## A Note for Witten Complex

### Shi Chen, B07202036, Department of Physics, NTU 6.Jan.2022

### 1 Clifford Actions and the Witten Deformation

Let M be a smooth compact manifold. For any e ∈ T M , and g^{T M} a Riemann
metrics. We can define the dual e^{∗} with g^{T 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(e^{0}) + c(e^{0})c(e) = −2he, e^{0}i
ˆ

c(e)ˆc(e^{0}) + ˆc(e^{0})ˆc(e) = 2he, e^{0}i
c(e)ˆc(e^{0}) + ˆc(e^{0})c(e) = 0

(2)

And we also have the formula from differential geometry d =

n

X

i=1

e^{i}∧ ∇e_{i}

d^{∗}= −

n

X

i=1

ι_{e}i∇ei

(3)

with ∇ the Levi-Civita connection. This give us
d + d^{∗}=

n

X

i=1

c(e^{i})∇_{e}_{i} (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:

D_{T}^{2} = D^{2}+ T

n

X

i=1

c(e_{i})ˆc(∇_{e}_{i}V ) + T^{2}|V |^{2} (6)

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 H^{0}(M ) be the corresponding Sobolev space. For each p ∈ zero(V ) (here
we assume it is discrite points) let U_{p} 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 \U_{p∈zero(V )}U_{p}, there is a constant C_{1}> 0 such
that on M \U_{p∈zero(V )}Up

|V |^{2}≥ C1

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

||DTs||^{2}_{0}= hD^{2}_{T}s, si ≥ (C1T^{2}− C2T )||s||^{2}_{0}

for any S ∈ Ω^{∗}(M ) with support in M \U_{p∈zero(V )}U_{p}. Then equation (7) is just
a consequence of it.

### 3 Harmonic Oscillator on Euclidean Space

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

g = (dy^{1})^{2}+ ... + (dy^{n})^{2}

Hence U_{p} 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

some A ∈ Gl(n) Let ei= _{∂y}^{∂}

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

D_{T}^{2} = −

n

X

i=1

( ∂

∂y^{i})^{2}+ T

n

X

i=1

c(ei)ˆc(eiA) + T^{2}hyAA^{∗}, yi

= −

n

X

i=1

( ∂

∂y^{i})^{2}− T Tr(√

AA^{∗}) + T^{2}hyAA^{∗}, yi

+ T Tr(√

AA^{∗}) +

n

X

i=1

c(e_{i})ˆc(e_{i}A)

(8)

The operator

K_{T} = −

n

X

i=1

( ∂

∂y^{i})^{2}− T Tr(√

AA^{∗}) + T^{2}hyAA^{∗}, yi

= −

n

X

i,j,k=1

( ∂

∂y^{i} + T yk

√

AA^{∗}ki)( ∂

∂y^{i} − yjT√
AA^{∗}ij)

(9)

is a rescaled harmonic oscillator. By the standard result of harmonic operator,
we knows that when T > 0, K_{T} is a non negative elliptic operator with ker K_{T}
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(e_{i})ˆc(e_{i}A) (11)

acting on Λ^{∗}(E_{n}^{∗}) is nonegative. Moreover, dim(ker L) = 1 with
ker L ⊂ Λ^{even}(E_{n}^{∗}) if det A > 0, and ker L ⊂ Λ^{odd}(E^{∗}_{n}) if det A < 0.

proof:

We write

A = U√
A^{∗}A

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

that √

A^{∗}A = W diags1, ..., snW^{∗}= W SW^{∗}

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(e_{i})ˆc(e_{i}A) =

n

X

i=1

c(e_{i})ˆc(e_{j}w_{ij}s_{j}W^{∗}) =

n

X

j=1

s_{j}c(e_{j}W^{∗}U^{∗})ˆc(e_{j}W^{j}) (14)

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

L =

n

X

i=1

s_{i}(1 + c(f_{i}U^{∗})ˆc(f_{i})) (15)
Now, we further define

η_{j}= c(f_{j}U^{∗})ˆc(f_{j})

Then by equation (2) again, we have η_{j} is a self-adjoint operator and η^{2}_{j} = 1.

Thus the lowest eigenvalue of η_{j} is −1. This give us L is a nonnegative
operator.

