# 1 Clifford Actions and the Witten Deformation

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## A Note for Witten Complex

### 1 Clifford Actions and the Witten Deformation

Let M be a smooth compact manifold. For any e ∈ T M , and gT M a Riemann metrics. We can define the dual e with gT M. Then we can defined the Clifford operator:

c(e)· = e∧ · − ιe·, ˆc(e)· = e∧ · + ιe· (1) Then we have the formula

c(e)c(e0) + c(e0)c(e) = −2he, e0i ˆ

c(e)ˆc(e0) + ˆc(e0)ˆc(e) = 2he, e0i c(e)ˆc(e0) + ˆc(e0)c(e) = 0

(2)

And we also have the formula from differential geometry d =

n

X

i=1

ei∧ ∇ei

d= −

n

X

i=1

ιeiei

(3)

with ∇ the Levi-Civita connection. This give us d + d=

n

X

i=1

c(ei)∇ei (4)

Now, for V ∈ Γ(T M ) and T ∈ R, Witten define the deformation of the operator

DT = d + d+ T ˆc(V ) (5)

It is easy to see that DT is self-adjoint and we have the Bochner type formula Bochner type formula

For any T ∈ R, we have the identity:

DT2 = D2+ T

n

X

i=1

c(ei)ˆc(∇eiV ) + T2|V |2 (6)

(2)

proof:

Just using equation (2), (4) and (5).



### 2 Estimation Outside the Zero of V

Let || · ||0 be the 0-th Sobolev norm on Ω(M ) induced by the inner product.

And H0(M ) be the corresponding Sobolev space. For each p ∈ zero(V ) (here we assume it is discrite points) let Up be a neighborhood of it. Then we have the estimation:

Proposition 1.

There exist constants C > 0, T0> 0 such that for any section s ∈ Ω(M ) with Supp(s)⊂ M \Up∈zero(V )Up and T ≥ T0, one has

||DTs||0≥ C√

T ||s||0 (7)

proof:

Since V is nowhere zero onM \Up∈zero(V )Up, there is a constant C1> 0 such that on M \Up∈zero(V )Up

|V |2≥ C1

From equation(6), we then have a constant C2> 0 such that:

||DTs||20= hD2Ts, si ≥ (C1T2− C2T )||s||20

for any S ∈ Ω(M ) with support in M \Up∈zero(V )Up. Then equation (7) is just a consequence of it.



### 3 Harmonic Oscillator on Euclidean Space

We first shrink the neighborhood Up enough and redefine the metric g, such that on each neighborhood Up, we have the metric is standard:

g = (dy1)2+ ... + (dyn)2

Hence Up can be identify with an open neighborhood of the n-dimensional Euclidea space En. We assume that V can be locally written as V = yA for

(3)

some A ∈ Gl(n) Let ei= ∂y

i be an oriented orthonormal basis of En. Then we can refine the Bochner formula by:

DT2 = −

n

X

i=1

( ∂

∂yi)2+ T

n

X

i=1

c(ei)ˆc(eiA) + T2hyAA, yi

= −

n

X

i=1

( ∂

∂yi)2− T Tr(√

AA) + T2hyAA, yi

+ T Tr(√

AA) +

n

X

i=1

c(ei)ˆc(eiA)

(8)

The operator

KT = −

n

X

i=1

( ∂

∂yi)2− T Tr(√

AA) + T2hyAA, yi

= −

n

X

i,j,k=1

( ∂

∂yi + T yk

AAki)( ∂

∂yi − yjT√ AAij)

(9)

is a rescaled harmonic oscillator. By the standard result of harmonic operator, we knows that when T > 0, KT is a non negative elliptic operator with ker KT being one-dimensional and generated by the Gaussian function:

exp(−T |yA|2

2 ) (10)

Furthermore, the nonzero eigenvalues of KT are all greater than CT for some fixed constant C > 0. For the remaining part, we have:

Lemma 2.

The linear operator

L = Tr(√ AA) +

n

X

i=1

c(ei)ˆc(eiA) (11)

acting on Λ(En) is nonegative. Moreover, dim(ker L) = 1 with ker L ⊂ Λeven(En) if det A > 0, and ker L ⊂ Λodd(En) if det A < 0.

proof:

We write

A = U√ AA

with U ∈ O(n)(singular value decomposition). Also, let W ∈ SO(n) be such

that √

AA = W diags1, ..., snW= W SW

(4)

then, one can easily deduces that Tr√

AA=

n

X

i=1

si (12)

and n

X

i=1

c(ei)ˆc(eiA) =

n

X

i=1

c(ei)ˆc(eiU W SW) (13) now, we define {U W }ij = wij, we can get

n

X

i=1

c(ei)ˆc(eiA) =

n

X

i=1

c(ei)ˆc(ejwijsjW) =

n

X

j=1

sjc(ejWU)ˆc(ejWj) (14)

Set fj= ejW. They are another oriented orthonormal basis of En, then by equation (12) and (13):

L =

n

X

i=1

si(1 + c(fiU)ˆc(fi)) (15) Now, we further define

ηj= c(fjU)ˆc(fj)

Then by equation (2) again, we have ηj is a self-adjoint operator and η2j = 1.

Thus the lowest eigenvalue of ηj is −1. This give us L is a nonnegative operator.

