negative eigenvalues). Hence for acurve having only simple crossings

index DA = t

sign (t) (13)

where denotes the crossing eigenvalue at t.

Note that the direct sum of two curves having only regular crossings againhas only regular crossings. The analogous result fails for simple crossings. In-deed A A has no simple crossings. We will see in this section that the propertyof having only simple crossings is generic so that (13) suffices. On the other handthe following theorem shows how to use formula (12) without perturbing A.

Theorem 4.2 The curve A 1l has only regular crossings for almost every R.

To prove these results we characterize the Fredholm index axiomatically andshow that the right hand sides of the formulae (12) and (13) satisfy these axioms.It is convenient to introduce some notation.

Denote the Banach space of bounded symmetric operators from W to H by

Lsym(W, H) = {A L(W, H) : A|W = A}

and let S(W, H) Lsym(W, H) be the open subset consisting of those operatorswith nonempty resolvent set, i.e. the operator

1l A : W H

is bijective for some real number . This means that the operator A, whenregarded as an unbounded operator on H with dense domain W = dom A isself-adjoint with compact resolvent. (See remark 3.2.)

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Denote by B = B(R, W , H ) the Banach space of all continuous1 maps A :R Lsym(W, H) which have limits

A = limt

A(t)

in the norm topology. Denote by B1 = B1(R, W , H ) B the Banach space ofthose A B which are continuously differentiable in the norm topology andsatisfy

AB1 = suptR

A(t) + A(t)

< .

Define an open subset

A = A(R, W , H ) B(R,W,H)

consisting of those A B for which the limit operators A : W H arebijective and A(t) S(W, H) for each t R. The set

A1 = A1(R,W,H) = A B1

is open in B1. The set A consists of all maps A : R L(W, H) which arecontinuous in the norm topology and satisfy (A-2), and (A-3). The set A1

consists of all maps A : R L(W, H) which satisfy (A-1), (A-2), (A-3) andin addition are continuously differentiable in the norm topology. Theorem 3.10implies that DA is a Fredholm operator for every A A1.

Given Ai A(R, Wi, Hi), i = 1, 2, the direct sum

A1 A2 A(R, W1 W2, H1 H2)

is defined pointwise. Given A, A, Ar A(R, W , H ) we say that A is the cate-nation of A and Ar if

A(t) =

A(t) for t 0Ar(t) for t 0

and A(t) = A(0) = Ar(t) for t > 0. In this case we write

A = A#Ar.

Note that the operation (A, Ar) A#Ar is only partially defined.

Theorem 4.3 There exist unique maps : A(R, W , H ) Z, one for everycompact dense injection of Hilbert spaces W H, satisfying the followingaxioms.

(homotopy) is constant on the connected components of A(R, W , H ).

1Continuous means continuous in the norm topology unless otherwise mentioned.

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(constant) If A is constant, then (A) = 0.

(direct sum) (A1 A2) = (A1) + (A2).

(catenation) If A = A#Ar, then (A) = (A) + (Ar).

(normalization) For W = H = R and A(t) = arctan(t), we have (A) = 1.

The number (A) is called the spectral flow of A.

Remark 4.4 Choose A A(R,W,H). Then there exists a constant > 0 suchthat if A0, A1 A1(R,W,H) satisfy

suptR

A(t) Aj(t)L(W,H)

then the path A = (1 )A0 + A1 lies in A1(R, W , H ) for 0 1.Moreover, if A0, A1 A

1(R, W , H ) are homotopic by a continuous homotopyin A(R,W,H) then they are homotopic by a C1-homotopy in A1(R, W , H ).Hence any homotopy invariant on A1(R,W,H) extends uniquely to a homotopyinvariant on A(R, W , H ).

We may write the set S = S(W, H) as an infinite disjoint union

S(W, H) =k=0

Sk(W, H)

where Sk = Sk(W, H) denotes the set of operators L S(W, H) with k-

dimensional kernel. The set Sk is a submanifold ofS of codimension k(k + 1)/2.The tangent space to Sk at a point L Sk is given by

TLSk =

L Lsym(W, H) : PLP = 0

where P : H H denotes the orthogonal projection onto the kernel of L. Inother words a curve A A is tangent to Sk at t = 0 if and only if A(0) Skand the crossing operator (A, 0) = 0. Since S1 has codimension 1 a curve hasonly simple crossings if and only it is transverse to Sk for every k. (Recall thatSk has codimension greater than 1 for k 2 and hence a curve is transverse toSk if and only if it does not intersect Sk.)