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

ηiηj= ηjηi

ˆ

c(f_{j})η_{j}= −η_{j}ˆc(f_{j})
ˆ

c(fi)ηj= ηiˆc(fj) if i 6= j

(16)

Then by induction, we have:

dim{x ∈ Λ^{∗}(E_{n}^{∗}) : (1 + ηj)x = 0 for 1 ≤ j ≤ n} = dim Λ^{∗}(E_{n}^{∗})
2^{n} = 1
Moreover, let ρ ∈ Λ^{∗}(E_{n}^{∗}) 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|_{Λ}even/odd(E_{n}^{∗})

Hence, we have ρ ∈ Λ^{even/odd}(E_{n}^{∗}) if and only if det(U ) = ±1.

Combining the result above, we finally get the result:

Propostion 3.

For T > 0, the operator

−

n

X

i=1

( ∂

∂y^{i})^{2}+ T

n

X

i=1

c(e_{i})ˆc(e_{i}A) + T^{2}hyAA^{∗}, yi

acting on Γ(Λ^{∗}(E_{n}^{∗})) 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 U_{x} of x ad an oriented coordinate system y such that on U_{x}, one
has

f (y) = f (x) −1

2(y^{1})^{2}− ... −1

2(y^{n}^{f}^{(x)})^{2}+1

2(y^{n}^{f}^{(x)+1})^{2}+ ... +1

2(y^{n})^{2} (17)
We call the integer n_{f}(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 , U_{x}∩ U_{y}= ∅ 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 f}de^{T f} (18)

One obviously have d^{2}_{T f} = 0, and hence we can define a deform de Rham
complex (Ω^{∗}(M ), dT f)

0 Ω^{0}(M ) ^{d}^{T f} Ω^{1}(M ) ^{d}^{T f} ... ^{d}^{T f} Ω^{dim(M )}(M ) ^{d}^{T f} 0 Then
we have the cohomology group:

H_{T f,dR}^{i} (M ; R) = ker dT f|_{Ω}i(M )

Imd_{T f}|Ω^{i−1}(M )

The first simple conclusion is that

Propostion 4.

For any i, we have

dim H_{T f,dR}^{i} (M ; R) = dim HdR^{i} (M ; R)

proof:

It is easy to see that α → e^{−T f}α gives an well-defined isomorphism from
H_{dR}^{i} (M ; R) to HT f,dR^{i} (M ; R).

Similarly to the undeformed one, we can develop the Hodge theory via the
deformed exteritor derivative d_{T f}, for α, β ∈ Ω^{∗}(M ),

hdT fα, βi = he^{−T f}de^{T f}α, βi = hα, e^{T f}d^{∗}e^{−T f}βi
Thus we have

d^{∗}_{T f} = e^{T f}d^{∗}e^{−T f} (19)
is the formal adjoint of dT f. Then we also have the operator:

D_{T f} = d_{T f} + d^{∗}_{T f} (20)

^{T f} = D^{2}_{T f} = dT fd^{∗}_{T f} + d^{∗}_{T f}dT f (21)
and the Hodge theory tells us:

dim(ker ^{T f}|_{Ω}i(M )) = dim H_{T f,dR}^{i} (M ; R) = dim HdR^{i} (M ; R) (22)
One can actually verifies that

dT f = d + T df ∧, d^{∗}_{T f} = d^{∗}+ T ιdf

Hence we get

D_{T f} = D + T ˆc(df ) (23)

is the special case of equation (5). For general D_{T}, the Laplacian D_{T}^{2} 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 = Id_{n}_{f}_{(x)×n}_{f}_{(x)}⊕ (-Id)_{(n−n}

f(x))×(n−nf(x)); hence, we have:

T f = −

n

X

i=1

( ∂

∂y^{i})^{2}− nT + T^{2}|y|^{2}+ T

n_{f}(x)

X

i=1

(1 − c(e_{i})ˆc(e_{i})) + T

n

X

i=nf(x)+1

(1 + c(e_{i})ˆc(e_{i}))

= −

n

X

i=1

( ∂

∂y^{i})^{2}− nT + T^{2}|y|^{2}+ 2T

nf(x)

X

i=1

ιe_{i}e^{∗}_{i} ∧ +

n

X

i=nf(x)+1

e^{∗}_{i} ∧ ιe_{i}

(24)

The kernel of the operator:

nf(x)

X

i=1

ιe_{i}e^{∗}_{i} ∧ +

n

X

i=nf(x)+1

e^{∗}_{i} ∧ ιe_{i}

on the Euclidean space can be easily found to be
dy^{1}∧ · · · ∧ dy^{n}^{f}^{(x)}
Hence we can get a refinement of Proposition 3 by
Proposition 5.