By applying equation (2), we actually can get relations:

ηiηj= ηjηi

ˆ

c(fjj= −ηjˆc(fj) ˆ

c(fij= ηiˆc(fj) if i 6= j

(16)

Then by induction, we have:

dim{x ∈ Λ(En) : (1 + ηj)x = 0 for 1 ≤ j ≤ n} = dim Λ(En) 2n = 1 Moreover, let ρ ∈ Λ(En) denote one of the unit sections of kerL, then one has:

ρ = (−1)n(

n

Y

i=1

ηi)ρ = (−1)n(det U )(

n

Y

i=1

c(fi)ˆc(fi))ρ Now, it is easy to see that the chiral element:

(−1)n(

n

Y

i=1

c(fi)ˆc(fi)) = ±Id|Λeven/odd(En)

Hence, we have ρ ∈ Λeven/odd(En) if and only if det(U ) = ±1.

 Combining the result above, we finally get the result:

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Propostion 3.

For T > 0, the operator

n

X

i=1

( ∂

∂yi)2+ T

n

X

i=1

c(ei)ˆc(eiA) + T2hyAA, yi

acting on Γ(Λ(En)) is nonnegative. Its kernel is of dimension one and is generated by

exp(−T |yA|2 2 ) · ρ

Moreover, all the nonzero eigenvalues of this operator are greater than CT for some fixed constant C > 0 (independent of T ).

### 4 Witten Deformation via Morse Function

Let f ∈ C(M ) be a Morse function on M . Then we have the Morse lemma:

Morse Lemma

For any critical points x ∈ M of the Morse function f , there is an open neighborhood Ux of x ad an oriented coordinate system y such that on Ux, one has

f (y) = f (x) −1

2(y1)2− ... −1

2(ynf(x))2+1

2(ynf(x)+1)2+ ... +1

2(yn)2 (17) We call the integer nf(x) the Morse index of f at x. Also, for later use, we assume that for any two different critical points x, y ∈ M of f , Ux∩ Uy= ∅ and equip M with a metric such that the coordinate isometric to an open subset of the Euclidean space. Let mi be the number of critical points such that nf = i.

Given a Morse function f (in this step, any function works). Witten suggested to deform the exterior differential operator d as follows:

dT f = e−T fdeT f (18)

One obviously have d2T f = 0, and hence we can define a deform de Rham complex (Ω(M ), dT f)

0 Ω0(M ) dT f1(M ) dT f ... dT fdim(M )(M ) dT f 0 Then we have the cohomology group:

HT f,dRi (M ; R) = ker dT f|i(M )

ImdT f|i−1(M )

The first simple conclusion is that

(6)

Propostion 4.

For any i, we have

dim HT f,dRi (M ; R) = dim HdRi (M ; R)

proof:

It is easy to see that α → e−T fα gives an well-defined isomorphism from HdRi (M ; R) to HT f,dRi (M ; R).

 Similarly to the undeformed one, we can develop the Hodge theory via the deformed exteritor derivative dT f, for α, β ∈ Ω(M ),

hdT fα, βi = he−T fdeT fα, βi = hα, eT fde−T fβi Thus we have

dT f = eT fde−T f (19) is the formal adjoint of dT f. Then we also have the operator:

DT f = dT f + dT f (20)

T f = D2T f = dT fdT f + dT fdT f (21) and the Hodge theory tells us:

dim(ker T f|i(M )) = dim HT f,dRi (M ; R) = dim HdRi (M ; R) (22) One can actually verifies that

dT f = d + T df ∧, dT f = d+ T ιdf

Hence we get

DT f = D + T ˆc(df ) (23)

is the special case of equation (5). For general DT, the Laplacian DT2 only preserve the Z2 grading, but in this case T f actually preserving the Z grading of Ω(M )

By the Morse lemma, we have the special case of equation (8) on the local coordinate for A = Idnf(x)×nf(x)⊕ (-Id)(n−n

f(x))×(n−nf(x)); hence, we have:

T f = −

n

X

i=1

( ∂

∂yi)2− nT + T2|y|2+ T

nf(x)

X

i=1

(1 − c(ei)ˆc(ei)) + T

n

X

i=nf(x)+1

(1 + c(ei)ˆc(ei))

= −

n

X

i=1

( ∂

∂yi)2− nT + T2|y|2+ 2T

nf(x)

X

i=1

ιeiei ∧ +

n

X

i=nf(x)+1

ei ∧ ιei



(24)

(7)

The kernel of the operator:

nf(x)

X

i=1

ιeiei ∧ +

n

X

i=nf(x)+1

ei ∧ ιei

on the Euclidean space can be easily found to be dy1∧ · · · ∧ dynf(x) Hence we can get a refinement of Proposition 3 by Proposition 5.

For any T > 0, the operator

n

X

i=1

( ∂

∂yi)2− nT + T2|y|2+ 2T

nf(x)

X

i=1

ιeiei ∧ +

n

X

i=nf(x)+1

ei ∧ ιei



acting on Γ(Λ(En)) is nonnegative. Its kernel is one-dimensional and is generated by

exp(−T |y|2

2 ) · dy1∧ · · · ∧ dynf(x)

Moreover, all the nonzero eigenvalues of this operator are greater than CT for some fixed constant C > 0.

### 5 Witten’s instanton complex

The key result of the deformation is the proposition below:

Proposition 6.

For any C > 0, there exists T0> 0 such that when T ≥ T0, the number of eigenvalues in [0, c] of T f|i(M ), 0 ≤ i ≤ n, equals to mi.

With he lemma, we can define F[0,c]T f,i⊂ Ω(M ) the mi dimensional vector space generated by the eigenspaces of T f|i(M ) associated with eigenvalues in [0, c].

