Morse Theory for Manifolds With Boundary

8/11/2019 Morse Theory for Manifolds With Boundary

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arXiv:1207

.3066v3

[math.GT

]27Nov2012

MORSE THEORY FOR MANIFOLDS WITH BOUNDARY

MACIEJ BORODZIK, ANDRAS NEMETHI, AND ANDREW RANICKI

Abstract. We develop Morse theory for manifolds with boundary. Beside standard andexpected facts like the handle cancellation theorem and the Morse lemma for manifoldswith boundary, we prove that under suitable connectedness assumptions a critical pointin the interior of a Morse function can be moved to the boundary, where it splits intoa pair of b oundary critical points. As an application, we prove that every cobordism ofconnected manifolds with boundary splits as a union of left product cobordisms and rightproduct cobordisms.

1. IntroductionFor some time now, Morse theory has been a very fruitful tool in the topology of mani-

folds. One of the milestones was the h-cobordism theorem of Smale [Sm2], and its Morse-theoretic exposition by Milnor[Mi1, Mi2]. Recently, Morse theory has become even morepopular, for two reasons. In the first instance, on account of its connections with Floerhomology, see e.g. [Sa, Wi, Ni, KM]. Secondly, the stratified Morse theory developed byGoresky and MacPherson[GM]. In the last 20 years Morse theory has also had an enormousimpact on the singularity theory of complex algebraic and analytic varieties.

Despite much previous interest in Morse theory, there still remain uncharted territories.Morse theory for manifolds with boundary is a particular example. The theory was initiatedby Kronheimer and Mrowka in [KM], and there is also a recent paper of Laudenbach [La]

devoted to the subject. Our paper is a further contribution.In this paper we prove some new results in the Morse theory for manifolds with boundary.Beside some standard and expected results, like the boundary handle cancellation theorem(Theorem 5.1) and the topological description of passing critical points on the boundary(using the notions of right and left half-handles introduced in Section 2) we discover anew phenomenon. An interior critical point can be moved to the boundary and there splitinto two b oundary critical points. In particular, if we have a cobordism of manifolds withboundary, then under a natural topological assumption we can find a Morse function whichhas only boundary critical points. We use this result to prove a structure theorem forconnected cobordisms of connected manifolds with connected non-empty boundary: sucha cobordism splits as a union of left and right product cobordisms. This is a topologicalcounterpart to the algebraic splitting of cobordisms obtained in [BNR]: an algebraic splitting

of the chain complex cobordism of a geometric cobordism can be realized topologically bya geometric splitting.The structure of the paper is the following. After preliminaries in Section1.1we study

in Section2 the changes in the topology of the level sets when crossing a boundary criticalpoint. Theorem 2.24 is the main result: passing a boundary stable (unstable) criticalpoint produces a left (right) half-handle attachment. In Section3we prove Theorem 3.1,

Date: November 28, 2012.2010 Mathematics Subject Classification. primary: 57R19, secondary: 58E05, 58A05.Key words and phrases. Morse theory, manifold with boundary, cobordism, bifurcation of singular points.The first author is supported by Polish MNiSzW Grant No N N201 397937. The second author is partially

supported by OTKA Grant K100796.

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2 MACIEJ BORODZIK, ANDRAS NEMETHI, AND ANDREW RANICKI

which moves interior critical points to the boundary. Then we pass to some more standardresults, namely rearrangements of critical points in Section 4. We finish the section withour most important up to now application, Theorem4.18, about the splitting of acobordism into left product and right product cobordisms. Finally, in Section5we discuss

the possibility of cancelling a pair of critical points.

Acknowledgements. The first author wishes to thank Renyi Institute for hospitality, thefirst two authors are grateful for Edinburgh Mathematical Society for a travel grant toEdinburgh and Glasgow in March 2012. The authors thank Andras Juhasz and AndrasStipsicz for fruitful discussions and to Rob Kirby for drawing their attention to [KM].

1.1. Notes on gradient vector fields. To fix the notation, let us recall what a cobordismof manifolds with boundary is.

Definition 1.1. Let 0 and 1 be compact oriented, n-dimensional manifolds with non-empty boundary M0 and M1. We shall say that (, Y) is a cobordismbetween (0, M0)and (1, M1), if is a compact, oriented (n + 1)-dimensional manifold with boundary

=Y 0 1, where Y is nonempty, 0 1=, and Y 0= M0, Y 1= M1.Remark 1.2. Strictly speaking, is a manifold with corners, so around a point xM0 M1 it is locally modelled by Rn1 R20. Accordingly, sometimes we write that 0,1 and Y, as manifolds with boundary, have tubular neighbourhoods in of the form0 [0, 1), 1 [0, 1), orY [0, 1), respectively. Nevertheless, in most cases it is safe (andmore convenient) to assume that is a manifold with boundary, i.e. that the corners aresmoothed along M0 and M1. Whenever p ossible we make this simplification in order toavoid unnecessary technicalities.

Example 1.3. Given a manifold with boundary (, M), we call (, M)[0, 1] a trivialcobordism, with = [0, 1], Y =M [0, 1], i= {i},Mi=M {i} fori= 0, 1.

We recall the notion of a Morse function. For this it is convenient to fix a Riemannianmetric g on .

Definition 1.4. A function F: [0, 1] is called a Morse function on the cobordism(, Y) ifF(0) = 0, F(1) = 1, F has only Morse critical points, the critical points arenot situated on 0 1, andF is everywhere tangent to Y.

There are two ways of doing Morse theory on manifolds. One can either consider thegradient flow ofF associated withFand the Riemannian metric (in the Floer theory, oneoften usesF), or, the so-called gradient-like vector field.Definition 1.5. (See [Mi2, Definition 3.1].) LetF be a Morse function on a cobordism(, Y). Let be a vector field on . We shall say that is gradient-like with respect to F,if the following conditions are satisfied:

(a) F >0 away from the set of critical points ofF;(b) ifp is a critical point ofFof indexk, then there exist local coordinates x1, . . . , xn+1

in a neighbourhood ofp, such that

F(x1, . . . , xn+1) =F(p) (x21+ + x2k) + (x2k+1+ + x2n+1)and

= (x1, . . . , xk, xk+1, . . . , xn+1) in U;(b) furthermore, ifp is a boundary critical point, then the above coordinate system can

be chosen so that Y ={xj = 0} and U={xj 0}for some j {1, . . . , n + 1}.(c) is everywhere tangent to Y ;


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MORSE THEORY FOR MANIFOLDS WITH BOUNDARY 3

The conditions (a) and (b) are the same as in the classical case. Condition (b) is aspecification of the condition (b) in the boundary case, compare Lemma 2.6.

Smale in [Sm1] noticed that for any gradient-like vector field for a function F thereexists a choice of the Riemannian metric such that =

F in that metric. The situation

is identical in the boundary case. This is stated explicitly in the following lemma, whoseproof is straightforward and will be omitted.

Lemma 1.6. Let U be a paracompact k-dimensional manifold and F: U R a Morsefunction without critical points. Assume that is a gradientlike vector field onU. Thenthere exists a metricg onUsuch that=F in that metric.

Similar statement holds ifUhas boundary andis everywhere tangent to the boundary.

Hence the two approaches gradients and gradient-like vector fields are equivalent.However, we shall need both approaches. In Section3we use gradients of functions and aspecific choice of a metric, because the argument becomes slightly simpler. In Section 5 wefollow[Mi2]very closely; as he uses gradient-like vector fields, we use them as well.

The next result shows that the condition from Definition 1.4 that

F is everywhere

tangent toYcan be relaxed. We shall use this result in Proposition 4.1.Lemma 1.7. Let be a compact, Riemannian manifold of dimension(n + 1)andYbe compact as well. Letg denote the metric. Suppose that there exists a functionF: R,and a relative open subsetU Y such thatFis tangent toYat each pointyU. Supposefurthermore, that for anyyY\ U we have

ker dFTyY.Then, for any open neighbourhood W ofY\ U, there exist a metrich on, agreeingwithg away fromW, such thathF(the gradient in the new metric) is everywhere tangentto Y.

Proof. Let us fix a point y Y\ U and consider a small open neighbourhood Uy ofy inW, in which we choose local coordinates x1, . . . , xn+1 such that Y Uy ={xn+1 = 0} andUy {xn+1 0}. In these coordinates we have dF =n+1i=1 fi(x) dxi for some smoothfunctions f1, . . . , f n+1. By the assumption, for each x Uy , there exists i n such thatfi(x)= 0. Shrinking Uy if needed, we may assume that for each xUy we havefi(x)= 0 forsomei. We may suppose that i = 1, hencef1(x)> 0. Let us choose a symmetric positivedefinite matrix Ay ={aij(x)}n+1i,j=1 so that a11 =f1(x) and for i > 1, a1i = ai1 = fi(x).Ay defines a metric hy on Uy such thathyF = (1, 0, . . . , 0)T Y in that metric.

Now let us choose an open subsetV of \ (Y\ U) such thatVyY\UUy is a coveringof . Let{V} {y}yY\Ube a partition of unity subordinate to this covering. Define

h= V g+

yY\Uyhy .

Thenh is a metric, which agrees withg away fromW. Moreover, as for each metrichy , andx Uy Y we havehyF(x) TxY by construction, the same holds for a convex linearcombination of metrics.

2. Boundary stable and unstable critical points

2.1. Morse function for manifolds with boundary. The whole discussion of Morsefunctions on manifolds with boundary would be pointless if we did not have the following.

Lemma 2.1. Morse functions exist. In fact, for any Morse function f: Y [0, 1] withf(M0) = 0, f(M1) = 1 there exists a Morse functionF: [0, 1] whose restriction to Yisf.


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Figure 1. Boundary stable (on the left) and unstable critical points.

Proof. Let f: Y [0, 1] be a Morse function on the boundary, such that f(M0) = 0 andf(M1) = 1. We want to extend fto a Morse function on .