For any T > 0, the operator

−

n

X

i=1

( ∂

∂y^{i})^{2}− nT + T^{2}|y|^{2}+ 2T

n_{f}(x)

X

i=1

ι_{e}_{i}e^{∗}_{i} ∧ +

n

X

i=n_{f}(x)+1

e^{∗}_{i} ∧ ιei

acting on Γ(Λ^{∗}(E_{n}^{∗})) is nonnegative. Its kernel is one-dimensional and is
generated by

exp(−T |y|^{2}

2 ) · dy^{1}∧ · · · ∧ dy^{n}^{f}^{(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}|_{Ω}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|Ω^{i}(M ) associated with eigenvalues in [0, c].

Since we have

d_{T f}T f = T fd_{T f} = d_{T f}d^{∗}_{T f}d_{T f}
and

d^{∗}_{T f}^{T f} = ^{T f}d^{∗}_{T f} = d^{∗}_{T f}dT fd^{∗}_{T f}

Hence we have d_{T f}(resp. d^{∗}_{T f}) maps F^{[0,c]}_{T f,i}to 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 ), d_{T f}):

(F^{[0,c]}_{T f} , dT f) : 0 F^{[0,c]}_{T f,0} ^{d}^{T f} F^{[0,c]}_{T f,1} ^{d}^{T f} ... ^{d}^{T f} F^{[0,c]}_{T f,n} ^{d}^{T f} 0
(25)

And the Hodge decomposition give us

β_{T f,i}^{[0,c]}:= dim(

ker d_{T f}|_{F}[0,c]

T f,i

Imd_{T f}|_{F}[0,c]

T f,i−1

)

is equal to dim(ker T f|_{Ω}i(M )), which is equal to β_{i} = dim(H^{i}(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
β_{i}≤ mi

(weak Morse inequality) and

β_{i}− β_{i−1}+ ... + (−1)^{i}β_{0}≤ m_{i}− m_{i−1}+ ... + (−1)^{i}m_{0}
(strong Morse inequality). Moreover

βn− βn−1+ ... + (−1)^{n}β0= mn− mn−1+ ... + (−1)^{n}m0

proof:

The weak Morse inequality is a direct consequence of β_{T f,i}^{[0,c]}= β_{i}. For strong
one, we first find that

m_{i}= dim(F^{[0,c]}_{T f,i}) = dim(ker d_{T f}|_{F}[0,c]

T f,i

) + dim(Imd_{T f}|_{F}[0,c]

T f,i

)

= β_{T 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)^{j}mi−j=

i

X

j=0

(−1)^{j}(βi−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)^{j}βi−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.

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|)^{2}exp(−T |yAp|^{2})dy^{1}∧ ... ∧ dy^{n},

ρ_{p,T} = γ(|y|)

√α_{p,T} exp(−T |yAp|^{2}
2 )ρ_{p}

(27)

Then ρp,T ∈ Ω^{∗}(M ) is of unit length with compact support contained in Up.
Let E_{T} be the direct sum of the vector space generated by ρ_{p,T}’s, where p runs
through the set of zero points of V . Since ρ_{p} is either even or odd, we have E_{T}
admit the decompostion E_{T} = E_{T ,even}⊕ ET ,odd. Let E_{T}^{⊥} be the orthogonal
complement of E_{T} in H^{0}(M ). Then we have

H^{0}(M ) = ET ⊕ E_{T}^{⊥}

orthogonally. Let pT and p^{⊥}_{T} denote the orthonoal projection operators from
H^{0}(M ) to ET and E_{T}^{⊥} respectively. Then we defined

D_{T ,1}= p_{T}D_{T}p_{T}, D_{T ,2}= p_{T}D_{T}p^{⊥}_{T},

DT ,3= p^{⊥}_{T}DTpT, DT ,4= p^{⊥}_{T}DTp^{⊥}_{T} (28)
Let H^{1}(M ) be the first Sobolev space with first Sobolev norm on Ω^{∗}(M ).