Since we have

dT fT f = T fdT f = dT fdT fdT f and

dT fT f = T fdT f = dT fdT fdT f

Hence we have dT f(resp. dT f) maps F[0,c]T f,ito F[0,c]T f,i+1 (resp. F[0,c]T f,i−1). Thus one has the following finite dimensional subcomplex of (Ω(M ), dT f):

(F[0,c]T f , dT f) : 0 F[0,c]T f,0 dT f F[0,c]T f,1 dT f ... dT f F[0,c]T f,n dT f 0 (25)

(8)

And the Hodge decomposition give us

βT f,i[0,c]:= dim(

ker dT f|F[0,c]

T f,i

ImdT f|F[0,c]

T f,i−1

)

is equal to dim(ker T f|i(M )), which is equal to βi = dim(Hi(M ; R)) One of the application of this complex is a proof of Morse inequality:

Morse Inequality

For any integer i such that 0 ≤ i ≤ n,, one has βi≤ mi

(weak Morse inequality) and

βi− βi−1+ ... + (−1)iβ0≤ mi− mi−1+ ... + (−1)im0 (strong Morse inequality). Moreover

βn− βn−1+ ... + (−1)nβ0= mn− mn−1+ ... + (−1)nm0

proof:

The weak Morse inequality is a direct consequence of βT f,i[0,c]= βi. For strong one, we first find that

mi= dim(F[0,c]T f,i) = dim(ker dT f|F[0,c]

T f,i

) + dim(ImdT f|F[0,c]

T f,i

)

= βT f,i[0,c]+ dim(ImdT f|F[0,c]

T f,i−1

) + dim(ImdT f|F[0,c]

T f,i

)

(26)

Hence we get

i

X

j=0

(−1)jmi−j=

i

X

j=0

(−1)ji−j+ dim(ImdT f|F[0,c]

T f,i−j−1

) + dim(ImdT f|F[0,c]

T f,i−j

))

=

i

X

j=0

(−1)jβi−j+ dim(ImdT f|F[0,c]

T f,i

)

Then the strong Morse inequalities follows.

 For the case c = 1, the resulting complex is called Witten’s instanton complex.

(9)

Now, the key point of the construction is to prove Propostion 6, which we start now. The first few step can be doing more general to the case of general deformation to vector field V = Ay. We first shrink all the neighborhood Up

isometric to open ball of radius 4a. And let γ : R → [0, 1] being the bunp function with γ(z) = 1 if |z| ≤ a and γ(z) = 0 if |z| ≥ 2a. Then we define

αp,T = Z

Ux

γ(|y|)2exp(−T |yAp|2)dy1∧ ... ∧ dyn,

ρp,T = γ(|y|)

√αp,T exp(−T |yAp|2 2 )ρp

(27)

Then ρp,T ∈ Ω(M ) is of unit length with compact support contained in Up. Let ET be the direct sum of the vector space generated by ρp,T’s, where p runs through the set of zero points of V . Since ρp is either even or odd, we have ET admit the decompostion ET = ET ,even⊕ ET ,odd. Let ET be the orthogonal complement of ET in H0(M ). Then we have

H0(M ) = ET ⊕ ET

orthogonally. Let pT and pT denote the orthonoal projection operators from H0(M ) to ET and ET respectively. Then we defined

DT ,1= pTDTpT, DT ,2= pTDTpT,

DT ,3= pTDTpT, DT ,4= pTDTpT (28) Let H1(M ) be the first Sobolev space with first Sobolev norm on Ω(M ).

Then we have Proposition 7.

There exists a constant T0> 0 such that

1. for any T ≥ T0 and 0 ≤ u ≤ 1, the operator

DT(u) = DT ,1+ DT ,4+ u(DT ,2+ DT ,3) : H1(M ) → H0(M ) is Fredholm;

2. the operator DT ,4: ET∩ H1(M ) → ET is invertible.

We first doing some estimate of DT ,ifor i = 2, 3, 4.

Lemma 8.

There exists constant T0> 0 such that for any s ∈ ET∩ H1(M ), s0∈ ET and T ≥ T0, one has

||DT ,2s||0≤ ||s||0

T ,

||DT ,3s0||0≤ ||s0||0

T ,

(10)

proof:

It is easy to see that DT ,3is the formal adjoint of DT ,2. Thus one needs only to prove the first estimate . Since each ρp,T, p ∈zero(V ), has support in Up, by equation (27) and proposition 3, we have that for any s ∈ ET∩ H1(M ),

DT ,2s = X

p∈zero(V )

ρp,T

Z

Up

p,T, DTsidvUp

= X

p∈zero(V )

ρp,T

Z

Up

hDTρp,T, sidvUp

= X

p∈zero(V )

ρp,T

Z

Up

hDT(γ(|y|)

√αp,T

exp(−T |yAp|2

2 )ρp), sidvUp

= X

p∈zero(V )

ρp,T

Z

Up

h(c(dγ(|y|))

√αp,T

exp(−T |yAp|2

2 )ρp), sidvUp

(29)

Since γ equals to one in an open neighborhood around zero(V ). dγ vanishes on this open neighborhood. Thus by equation (29), one can easily find that there exist constants T0> 0, C1> 0, C2> 0 such that when T ≥ T0, for any s ∈ ET∩ H1(M ),

||DT ,2s||0≤ C1Tn/2exp(−C2T )||s||0 (30) and the lemma is just the limited case.