First, let us choose a small tubular neighbourhood U ofY and a diffeomorphism U=Y [0, ) for some >0. Let F: U[0, 1] be given by the formula(2.2) U=Y [0, )(x, t) F(x, t) =f(x) f(x)(1 f(x))t2.The factorf(x)(1f(x)) ensures that Fattains values in the interval [0, 1] and F1(i)ifori {0, 1}. It is obvious that there exists a smooth function F: [0, 1], which agreeson Y [0, /2) with F, and it satisfies the Morse condition on the whole . The gradientF is everywhere tangent to Y . Remark 2.3. The above construction yields a function with the property that all itsboundary critical points are boundary stable (see Definition 2.4below). This is due to thechoice of sign1 in front of f(x)(1f(x))t2 in (2.2). If we change the sign to +1, weobtain a function with all boundary critical points boundary unstable.

We fix a Morse function F: [0, 1] and we start to analyze its critical points. Letzbe such a point. Ifz \ Y, we shall call it an interior critical point. IfzY, it will becalled aboundarycritical point. There are two types of boundary critical points.

Definition 2.4. Let z be a boundary critical point. We shall call it boundary stable, ifthe tangent space to the unstable manifold ofz lies entirely in TzY, otherwise it is calledboundary unstable.

The index of the boundary critical point z is defined as the dimension of the stable

manifold Wsz . Ifz is boundary unstable, this is the same as the index ofz regarded as acritical point of the restriction f of F on Y. If z is boundary stable, we have indFz =indfz + 1. In particular, there are no boundary stable critical point with index 0, norboundary unstable critical points of index n + 1.

Remark 2.5. We point out that we use the flow ofFand not ofFas Kronheimerand Mrowka[KM]do, hence our definitions and formulae are slightly different from theirs.

We finish this subsection with three standard results.

Lemma 2.6 (Boundary Morse Lemma). Assume that F has a critical point z Y suchthat the Hessian D2F(z) at z is non-degenerate, andF is everywhere tangent to Y.


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Then there are local coordinates (x1, . . . , xn+1) in an open neighbourhood U z such thatU ={x21+ +x2n+1 2} {x1 0} andU Y ={x1 = 0} for some > 0, andF inthese coordinates has the formx21 x22 x2n+1+ F(z).

Proof. We choose a coordinate system y1, . . . , yn+1 in a neighbourhood U ofz suchthat z = (0, . . . , 0), Y ={y1 = 0}, U={y1 0}, and the vector field y1 is orthogonal toY. We might assume F(z) = 0. The tangency ofF to Y implies that at each point ofY

(2.7) F

y1(0, y2, . . . , yn+1) = 0.

By Hadamards lemma one writes

(2.8) F =y1K1(y1, . . . , yn+1) +

n+1j=2

yjKj(y1, y2, . . . , yn+1),

for some functions K1, . . . , K n+1. We can assume that for j > 1, Kj does not dependon y1. Indeed, if it does depend, we might write Kj(y1, . . . , yn+1) = Kj(0, y2, . . . , yn+1) +

y1L1j(y1, . . . , yn+1), and then replace Kj byKj(0, y2, . . . , yn+1) and K1 byK1+ yjL1j .The condition (2.7) implies now that K1(0, y2, . . . , yn+1) = 0, hence K1= y1H11(y1, . . . , yn+1).

By Hadamards lemma applied to K2, . . . , K n+1 we get

(2.9) F=y21H11(y1, . . . , yn+1) +

nj,k=2

yjykHjk(y2, . . . , yn+1).

The non-degeneracy ofD2F(z) means that H11(z)= 0. Then, after replacing y1H11

byx1, we can assume that H11=1. Finally, the sum in (2.9) can be written as

j2 jx2j

(j =1) by the classical Morse lemma[Mi1,Lemma 2.2]. The next result is completely standard by now.

Lemma 2.10. Assume thatF is a Morse function on a cobordism(, Y) between(0, M0)and(1, M1). IfFhas no critical points then(, Y)=(0, M0) [0, 1]. Furthermore, wecan choose the diffeomorphism to map the level setF1(t) to the set0 {t}.Proof. The proof is identical to the classical case, see e.g. [Mi2,Theorem 3.4].

2.2. Half-handles. For anyk we consider thekdimensional discDk ={x21 + +x2k 1}.In the classical theory, an ndimensional handle of index k is the ndimensional manifoldH=Dk Dnk with boundary

H =

Dk Dnk

Dk Dnk

= B0 B0 .Given annmanifold with boundary (, ) and a distinguished embedding : B0,

the effect of a classical handle attachment is the n-dimensional manifold with boundary(, ) = ( H, ( \ B0) B0),

where we glue along (B0) identified with B0. The boundary is the effect of surgery

on (B0). We now extend this construction to relative cobordisms of manifolds withboundary, using half-handles. Since our ambient space is (n+ 1)dimensional, (n+ 1)is the dimension of the handles, and they induce ndimensional handle attachments on Y.

In order to do this, for any k 1 we distinguish the following subsets of Dk: thehalf-disc Dk+ := D

k {x1 0}, and its boundary subsets Sk1+ := Dk {x1 0},Sk20 := D

k {x1= 0}and Dk10 :=Dk {x1= 0}. Clearly, Sk20 is a boundary the two(k 1)discs Sk1+ and Dk10 . We will callx1 the cutting coordinate.


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B

B B0

B0

C

Figure 2. A right half-handle of index 1. The picture on the left is thehandle, the two other pictures explain the notation. The two half-circlesformB,Cis the bottom rectangle.

Definition 2.11. Let 0 k n. An (n + 1)-dimensional right half-handle of indexk is the(n+ 1)-dimensional manifold Hright=D

k Dn+1k+ , with boundary subdivided into threepiecesHright= B C N, where

B:= D

k

Dn+1

k

+ , C :=D

k

Dn

k

0 , N :=D

k

Sn

k

+ .One has the following intersections too

B0:= C B= Dk Dnk0 , N0 := C N=Dk Snk10 .Hence the handle H is cut along C into two pieces, one of them is the half-handle Hright.Note that (C, B0) is andimensional handle of index k .

Symmetrically, we define the left half-handles by cutting the handle H along the leftcomponent disc Dk.

Definition 2.12. Fix k with 1 k n+ 1. An (n+ 1)dimensional left half-handle ofindex k is the (n+ 1)dimensional diskHleft := D

k+ Dn+1k with boundary subdivided

into three pieces Hleft= B C N, whereB :=Sk1+ Dn+1k, C:=Dk10 Dn+1k, N :=Dk+ Dn+1k.

Furthermore, we specifyB0:= CB=Sk20 Dn+1k andN0:= NC=Dk10 Dn+1k.Ahalf-handlewill from now on refer to either a right half-handle or left half-handle. We

pass to half-handle attachments. We will attach a half-handle along B. The definitionsof the right half-handle attachment and the left half-handle attachment are formally verysimilar, but there are significant differences in the properties of the two operations.

Definition 2.13. Let (, Y; 0, M0, 1, M1) be an (n + 1)-dimensional relative cobordism.Given an embedding : (B, B0)(1, M1) define the relative cobordism (, Y; 0, M0, 1, M1)obtained from (, Y; 0, M0, 1, M1) byattaching a (right or left) half-handle of indexk by

= BH, Y= YB0 C,1= (1 \ B) N, M1= (M1\ B0) N0.

See Figure4 and Figure5for right, respectively left half-handle attachments.

We point out that in the case of the right half-handle attachment, any embedding ofB0into M1 determines (up to an isotopy) an embedding of pairs (B, B0)(1, M1). Indeed,as (B, B0) =D

k(Dn+1k+ , Dnk0 ), a map :B0M1extends to a map : B1in acollar neighbourhood ofM1 in 1. (This is not the case in the left half-handle attachment.)

In particular, in the case of right attachments, we specify only the embeddingB0M1.


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B

B0B0 C

Figure 3. A left half-handle of dimension 3 and indexk = 2. The two linesare B0, the bottom rectangle is C. B is the surface between the two halfcircles on the picture.

Example 2.14. (a) We exemplify the left half-handle attachment for k = 1. In this caseB0is empty. If we are given an embedding ofB={1}Dn into 1\M1, we glue [0, 1]Dn to alongB. Then we setY= Y{0}Dn, 1= (1\B)[0, 1]B andM =M{0}B .

(b) The right half-handle attachment of index 0 is the disconnected sum Dn+1+ withboundary D

n+1+ = S

n+ D

n0 . We think of first discS

n+ as a part of 1, while the second

disc as a part ofY, and M1 = M1 Sn10 .Remark 2.15. In the next subsection we shall see that crossing a boundary stable criticalpoints corresponds to left half-handle attachment, while a boundary unstable critical pointcorresponds to a right half-handle attachment. Theorem3.1can be interpreted informally,as splitting a handle into a right half-handle and left half-handle. This also motivates thename half-handle.

2.3. Elementary properties of half-handle attachments. The following results aretrivial consequences of the definitions.

Lemma 2.16. Let be the result of a right half-handle attachment to along(B, B0)(1, M1). LetB beB pushed slightly offM1 into the interior of1. Let be a result ofattaching a (standard) handle of indexk to alongB. Then and are diffeomorphic.

Proof. This is obvious, since when we forget about C and B0, the pair (Hright, B) is astandard (n + 1)-dimensional handle of indexk .

In particular, the effect of a right half-handle attachment on is the same as the effectof a standard handle attachment of the same index.

The situation is completely different in the case of left half-handle attachments.

Lemma 2.17. If is a result of a left half-handle attachment, then is diffeomorphic to.

Proof. By definition the pair (Hleft, B) is diffeomorphic to the pair (Dn

[0, 1], Dn

{0}

).Attaching Hleftalong B to does not change the diffeomorphism type of .

The effect on Yof a right/left half-handle attachments are almost the same.

Lemma 2.18. If(, Y; 0, M0, 1, M1)is the result of left (respectively, right) half-handle

attachment to (, Y; 0, M0, 1, M1) along (B, B0) (1, M1), then Y is a result of aclassical handle attachment of indexk 1 (respectivelyk) alongB0.Proof. This follows immediately from Definition2.13.