Proof of Theorem 4.3: For A A1(R,W,H) the number (A) can be defined

as the intersection number of the curve A with the cycle S1 (appropriatelyoriented) as in [18]. This requires showing that the set of curves transverseto all Sk is open and dense in A1 and that the set of homotopies which aretransverse to all Sk is dense. The assertion about curves can be proved usingthe transversality theory in [1]. In the notation of [1] ne uses the representationof maps : A C1(X, Y) with A = A1(R,W,H), X = R, Y = S(W, H),

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W = Sk(W, H) and the inclusion. Homotopies are handled similarly but apreliminary smoothing argument is required. By Remark 4.4 the definition of(A) as the intersection number with S1 extends to A A(R,W,H).

If the curve A A1(R, W , H ) is transverse to each Sk then the intersectionnumber of A with S1 is given by the explicit formula

(A) =t

sign (t)

where the right hand side is as in (13). The transversality argument shows thatthis intersection number satisfies the homotopy axiom. The other axioms areobvious. The proof that the axioms characterize the spectral flow requires thefollowing

Theorem 4.5 For every A A(R,W,H) there exists an integer m and a pathof matrices B A(R,Rm,Rm) such that A B is homotopic to a constant path.

Proof: We prove the theorem in three steps.

Step 1: If A(t) is bijective for every t then the theorem holds with m = 0.

Homotop A to a constant A(0) using the formula A(t) = A(tan(arctan t)).

Step 2: If A A1(R, W , H ) has m simple crossings then there exists a curveb A1(R,R,R) such that A b is homotopic to a curve with m 1 simplecrossings.

Assume A has a simple crossing at t = t0 and let (t) (A(t)) be thecrossing eigenvalue for t near t0. For > 0 sufficiently small choose a C

1-curve of eigenvectors : (t0 , t0 + ) H such that (t)(t) = A(t)(t) and(t)H = 1. Define (t) : H R by

(t) = (t), .

Moreover, choose a smooth cutoff function : R [0, 1] such that (t) = 1for |t t0| /2 and (t) = 0 for |t t0| . Finally, choose a C1-functionb : R R such that

b(t) = (t) for |t t0| /2,

b(t) = 0 for t = t0, and b(t) is constant for |t| . Consider the curve A A1(R, W R, H R) defined by

A(t) =

A(t) (t)(t)

(t)(t) b(t)

.

Then A0 = A b and for > 0 the curve A has no crossing at t = t0.

Step 3: The general case.

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By transversality homotop A to a curve in A1(R, W , H ) with only simplecrossings. Now use step 2 inductively to construct a curve B A1(R,Rm,Rm)such that A B is homotopic to a curve without crossings. Finally use step 1.2

Proof of Theorem 4.3 continued: Denote by (A) the spectral flow asdefined by intersection numbers. Let : A(R, W , H ) Z be any putativespectral flow which satisfies the axioms of Theorem 4.3. We prove that = .To see this note that a curve of matrices B A(R,Rm,Rm) is homotopic toa curve of diagonal matrices. Hence it follows from the homotopy, direct sum,and normalization axioms that

(B) = (B) = 12

sign B+ 12 sign B.

Now let A A(R,W,H) be any curve and choose B A(R,Rm

,Rm

) as inTheorem 4.5. Then it follows from the homotopy and constant axioms that(A B) = 0. Hence (A) = (B) = (B) = (A). This proves thetheorem. 2

We have not used the catenation axiom to prove uniqueness of the spectralflow. Hence the catenation axiom follows from the other axioms. Here is adirect proof of this observation.

Proposition 4.6 The catenation axiom follows from the homotopy, direct sum,and constant axioms.

Proof: Let A, Ar A(R,W,H) such that A(t) = Ar(t) for t 1. Callthis constant operator L. We construct a homotopy A A(R, W W, H H)

such thatA0 = A Ar, A1 = A#Ar L.

This homotopy is given by A(t) = A(t) Ar(t) for t 0 and

A(t) = R

2

A(t) 0

0 Ar(t)

R

2

for t 0. Here R() is the Hilbert space isomorphism of H H defined by

R() =

cos sin sin cos

.

Note in particular that R() commutes with L L. Thus we have proved that

A0 = A Ar and A1 = A#Ar L are homotopic. Hence

(A) + (Ar) = (A Ar)

= (A#Ar L)

= (A#Ar) + (L)

= (A#Ar).