 By the lemma, DT ,2and DT ,3are compact operators ( They are actually bounded of finite rank), and we already know DT is Fredholm. Hence we have prove the first part of propostion 7.

For the second part, one just needs to show that there exist constants T0> 0, C3> 0 such that for any T ≥ T0 and s ∈ ET∩ H1(M ),

||DT ,4||0≥ C3||s||0. Notice since for s ∈ ET∩ H1(M ) one has

DTs = DT ,2s + DT ,4s,

Then by lemma 8., we just need to show that for some C4≥ 0.

||DTs||0≥ C4||s||0

when T > 0 is large enough.

Lemma 9.

There exist constants T0> 0 and C > 0 such that for any s ∈ ET ∩ H1(M ) and T ≥ T0,

||DTs||0≥ C√

T ||s||0 (31)

(11)

proof:

We denote Up(b) for the open ball around p of radius b.

Step 1. Assume Supp(s) ∈ ∪p∈zero(V )Up(4a), Then we can assume that we are in a union of Euclidean spaces Ep’s containing Up0s, p ∈ zero(V ) and can thus applied the results of section 3. Thus, for any T > 0, p ∈ zero(V ), set

ρ0p,T = (T π)n/4

q

| det(Ap)| exp(−T |yAp|2

2 ) · ρp (32)

And we can set p0T for all s with Supp(s) ∈ ∪p∈zero(V )Up(4a) by p0Ts = X

p∈zero(V )

ρ0p,T Z

Ep

0p,T, sidvEp (33)

Since pTs = 0, we can rewrite p0T by

p0Ts = X

p∈zero(V )

ρ0p,T Z

Ep

h(1−γ(|y|))(T π)n/4

q

| det(Ap)| exp(−2T |yAp|2

2 )·ρp, sidvEp

As γ equals to 1 near each p, hence there exists C5> 0 such that when T ≥ 1,

||p0Ts||20≤ C5

√T||s||20 (34)

By proposition 3, we have

DTp0Ts = 0

By propostion 3 again, we actually have C6, C7≥ 0 such that

||DTs||20= ||DT(s − p0Ts)||20≥ C6T ||s − p0Ts||20≥ C6T

2 ||s0||20− C7

T ||s||20 (35) Hence there exist T1> 0 such that T ≥ T1 imply

||DTs||0

√C6T 2 ||s||0

Step 2. Suppose Supp(s) ⊂ M \ ∪p∈zero(V )Up(2a). By proposition 1, we have T2> 0, C8> 0 such that for T ≥ T2, we have

||DTs||0≥ C8

T ||s||0 (36)

Step 3. Let ˜γ ∈ C be such that on each Up, ˜γ(y) = γ(|y|/2), and that

˜ γ|M \∪

p∈zeroUp(4a) = 0. Then for any s ∈ ET∩ H1(M ), we have

˜

γs ∈ ET∩ H1(M )

(12)

Then, we get, there exist C9> 0 such that

||DTs||0≥1

2(||(1 − ˜γ)DTs||0+ ||˜γDTs||0)

=1

2(||DT((1 − ˜γ)s) + [D, ˜γ]s||0+ ||DT(˜γs) + [D, ˜γ]s||0)

√ T

2 (C8||(1 − ˜γ)s||0+p

C6||˜γs||0) − C9||s||0

C10

T ||s||0− C9||s||0

where C10 = min{√

C6/2, C8/2}. Complete the proof.

 Now we can come back to our concrete cases,for D2T f operator. In this case, for x ∈ M a critical point of f , we have:

αx,T = Z

Ux

γ(|y|)2exp(−T |y|2)dy1∧ dy2∧ · · · ∧ dyn ρx,T = γ(|y|)

√αx,T

exp(−T |y|2)dy1∧ dy2∧ · · · ∧ dynf

(37)

Propostion 10.

For any T > 0, we have

DT ,1= 0; (38)

proof:

For any s ∈ H0(M ), we have:

pTs = X

x∈crit(f )

x,T, siH0(M )ρx,T (39)

and we also have:

DT fx,T, siH0(M )ρx,T ∈ Ωnf−1(M ) ⊕ Ωnf+1(M ) and has compact support in Ux. Thus DT ,1= 0.

 Now for any positve constant c > 0, let ET(c) denote the direct sum of

eigenspaces of DT f associated with the eigenvalues lying in [−c, c]. Clearly, ET(c) is a finite dimensional subspace of H0(M ). Let P (c) denote the orthogonal projection to ET(c). Then

(13)

Lemma 11.

There exist C1> 0, T3> 0 such that for any T ≥ T3 and any σ ∈ ET,

||PT(c)σ − σ||0≤ C1

T ||σ||0 (40)

proof:

Let δ = {λ ∈ C : |λ| = c} be the counter-clockwise oriented circle. By our pprevios estimate, once can deduce that there is a T0 such that for any λ ∈ δ, T > T0 and s ∈ H1(M ),

||(λ − DT fs)||o≥1

2||λpTs − DT ,2pTs||0+1

2||λpTs − DT ,3pTs − DT ,4pTs||0

≥1 2((c − 1

T)||pTs||0+ (C√

T − c − 1

T)||pTs||0)

(41) Hence, there exist T1> 0 and C2> 0 such that for any T ≥ T1 and

s ∈ H1(M ),

||(λ − DT f)s||0≥ C2||s||0 (42) Thus for any T > T1 and λ ∈ δ, we have

λ − DT f : H1(M ) → H0(M )

is invertible. Thus the resolvent (λ − DT f)−1 is well-defined. From spectral theorem, we have:

PT(c)σ − σ = 1 2π√

−1 Z

δ

((λ − DT f)−1− λ1)σdλ. (43)

and by Proposition 10, we have

((λ − DT f)−1− λ−1)σ = λ−1(λ − DT f)−1DT ,3σ (44) Then, we get:

||(λ − DT f)−1DT ,3σ||0≤ C2−1||DT ,3σ||0≤ 1

C2T||σ||0 (45) for T > T1.