The effects of half handle attachment on are also easily described. The next lemma isa direct consequence of the definitions; its proof is omitted. We refer to Figures 4and5.


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1

B B

Y

M1 M1

B0B0

Y

1

M1 M1

Figure 4. Right half-handle attachment. Herek = 1, n= 2. On the right,the two black points represent a sphere S0 with a neighbourhoodB0 in M1and B in 1. On the picture on the right the dark green coloured part ofthe handle belongs to 1, the dashed lines belong to 1 and are drawn onlyto make the picture look more three-dimensional.

Lemma 2.19. (a) If(, Y; 0, M0, 1, M1)is the result of left half-handle attachment to

(, Y; 0, M0, 1, M1) along(B, B0)(1, M1), then

1=1\ B.(b) If (, Y; 0, M0, 1, M

1) is the result of index k right half-handle attachment to

(, Y; 0, M0, 1, M1) alongB0M1, then1=1 B0 N,

where hereN is anndimensional diskDk Dnk andB0= Sk1 Dnk.These facts also emphasize that right half-handle attachments and left half-handle at-

tachments are somehow dual operations on . This can be seen also at the Morse functionlevel: changing a Morse function F toFchanges all right half-handles to left-half handlesand conversely, see Section2.4and2.5below. But the above lemma shows another aspect aswell: a right half handle attachment consists on gluing a disk, a left half-handle attachmentconsists of removing a disk. Indeed, in the case of right attachment, (1, M

1) = ( 1

Dk Dnk, 1) associated with an embedding : Dk Dnk M1. On the other hand,by definition, for an embedding : (Dk1 Dn+1k, Dk1 Dn+1k) (1, M1) thepair

(1, M1) = (closure of(1 \ Dk1 Dn+1k), 1)

is obtained from (1, M1) by ahandle detachmentof index k 1. Hence we obtain:Corollary 2.20. The effect on(1, M1)of a right halfhandle attachment of indexk (at thelevel of cobordism) is a handle attachment of indexk at the level of(1, M1). Conversely,the effect on(1, M1)of a left half-handle attachment of indexk (at the level of cobordism)

is a handle detachment of indexk 1at the level of(1, M1). In particular, M1 is obtainedfromM1 as the result of ak surgery in the first case, and(k 1) surgery in the second.The duality can be seen also as follows: we can cancel any handle attachment by a

suitably defined handle detachment, and conversely.The following definition introduces a terminology which is rather self-explanatory. We

include it for completeness of the exposition.

Definition 2.21. We shall say that a cobordism (, Y) between (, M) and (, M) isa right (respectively left) half-handle attachmentof index k , if (, Y, , M) is a result ofright (respectively left) half-handle attachments of index k (in the sense of Definition2.13)to ( [0, 1], M [0, 1], {0}, M {0}, {1}, M {1}).


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1

B

Y

M1 M1

B0B0

Y1

M1 M1

Figure 5. Left handle attachment with k = 2 and n = 2. This time thesphere on the left (denoted by two points) bounds a disk in 1.

We conclude this section by studying homological properties of the handle attachment.These properties will be used in [BNR]. The proofs are standard and are left to the reader.

Let (H+, C , B , N ) be a half-handle of indexk.

Lemma 2.22. If (Hright, C , B , N ) is a right half-handle, then the pair (C, B0) is a strong

deformation retract of(Hr+, B), while(Dk, Dk) is a strong deformation retract of(C, B0).

In particular, Hj(Hr+, B)

=Hj(C, B0) = Z forj = k, and it is zero otherwise.The situation is completely different for left half-handles.

Lemma 2.23. If(Hleft, C , B , N ) is a left half-handle, then the pair(Hleft, B) retracts ontothe trivial pair (point, point). In particular, all the relative homologies H(Hleft, B) van-ish. On the other hand, (Dk10 , S

k20 ) is a strong deformation retract of (C, B0), hence

Hj(C, B0) = Z forj =k 1, and it is zero otherwise. Therefore, the inclusion (C, B0)(Hleft, B) induces a surjection on homologies.

2.4. Boundary critical points and half-handles. Consider a Morse function F on acobordism (, Y) and assume that it has a single boundary critical point z of index k withcritical value c.

Theorem 2.24. If z is boundary stable (unstable), then the cobordism is a left (right)half-handle attachment of indexk respectively.

Proof. We can assume thatc = F(z) = 0. Let be a positive number small enough and letUbe a half ball around z of radius 2 such that in Uwe can choose Morse coordinatesx1, . . . , xn+1 (cf. Lemma2.6) with

U={x21+ + x2n+1 42} {x1 0},and Y Udefined by{x1= 0}, and

F(x1, . . . , xn+1) =

a2 + b2,

where ifz is boundary stable we set

(2.25) a2 =x21+ x22+ + x2k, b2 =x2k+1+ + x2n+1 (k 1),

and ifz is boundary unstable

(2.26) a2 =x22+ + x2k+1, b2 =x21+ x2k+2+ + x2n+1 (k 0).We also assume that x1, . . . , xn+1 is an Euclidean orthonormal coordinate system.

Next, we consider > 0 such that , and we define the spaceH bounded by thefollowing conditions (see Figure6)

H :={a2 + b2 [2, 2], a2b2 (4 4)/4, x1 0}.


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a2

b2

a2

b2

= 4

4

4

a2 + b2 =2

a2 + b2 =2

H

Figure 6. A schematic presentation ofH,B,P,Kfrom the proof of The-orem2.24. To each p oint (a2, b2) inHon the picture, correspond all thosepoints (x1, . . . , xn+1) for which (2.25) or (2.26) holds and x1 0.

Observe that HU.Let us now define the following parts of the boundary ofHB = H {a2 + b2 =2} F1(2),P =H {a2 + b2 =2} F1(2),K=H {a2b2 = (4 4)/4},C=H {x1 = 0} Y.(2.27)

We have

B

P

K

C=

H(on Figure6 we do not see

C, because this would require one

more dimension). Ifz is boundary unstable andk = 0 in (2.26) then the term a2 is missing

andB=. OtherwiseB=.Lemma 2.28. The flow ofF is tangent toK.Proof. Assume the critical point is boundary stable. Then the time tflow acts by

(x1, . . . , xn+1)(e2tx1, . . . , e2txk, e2txk+1, . . . , e2txn+1),hencea2 e4ta2 and b2 e4tb2, and the hypersurface a2b2 = const is preserved. Lemma 2.29. The inclusion of pair of spaces

F1(2) B H, Y F1(2) B H (, Y)admits a strong deformation retract (inducing an isomorphism of these pairs).

Proof. By Lemma2.10we can assume that (, Y) is (F1([2, 2]), Y F1([2, 2]).First we assume thatB is not empty, and it is given by the equation (2.27) in U. Set

= closure of (F1(2) \B) and let T be the part of the boundary of given byT= closure of ( \ F1(2) ).

ObviouslyTB, see Figure 7. Let us choose a collar of T in , that is a subspaceU diffeomorphic to T[0, 1], T identified with T {0} and T[0, 1] F1(2). Let T be the space identified with T {1} by this diffeomorphism.

Similarly, let += closure of (F1(2) \P), andT+ = closure of (+ \ F1(2)). We

also define 0 as the closure of \

H. Clearly Fhas no critical points in 0 andF is


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BUT

T

+

HY

T

T+

T

V

Figure 7. Notation used in Lemma 2.29. Please note that the left pictureis drawn on , while the right one is on .

F1(2)Y

HV

F1(2)

VH

V = V(V)

HH

Figure 8. Contractions H and V from the proof of Lemma 2.29. ThesetV is now drawn as a rectangle.

everywhere tangent to 0\ ( +) = (Y 0)

Kby Lemma2.28. In particular, by

Lemma2.10,the flow ofF on 0 yields a diffeomorphism between and +, mappingT to T+. We defineV as the closure of the set of points v such that a trajectorygoing through v hits U. Lemma2.10 implies that there is a diffeomorphism V= T[0, 1] [2, 2] such that for (x,t,s) V we have F(x,t,s) = s. Finally, we also defineV:={(x,t,s)V : s 2(1 2t)}.

We define the contraction in two steps: vertical and horizontal. The vertical contraction

is defined as follows. For vH V we define V(v) =v. For a point v0 \ V we takefor V(v) the unique point s such that a trajectory ofFgoes from s to v . Finallyifv = (x,t,s)V\ V we define V(v) = (x,t,2(1 2t)).

By construction, the image of V isH V F1(2). Next, we define H.It is an identity onH F1(2), and maps (x,t,s)V to (x, t (2 + s)/(22), 2)

ifs 2(2t 1), and to (x, 0, s 22t) otherwise. Note that the expressions agree for any(x,t,s) with s= 2(2t

1) and these points are sent to (x, 0,

2). Both Hand V are

continuous retractions, by smoothing corners we can modify them into smooth retractions;also they can be extended in a natural way to strong deformation retracts. By construction,the retracts preserve Y too. See also Figure8.

IfBis empty, thenHis necessarily a unstable (right) half-handle of index 0,F1([2, 2])is a disconnected sum ofHand the manifold F1(2) [2, 2].

Continuation of the proof of Theorem2.24. We want to show thatHis a half-handle.By subsection2.2we have the following description in local coordinates of the left half-

handle (2.30) and right half-handle (2.31) with cutting coordinate x1:

(2.30) Hleft={x21+ + x2k 1} {x2k+1+ + x2n+1 1} {x1 0}


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(2.31) Hright={x22+ + x2k+1 1} {x21+ x2k+2+ + x2n+1 1} {x1 0}.We consider the subsets R and S ofR2 and a diffeomorphism : RS continuous on

the boundary, where

R={(u, v)R2

: u 0, v 0, uv (4

4

)/4,u + v[2

, 2

]},S:{(u, v)R2 : u[0, ], v[0, ]}.

Assume that maps the edge ofRgiven by{u + v =2}to the edge{u= }ofSandthe images of coordinate axes are the corresponding coordinate axes. (Note that R can beseen on Figure6 if we replace a2 byuand b2 byv .)