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The first equality follows from the direct sum axiom, the second from the ho-motopy axiom, and the last from the constant axiom. 2

Lemma 4.7 Assume A : R L(W, H) satisfies (A-1),(A-2), (A-3) and hasonly regular crossings. Then the number of crossings is finite and the spectralflow of A is given by

(A) =t

sign(A, t)

where the right hand side is as in (12).

To prove this result we require Katos selection theorem for the eigenvalues

of a one parameter family of self-adjoint operators.

Theorem 4.8 (Kato Selection Theorem) For A A1(R,Rn,Rn) there ex-ists a C1 curve of diagonal matrices

R Rnn : t (t) = diag(1(t), . . . , n(t))

such that(t) A(t) for everyt. Here the sign denotes similarity. Moreover,

( , t) (A , t)

for allt and .

This is a reformulation of Theorem II.5.4 and Theorem II.6.8 in [14]. The

existence of a continuous family (t) which is pointwise similar to A(t) is easyto prove. Simply use the ordering of the real line to select the diagonal entries.In general one cannot choose the similarity continuously. There might not exista continuous family of bijective matrices Q(t) with A(t) = Q(t)(t)Q(t)1. Tofind a differentiable function (t) is much harder.

The functions j(t) are the eigenvalues of A(t) counted with multiplicity.

The theorem also asserts that the derivatives j(t) for those j with j(t) = are the eigenvalues of the crossing operator (A , t), again counted withmultiplicity.

Corollary 4.9 Assume A : R L(W, H) satisfies (A-1) and (A-2). Lett0 Rand c > 0 such that c / (A(t0)). Then there exists a constant > 0 and a

C1

function (t) of diagonal matrices defined for t0 < t < t0 + such that

( , t) (A , t)

for t0 < t < t0 + and c < < c. (This implies that the diagonal entriesof (t) are the eigenvalues of A(t) betweenc andc, counted with multiplicity.)

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Proof: We prove this by reducing it to the finite dimensional situation. Firstchoose > 0 such that c / (A(t)) for t0 < t < t0 + . Let E(t) denotethe sum of the eigenspace of A(t) corresponding to eigenvalues between cand c. Choose C1-functions j : (t0 , t0 + ) H such that the vectors1(t), . . . , N(t) form an orthonormal basis of E(t) for every t. Define (t) :H RN by the formula

(t)x =

Nj=1

xjj(t).

By the finite dimensional case the theorem holds for the symmetric matrix

B(t) = (t)A(t)(t).

Differentiating the definition of B gives

B = A + A + A

and henceQBQ = PAP

where P and Q denote the spectral projections for A and B corresponding tothe same eigenvalue . 2

Proof of Lemma 4.7: A curve A : R L(W, H) which satisfies (A-1), (A-2),and (A-3) has only regular crossings if and only if it is transverse to

S1 =k1

Sk.

in the sense that its derivative A(t) at a crossing t does not lie in the tangentcone at L = A(t)

TLS1 =

L Lsym(W, H) : 0 (PLP|kerL)

.

Choose t0 R with A(t0) S1. Then 0 is an eigenvalue of A(t0) with finitemultiplicity m. Choose c > 0 such that there is no other eigenvalue of A(t0)in the interval c c. Now choose > 0 such that c / (A(t)) fort0 t t0 + . By Corollary 4.9 there exist m continuously differentiablecurves

1, . . . , m : [t0 , t0 + ] (c, c)

representing the eigenvalues of A(t) in (c, c). Since t0 is a regular crossingof A it follows that j(t0) = 0 for every j. This proves that the crossings areisolated.

Shrinking if necessary we may assume that j(t) = 0 for 0 < |t t0| .This proves that

sign j(t0 + ) = sign j(t0 ) = sign j(t0).

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Hence

sign ((A, t0)) = # {j : 0 < j(t0 + ) < } # {j : 0 < j(t0 ) < } .

The right hand side is unchanged by small perturbation and agrees with thespectral flow across the interval [t0 , t0 + ] for a nearby curve in A withsimple crossings. This proves that the intersection number of A with S1 at t0is the signature of the crossing operator. 2

Proof of Theorem 4.2: By Corollary 4.9 cover the set {(t, ) : (A(t))}by countably many graphs of smooth curves t j(t) each defined on aninterval [aj , bj ]. By Sards theorem the complement of the set of common regularvalues has measure zero. By Corollary 4.9 R is a common regular value ofthe functions j if an only if A 1l has only regular crossings. 2

Proof of Theorem 4.1: We prove that minus the Fredholm index satisfiesthe axioms of Theorem 4.3:

A1(R, W , H ) Z : A index DA.