 Now, we are ready to prove the most important Proposition 6.

(14)

proof of proposition 6:

By lemma 11, there is T5> 0 such that for T > T5 PT(c)ρx,T are linear independent for xcrit(f ). Thus, for such T , we have:

dim ET(c) ≥ ET (46)

Now, if dim ET(c) > ET, then there should exist a nonzero element s ∈ ET(c) uch that s is perpendicular to PT(c)ET. That is

hs, PT(c)ρx,Ti0= 0 (47)

for all x ∈ crit(f ). Then we deduced that pTs =X

x

hs, ρx,Tx,T −X

x

hs, PT(c)ρx,TiPT(c)ρx,T

=X

x

hs, ρx,Ti(ρx,T − PT(c)ρx,T) +X

x

hs, ρx,T − PT(c)ρx,TiPT(c)ρx,T

(48) By lemma 11, there is C3> 0 and T ≥ T5 such that:

||pTs||0≤C3

T ||s||0 (49)

Thus, there exists a constant C4> 0 such that when T > 0 is large enough,

||pTs||0≥ C4||s||0 (50) Then we find that for T large enough, we have

CC4

T ||s||0≤ ||DT fpTs||0

= ||DT f − DT fpTs||

= ||DT fs − DT ,3s||0

≤ ||DT fs||0+ ||DT ,3s||0

≤ ||DT fs||0+ 1 T||s||0

(51)

Hence we get

||DT fs||0≥ CC4

T ||s||0− 1 T||s||0

Hence if T large enough, the asumption s ∈ ET(c) nonzero is contradictive.

Hence we get:

dim ET(c) = dim ET =

n

X

i=0

mi

Moreoveer, ET(c) is generated by PT(c)ρx,T for all x ∈ crit(f ). Now, let Qi

denote the orthogonal projection operator from H0(M ) onto the L2space of

(15)

i(M ). Since T f preserves the Z−grading structure, we have for any eigenvectors s of DT f associated with an eigenvalue µ ∈ [−c, c],

T fQis = QiT fs = µ2Qis

Hence Qis is an eigenvector of T f with eigenvalue µ2. Moreover, by Lemma 11, we also have:

||Qnf(x)PT(c)ρx,T − ρx,T|| ≤ C1

T (52)

One can see that when T > 0 large enough, the forms Qnf(x)PT(c) is linear independent for all x, hence we have:

dim QiET(c) ≥ mi (53)

But we also have:

i=n

X

i=0

dim QiET(c) ≤ dim ET(c) =

n

X

i=0

mi (54)

Conbining equation (53) and (54), we get the desired result:

dim QiET(c) = mi (55)



### 6 Thom-Smale Complex

Let f ∈ C(M ) be a Morse function on an n−dimensional closed oriented manifold M . Let gTM be a metric on T M , and let

∇f = (df )

be the gradient vector field of f . Then we can defines a one parameter subgroup of the diffeomorphism group (ψt)t∈R of M :

dy

dt = −∇f (y) (56)

If x ∈ crit(f ), then we set:

Wu(x) = {y ∈ M : lim

t→−∞ψt(y) = x}

Ws(x) = {y ∈ M : lim

t→+∞ψt(y) = x} (57)

Be the unstable and stable cells at x respectively.

Assume that the vector field ∇f satisfies the Smale transversality conditions:

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For any x, y ∈ crit(f ) and x 6= y, we have Wu(x) and Ws(y) intersect transversally

In particular, since we know that the dimension of Wu(x) (resp. Ws(x)) should be nf(x)(resp. n − nf(x)). Hence if nf(y) = nf(x) − 1, then

Wu(x) ∩ Ws(y) consists of finite set Γ(x, y) of integral curves γ of the vector field −∇f , with γ−∞= x and γ= y along with Wu(x) and Ws(y)

transversally.

By [S.Smale Theorem A], there always exists a metric g such that his transversality conidtions holds.

Now, we fixed an orientation on each Wu(x). Let x, y ∈ crit(f ) with nf(y) = nf(x) − 1. Take γ ∈ Γ(x, y), Then the tangent space TyWu(y) is orthogonal to the tangent space TyWs(y) and is oriented.

For any t ∈ (−∞, ∞), the orthogonal space TγtWs(y) to TγtWs(y) in Tγt(M ) carries a natural orientation, whith is induced form the orientaion on TyWu(y).

On the other hand, the orthognal space Tγ0tWu(x) to −∇f in TγtWu(x) can be oriented in such a way that s is an oriented basis of Tγ0tWu(x) if

(−∇f (γt), s) is an orented basis of TγtWu(y).

Since Wu(x) and Ws(y) intersect transversally along γ, for any t ∈ (−∞, ∞), TγtWs(y) and Tγ0tWu(x) can be identified, and hence can compare the induced orentations on them. Then we defined

nγ(x, y) =

(1 if the orientation are the same.