Then we have Proposition 7.

There exists a constant T_{0}> 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) : H^{1}(M ) → H^{0}(M )
is Fredholm;

2. the operator D_{T ,4}: E_{T}^{⊥}∩ H^{1}(M ) → E_{T}^{⊥} is invertible.

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

Lemma 8.

There exists constant T_{0}> 0 such that for any s ∈ E_{T}^{⊥}∩ H^{1}(M ), s^{0}∈ ET and
T ≥ T_{0}, one has

||DT ,2s||0≤ ||s||0

T ,

||DT ,3s^{0}||0≤ ||s^{0}||0

T ,

proof:

It is easy to see that D_{T ,3}is the formal adjoint of D_{T ,2}. Thus one needs only
to prove the first estimate . Since each ρ_{p,T}, p ∈zero(V ), has support in U_{p}, by
equation (27) and proposition 3, we have that for any s ∈ E_{T}^{⊥}∩ H^{1}(M ),

DT ,2s = X

p∈zero(V )

ρp,T

Z

Up

hρp,T, DTsidvU_{p}

= X

p∈zero(V )

ρp,T

Z

Up

hDTρp,T, sidvU_{p}

= X

p∈zero(V )

ρp,T

Z

U_{p}

hDT(γ(|y|)

√αp,T

exp(−T |yAp|^{2}

2 )ρp), sidvU_{p}

= X

p∈zero(V )

ρp,T

Z

U_{p}

h(c(dγ(|y|))

√αp,T

exp(−T |yAp|^{2}

2 )ρp), sidvU_{p}

(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 ∈ E_{T}^{⊥}∩ H^{1}(M ),

||DT ,2s||0≤ C1T^{n/2}exp(−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 ∈ E_{T}^{⊥}∩ H^{1}(M ),

||DT ,4||0≥ C3||s||0.
Notice since for s ∈ E_{T}^{⊥}∩ H^{1}(M ) one has

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

Then by lemma 8., we just need to show that for some C_{4}≥ 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 ∈ E^{⊥}_{T} ∩ H^{1}(M )
and T ≥ T0,

||DTs||0≥ C√

T ||s||0 (31)

proof:

We denote U_{p}(b) for the open ball around p of radius b.

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

ρ^{0}_{p,T} = (T
π)^{n/4}

q

| det(Ap)| exp(−T |yA_{p}|^{2}

2 ) · ρp (32)

And we can set p^{0}_{T} for all s with Supp(s) ∈ ∪p∈zero(V )Up(4a) by
p^{0}_{T}s = X

p∈zero(V )

ρ^{0}_{p,T}
Z

E_{p}

hρ^{0}_{p,T}, sidvE_{p} (33)

Since pTs = 0, we can rewrite p^{0}_{T} by

p^{0}_{T}s = X

p∈zero(V )

ρ^{0}_{p,T}
Z

Ep

h(1−γ(|y|))(T
π)^{n/4}

q

| det(Ap)| exp(−2T |yAp|^{2}

2 )·ρp, sidvE_{p}

As γ equals to 1 near each p, hence there exists C_{5}> 0 such that when T ≥ 1,

||p^{0}_{T}s||^{2}_{0}≤ C5

√T||s||^{2}_{0} (34)

By proposition 3, we have

DTp^{0}_{T}s = 0

By propostion 3 again, we actually have C_{6}, C_{7}≥ 0 such that

||DTs||^{2}_{0}= ||DT(s − p^{0}_{T}s)||^{2}_{0}≥ C6T ||s − p^{0}_{T}s||^{2}_{0}≥ C6T

2 ||s0||^{2}_{0}− C7

√

T ||s||^{2}_{0} (35)
Hence there exist T_{1}> 0 such that T ≥ T_{1} imply

||DTs||_{0}≥

√C_{6}T
2 ||s||0

Step 2. Suppose Supp(s) ⊂ M \ ∪_{p∈zero(V )}U_{p}(2a). By proposition 1, we have
T_{2}> 0, C_{8}> 0 such that for T ≥ T_{2}, 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 ∈ E_{T}^{⊥}∩ H^{1}(M ), we have

˜

γs ∈ E_{T}^{⊥}∩ H^{1}(M )

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 C_{1}0 = min{√

C_{6}/2, C_{8}/2}. Complete the proof.