The homotopy and direct sum axioms are obvious. The constant axiom fol-lows from Corollary 3.13, the normalization axiom from Theorem 2.1, and thecatenation axiom from Proposition 4.6. Hence

index DA = (A)

for A A1(R,W,H). To prove this formula in general approximate a curve

which satisfy (A-1), (A-2), and (A-3) (but is only continuously differentiable inthe weak operator topology) by a curve in A1(R,W,H). Finally use Lemma 4.7.2

We include here some observations about catenation. Assume A, Ar Asuch that A(t) = Ar(t) for t 0. Form the shifted catenation

A = A#Ar

by A(t) = A(t + ) for t 0 and A(t) = Ar(t ) for t 0.

Proposition 4.10 If the operators DA andDAr are onto (resp injective), thenthe operator DA is onto (resp injective) for sufficiently large.

Proof: We consider the injective case; the onto case follows by duality. Byassumption there exist constants c > 0 and cr > 0 such that

W c DAH , W cr DArH ,

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for every W. Choose a nondecreasing cutoff function : R R such that(t) = 1 for t T, (t) = 0 for t T and (t) 1/T. Then for T

W (1 )W + W

c DA((1 ))H + cr DAr()H

c (1 )DAH + cr DArH + (c + cr)1

TH

(c + cr)

DAH +

1

TH

.

This estimate shows that DA is injective for T sufficiently large. Note that inthe estimate

W cDAH

the constant c is independent of > T. 2

The operator A = A#Ar is constant and bijective on the time interval t . The proof of Proposition 3.12 shows that every in the kernel ofDA satisfies an estimate

d2

dt22H

22H

on this interval. Hence for large , (0) must be small. This shows that elementsof the kernel of DA are roughly of the form = #r where ker DAand r ker DAr . Conversely the catenation of two such elements and rcan be approximated by an element in the kernel of DA. The argument usesProposition 4.10. This provides an alternative proof for the catenation axiom.

5 The Maslov index

Let (E, ) be a symplectic vector space and denote by L = L(E, ) the manifoldof Lagrangian subspaces of E. The Maslov index as defined in [21] assigns toevery pair of Lagrangian paths , : [a, b] L(E, ) a half integer (, ).In this section we enumerate the properties of the Maslov index that will beneeded in the sequel.

Any two symplectic vector spaces of the same dimension are symplectomor-phic. The Maslov satisfies the naturality property

(, ) = (, ) (14)

for a symplectomorphism : (E, ) (E, ). Hence we will give our defini-tions for the standard symplectic vector space

E = R2n, = 0 =

nj=1

dxj dyj.

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The only other example which we need is

E = R2n R2n, = (0) 0.

In the latter case the graph of a symplectomorphism of (R2n, 0) is an elementof L. The Maslov index has the following properties.

(naturality) Equation (14) holds when is time dependent.

(homotopy) The Maslov index is invariant under fixed endpoint homotopies

(zero) If (t) (t) is of constant dimension then (, ) = 0.

(direct sum) If E = E1 E2 then

(1 2, 1 2) = (1, 1) + (2, 2).

(catenation) For a < c < b

(, ) = (|[a,c], |[a,c]) + (|[c,b],

|[c,b]).

(localization) If (E, ) = (R2n, 0), (t) = Rn 0 and (t) = Gr(A(t)) for a

path A : [a, b] Rnn of symmetric matrices then the Maslov index of is given by the spectral flow

(, ) = 12 signA(b) 12 signA(a). (15)

Remark 5.1 These axioms characterize the Maslov index (see [21]). By thelocalization property the spectral flow of a path of finite dimensional symmetricmatrices is a special case of the Maslov index. However, we only define thespectral flow in the case where the matrices A(a) and A(b) are invertible whereasthe Maslov index is defined for any path. The reason for the former is that theoperator DA is not Fredholm unless A

are invertible. The reason for the latteris that it is often necessary to consider Lagrangian pairs with (a) = (a).

For t [a, b] and (t) = V constant the crossing form (, V , t) is aquadratic form on (t) V defined as follows. Let W be a fixed Lagrangiancomplement of (t). For v (t) V and s t small define w(s) W byv + w(s) (s). The form

(, V , t)(v) = dds s=t (v, w(s))is independent of the choice of W. In general the crossing form (, , t) isdefined on (t) (t) and is given by

(, , t) = (, (t), t) (, (t), t).