We lift to a diffeomorphism betweenHand Hright(respectivelyHleft) as follows. Firstlet us write (u, v) = (1(u, v), 2(u, v)). As maps axes to axes, we have 1(0, v) = 0and 2(u, 0) = 0. Furthermore 1, 2 0. By Hadamards lemma there exists smoothfunctionsand such that

(u, v) = (u(u, v)2, v(u, v)2).

We define now

(x1, . . . , xn+1) = ((a, b)x1, . . . , (a, b)xk, (a, b)xk+1, . . . , (a, b)xn+1)

ifz is boundary stable, and

(x1, . . . , xn+1) = ((a, b)x1, (a, b)x2, . . . , (a, b)xk, (a, b)xk+1, . . . , (a, b)xn+1)

ifz is boundary unstable. Here a and b are given by (2.25) or (2.26). By the very construc-

tion, maps ( H,B,C) diffeomorphically to the triple (H , B , C ), whereH={a2 [0, 2], b2 [0, 2], x1 0}B={a2 =2, b2 [0, 2], x1 0}C={a2 [0, 2], b2 [0, 2], x1= 0}.

After substituting for a and b the values from (2.25) or (2.26) (depending on whether zis boundary stable or unstable), we recover the model (2.31) of a right half-handle ifz isboundary unstable; or the model (2.30) of a left half-handle (both of index k).

The fact that each half-handle can be presented in a left or right model will be now usedto show the following converse to Theorem2.24.

Proposition 2.32. Let (, Y) = (0 [0, 1], M0 [0, 1]) be a product cobordism between(0, M0) and(1, M1)=(0, M0). Let us be given a half-handle(H,C,B) of indexk andan embedding ofB0= C B into M1 (respectively an embedding of(B, B0)into(1, M1)),and let (, Y) be the result of a right half-handle attachment along B0 (respectively, aleft half-handle attachment along(B, B0)) of indexk. Then, there exists a Morse functionF: (, Y) R, which has a single boundary unstable critical point (respectively, a singleboundary stable critical point) of indexk onHand no other critical points. In particular,Fis a Morse function on a cobordism(, Y).

Proof. We shall prove the result for right half-handle attachment, the other case is com-pletely analogous. In this case we haveB0embedded intoM1and we extend this embeddingto an embedding ofB into 1(see Definition2.13). We shall be using the notation from theproof of Theorem2.24, with 2 = 1 and 2 = 4. We have = BH, where we identify(H,B,C ) with ( H,B,C) using the diffeomorphism . Let now = closure of (1\ B)and T = closure of (\ 1).

By Lemma2.29used back to front we deduce that is diffeomorphic to , where

(2.33) = [1, 1].


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The gluing in (2.33) is as follows: we glue {1} to 1 and T [1, 1] toK(in theproof of Lemma2.29we proved thatK is T [1, 1]). Now let us define a function

F(x) =

t ifx = (v, t)

0 {

t}

0

[0, 1] =

2 + t ifx = (v, t) {t} [1, 1]2

kj=1

x2j +n+1

j=k+1x2j ifx = (x1, . . . , xn+1)H.

Checking continuity ofF is straightforward. Moreover F is piecewise smooth and smoothaway from B. Let us choose a vector field e on B, normal to B and pointing towards H.ThenF, e is positive on both sides ofB. By a standard analytic argument F can beapproximated by a smooth function equal to Faway from some neighbourhood ofB , whichhas no new critical point.

2.5. Left and right product cobordisms and traces of handle attachments. In

this subsection we relate half-handle attachments to handle attachments and detachmentsin the sense subsection 2.3. This also creates a dictionary between surgery theoreticalnotions (traces of handle attachments and detachments) and Morse theoretical (additionsof half-handles). To begin with, let (, Y) be a cobordism between (0, M0) and (1, M1).

Definition 2.34. We shall say that is a left product cobordism if =0[0, 1]. Similarly,if =1 [0, 1], then we shall say that is a right product cobordism.Proposition 2.35. (a) If(, Y) is a cobordism between(0, M0) and(1, M1) consistingonly of left half-handle attachments, then it is a left-product cobordism. Conversely, if itconsists only on right half-handle attachments, then it is a right product cobordism.

(b) LetF: [0, 1] be a Morse function in the sense of Definition1.4. Assume thatFhas no critical points in the interior of. If all critical points on the boundary are boundary

stable, thenF is a left-product cobordism. If all critical points are boundary unstable, thenF is a right product cobordism.

Proof. The two statements (a) and (b) are equivalent via Theorem 2.24 and Proposi-tion 2.32. The stable-unstable (right-left) statements are also equivalent by replacing theMorse function F byF. The stable case follows from Lemma 2.17.

The next results of this subsection will be not used in this paper, but we insert thembecause they bridge surgery techniques and applications, e.g. with [Ra] or[BNR].

In order to clarify what we wish, let us recall that by Theorem 2.24if a Morse functionFdefined on a cobordism (, Y) has only one critical point of boundary type then (, Y)is a half-handle attachment. Proposition2.32 is the converse of this, the (total) space of

a half-handle attachment can be thought as a cobordism with a Morse function on it withonly one critical point.

We wish to establish the analogues of these statements at the level of . In Subsection2.3we proved that the output of a right/left half-handle attachment at the level of inducesa handle attachment/detachment. The next lemma is the converse of this statement. (Infact, the output cobordism provided by it can be identified with the cobordism constructedin Proposition2.32.)

Lemma 2.36. Assume that (1, M1) is a result of a handle attachment (respectively de-tachment) to (0, M0). Then, there exists a cobordism (, Y; 0, M0, 1, M1) such that=1 [0, 1] (respectively=0 [0, 1]).


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0

1 0

0

1=1 [0, 1]

Y

Y

Y

Figure 9. Lemma2.36. On the left a 1-handle is attached to 0. On theright there is a cobordism between 0 and 1, which is a right productcobordism.

Proof. Assume that (1, M1) arises from a handle attachment to (0, M0), i.e. 1 = 0Dk Dnk. Let us define = 1 [0, 1]. The boundary can be split as

=0 Dk Dnk {0} (M1 [0, 1]) (1 {1})= 0 {0} Y 1 {1},

where Y = Dk Dnk (M1 [0, 1]). Its Dk Dnk part can be pushed inside transforming (diffeomorphically) into a cobordism, see Figure 9.

Analogous construction can be used in the case of a handle detachment. If (1, M1) is

a result of a handle detachment from (0, M0), then the trace of the handle detachment isthe cobordism between (0, M0) and (

1, M

1) such that

(, Y) = (0 [0, 1], M0 [0, 1] Dk Dnk).

Definition 2.37. The cobordism (, Y; 0, M0, 1, M1) determined by the Lemma 2.36is called the trace of a handle attachment of (0, M0) (respectively the trace of a handledetachment).

3. Splitting interior handles

We prove here the theorem about moving critical points to the boundary.

Theorem 3.1. Assume that on a cobordism(, Y)between(0, M0)and(1, M1)we havea Morse functionFwith a single critical pointz of indexk {1, . . . , n} in the interior of situated on the level set1/2 = F

1(F(z)). If

(3.2) the connected component of 1/2 containing z has non-empty intersection with Y ,then there exists a functionG : [0, 1], such that

G agrees withF in a neighbourhood of0 1; G is everywhere tangent to Y; G has exactly two critical pointszs andzu, which are both on the boundary and of

indexk. The pointzs is boundary stable andzu is boundary unstable. There exists a Riemannian metric such that there is a single trajectory ofG from

zs to zu insideY.

Remark 3.3. A careful reading of the proof shows that we can in fact construct a smoothhomotopy Gt such that F = G0, G = G1 and there exists t0 (0, 1) such that Gt has


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Figure 10. The trajectories of the gradient vector field ofD for values ofa >0, a = 0 and a t0 and adegenerate critical point on the boundary for t= t0. See Remark3.15.

The proof of Theorem3.1 occupies Sections3.2 to 3.4. We make a detailed discussion ofCondition (3.2) in Section3.5.

3.1. About the proof. The argument is based on the following two-dimensional picture.Consider the set Z={(x, y)R2 : x 0} and the function D : Z Rgiven by

D(x, y) =y3 yx2 + ay,

wherea Ris a parameter. Observe that the boundary ofZgiven by{x= 0} is invariantunder the gradient flow ofD (see Figure10). The proof of the following lemma is completelystraightforward and will be omitted.

Lemma 3.4. For a > 0, D has a single Morse critical point in the interior of Z. Fora 0, 1 and an open half-discU , intersectingY along a disk, and coordinates x,y,u1, . . . , un1 such that in these coordinatesU is givenby

0 x


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y

xU1

S1

S2

U21= S1 S2

z (3, 0)

Figure 11. Sets U1, U21, S1 and S2 in two dimensions (coordinates xand y).

U Y is given by{x= 0}, and in these coordinatesF is given by

y3 yx2 + y+12

+n1j=1

ju2j ,

where1, . . . , n1 {1}are choices of signs. In particular#{j : j =1}= k 1, wherek= indzF.

Assuming the proposition, we prove Theorem3.1. Let us introduce some abbreviations.

(3.6) u= (u1, . . . , un1), u2 =n1j=1

ju2j , ||u||2 =

n1j=1

u2j .

We fix a small real number >0 such that and two subsets U1U2 ofU byU1 ={|y| , x 3} {(x 3)2 + y2 2}U2 ={|y| 2, x 3} {(x 3)2 + y2 42}.

The difference U21 :=U2\ U1 splits into two subsets S1 S2 (see Figure11), whereS1= U21 {x 3}, S2= U21 {x 3}.