−1 if the orientation are different. (58) Then we can defined our complex:

Ci(Wu) = M

nf(x)=i

R[Wu(x)] (59)

and the boundary map

∂Wu(x) = X

nf(y)=nf(x)−1

X

γΓ(x,y)

nγ(x, y)Wy(y). (60)

The basic result,is Theorem 12.

(C(Wu), ∂) is a chain complex. Moreover, we have a canonical identification between its homology group H(C(Wu), ∂) to singular homology group H(M )

We now consider its dual complex (C(Wu), ∂) and we are going to construct an isomorphism from this dual complex to the singular cohomology group.

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### The de Rham map of the Thom-Smale Complex

We first state the result of Lauudenbach.

Proposition 13.

1. If x ∈ crit(f ), the nthe clousure ¯Wu(x) is an nf(x) dimensional submanifold of M with conical singularities.

2. ¯Wu(x)\Wu(x) is stratified by unstable manifolds of critical points of index strictly less than nf(x).

By this proposition, we have a well defined integration:

Z

W¯u(x)

α

for α ∈ Ω(M ). Moreover, if α ∈ Ωi(M ), this integral is not zero only if nf(x) = i. Hence we get a Z−graded map from Ω(M ) to H(C(Wu), ∂):

P: α 7−→ X

x∈crit(f )

[Wu(x)] Z

W¯u(x)

α (61)

where [Wu(x)] is the dual basis of [Wu(x)].

We are going to prove the theorem:

Theorem 14

P is an quasi-isomorphism

By Stokes theorem and proposition 13, it is easy to see that Pis a chain map. And we are going to prove this theorem via Witten’s instanton complex.

### 7 Proof of the Isomorphism via Witten’s Instanton Complex

In the following, we always assume T is sufficient large such that Proposition 6 is valid.

We first endow C(Wu) with an inner product such that [Wu] become an orthonormal basis. And we now define a linear map JT : C(Wu) → Ω(M )

JTWu(x)= ρx, T (62)

Clearly, JT is an isometry preserving the Z-gradings.

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Noew, we let PT denote the orthogonal projection from Ω(M ) on FT f[0,1].(Actually, it is the PT(1) we defined before.)Futhermore, we definedeT : C(Wu) → FT f[0,1]:

eT = PTJT. (63)

Then we have an estimation:

Theorem 15

There exists c > 0 such that as T → ∞, for any s ∈ C(Wu),

(eT − JT)s = O(e−cT)||s||0 (64) uniformly on M . In particular, eT is an isomorphism

proof:

Let δ = U (1) ∈ C be the counter-clockwise oriented circle. By equation (43), we have for any x ∈ crit(f ) and T > 0 large enough,

(eT − JT)Wu(x)= PTρx,T − ρx,T

= 1

2π√

−1 Z

δ

((λ − DT f)−1− λ−1x,T

= 1

2π√

−1 Z

δ

(λ − DT f)−1DT fρx,T

λ dλ

(65)

For any p ≥ 0, let || · ||p denote the p-th Soolev norm on Ω(M ).

By the construction of ρx,T, for small neighborhood of x, we have:

DT fρx,T = 0; (66)

Hence by definition, for any positive p, there is vp> 0 such that as T → ∞,

||DT fρx,T||p= O(e−cpT) (67) Take p > 1, Since D is a first order elliptic operator, by G˚arding’s inequality, we have C, C1, C2> 0 such that for s ∈ ΩM ):

||s||q ≤ C1(||Ds||q−1+ ||s||0)

≤ C1(||(λ − DT f)s||q−1+ C2T ||s||q−1+ ||s||0)

≤ CTq(||(λ − DT f)s||q−1+ ||s||0)

(68)

and by equation (42), there also exist C0> 0 such that for λ ∈ δ, s ∈ Ω(M ) and T large enough

||(λ − DT f)−1s||0≤ C0||s||0 (69)

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Combine equation (68) and (69), we get:

||(λ − DT f)−1s||q ≤ CTq(||s||q−1+ C0||s||0) ≤ C00||s||q−1 (70) Hence, there exist cq > 0 such that for T large enough, we have:

||(λ − DT f)−1DT fρx,T||q = O(e−cT) (71) uniformmly on λ ∈ δ. Then by Sobolev inequality, we get there exist c > 0 such that:

|(λ − DT f)−1DT fρx,T| = O(e−cT) (72) uniformlly on λ. Hence prove the first assertion. Since JT is an isometry, in particular, we have eT is an isomorphism for T large enough.

 Now, we define the deform Pby P∞,T : FT ,f[0,1]→ C(Wu). By:

P∞,T : α 7→ PeT fα (73)

Being composition of chain map, P∞,T is again a chain map. We also define two operator F , N on C(Wu) by

F [Wu(x)]= f (x)[Wu(x)]

N [Wu(x)]= nf[Wu(x)] (74) Then we have an estimate:

Theorem 16.