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A crossing is a time t [a, b] such that (t) (t) = {0}. A crossing iscalled regular if (, , t) is nondegenerate. It is called simple if in addition(t) (t) is one dimensional. For a pair with only regular crossings theMaslov index is defined by

(, ) = 12 sign (, , a) +

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6 The Morse index

6.1 Sturm oscillation

Consider the operator family A(t) : W H defined by

A(t) = d2

ds2 q(s, t)

withH = L2([0, 1]), W = W2,2([0, 1]) W1,20 ([0, 1]).

Here s is the coordinate on [0, 1]. Assume that q is C1 on the closed strip[0, 1] R and independent oft for |t| T. Let = (s, t) be the solution of theinitial value problem

2

s2+ q = 0, (0, t) = 0,

s(0, t) = 1.

Define q(s) = q(s, T) and (s) = (s, T) and assume that

(1) = 0.

This means that 0 is not in the spectrum of A.

Proposition 6.1 The spectral flow of A(t) is given by

(A) = () (+)

where () denotes the number of zeros of the function (s) in the interval0 < s 1.

Proof: First note that all eigenvalues of A(t) have multiplicity 1. By The-orem 4.2 we may assume that all crossings are simple. We investigate thebehaviour of the quotient /s along the boundary s = 1. Differentiating theidentity 1

0

(s)

2 q2

ds = (1, t)s(1, t)

with respect to t and integrating by parts we obtain

1

0 (tq)2

ds = t(1, t)s(1, t) (1, t)ts(1, t).

At a crossing t the left hand side is the crossing operator (up to a scalar multiple):

10

(tq)2 ds1

02 ds

= tr(PAP) = (t).

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Here (t) is the crossing eigenvalue. Hence

sign (t) = signd

dt

(1, t)

s(1, t).

Since s (s, t) solves a second order equation and s cannot vanish si-multaneously. Hence the loop of lines R((s, t), s(s, t)) R2 around theboundary of [T, T] [0, 1] is contractible. Set t = T and let s run from 0 to1: The intersections of the line R((s, T), s(s, T)) with the vertical R(0, 1)count the zeros of

() =

(s,T)=0

signd

ds

(s, T)

s(s, T)=

(s,T)=0

1.

The intersections of the line R((1, t), s(1, t)) with the vertical R(0, 1) countthe crossings of A(t)

(A) =

(1,t)=0

sign (t)

=

(1,t)=0

signd

dt

(1, t)

s(1, t)

=

(s,T)=0

signd

ds

(s, T)

s(s, T)

(s,T)=0

signd

ds

(s, T)

s(s, T)

= () (+).

2

Corollary 6.2 (The Sturm oscillation theorem) The n-th eigenfunction ofthe problem

2u

s2+ qu + u = 0, u(0) = u(1) = 0

has n 1 interior zeros.

Proof: Consider the spectral flow for the operator family

A(t) = d2

ds2 q b(t)

where b : R R is a smooth function such that b(t) = b

< 1 for t 1 andn1 < b(t) = b+ < n for t 1. 2

This proof also shows that if the operator A : W H defined by Au =d2u/ds2 qu is invertible then its Morse index (the number of negative eigen-values) is the number of zeros of the fundamental solution (s) in the interval0 < s < 1. This is a special case of the Morse index theorem proved below.

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6.2 The Morse index theorem

In suitable coordinates the Jacobi equation in differential geometry has the form

Au = d2u

ds2 Q(s)u = 0. (17)

Here s [0, 1], u(s) Rn, and Q(s) is a symmetric matrix representiong thecurvature. This generalizes the previous example from 1 dimension to n. Wecall s0 (0, 1] a conjugate point of A iff there is a non-trivial solution u ofequation (17) satisfying u(0) = u(s0) = 0. The dimension of the vector spaceof all solutions u of (17) satisfying u(0) = u(s0) = 0 is called the multiplicityof the conjugate point. Denote by (A) the number of conjugate points of Ain the interval 0 < s 1 counted with multiplicity. Let (s) Rnn be the

fundamental solution of (17) defined by

d2

ds2+ Q = 0, (0) = 0,

d

ds(0) = 1l,

for 0 s 1. Then

(s) = range

(s)

(s)

is a Lagrangian plane for every s.

Proposition 6.3 Assume det (1) = 0. Then the number (A) of conjugatepoints is related to the Maslov index of by

(, V) = (A)

n

2

where V = 0 Rn.

Proof: Suppose s0 is a crossing of multiplicity m0. By Remark 5.2 the crossingform is given by

(, V , s0)(v) = (s0)u0, (s0)u0, v = (0, (s0)u0), (s0)u0 = 0.