For a point v = (x,y,u1, . . . , un1)U, let us define:

s(v) =

1 ifvU1,0 ifvU\ U2,2

|y|

ifv

S1,

2 (x3)2+y2 ifvS2.The above formula defines a continuous functions : U2 [0, 1]. It is smooth away ofS1 S2. We can perturb it to a C function s, with the following properties:

(S1) s1(1) =U1,s1(0) ={|y| 2 2} {(x 3)2 + y2 42 3};(S2) suj = 0 for any j= 1, . . . , n 1;(S3) sx = 0, and

sy < 2 at all points ofS1. Furthermore y sy


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Observe thats satisfies (S1)(S4) at every point, where it is smooth, the only issue is thaton S1 S2,s fails to be C2.

Now let us choose smooth decreasing function : [0, 2][0, 1], which is equal to 0 on[ 34

2, 2] and (0) = 1. We define now a new function b :U2

[0, 1] by the formula

(3.7) b(x, y, u) =s(x, y, u) (||u||2).Let us finally define the function G : [0, 1] by

(3.8) G(w) =

F(w) ifwU2y3 yx2 + y (+ 1)b(x, y, u)y+ 12 + u2 ifw= (x, y, u)U2,

where >0 is a very small number. Later we shall show that it is enough to take < 2/2.In the following lemmas we shall prove that G satisfies the conditions of Theorem3.1.

Lemma 3.9. The functionG is smooth.

Proof. It is a routine checking and we leave it for the reader. In the next two lemmas we show that G has no critical points in U21.

Lemma 3.10. G has no critical points onU21 {y= 0}.Proof. If (x, 0, u1, . . . , un1)U21 then x >3. Consider the derivative over y ofG:

(3.11) G

y = 3y2 x2 + 1 (+ 1)b (+ 1)(u21+ + u2n1)

s

yy.

Takingy= 0 we getx2 + 1 (+ 1)b. Since b[0, 1] and x >3, one gets Gy


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Lemma 3.14. G has two critical points onU1 at

zs := (0,

/3, 0, . . . , 0)

zu := (0, /3, 0, . . . , 0).Both critical points are boundary, both of Morse indicesk,zs is stable, whilezu is unstable.Proof. The derivative ofG vanishes only at zs and zu. Indices are immediately computedfrom (3.13). The pointzs is boundary stable, because for zs the expressionyx2 is negativeand the boundary is given byx = 0, hence it is attracting in the normal direction. Similarlywe prove for zu. See also Figure10 for the two-dimensional picture.

Remark 3.15. If we define Gt =y3 yx2 + y t(+ 1)b y + 12+ u2 fort[0, 1], then the

same argument as in Lemmas3.10and 3.12shows thatGt has no critical points in U2 \ U1.As for critical points in U1, observe that on U1 we have

Gt = y3 yx2 + (1 t(1 + ))y+ 1

2+ u2.

Let t0 = 11+ . If t > t0, the function Gt has two critical points on the boundary Y, whilefor t < t0, Gt has a single critical point in the interior U1\ Y. If t = t0, Gt has a singledegenerate critical point on Y. In this way we construct an isotopy between F andG.

Let us now choose a Riemannian metric g on

U1 := U1 {||u||< }by the condition that (x,y,u1, . . . , un1) be orthonormal coordinates. Clearly, any metricg on can be changed near U1 so as to agree with g

on U1. In this metric the gradient ofG is

(2xy, 3y2 x2 , 21u1, . . . , 2n1un1).We want to show that there is a single trajectory starting from zs and terminating at zu.

Clearly, there is one trajectory from zs

to zu

which stays in U1 (having y = 0 and u= 0).In order to eliminate the others, we need the following lemma.

Lemma 3.16. Letbe a trajectory ofG starting fromzs. Letw be the point, where hitsU1 for the first time. Ifis sufficiently small, thenG(w)> G(zu).

Proof. Assume that(t) is such trajectory. Assume that among numbers i, we havei=1for i k 1 and i = 1 otherwise. Aszs is a critical point of the vector fieldG with anon-degenerate linear part, we conclude that the limit

limt

(t)||(t)|| =:v = (x0, y0, u01, . . . , u0,n1)

exists. The vectorv is the tangent vector to the curve at the point zs, and it lies in the

unstable space. Hence x0 = 0 as (1, 0, . . . , 0) is a stable direction; similarly u01 = =u0,k1= 0. Therefore, untilhits the boundary ofU1 for the first time, we have

x= u1= = uk1= 0.Set also g(y) = y3 y. One has the following cases, depending the position ofw, where hits U1 for the first time: (a) y =, (b) y = , or (c)||u||2 = 2. The case (a)cannot happen sinceG is increasing along the trajectory, henceG(w)> G(zs), a fact which

contradicts g() < g(

/3) valid for 2 < 2. In case (b), G(w) > G(zu) follows from

g()> g(

/3). Finally, assume the case (c). Then, as u01= = u0,k1= 0, we obtainu2 =||u||2 = 2. ThenG(w)G(zs) 2, because the contribution to G from y3 yincreases along . Hence G(w)> G(zu) follows again since .


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Z1 Z2

y

x

Figure 12. Sets Z1 and Z2 from Section 3.3. There are drawn also thestable and unstable manifolds of

A.

Given the above lemma it is clear that if a trajectory leaves U1, thenG becomes biggerthanG(zu). AsG increases along any trajectory, it is impossible that such trajectory limitsin zu. The proof of Theorem3.1, up to Proposition3.5, is accomplished.

3.3. An auxiliary construction. We provide now a construction, a crucial ingredient inthe proof of Proposition3.5,see next Section. Set

Z={(x, y)R2 :x 0},and define the two functions

(3.17) A(x, y) = x3

3

3

3xy2 x

3+

2

3

3, B(x, y) =y3 yx2 + y.

They are the real/imaginary parts of a complex function; indeed (with i = 1)

A+ iB=

x

3 iy

3

x3

iy

+ 2

3

3.

It follows from the CauchyRiemann equations (or straightforward computations), that thelevel sets ofA and B are orthogonal, away from the point (1, 0). If we choose a Riemannianmetric such that (x, y) are orthonormal coordinates, then A is constant on trajectories ofB (i.e. A is a first integral ofB). Let us fix >0 smaller than 2

3

3. Consider two sets

(3.18) Z1={(x, y)Z, x 1, A(x, y)}.We have the following result.

Lemma 3.19. The map (x, y) = (A(x, y), B(x, y)) maps Z1 and Z2 diffeomorphicallyonto E1 andE2 respectively, where

E1=

(a, b)R2 : a

, 2

3

3

, E2=

(a, b)R2 : a

.

Moreover, maps trajectories ofB onto vertical lines.Proof. This follows from the covering theory of one-variable complex functions. Alterna-tively, one readily checks that : Z1 V1 and : Z2 V2 are bijections. As D isnon-degenerate on Z1 Z2, is a diffeomorphism between the two pairs of sets. Now A isa first integral of the flowB, so each trajectory ofB lies in the set A1(c).


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V

V1

D1 z

V2

D2

W

Y

h3

V

(1, 0, . . . , 0)

11 (W) 12 (h4(D2)) = h

3(D2)

V1 V2x= 0

1 2

C1 = 1(V1) C2 = 2(V2)

W D1

a= a=

D2h4

h4(D2)

12

a= 23

3a= 2

3

3a= 2

3

3

E1

Figure 13. Notation used in Section3.4. The top line is the picture on ,the middle line is in coordinates such thatFis equal toy3 yx2 + y +12+ u2.The bottom line is in coordinates such thatF =b + 12 . There is no mistake,the linea = appears twice on the picture, in coordinates on C1 and on C2.

3.4. Proof of Proposition 3.5. First, as z is a critical point of index k {1, . . . , n},by Morse lemma we can find a neighbourhoodV of z and a chart h1 :V Rn+1, withcoordinates (x, y, u) such that

F h11 (x, y, u) =xy+ u2 +1

2.

Let us define a map h2(x, y, u) = (x, y, u), where x = y2 + 1 x2. By the inverse func-

tion theorem, h2 is a local diffeomorphism near (1, 0, . . . , 0). ShrinkingV if needed, andconsideringh3= h

12 h1, we obtain h3(z) = (1, 0, . . . , 0) and

F h13 (x,y,u) =y3 yx2 + y+ u2 +1

2=B(x, y) + u2 +

1

2.

Let us pick now >0 such that the cylinder

V ={|x 1|< , |y|< , ||u||< } Rn+1

lies entirely inh3(V). By shrinkingVwe may in fact assume that h3(V) =V. If 0< 1is sufficiently small then A(x, 0)< implies|x 1|< . Choose such a , and set

V1 := V {x


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(compare (3.18)). By Lemma3.19the map

(3.20) 1(x, y, u) = (A(x, y), B(x, y) + u2, u),

is a diffeomorphism (being the composition of and a triangular map). Set C1:= 1(V1)

andV1:= h13 (V1). Finally, leth= 1 h3.

Using Lemma3.19again we obtain that

F h1(a, b, u) =b+12

.

Let > be sufficiently closed to satisfying the inclusion

D1 := [, ] (, ) (, )n1 C1.LetD1= h1(D1)V1, see Figure13.Lemma 3.21. If andare small enough, there is an open/closed ball

W in, containing

D1, such thath extends to a diffeomorphism betweenW and [, 233 ] (, ) (, )n1withF h1(a, b, u) =b + 12 , sending points witha= 233 to Y.In the proof we shall use the following result.

Lemma 3.22. There exists a smooth curve: [, 23

3], such that( 2

3

3)Y, (t)

1/2, (t)D1 if and only if t [, ] and h((t)) = (t, 0, . . . , 0) and omitsV\V1.Furthermore, is transverse to Y.

Proof of Lemma3.22. Letp= h1(, 0, . . . , 0)1/2. Let B1/2 be an open ball withcentrez and pB. Let be the connected component of 1/2 containingp. We considertwo cases.

If \ B is connected, it is also path connected. By (3.2), there exists a path \ B

joining p with a p oint on the b oundary. We can assume that is transverse to Y. Wechoose = h1([, ] {0, . . . , 0}) (and we smooth a possible corner at p). It is clearthat omitsV\V1 and that we can find a parametrization ofby the interval [, 233 ].