There exists c > 0 such that as T → ∞, P∞,TeT = eT F

T)N/2−n/4(1 + O(e−cT)) (75) In particular, P∞,T is an isomorphism for T > 0 large enough.

proof:

Take x ∈ crit(f ), s = Wu(x). By definition, we have:

P∞,TeTs = X

y,nf(y)=nf(x)

eT f (y)Wu(y) Z

W (y)¯

eT (f −f (y))eTs (76)

By the definition of unstable manifold, we must have:

f − f (y) ≤ 0 (77)

(20)

on ¯Wu(y). Apply Theorem 15, we have:

Z

W (y)¯

eT (f −f (y))eTs = Z

W (y)¯

eT (f −f (y))JTs + O(e−cT) (78)

for some c > 0. Since supp(JTs) ∈ Ux, we can using the definition of ρx,T to give us:

Z

W (x)¯

eT (f −f (x))eTs = (π

T)nf(x)/2−n/4(1 − O(e−cT)) (79) By Proposition 13, we have ¯W (y)\Wu(y) is a union of certain ¯Wu(y0), with nf(y0) < nf(y). Thus we find that for y ∈ crit(f )m with y 6= x and

nf(y) = nf(x), we then have

x /∈ ¯Wu(y) (80)

Hence, the by the definition of ρx,T again, we have:

JTs = O(e−c0T) (81)

on ¯Wu(y) for some c0> 0. Hence we get:

Z

W (y)¯

eT (f −f (y))eTs = O(e−cT) (82)

Combine all the result, we are done.

 Finally, we get our proof for theorem 14:

proof of theorem 14:

Since we have already seen that eT f : FT f[0,1]→ Ω is an quasi-isomorphism by Proposition 4. And now P∞,T = P◦ eT f is an quasi-isomorphism (actually isomorphism) for T large enough, too. Hence Pis an quasi-isomorphism.



### 8 The Product Structure of The Thom-Smale Complex (Notation)

As we all know, the cohomology is a graded ring with cup product as its product structure. In this section, we will follows the discussion of C. Viterbo to encode the product structure on the Thom-Smale Complex. For this, we first clear the notation on his paper.

Let f being a Morse function on a smooth compact manifold M (In his paper, the result can be generalized to non-compact cases in certain ways, but we assume the compactness for simplicity.) And also assume all good property as

(21)

before. For all critical points x, y ∈ M of f . We define P (x, y) being all gradient flow with start points x and end points y.

P (x, y) = {γ : R → M | ˙γ(s) = df (s), lim

s→−∞γ(s) = x, lim

s→∞γ(s) = y} (83) There is a natural R action on P (x, y) defined by (t.γ)(s) = γ(s + t); Hence, we can defined:

P (x, y) = P (x, y)/Rˆ (84)

Clearly, this is the set of all gradient flow from y to x regardless of the midpoint. Finally, we can define the set:

M (x, y) = {γ(0) ∈ M |γ ∈ P (x, y)} (85) Be the set of points on P (x, y), this is the same as the intersection

Wu(x) ∩ Ws(y) we defined last time up to the points x, y. By the uniqueness theorem, we know that P (x, y) is diffeomophic to M (x, y).

Let i(x) denote the Morse index of the critical point x, we then have:

dim P (x, y) = i(y) − i(x)

dim ˆP (x, y) = i(y) − i(x) − 1 (86) Then the Thom-Smale complex W(f ) is defined to be the free R-module with one generator for each critical points and graded by its Morse index (when M is not compact, it is useful to define W(f ; a, b) to restrict the discussion to the region f−1[a, b]), and we can define the coboundary map by:

δ :Wk(f ) → Wk+1(f )

δ(x) = X

y∈crit(f ),i(y)=i(x)+1

n(x, y) · y (87)

Where the coefficient n(x, y) is the as before (the intersection number of Wu(x) and Ws(y)). We fix the orientation of Wu(x) arbitrarily and define the orientation of Ws(x) by requiring that Ws(x) ∩ Wu(x) = (+1) · x. We have known that this cohomology is isomorphic the the de Rham cohomology. Now, we are going to define the “cap” product structure on the cohomology:

H(M ) ⊗ HT S (M, f ) → HT S (M, f )

### (M )-module Structure on the Cohomology of (W, δ)

Theorem 17.

Let omega be a closed d-form on M , and let π(ω) be the map:

π(ω) :Wk(f ) →Wk+d(f )

x → X

y∈crit(f ),i(y)=i(x)+d

( Z

M (x,y)

ω) · y (88)

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Then we have π(ω) commutes with δ, inducing a map in cohomology:

P (ω) : HT Sk (M, f ) → HT Sk+d(M, f ) (89) Moreover, this map depends only on the cohomology class of ω in H(M ), and we have the associativity;

P (ω)P (ω0) = P (ω ∪ ω0) (90) As a result, P defines an H(M ) module structure on HT S (M, f )

We will need some lemma to prove this theorem. First is to explore the structure of the set M (x, y):

Lemma 18.

The closure of M (x, z) may be described as M (x, z) =¯ [

M (x, y1) ∪ M (y1, y2) ∪ ... ∪ M (yq, z) (91) The union being over all sequences y1, ..., yq of critical points such that

M (x, y1), M (y1, y2), ..., M (yq, z) are all non empty. Moreover, for any such sequence (y1, ..., yq), there is a map

G : ˆP (x, y1) × ... × ˆP (yq, z) × ∆q+1→ ¯M (x, z) (92) where

q+1= {(λ0, ..., λq) ∈ [−∞, +∞][q + 1]|1 + λj ≤ λj+1} (93) and

1. The image of G is a neighborhood of M (x, y1) ∪ ... ∪ M (yq, z) in ¯M (x, z) 2. The restriction of G to ˆP (x, y1) × ... × ˆP (yq, z) × ∆q+1◦ is a

diffeomorphism onto its image.

3.

G(a0, ..., aq, −∞, ..., ∞, µj, ..., µj+p, +∞, ..., +∞) = G(aj, ..., aj+p, µj, ..., µj+p) (94) proof:

Since the paper itself does not contains the complete proof, either, and the theorem is very intuitive, we omit the proof.