Now we can come back to our concrete cases,for D^{2}_{T f} operator. In this case,
for x ∈ M a critical point of f , we have:

αx,T = Z

Ux

γ(|y|)^{2}exp(−T |y|^{2})dy^{1}∧ dy^{2}∧ · · · ∧ dy^{n}
ρ_{x,T} = γ(|y|)

√αx,T

exp(−T |y|^{2})dy^{1}∧ dy^{2}∧ · · · ∧ dy^{n}^{f}

(37)

Apart from the estimate of lemma8 and lemma9, we have more information about this operator:

Propostion 10.

For any T > 0, we have

DT ,1= 0; (38)

proof:

For any s ∈ H^{0}(M ), we have:

pTs = X

x∈crit(f )

hρx,T, si_{H}0(M )ρx,T (39)

and we also have:

DT fhρx,T, si_{H}0(M )ρx,T ∈ Ω^{n}^{f}^{−1}(M ) ⊕ Ω^{n}^{f}^{+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 H^{0}(M ). Let P (c) denote the
orthogonal projection to ET(c). Then

Lemma 11.

There exist C_{1}> 0, T_{3}> 0 such that for any T ≥ T_{3} and any σ ∈ E_{T},

||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 ∈ H^{1}(M ),

||(λ − DT fs)||_{o}≥1

2||λpTs − D_{T ,2}p^{⊥}_{T}s||_{0}+1

2||λp^{⊥}_{T}s − D_{T ,3}p_{T}s − D_{T ,4}p^{⊥}_{T}s||_{0}

≥1 2((c − 1

T)||p_{T}s||_{0}+ (C√

T − c − 1

T)||p^{⊥}_{T}s||_{0})

(41)
Hence, there exist T_{1}> 0 and C_{2}> 0 such that for any T ≥ T_{1} and

s ∈ H^{1}(M ),

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

λ − DT f : H^{1}(M ) → H^{0}(M )

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

P_{T}(c)σ − σ = 1
2π√

−1 Z

δ

((λ − D_{T f})^{−1}− λ^{−}1)σdλ. (43)

and by Proposition 10, we have

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

||(λ − DT f)^{−1}DT ,3σ||0≤ C_{2}^{−1}||DT ,3σ||0≤ 1

C_{2}T||σ||0 (45)
for T > T_{1}.

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

proof of proposition 6:

By lemma 11, there is T_{5}> 0 such that for T > T_{5} P_{T}(c)ρ_{x,T} are linear
independent for xcrit(f ). Thus, for such T , we have:

dim E_{T}(c) ≥ E_{T} (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,Tiρx,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 C_{3}> 0 and T ≥ T_{5} such that:

||pTs||0≤C3

T ||s||0 (49)

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

||p^{⊥}_{T}s||0≥ C4||s||0 (50)
Then we find that for T large enough, we have

CC4

√

T ||s||0≤ ||DT fp^{⊥}_{T}s||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 H^{0}(M ) onto the L^{2}space of

Ω^{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 f}Qis = Qi^{T f}s = µ^{2}Qis

Hence Q_{i}s is an eigenvector of T f with eigenvalue µ^{2}. Moreover, by Lemma
11, we also have:

||Q_{n}_{f}_{(x)}PT(c)ρx,T − ρx,T|| ≤ C1

T (52)

One can see that when T > 0 large enough, the forms Q_{n}_{f}_{(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 Q_{i}E_{T}(c) ≤ dim E_{T}(c) =

n

X

i=0

m_{i} (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 g^{T}M 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:

W^{u}(x) = {y ∈ M : lim

t→−∞ψt(y) = x}

W^{s}(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:

For any x, y ∈ crit(f ) and x 6= y, we have W^{u}(x) and W^{s}(y) intersect
transversally

In particular, since we know that the dimension of W^{u}(x) (resp. W^{s}(x))
should be nf(x)(resp. n − nf(x)). Hence if nf(y) = nf(x) − 1, then

W^{u}(x) ∩ W^{s}(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 W^{u}(x) and W^{s}(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 W^{u}(x). Let x, y ∈ crit(f ) with
nf(y) = nf(x) − 1. Take γ ∈ Γ(x, y), Then the tangent space TyW^{u}(y) is
orthogonal to the tangent space TyW^{s}(y) and is oriented.