Since (s0) is injective on the kernel of (s0) the crossing form is negativedefinite and of rank m0. This shows that all crossings are regular. Moreover,s0 = 0 is a crossing with crossing index m0 = n. The Proposition is proved bysumming over the crossings:

(, V) = 12

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Let H and W be as before but tensored with Rn:

H = L2([0, 1],Rn), W = W2,2([0, 1],Rn) W1,20 ([0, 1],Rn),

and replace the operator of (17) by a one-parameter family

A(t) = d2

ds2 Q(s, t).

Assume that Q is C1 on the closed strip [0, 1] R and independent of t for|t| T. Assume that 1 is not a conjugate point for either of the operators A.This says that these operators are invertible.

Proposition 6.4 The spectral flow of A(t) is given by

(A) = (A) (A+).

Proof: Let = (s, t) Rnn be the fundamental solution defined by

2

s2+ Q = 0, (0, t) = 0,

s(0, t) = 1l.

Then the operator A(t) is injective if and only if det (1, t) = 0. The kernel ofA(t) consists of all functions of the form

u(s) = (s, t)u0, (1, t)u0 = 0.

Think of the crossing operator (A, t) as a quadratic form on the kernel of A(t):

(A, t)(u) = 10

u(s), tQ(s, t)u(s) ds.

The next lemma shows that that this agrees with the crossing form of theLagrangian path t (1, t) with the vertical V = 0 Rn evaluated at u0 where

(s, t) = range

(s, t)

s(s, t)

.

Hence

(A) = ((1, ), V)

= (

+

, V) (

, V)= (A) (A+).

The second equality follows from the fact that the loop of Lagrangian subspaces(s, t) around the boundary of the square [0 , 1] [T, T] is contractible. Thelast equality follows from Proposition 6.3. 2

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Lemma 6.5 LetV = 0 Rn. Then

(A, t)(u) = ((1, ), V , t)(v)

for u(s) = (s, t)u0 and v = (0, s(1, t)u0) with (1, t)u0 = 0.

Proof: Differentiate the identity10

(s)

Ts TQ

dx = (1, t)Ts(1, t)

with respect to t and integrate by parts to obtain

1

0

T(tQ) ds = t(1, t)Ts(1, t) ts(1, t)

T(1, t).

Now multiply on the left an right with u0 where (1, t)u0 = 0. The result is10

u(s), tQ(s, t)u(s) ds = s(1, t)u0, t(1, t)u0.

The left hand side is (A, t)(u) and the right hand side is ((1, ), V , t)(v).2

Corollary 6.6 (The Morse index theorem) Assume that the operator A :W H defined by (17) is invertible. Then its Morse index (the number ofnegative eigenvalues) is the number (A) of conjugate points.

Proof: Consider the spectral flow for the operator family

A(t) = d2

ds2 Q b(t)

where b : R R is a smooth function such that b(t) = b < 1 for t 1 andb(t) = 0 for t 1. 2

7 Cauchy-Riemann operators

Denote by

J0 = 0 1l1l 0 .the standard complex structure on R2n = Cn. Consider the perturbed CauchyRiemann operator

S, =

t J0

s+ S (18)

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where : [0, 1] R R2n satisfies the non-local boundary condition

((0, t), (1, t)) (t). (19)

Here (t) R2n R2n is a path of Lagrangian subspaces and S(s, t) R2n2n

is a family of matrices. We impose the following conditions:

(CR-1) The function : R L(R2n R2n, (0) 0) is of class C1. More-over, there exist Lagrangian subspaces and a constant T > 0 with(t) = + for t T and (t) = for t T.

(CR-2) The function S : [0, 1] R R2n2n is continuous. Moreover, thereexist symmetric matrix valued functions S : [0, 1] R2n2n such that

limt sup0s1 S(s, t) S

(s) = 0.

(CR-3) Let : [0, 1] Sp(2n) be defined by

s+ J0S

= 0, (0) = 1l.

Then the graph of (1) is transverse to .

The operator S, has the form DA = d/dt A(t) but in contrast to section 3the domain of the operator A(t) : W(t) H depends on t so Theorem 3.10 doesnot apply directly. We overcome this difficulty below by changing coordinates.Condition (CR-1) asserts that the domain of the operator A(t) is independent

of t for |t| T. The Lagrangian boundary condition and the symmetry of Simply that the limit operators A are self-adjoint. Condition (CR-3) assertsthat these operators are invertible. Abbreviate

L2 = L2([0, 1] R,R2n),

W1,2 =

W1,2([0, 1] R,R2n) : ((0, t), (1, t)) (t)

.

In the case (t) of constant boundary conditions these are the spacesH = L2 and W = W1,2 of section 3.