If \ B is not connected, then as is connected, by a homological argument we haven= 1 and k = 1. Since is connected and has boundary, then is an interval and B isan interval too. Then \ B consists of two intervals, each intersecting Y. One of theseintervals containsp. Sop is connected with Yby an interval, which omits B . We concludethe proof by the same argument as in the above case, when \ B was connected.

Proof of Lemma3.21. Given Lemma 3.22, let us choose a tubular neighbourhood X of in F1(1/2)\(

V\

V1). Shrinking X if needed we can assume that it is a disk and

X1 := XV =D1F1(1/2). Now letbe a vector field onD1given by (Dh)1(1, 0, . . . , 0),where Dhdenotes the derivative ofh. This vector field is everywhere tangent to X1 and(3.23) | D1 =

d

dt(t)

by the very definition of . We extend to a smooth vector field on the whole X, suchthat (3.23) holds on the whole . For any point z, the trajectory of (which is ) hitseventually Yand, on the other end, it hits the right wallR= h1({} {0} (, )n1).(compare Figure14; note that the horizontal coordinate there increases from right to left

for consistency with Figure13). Since is transverse to

R and to Y, by implicit function


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X1 =D1 F1(1/2)X

Y

R

z= z(tz)

zz()

Figure 14. Proof of Lemma3.21. Construction of the vector field. Pic-ture on F1(1/2). The parallel vector field from the region on the right isextended to the whole Xso that it is tangent to .

theorem trajectories close to also start atR and end up at Y. ShrinkingX if necessarywe may assume that each point ofX lies on the trajectory ofwhich connects a point ofR to some point ofY, and all the trajectories are transverse to both Y andR.

We can now rescale (that is multiply by a suitable smooth function constant on tra-

jectories) so that all the trajectories go fromRtoY in time 23

3 , i.e. the same time as

does. The rescaled vector field allows us to introduce coordinates on X in the followingway. For any pointzX, let z be the trajectory of, going through z. We can assumethat z()

R. Let tz =

1z (z), i.e. the moment when z passes through z. Since we

normalizedz, we know thattz[, 233 ] and tz = 23

3if and only ifzY X.

Let uz be such that h(z()) = (, 0, uz). The vector uz might be thought off as acoordinate onR. We define now

h(z) = (tz, 0, uz).

This maps clearly extends hto the whole X.

Now letWbe a tubular neighbourhood ofX in \ (V\V1). We use the flow ofF toextend coordinates from X toW. More precisely, shrinkingW if needed we may assumethat for each wW the trajectory ofF intersects X. This intersection is necessarilytransverse and it is in one point, which we denote by zwX. We define now

h(w) = (tzw , F(w) F(zw), uzw).As h is a local diffeomorphism on X (because

F is transverse to X), it is also a local

diffeomorphism near X. We put W = h(W). Clearly b oth definitions ofh onV andWagree. We may now decrease and shrink W so that

W = [, 2

3

3] (, ) (, )n1.

We haveFh1(a, b, u) =b + 12 . We now extendh3over Wby the formulah3= 1h. Consider now

V2:= V {x >1, A(x, y)}.Let 2 : V Rn+1 be given by 2(x, y, u) = (a, b, u) = (A(x, y), B(x, y) + u2, u), providedby the same formula as 1 in (3.20) but the image now satisfies a , cf. Lemma3.19.


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LetC2= 2(V2), and let us choose sufficiently small such that

D2:= [, ] (, ) (, )n1 C2.

We shall denote h= 2

h3 andD2 = h

1(D2).

Let us now fix M >0 large enough and consider a map h4 : Rn+1 Rn+1 of the formh4(a,b,u) = ((a), b, u),

where : [, ]=[, M] is a strictly increasing smooth function, which is an identity near.Consider the map h3 :

12 h4 h :D2 Rn+1. Since h is an identity for a close to, this

map agrees with h3 fora close to. Furthermore F h1(a, b, u) =F h1 h14 (a, b, u) =b + 12 by a straightforward computation. On the other hand, the pointh

1(, 0, . . . , 0)D2is mapped by h3 to (M, 0, . . . , 0) Rn+1, where Mcan be arbitrary large, e.g. M >3.

Having gathered all the necessary maps, we now conclude the proof. Let

U=

W (

V\ h13 (V1 V2)

D2.

The map h3 :U [0, ) Rn is given by h3 onW and onV\h13 (V2), and by h3onD2. This map is clearly a diffeomorphism onto its image, so it is a chart near z.By construction F h13 is equal to y3 yx2 +y +u2 + 1/2 and h3(W) contains thesegment with endpoints (0, 0, . . . , 0) and (3, 0, . . . , 0). Since it is an open subset, it contains[0, 3 +) (, )(, )n1 for > 0 small enough. The inverse image of this cubegives the required chart.

This ends the proof of Theorem 3.1 which moves a single interior critical point to theboundary. Section4generalizes this fact for more critical points; one of the needed toolswill be the rearrangements of the critical values/points.

3.5. Condition (3.2) revisited. We will provide two characterizations of Condition (3.2).One is valid for arbitrary n 1, the other one holds only in the case n = 1. We shall

keep the notation from previous subsections, in particular (, Y) is a cobordism between(0, M0) and (1, M1),F: [0, 1] is a Morse function with a single critical pointz in theinterior of , and F(z) = 1/2. Let 1/2 = F

1(1/2) and be the connected componentof 1/2 such that z.Proposition 3.24. If0,1 and have no closed connected components, then

Y=.In particular, in Theorem 3.1we can assume that0, 1 and have no closed connectedcomponents instead of (3.2).

Proof. Letp = h1(, 0, . . . , 0)D1 and let B be an open ball in near z , such thatpB. It is enough to show that pcan be connected to Yby a path in 1/2, which omitsB (compare Lemma3.22).

Let us choose a Riemannian metric on . Let Wsz be the stable manifold ofz and let

T be intersection ofWsz and 0. This is a (k1)dimensional sphere. The flow ofFinduces a diffeomorphism : 1/2\ B=0\ B0, where B0 is a tubular neighbourhood ofT in 0 (here we tacitly use the fact that and are small enough), see Figure 15. Letp0= (p). Let

0 be the connected component of 0, which contains B0.

Now we will analyze several cases. Recall that k = indzF {1, . . . , n}. First we assumethat k < n. Then 0\ T is connected, so p0 can be connected to the boundary of 0 which is non-empty by assumptions of the proposition by a path 0. Now the inverseimage 1(0) is the required path.

Ifk =n > 1 then we reverse the cobordism and look atF, hence this case is coveredby the previous one (since k = n will be replaced by k = 1< n).


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T

p00

Figure 15. Notation on 0.

M0M0 B0B0 T T

p0p0

Figure 16. Proof of Proposition 3.24. Case k= 1 and n = 1 and T lies intwo components of . 0 is the horizontal segment. The pointsp

0 and p

0

are the two possible positions of the point p0.

M0M0 M0M0B0 B0

TT p0p0

Figure 17. Proof of Proposition 3.24. Case k= 1 and n = 1 and T lies intwo components of 0. The points p

0 and p

0 are the two possible positions

of the point p0. Both can be connected to the boundaryM0.

Finally, it remains to deal with the situation k = n = 1. Then dim0 = 1. T consistsof two points. Assume first that they lie in a single connected component 0 of 0. Weshall show that this is impossible. As 0 is connected with non-trivial boundary, it is aninterval. The situation is like on Figure16. Now asFhas precisely one Morse critical pointof index 1, 1 is the result of a surgery on 0. This surgery consists of removing two innersegments from 0 and gluing back two other segments, which on Figure 16are drawn asdashed arc. But then 1 has a closed connected component, which contradicts assumptionsof Theorem3.1.

Therefore,Tlies in two connected components of 0. The situation is drawn of Figure17,and it is straightforward to see that p0 can be connected to M0 by a segment omittingB0.

The proof of Proposition3.24suggests that the case n= 1 is different than case n >1.We shall provide now a full characterization of a failure to ( 3.2).


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Proposition 3.25. Assume thatk = n = 1and is connected. If (3.2)does not hold, then is a pair of pants, 0 is a circle and1 is a disjoint union of two circles; or converse:1= S

1 and0 is a disjoint union of two circles. In particular,Y =.

Proof. A one-handle attached to a surface changes the number of boundary componentsby1. Let us assume that 1 has less components than 0, if not we can reverse thecobordism. As is connected, 0 has two components and 1 only one. Let A0 0be the attaching region, i.e. the union of two closed intervals to which the one-handle isattached. With the notation of Section 3.5 we have (, z)= (0/A0, A0/A0), where thequotient denotes collapsing a space to a point. In particular z can not be joined toY by apath in if and only if 0 is disjoint from Y. Hence 0 is closed, that is, it is a union oftwo circles.

4. Rearrangements of boundary handles

4.1. Preliminaries. Let (, Y) be a cobordism between twon-dimensional manifolds with

boundary (0, M0) and (1, M1). LetFbe a Morse function, with critical points w1, . . . , wkInt and y1, . . . , ylY. In the classical theory (that is, when Y =), the ThomMilnorSmale theorem (see [Mi2, Section 4]) says that we can alter F without introducing newcritical points such that if ind wi 0 for all (x, y)[0, 1] [0, 1];(PS2) there exists >0, such that (x, y) =x for all x[0, ] [1 , 1] and y[0, 1];(PS3) for any s(, ) we have (F(p1) + s, 0) =a1+ s and (F(p2) + s, 1) =a2+ s.


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For any we define G() = (F(), ()). From the properties (PS3), (M1) and(M2) we see that nearpi,G differs fromFby a constant. The property (PS2) ensures thatG agrees with F in a neighbourhood of 0 and 1. Let us show thatG does not vanishaway from pi. By a chain rule we have

(4.4) G= x

F+ y

.

Since is constant on all trajectories ofF, the scalar productF, = 0. Then theproperty (PS1) guarantees thatG, F> 0 away from p1 and p2.