For any t ∈ (−∞, ∞), the orthogonal space T_{γ}^{⊥}_{t}W^{s}(y) to Tγ_{t}W^{s}(y) in Tγ_{t}(M )
carries a natural orientation, whith is induced form the orientaion on TyW^{u}(y).

On the other hand, the orthognal space T_{γ}^{0}_{t}W^{u}(x) to −∇f in Tγ_{t}W^{u}(x) can
be oriented in such a way that s is an oriented basis of T_{γ}^{0}_{t}W^{u}(x) if

(−∇f (γt), s) is an orented basis of Tγ_{t}W^{u}(y).

Since W^{u}(x) and W^{s}(y) intersect transversally along γ, for any t ∈ (−∞, ∞),
T_{γ}^{⊥}_{t}W^{s}(y) and T_{γ}^{0}_{t}W^{u}(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(W^{u}) = M

n_{f}(x)=i

R[W^{u}(x)] (59)

and the boundary map

∂W^{u}(x) = X

n_{f}(y)=n_{f}(x)−1

X

γΓ(x,y)

nγ(x, y)W^{y}(y). (60)

The basic result,is Theorem 12.

(C_{∗}(W^{u}), ∂) is a chain complex. Moreover, we have a canonical identification
between its homology group H_{∗}(C_{∗}(W^{u}), ∂) to singular homology group H_{∗}(M )

We now consider its dual complex (C^{∗}(W^{u}), ∂) and we are going to construct
an isomorphism from this dual complex to the singular cohomology group.

### 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 ¯W^{u}(x) is an nf(x) dimensional
submanifold of M with conical singularities.

2. ¯W^{u}(x)\W^{u}(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 ∈ Ω^{i}(M ), this integral is not zero only if
nf(x) = i. Hence we get a Z−graded map from Ω^{∗}(M ) to H_{∗}(C_{∗}(W^{u}), ∂):

P_{∞}: α 7−→ X

x∈crit(f )

[W^{u}(x)]^{∗}
Z

W¯^{u}(x)

α (61)

where [W^{u}(x)]^{∗} is the dual basis of [W^{u}(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 P_{∞}is 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^{∗}(W^{u}) with an inner product such that [W^{u}]^{∗} become an
orthonormal basis. And we now define a linear map J_{T} : C^{∗}(W^{u}) → Ω^{∗}(M )

JTW^{u}(x)^{∗}= ρx, T (62)

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

Noew, we let P^{T} denote the orthogonal projection from Ω^{∗}(M ) on
F_{T f}^{[0,1]}.(Actually, it is the PT(1) we defined before.)Futhermore, we
definede_{T} : C^{∗}(W^{u}) → F_{T 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^{∗}(W^{u}),

(e_{T} − J_{T})s = O(e^{−cT})||s||_{0} (64)
uniformly on M . In particular, e_{T} 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,

(e_{T} − J_{T})W^{u}(x)^{∗}= P_{T}ρ_{x,T} − ρ_{x,T}

= 1

2π√

−1 Z

δ

((λ − DT f)^{−1}− λ^{−1})ρx,Tdλ

= 1

2π√

−1 Z

δ

(λ − D_{T f})^{−1}DT 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 v_{p}> 0 such that as T → ∞,

||D_{T f}ρ_{x,T}||_{p}= O(e^{−c}^{p}^{T}) (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 ∈ Ω^{∗}M ):

||s||q ≤ C1(||Ds||q−1+ ||s||0)

≤ C_{1}(||(λ − D_{T f})s||_{q−1}+ C_{2}T ||s||_{q−1}+ ||s||_{0})

≤ CT^{q}(||(λ − DT f)s||q−1+ ||s||0)

(68)

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

||(λ − DT f)^{−1}s||0≤ C^{0}||s||0 (69)

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

||(λ − DT f)^{−1}s||q ≤ CT^{q}(||s||q−1+ C^{0}||s||0) ≤ C^{00}||s||q−1 (70)
Hence, there exist cq > 0 such that for T large enough, we have:

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

|(λ − D_{T f})^{−1}D_{T f}ρ_{x,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 P_{∞}by P_{∞,T} : F_{T ,f}^{[0,1]}→ C^{∗}(W^{u}). By:

P_{∞,T} : α 7→ P_{∞}e^{T f}α (73)

Being composition of chain map, P_{∞,T} is again a chain map. We also define
two operator F , N on C^{∗}(W^{u}) by

F [W^{u}(x)]^{∗}= f (x)[W^{u}(x)]^{∗}

N [W^{u}(x)]^{∗}= nf[W^{u}(x)]^{∗} (74)
Then we have an estimate:

Theorem 16.