Theorem 7.1 The operator S, : W1,2 L

2 is Fredholm. Its index is givenby

index S, = (Gr(), ) (Gr(+), +) (, ) (20)

Proof: We prove the theorem in five steps.

Step 1: Let(s, t) Sp(2n) be defined by

s+ J0

S+ ST

2 = 0, (0, t) = 1l. (21)

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Then

(Gr((1, )), ) = (, ) + (Gr(+), +) (Gr(), ). (22)

By condition (CR-2) we have

(s) = limt

(s, t).

By condition (CR-3) the path s (s, T) has the same Maslov index as for T sufficiently large. Hence step 1 follows by considering the loops ofLagrangrangian subspaces (s, t) = (t) and (s, t) = Gr((s, t)) around theboundary of the rectangle [0, 1] [T, T]. In view of step 1 it suffices to provethat

index S, = (Gr((1, )), ). (23)

Step 2: The theorem holds when (t) = V V, S(s, t) = S(s, t)T is sym-metric and continuouly differentiable, and the path t (1, t) has only simplecrossings.

By Theorem 3.10 the operator is Fredholm and by Theorem 4.1 the Fredholmindex is given by the spectral flow for the self-adjoint operator family

A(t) = J0d

ds S(s, t)

on H = L2([0, 1],R2n) with dense domain

W = W1,20 ([0, 1],Rn) W1,2([0, 1],Rn).

We examine the crossing operator (A, t) at acrossing t. The operator A(t) isinjective if and only if (1, t)V V = 0 where (s, t) Sp(2n) is defined as instep 1. The kernel of A(t) consists of all functions of the form

(s) = (s, t)v, v = (0, y), B(1, t)y = 0.

Here B(1, t) is the right upper block in the block decomposition (16) of (1, t).Think of the crossing operator (A, t) as a quadratic form on the kernel of A:

(A, t)() =

10

(s), tS(s, t)(s) ds.

We shall prove in the lemma below that this form agrees with ((1 , ), t).Hence the operator family A(t) has only regular crossings and

index S,VV = (A) = ((1, )) = (Gr((1, )), V V).

The last equality follows from Remark 5.4. Now use step 1.

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Lemma 7.2 We have

(A, t)() = ((1, ), t)(y)

for (s) = (s, t)v with v = (0, y) V and B(1, t)y = 0.

Proof: Differentiate the identity

S = J0s

with respect to t, multiply on the left by T and integrate by parts to obtain10

T(tS) ds =

10

TJ0st ds

10

TSt ds

=10

TJ0st ds +10

(s)TJ0t ds

= (1, t)TJ0t(1, t).

Now multiply on the left and right by v = (0, y) with B(1, t)y = 0 to obtain

(A, t)() = (1, t)v, J0t(1, t)v

= D(1, t)y, tB(1, t)y

= ((1, ), t)(y).

Here D(1, t) denote the lower right block in the decomposition (16) of (1, t).The last equality follows from Remark 5.4. This proves the lemma.

Step 3: The theorem holds when (t) = V V.

Choose any smooth cutoff function : R [0, 1] such that (t) = 0 fort T and (t) = 1 for t T and replace S by

S(s, t) = (t)S+(s) + (1 (t))S(s).

Then the right hand side of (18) is unchanged. Moreover, the multiplicationoperator induced by S S satisfies the assumptions of Lemma 3.16. Hence,by Corollary 3.17, the operator S,VV is Fredholm and has the same index asS,VV. Now choose a small perturbation to obtain a symmetric C

1-functionS such that the associated symplectic path t (1, t) has only simple cross-

ings. Finally use step 2.Step 4: The theorem holds in the case of local boundary conditions

(t) = 0(t) 1(t)

with j(t) L(R2n, 0).

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Identify R2n = Cn and choose a unitary transformation (s, t) U(n) =O(2n) GL(n,C) of class C1 such that (s, t) is independent of t for |t| T.Then

S, = S,

where

S = 1

t 1J0

s+ 1S

and = 0 1 with

0(t) = (0, t)10(t),

1(t) = (1, t)

11(t).

Since 1 = T the matrix S(s, t) is symmetric for |t| T. The correspondingsymplectic matrices (s, t) Sp(2n) defined by (21) with S replace by S are

given by(s, t) = (s, t)1(s, t)(0, t).

Now denote (t) = (0, t) (1, t) Sp(R2n R2n, (0) 0). Then

Gr((1, t)) = (t)1Gr((1, t)), (t) = (t)1(t).

and, by the naturality axiom for the Maslov index,

(Gr((1, )), ) = (Gr((1, )), ).