We need to show thatGis everywhere tangent to Y. If one of the points is interior, by(M4) vanishes on Y, henceG is parallel on Y toFand we are done. Next assumethat both critical points are on the b oundary. Let us choose an open subsetU ofY suchthat|U= 0 andK1 K2U. This is possible, because of the properties (M1) and (M2).Then let us choose a neighbourhood W in ofY\ U, disjoint from K1 and K2. Observethat dG(F) =G, F> 0. AsF T Yone has ker dGT Y, so by Lemma1.7wecan change the metric in W so thatG is everywhere tangent to Y . Proof of Lemma4.3. Let us define T1 =K1 0 and T2 =K2 0. Assume thatT1 andT2 are not empty. For each \ K1 K2, let () be the intersection of the trajectoryof underF with 0. This gives a map : \ (K1 K2)0 \ (T1 T2).

Let us define first on 0 by the following conditions: 1 in a neighbourhood ofT2, 0 in a neighbourhood ofT1. Furthermore, if either T1 or T2 is disjoint from theboundary M0 we extend to a constant function on M0. Finally, we extend to the whole by picking () =(()) ifK1 K2, and |Ki() =i 1, i= 1, 2.

IfT1 =, then indFp1 = 0 and the proof of the rearrangement theorem is completelystraightforward.

4.3. MorseSmale condition on manifolds with boundary. In the classical theory,the MorseSmale condition imposed on a Morse function F: M

R means that for each

pair of two critical points p1, p2 of M the intersection of stable manifold Wsp1 with theunstable manifold ofWup2 is transverse. (Note that this MorseSmale condition also dependson the choice of Riemannian metric onM.) Following[KM,Definition 2.4.2], we reformulatethe MorseSmale condition in the following way

Definition 4.5. The functionF is calledMorseSmale if for any two critical points p1 andp2, the intersection of Int Wsp1 with Int Wup2 is transverse (as the intersection in the(n + 1)-dimensional manifold ) and the intersection ofY Wsp1 withY Wup2 is transverse(as an intersection in the n-dimensional manifold Y).

It is clear, that regular functions form an open-dense subset of all C2 functions, whichsatisfy the Morse condition from Definition 1.4.

Assume now thatFis MorseSmale. Given two critical points ofF, p1 andp2, we wantto check whetherWsp1Wup2 =. This depends not only on the index, but also on whetherany of the two points is boundary stable. We show this in a tabulated form in Table 1,where indp1 = k and indp2 = l. In studying the intersection, we remark thatW

sp1 Wup2

is formed from trajectories, so if for dimensional reasons we have dim Wsp1Wup2 < 1, itimmediately follows that this intersection is empty.

4.4. Global rearrangement theorem. Let us combine the rearrangement theorems fromSection4.2with the computations in Table 1.

Proposition 4.6. LetFbe a Morse function on a cobordism(, Y) between(0, M0) and(1, M1). Letw1, . . . , wm be the interior critical points ofFandy1, . . . , yk+l be the boundary


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Lemma 4.9. Each functionFcan be made technically good without introducing new criticalpoints.

Proof. By Proposition 4.6 we can rearrange the critical points of F, proving (TG1) and(TG2). The properties (TG3) and (TG4) can be guaranteed, using the handle cancellationtheorem (e.g. [Mi2,Theorem 5.4]. We refer to the beginning of Section5for an explanation,that one can use the handle cancellation theorem if the manifold in question has boundary.

Theorem 4.10. Let (, Y) be a cobordism between (0, M0) and (1, M1). Let F be atechnically good Morse function on that cobordism, which has critical points y1, . . . , yk onthe boundary Y, z1, . . . , zl+m in the interior Int, of which zm+1, . . . , zl+m have index 0orn + 1 and the indices ofz1, . . . , zm are in{1, . . . , n}. Suppose furthermore the followingproperties are satisfied:

(I1) 0 and1 have no closed connected components;(I2) has no closed connected component.

Then, there exist a Morse function G : [0, 1], on the cobordism (, Y), with criticalpointsy1, . . . , ykY, zm+1, . . . , zl+m andzs1, zu1 , . . . , zsm, zum such that: indG yi = indFyi for i = 1, . . . , k and for j = m + 1, . . . , m+l we have indG zj =

indFzj ; forj = 1, . . . , m, indG zsj = indG zuj = indFzj ; for j = 1, . . . , m, zsj and zuj are on the boundary Y, furthermore zsj is boundary

stable, zuj is boundary unstable andG(zsj )< G(z

uj).

In other words, we can move all critical points to the boundary at once. To proveTheorem 4.10 we use Theorem 3.1 independently for each critical point z1, . . . , zm. Weneed to ensure that Condition 3.2 holds. This is done in Proposition 4.11 stated below.Given these two ingredients the proof is straightforward.

4.6. Topological ingredients needed in the proof of Theorem 4.10.

Proposition 4.11. Let Fbe a technically good Morse function on the cobordism (, Y).Assume that0, 1 and have no closed connected components. Letc, d be as in Defini-tion4.8. Then

(a) Ifn >1, then for anyy[c, d], the inverse imageF1(y) has no closed connectedcomponent.

(b) If n = 1, then after possibly rearranging the critical values of the interior criticalpoints of index 1, for any interior critical point z of F of index 1, z can beconnected with Y by a curve lying entirely inF1(F(z)); and furthermore all thecritical points are on different levels.

Remark 4.12. The distinction between cases n > 1 and n = 1 is necessary. Point (a) ofProposition4.11is not necessarily valid ifn = 1, see Figure18for a simple counterexample.

First let us prove several lemmas, which are simple consequences of the assumptions ofProposition4.11. We use assumptions and notation of Proposition4.11.

Lemma 4.13. Let x, y [0, 1] with x < y. If is a connected component of F1[x, y]then either Y =, or for anyu[x, y] [c, d] we haveF1(u) Y=.Proof. Assume that for some u [x, y][c, d] the intersection F1(u)Y = and Y=. Then either Y F1[0, u] or Y F1[u, 1] is not empty. Assume thefirst possibility (the other one is symmetric) and let Y = Y F1[0, u]. Letf=F|Y


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0= F1(0)

1= F1(1)

1

0

1/2

Y

F

Figure 18. The statement of Proposition 4.11(a) does not hold ifn = 1.Here,F is the height function. The level setF1(1/2), drawn on the picture,has two connected components, one of which is closed.

be the restriction. Then Y is compact and fhas a local maximum on Y. This maximumcorresponds to a critical point offof index n, so either a boundary stable critical point ofFof index n+ 1, or a boundary unstable critical point of index n. But the correspondingcritical value is smaller than u, so smaller than d, which contradicts property (TG2).

Lemma 4.14. For any x [c, 1] the set F1[0, x] cannot have a connected componentdisjoint fromY.

Proof. Assume the contrary, and let be a connected component ofF1[0, x] disjoint fromY. Let 1 be the connected component of containing

. Suppose that 1 Y =. Theneither 1 (0 1) =or 1 (0 1)=. In the first case 1 is a closed connectedcomponent of , in the second either 1 0, or 1 1 is not empty, so either 0 or 1has a closed connected component. The contradiction implies that 1 Y=.

By Lemma4.13 we have then F1(x) 1 Y=, hence := (F1[0, x] 1) \ is not empty and is disjoint from . As and both belong to 1 which is connected,there must be a critical point z 1 of index 1, which joins to . We have F(z)> xand , belong to two different connected components ofF1[0, F(z)) and to a singleconnected component ofF1[0, F(z)]. The connected component ofF1[0, F(z)) containing has empty intersection withY(by Lemma4.13) hencez must be an interior critical pointof index 1. We also remind that all critical points ofF on are interior critical points,because Y =.

Let Ws be the stable manifold ofz of the vector fieldF. Then Ws 1 must be aconnected curve, with non-empty intersection with . One of its boundaries is either on0 or it is a critical point of F in , necessarily interior and by the MorseSmalecondition, its index is 0. In the first case, 0 is not empty and since it is disjointfrom Y, 0 has a closed connected component. In the other case, we have in 1 a singletrajectory between a critical point of index 0 and a critical point of index 1. This contradicts(TG3).

Lemma 4.15. Assume that for some y [c, d] 1 and 2 are two disjoint connectedcomponents ofF1(y). IfFhas no interior or boundary unstable critical points of indexn


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with critical value in[c, y), then1 and2 belong to two different connected components ofF1[0, y].

Proof. Forx < yand close toy the sets 1and 2lie in two different connected components

ofF1

(x, y]. Letx0

0 be the smallestx with that property. Assume x0> 0. Thenx0is acritical value ofF. The number of connected components ofF1(x) increases as x crossesx0. Thus the corresponding critical point is either an interior critical point of index n, or aboundary unstable critical point of indexn. But then x0 c because of (TG2), so we havex0[c, y] which contradicts the assumptions of the lemma.

It follows that x0 = 0. As F has no critical points on 0, it follows that 1 and 2belong to different components ofF1[0, y].

Lemma 4.16. Lety [c, d] be chosen so that there are no interior or boundary unstablecritical points of indexn with critical values in[c, y). ThenF1(y) has no closed connectedcomponents.

Proof. Assume that is a closed connected component ofF1(y). Let =F1(y)\

,it is not empty by Lemma4.13(applied for x = 0, y = 1), for otherwise Y =. Let be the connected component ofF1[0, y] containing . By Lemma4.15, and aredisjoint, in particular F1(y) Y = Y =. By Lemma4.13, Y =. But thiscontradicts Lemma4.14.

Remark 4.17. There exists a symmetric formulation of the last three lemmas, which canbe obtained by considering the function 1 F instead ofF. For instance, in Lemma4.16,the symmetric assumption is that there are no interior or boundary stable critical points ofindex 1 in (y, d]. The statement is the same.

Proof of Proposition4.11. Casen >1. Let x [c, d] be a non-critical value such that allthe critical points of F with index n have critical value greater than x, and all critical

points with index smaller than n have critical values smaller than x. Such x exists becauseof (TG1). If y x, then Lemma 4.16 guarantees that F1(y) has no closed connectedcomponents. Ify > x, then F1[y, d] has no critical points of index 1 (as n > 1), so weapply the symmetric counterpart of Lemma4.16.