There exists c > 0 such that as T → ∞,
P∞,TeT = e^{T F}(π

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 = W^{u}(x)^{∗}. By definition, we have:

P∞,TeTs = X

y,n_{f}(y)=n_{f}(x)

e^{T f (y)}W^{u}(y)^{∗}
Z

W (y)¯

eT (f −f (y))eTs (76)

By the definition of unstable manifold, we must have:

f − f (y) ≤ 0 (77)

on ¯W^{u}(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)^{n}^{f}^{(x)/2−n/4}(1 − O(e^{−cT})) (79)
By Proposition 13, we have ¯W (y)\W^{u}(y) is a union of certain ¯W^{u}(y^{0}), with
n_{f}(y^{0}) < n_{f}(y). Thus we find that for y ∈ crit(f )m with y 6= x and

n_{f}(y) = n_{f}(x), we then have

x /∈ ¯W^{u}(y) (80)

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

JTs = O(e^{−c}^{0}^{T}) (81)

on ¯W^{u}(y) for some c^{0}> 0. Hence we get:

Z

W (y)¯

eT (f −f (y))e_{T}s = 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 e^{T f} : F_{T f}^{[0,1]}→ Ω^{∗} is an quasi-isomorphism by
Proposition 4. And now P_{∞,T} = P_{∞}◦ e^{T f} is an quasi-isomorphism (actually
isomorphism) for T large enough, too. Hence P∞is 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

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

W^{u}(x) ∩ W^{s}(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 :W^{k}(f ) → W^{k+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
W^{u}(x) and W^{s}(y)). We fix the orientation of W^{u}(x) arbitrarily and define the
orientation of W^{s}(x) by requiring that W^{s}(x) ∩ W^{u}(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 ) ⊗ H_{T S}^{∗} (M, f ) → H_{T S}^{∗} (M, f )

### 9 An H

^{∗}

### (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) :W^{k}(f ) →W^{k+d}(f )

x → X

y∈crit(f ),i(y)=i(x)+d

( Z

M (x,y)

ω) · y (88)

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

P (ω) : H_{T S}^{k} (M, f ) → H_{T S}^{k+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 (ω^{0}) = P (ω ∪ ω^{0}) (90)
As a result, P defines an H^{∗}(M ) module structure on H_{T 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, y_{1}) ∪ M (y_{1}, y_{2}) ∪ ... ∪ M (y_{q}, z) (91)
The union being over all sequences y_{1}, ..., y_{q} of critical points such that

M (x, y_{1}), M (y_{1}, y_{2}), ..., M (y_{q}, z) are all non empty. Moreover, for any such
sequence (y_{1}, ..., y_{q}), 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(a_{0}, ..., a_{q}, −∞, ..., ∞, µ_{j}, ..., µ_{j+p}, +∞, ..., +∞) = G(a_{j}, ..., a_{j+p}, µ_{j}, ..., µ_{j+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.

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 G^{∗}dφ to doing
the integral. By the Stoke’s formula for manifolds with corners, we see that
G^{∗}φ must be integrated only on:

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

P (yˆ _{1}, y_{2}) × ... × ˆP (y_{q}, z) × ∆^{◦}_{q} (98)
or

P (x, yˆ _{1}) × ... × ˆP (y_{q−1}, y_{q}) × ∆^{◦}_{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)

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) : H_{T S}^{k} (M, f ) → H_{T S}^{k+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)

ω_{1}∧ ω2=X

y

Z

M (x,y)

ω_{1}
Z

M (y,z)

ω_{2} (102)

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 (a_{1}, −∞, a_{2}, ∞) to {y}. Now, we can pullback ω_{1}and ω_{2}on

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
φ_{1}∈ H(C ˆP (x, y) × C ˆP (y, z), D_{1})

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) × {(a_{1}, +∞)} 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.