Now choose such that (t) = V V and use step 3.

Step 5: The general case.

Define the operator T : L2([0, 1] R,R2n) L2([0, 1] R,R2n R2n) whichsends to the pair T = = (0, 1) where

0(s, t) = ((1 s)/2, t/2), 1(s, t) = ((1 + s)/2, t/2).

If W1,2 then satisfies the local boundary conditions

(0, t) = 0(t), (1, t) (t) = 1(t).

Moreover, the operator T S, T1 is given by

(0, 1) (t0 + J0t + S00, t1 J0t + S11)

where S0(s, t) = S((1 s)/2, t/2)/2 and S1(s, t) = S((1+ s)/2, t/2)/2. This is aCauchy-Riemann operator with respect to the complex structure J = (J0)J0which is compatible with = (0) 0. The corresponding fundamentalsolution (s, t) = 0(s, t) 1(s, t) is given by

0(s, t) = ((1 s)/2, t/2)(1/2, t/2)1,

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1

(s, t) = ((1 + s)/2, t/2)(1/2, t/2)1.

Hence (1, t) = Gr((1, t/2)) and it follows that

(Gr((1, )), ) = ((1, )0, 1)

By step 4 the operator T S, T1 is Fredholm and its index is given by

index T S, T1 = ((1, )0, 1).

Hence S, is a Fredholm operator and

index S, = (Gr((1, )), ).

By step 1 this proves the theorem. 2

Remark 7.3 (Periodic boundary conditions) Assume (t) = for all t.Then condition (5) above means that

det(1l (1)) = 0.

In this case the Fredholm index is related to the Conley-Zehnder index by

index S, = CZ () CZ (

+).

This result was proved in [25]. With these boundary conditions the operatorS plays a central role in Floer homology for symplectomorphisms. The mod 2index is the relative fixed point index () = sign det(1l+(1))sign det(1l

(1)):(1)index S = ().

As a result the Euler characteristic of Floer homology for a symplectomorphismis the Lefschetz number [11], [6].

Remark 7.4 (Local boundary conditions) Assume S = 0 and

(t) = 0(t) 1(t)

where 0(t), 1(t) L(R2n, 0). Then condition (5) above means that

0(T) 1(T) = 0.

By Remark 5.3 the Fredholm index is given by

index = (0, 1).

This was proved by Floer [8] using results by Viterbo [28]. With these bound-ary conditions the operator plays a central role in Floer homology for La-grangian intersections. The mod 2 index is the relative intersection number

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(0

, 1

). Choose orientations of 0

and 1

and define (0

, 1

) = 1 ac-cording to whether the induced orientations on R2n = 0(T) 1(T) agree.Then

(1)index = (0, 1).

As a result the Euler characteristic of Floer homology for a pair of Lagrangiansubmanifolds is the intersection number [9].

Remark 7.5 (Dirichlet boundary conditions) Assume S = ST is symmet-ric and (t) = V V where V = 0 Rn is the vertical. Then condition (5)above means that (1)V V = 0. By Remark 5.4 the Fredholm index is givenby

index S, = () (+).

In this case the results of section 3 apply and the operator A = A(t) is given by

A = J0 S

for = (, ) with boundary condition (0) = (1) = 0. This operator appearsas the second variation of the symplectic action on phase space. The signatureof A is undefined since the index and the coindex are both infinite. In [23] weinterpret the Maslov index as the signature of A via finite dimensional approx-imation:

sign A = 2().

Hence the index theorem can be written in the form

index S, = 12 sign A 12sign A

+.

This is consistent with the finite dimensional formula in Remark 2.2.

Remark 7.6 (Totally real boundary conditions) The operator S, con-tinues to be Fredholm when (t) is only totally real with respect to the complexstructure (J0) J0 on R2

n R2n. To see this in the case of local boundaryconditions = 01 choose a family of symplectic forms (s, t) on R2

n whichare compatible with J0 and satisfy

0(t) L(R2n, (0, t)), 1(t) L(R

2n, (1, t)).

Now choose a unitary frame

(s, t) : (R2n, J0, 0) (R2n, J0, (s, t))

and consider the operator S, in the new coordinates = 1. Then the

operator has the above form with Lagrangian boundary conditions. Its index isindependent of the choice of since the space of all symplectic forms on R2n

which are compatible with J0 is contractible. The general case can be reducedto that of local boundary conditions as in Step 5 of the proof of Theorem 7.1.

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Robbin and Salamon, Spectral Flow