 The next lemma is to develop a new Stokes’ formula for our application.

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Lemma 19.

Let x, z with i(x) − i(z) = k + 1. Define

∂M (x, z) = X

i(y)=i(x)+1

n(x, y)M (y, z) + X

i(y)=i(z)−1

n(y, z)M (x, y) (95)

Then we have for any k-form φ:

Z

∂M

φ = Z

M (x,z)

dφ (96)

proof:

We first using a partition of unity restrict our attention to a neighborhood of M (x, y1) ∪ ... ∪ M (yq, z). Since G−1 is a diffeomorphism of such a neigborhood into ˆP (x, y1) × ... × ˆP (yq, z) × ∆q+1, we may pull back the form Gdφ to doing the integral. By the Stoke’s formula for manifolds with corners, we see that Gφ must be integrated only on:

P (x, yˆ 1) × ... × ˆP (yq, z) × {−∞}∆q[ ˆP (x, y1) × ... × ˆP (yq, z)∆q× {∞} (97) By the third property of G, this integration is on:

P (yˆ 1, y2) × ... × ˆP (yq, z) × ∆q (98) or

P (x, yˆ 1) × ... × ˆP (yq−1, yq) × ∆q (99) which have dimension < k unless i(x) − i(y1) = 1 (resp. i(yq) − i(z) = 1).

Thus the only integrals of φ that appear are these on M (y, z) (resp. M (x, y)) with i(x) − i(y) = 1 (resp. i(y) − i(z) = 1) and the integral appears once for each elemet P (x, y) counted with the proper sign. This concludes the proof.

 . With the help of this Stokes formula, we can now prove the key lemma:

Lemma 20.

π(dω) = δπ(ω) + π(ω)δ (100)

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proof:

π(dω)x =X

t

( Z

M (x,t)

dω)t

=X

t

( Z

∂M (x,t)

ω)t

=X

t

( X

i(y)=i(x)+1

n(x, y) Z

M (y,t)

ω + X

i(z)=i(t)−1

n(z, t) Z

M (x,z)

ω)t

=X

t

X

i(y)=i(x)+1

n(x, y)(

Z

M (y,t)

ω)t +X

t

X

i(z)=i(t)−1

n(z, t)(

Z

M (x,z)

ω)t

= π(ω)δx + δπ(ω)x

= (δπ(ω) + π(ω)δ)x Corollary.

If ω is closed, π(ω) induces a map P (ω) : HT Sk (M, f ) → HT Sk+d(M, f ) which depends only on the cohomology class of ω.

proof:

If dω = 0, δπ(ω) = −π(ω)δ, hence (−1)deg· π(ω) is a chain map, hence define a map in cohomology.

If ω = dφ, π(dφ) = δπ(φ) + π(φ)δ, hence φ(dφ) sends cocycles to coboundaries:

it induces the zero map in cohomology.

 Now, we prove the final part of the theorem

Lemma 21.

Let ω1, ω2 be closed forms, Then we have P (ω1∧ ω2) = P (ω1)P (ω2) proof:

We have

P (ω1∧ ω2)x =X

z

( Z

M (x,z)

ω1∧ ω2)z

P (ω1)P (ω2)x =X

z

(X

y

( Z

M (x,y)

ω1

Z

M (y,z)

ω2))z

(101)

Hence, what we need to prove is the equality:

Z

M (x,z)

ω1∧ ω2=X

y

Z

M (x,y)

ω1 Z

M (y,z)

ω2 (102)

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We only have term with i(y) = i(x) + k1= i(z) − k2 in the left hand side.

By the technique of partition of unity, we may assume that ω1 and ω2 vanish outside neighborhood of ¯M (x, y) and ¯M (y, z). Then by lemma 18, we can pullback the problem on the image of ˆP (x, y) × ˆP (y, z) × ∆2. Now, considering the cone sapce C ˆP (x, y) × C ˆP (y, z), where

C ˆP (x, y) = ˆP (x, y) × [−∞, ∞]/ ˆP (x, y) × {+∞} And we have a map:

C ˆP (x, y) × C ˆP (y, z) → P (x, y) × ˆˆ P (y, z) × ∆2(a1, t1, a2, t2) → (a1, a2, t1, t1+ t2) (103) and maps (a1, −∞, a2, ∞) to {y}. Now, we can pullback ω1and ω2on

C ˆP (x, y) × C ˆP (y, z), and we get two forms φ1, φ2.

By the fact that ω1 vanishes away from M (x, y), we get that in fact φ1∈ H(C ˆP (x, y) × C ˆP (y, z), D1)

φ2∈ H(C ˆP (x, y) × C ˆP (y, z), D1)

(104)

where Di= {(a1, t1, a2, t2)|ti≥ C} for some C. If we let A = ˆP (x, y) and B = ˆP (y, z), then we have:

φ1∈ H(CA × CB, A × CB)

φ2∈ H(CA × CB, CA × B) (105)

Then by the fact of algebraic topology, we have the formula:

Z

CA×CB

φ1∧ φ2= Z

CA

φ1

Z

CB

φ2 (106)

And since C ˆP (x, y) × {(a1, +∞)} goes to M (x, y), we thus have:

Z

C ˆP (x,y)

φ1= Z

M (x,y)

ω1 (107)

The lemma follows.

 Finally, we are going to prove that his product structure coincide with the original cup product one.

Proposition 22.

With the above assumptions. Under the identification of the Thom-Smale cohomology and the de Rham cohomology. The product structure above is just the usual cup product.

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