Proof of Proposition4.11. Casen= 1. The property (TG2) implies that the only criticalpoints of F|[c,d] are the interior critical points of index 1. Let us call them z1, . . . , zm.Since they are all of the same index, by Proposition4.6 we are able to rearrange the valuesF(z1), . . . , F (zm) at will. Let us fix c1, . . . , cmwith the property thatc < c1


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4.7. Splitting of cobordisms. We have now all the ingredients needed to prove our maintheorem about splitting cobordisms. We slightly change the notation in this subsection, thecobordism will be between (, M) and (, M).

Theorem 4.18. Let (, Y) be a cobordism between (, M) and (, M

). If the following

conditions are satisfied

and have no closed connected components; has no closed connected component;

Then the relative cobordism can be expressed as a union

= 0 1/2 1 3/2 n+1/2 n+1such thats = s s+1/2 Ys with0= , n+3/2= , Y =Y0 Yn+1. In otherwords (s, Ys) is a cobordism between (s, Ms) and (s+1/2, Ms+1/2), where Ms = s =s Ys. Furthermore

(0, Y0) is a cobordism given by a sequence of index0 handle attachments;

fork = 1, . . . , n+ 1,(k

1/2, Yk

1/2)is a left product cobordism, given by a sequence

of indexk left half-handle attachments; fork= 1, . . . , n, (k, Yk) is a right product cobordism, given by a sequence of index

k right half-handle attachments; (n+1, Yn+1) is a cobordism provided by a sequence of index(n + 1)handle attache-

ments.

Proof. Let us begin with a Morse function F on the cobordism which has only boundarystable critical points (see Remark 2.3). Assume that w1, . . . , wm are the interior criticalpoints and y1, . . . , yk are the boundary critical points. After a rearrangement of criticalpoints and the cancellation of pairs of critical points as in Lemma 4.9 we can make Ftechnically good. After applying Theorem4.10we get that Fcan have only 0 handles andn + 1 handles as interior handles. Let us write = 1/(4n+ 6) and choose c0 =, c

s1 = 3,

cu1 = 5, . . . ,csk = (4k 1),cuk = (4k+ 1), . . . , csn+1= 1 3,cn+1= 1 . We rearrangethe function F according to Proposition4.6. Then we define for k = 0, 1/2, 1, . . . , n + 1 themanifold k =F

1[4k, (4k+ 2)], Yk = k Y and k = F1(4k).By construction, each part (k, Yk) contains critical p oints only of one type: fork = 0

and n+ 1 they are interior critical points, for k = 1, . . . , nthey are boundary unstable ofindex k and for k = 1/2, . . . , n+ 1/2, they are boundary stable of index k+ 1/2 and weconclude the proof by Proposition2.35.

Remark 4.19. If the cobordism is a product on the boundary, i.e. Y =M [0, 1], we canchoose the initial Morse function to have no critical points on the boundary. Then all thecritical points ofFcome in pairs,zsj andz

uj withz

sj boundary stable,z

uj boundary unstable

and there is a single trajectory ofFgoing from zsj do zuj.The strength of Theorem4.18 is that it is much easier to study the difference between

the intersection forms on (k, Mk) and on (k1/2, Mk1/2). We refer to [BNR] for anapplication of this fact.

5. The cancellation of boundary handles

In this section we assume that F is a Morse function on the cobordism (, Y) satisfyingthe KronheimerMrowkaMorseSmale regularity condition (Definition 4.5). We assumethatFhas precisely two critical points z andw, with ind z = k and ind w= k + 1 and thatthere exists a single trajectory ofF going from z to w. Ifz and w are both interiorcritical points, then[Mi2,Theorem 5.4] implies that (, Y) is a product cobordism. In fact,


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Milnors proof modifies Fonly in a small neighbourhood of, which avoids the boundaryY. Hence, it does not matter, that in our case the cobordism has a boundary.

We want to extend this result to the case of boundary critical points. In some cases ananalogue of the Milnors theorem holds, in other cases we can show that it cannot hold.

5.1. Elementary cancellation theorems.

Theorem 5.1. Letz andw be a boundary critical points of indexk andk + 1, respectively.Assume that is a single trajectory joiningz andw. Furthermore, assume that bothz andw are boundary stable, or both boundary unstable. Then(, Y) is a product cobordism.

As usual, it is enough to prove the result for boundary unstable critical points, the othercase is covered if we change F to 1 F. Note also, cf. Section 5.2, the assumption thatboth critical points are boundary stable, or both boundary unstable is essential.

A careful reading of Milnor[Mi2,pages 4666] shows that the proof there applies to thissituation with only small modifications. Below we present only three steps of that proof,adjusted to our situation. We refer to [Mi2]for all the missing details.

Letbe the gradient vector field ofF. The proof relies on the following proposition (seethe Preliminary Hypothesis 5.5 in [Mi2], proved on pages 5566).

Proposition 5.2. There exist an open neighbourhoodUofand a coordinate mapg : UR0 Rn and a gradient-like vector field agreeing withaway fromUsuch that

g(Y) {x1= 0}, andg(U) {x1 0}; g(z) = (0, 0, . . . , 0); g(w) = (0, 1, . . . , 0); g = = (x1, v(x2), x3, . . . , xk, xk+1, . . . , xn+1), wherev is a smooth function

positive in(0, 1), zero at0 and1 and negative elsewhere. Moreover dvdx2

= 1 near0and1.

Furthermore, U can be made arbitrary small (around).

Given the proposition, we argue in the same way, as in the classical case, cf. [Mi2,pages5055]: we improve the vector field in Uso that it becomes a gradient like vector field of afunctionF, which has no critical points at all. Then the cobordism is a product cobordism.

The proof of Proposition5.2is a natural modification of the Milnors proof. After ap-plying arguments as in [Mi2, pages 55-58] the proof boils down to the following result.

Proposition 5.3 (compare [Mi2, Theorem 5.6]). Let a+ b = n, a 1 and b 0 andwrite a pointx R0 Ra1 Rb as(xa, xb) withxa R0 Ra1 andxb Rb. Assumethath : (R0 Rn1, {0} Rn1)(R0 Rn1, {0} Rn1) is an orientation preservingdiffeomorphism such that h(0) = 0. Suppose that h(R0 Ra1 {0}) intersects{0} {0} Rb only at the origin and the intersection is transverse and the intersection index is+1. Then, given any neighbourhoodN of0

R

0 R

n

1, there exists a smooth isotopyh

tfor t[0, 1] of diffeomorphisms from (R0Rn1, {0} Rn1) to itself withh0 =h suchthat

(I) ht(x) =h(x) away fromN;(II) h1(x) =x in some small neighbourhoodN1 of0 such thatN1N;

(III) h1(R0 Ra1 {0}) {0} {0} Rb ={0} R0 Rn.Remark 5.4. The transversality assumption from the assumption of Proposition 5.3 isequivalent to the flow of being MorseSmale.

The proof of Proposition 5.3in Milnors book is given on pages 5966. We prove hereonly the analogue of[Mi2,Lemma 5.7]. For all other results we refer to Milnors book.


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Lemma 5.5. Let h be as in hypothesis of Proposition 5.3. Then there exists a smoothisotopyht : R0 Rn1 R0 Rn1, withh0 the identity map andht = h, such that foreacht we haveht(R0 Ra1) Rb = 0.Proof. We follow the proof of [Mi2, Lemma 5.7]. We shall construct the required isotopy intwo steps. First we isotopeh byht(x) = 1t h(tx). Thenh1= h and h0is a linear map. If thisis an identity, we are done. Otherwise h0 is just a nondegenerate linear map and clearly itmaps R0 Ra1 Rb diffeomorphically onto itself. It means that under the decompositionR

n = R Ra1 Rb, h0 has the following block structure

h0=

a11 0 0 A B C D

,where a11 > 0, and stars denote unimportant terms. As h0 is orientation-preserving,det h0 >0. We can apply a homotopy of linear maps which changes the first column ofh0to (a11, 0, . . . , 0) and preserves all the other entries ofh0. We do not change the determinantand the condition h0(R0

R

a1)R

b = 0 is preserved (it means thata11det A > 0). Let

h00=A B

C D

.

Obviously det h0= a11det h00, so det h00> 0. We use the same reasoning as in the Milnorsproof to find a homotopy ofh00 to the identity matrix, finishing the proof.

5.2. Non-cancellation results. The two results below have completely obvious proofs,we state them to contrast with Theorem 5.1.

Lemma 5.6. Assume that a Morse functionF on the cobordism (, Y) between(0, M0)and(1, M1) has two critical pointsz andw. Supposez is an interior critical point andwis a boundary critical point. Then(, Y) is not a product cobordism.

Proof. F restricted to Yhas a single critical point, so the cobordism between M0 and M1cannot be trivial.

Lemma 5.7. Suppose thatF has two critical pointsz andw. Assume thatz is boundarystable andw is boundary unstable. Then(, Y) is not a product.

Proof. If it were a product, we would have H(, 0) = 0. We shall show that this is notthe case.

IfF(z) =F(w), then there are no trajectories between z and w, so by Proposition4.1wecan ensure thatF(z)< F(w). So we can always assume that F(z)=F(w). For simplicityassume thatF(z)< F(w). Let c be a regular value such thatF(z)< c < F(w).

By Lemma2.17F1[0, c]0 [0, c]. ThenH(, 0)=H(, F1[0, c]). Now arisesfrom F1[0, c] by a right half-handle addition, hence H(, F1[0, c])= H(H, B), where(H, B) is the corresponding right half-handle. ButH(H, B) is not trivial by Lemma 2.22(or Lemma2.16).

References

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Institute of Mathematics, University of Warsaw, ul. Banacha 2, 02-097 Warsaw, Poland

E-mail address: [email protected]

A. Renyi Institute of Mathematics, 1053 Budapest, Realtanoda u. 13-15, Hungary.


School of Mathematics, University of Edinburgh, Edinburgh EH9 3JZ, Scotland UK.


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Morse Theory for Manifolds With Boundary