1 Clifford Actions and the Witten Deformation

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(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∧ ∇e_i

d^∗= −

n

X

i=1

ι_ei∇ei

(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:

D_T² = D²+ T

n

X

i=1

c(e_i)ˆc(∇_eiV ) + T²|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 H⁰(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 C₁> 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||²₀= hD²_Ts, si ≥ (C1T²− C2T )||s||²₀

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 = (dy¹)²+ ... + (dyⁿ)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_T² = −

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(e_i)ˆc(e_iA)

(8)

The operator

K_T = −

n

X

i=1

( ∂

∂yⁱ)²− T Tr(√

AA^∗) + T²hyAA^∗, yi

= −

n

X

i,j,k=1

( ∂

∂yⁱ + T yk

√

AA^∗ki)( ∂

∂yⁱ − 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_iA) (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_iA) =

n

X

i=1

c(e_i)ˆc(e_jw_ijs_jW^∗) =

n

X

j=1

s_jc(e_jW^∗U^∗)ˆc(e_jW^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_iU^∗)ˆc(f_i)) (15) Now, we further define

η_j= c(f_jU^∗)ˆc(f_j)

Then by equation (2) again, we have η_j is a self-adjoint operator and η²_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_jˆc(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^∗) 2ⁿ = 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|_Λeven/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

( ∂

∂yⁱ)²+ T

n

X

i=1

c(e_i)ˆc(e_iA) + T²hyAA^∗, 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(y¹)²− ... −1

2(y^nf(x))²+1

2(y^nf(x)+1)²+ ... +1

2(yⁿ)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 d²_{T f} = 0, and hence we can define a deform de Rham complex (Ω^∗(M ), dT f)

0 Ω⁰(M ) ^{dT f} Ω¹(M ) ^{dT f} ... ^{dT f} Ω^{dim(M )}(M ) ^{dT f} 0 Then we have the cohomology group:

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

Imd_{T f}|Ωⁱ⁻¹(M )

The first simple conclusion is that

Propostion 4.

For any i, we have

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

proof:

It is easy to see that α → e^{−T f}α gives an well-defined isomorphism from H_dRⁱ (M ; R) to HT f,dRⁱ (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} = D²_{T f} = dT fd^∗_{T f} + d^∗_{T f}dT f (21) and the Hodge theory tells us:

dim(ker ^{T f}|_Ωi(M )) = dim H_{T f,dR}ⁱ (M ; R) = dim HdRⁱ (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_T² 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_nf(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

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

( ∂

∂yⁱ)²− nT + T²|y|²+ 2T

nf(x)

X

i=1

ιe_ie^∗_i ∧ +

n

X

i=nf(x)+1

e^∗_i ∧ ιe_i



(24)

The kernel of the operator:

nf(x)

X

i=1

ιe_ie^∗_i ∧ +

n

X

i=nf(x)+1

e^∗_i ∧ ιe_i

on the Euclidean space can be easily found to be dy¹∧ · · · ∧ dy^nf(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

n_f(x)

X

i=1

ι_eie^∗_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 ) · dy¹∧ · · · ∧ dy^nf(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}|_Ωi(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|Ωⁱ(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} ^{dT f} F^[0,c]_{T f,1} ^{dT f} ... ^{dT f} F^[0,c]_{T f,n} ^{dT 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|_Ωi(M )), which is equal to β_i = 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 β_i≤ mi

(weak Morse inequality) and

β_i− β_i−1+ ... + (−1)ⁱβ₀≤ m_i− m_i−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 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)^jmi−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)^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.

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,

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 H⁰(M ). Then we have

H⁰(M ) = ET ⊕ E_T^⊥

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

D_{T ,1}= p_TD_Tp_T, D_{T ,2}= p_TD_Tp^⊥_T,

DT ,3= p^⊥_TDTpT, DT ,4= p^⊥_TDTp^⊥_T (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 D_{T ,4}: E_T^⊥∩ H¹(M ) → E_T^⊥ 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 ∈ E_T^⊥∩ H¹(M ), s⁰∈ ET and T ≥ T₀, one has

||DT ,2s||0≤ ||s||0

T ,

||DT ,3s⁰||0≤ ||s⁰||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^⊥∩ H¹(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^⊥∩ H¹(M ),

||DT ,2s||0≤ C1T^n/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 ∈ E_T^⊥∩ H¹(M ),

||DT ,4||0≥ C3||s||0. Notice since for s ∈ E_T^⊥∩ 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 ∈ E^⊥_T ∩ H¹(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_p⁰s, p ∈ zero(V ) and can thus applied the results of section 3. Thus, for any T > 0, p ∈ zero(V ), set

ρ⁰_p,T = (T π)^n/4

q

| det(Ap)| exp(−T |yA_p|²

2 ) · ρp (32)

And we can set p⁰_T for all s with Supp(s) ∈ ∪p∈zero(V )Up(4a) by p⁰_Ts = X

p∈zero(V )

ρ⁰_p,T Z

E_p

hρ⁰_p,T, sidvE_p (33)

Since pTs = 0, we can rewrite p⁰_T by

p⁰_Ts = X

p∈zero(V )

ρ⁰_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 C₅> 0 such that when T ≥ 1,

||p⁰_Ts||²₀≤ C5

√T||s||²₀ (34)

By proposition 3, we have

DTp⁰_Ts = 0

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

||DTs||²₀= ||DT(s − p⁰_Ts)||²₀≥ C6T ||s − p⁰_Ts||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 )}U_p(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 ∈ E_T^⊥∩ H¹(M ), we have

˜

γs ∈ E_T^⊥∩ H¹(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 C₁0 = min{√

C₆/2, C₈/2}. Complete the proof.

 Now we can come back to our concrete cases,for D²_{T 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 ρ_x,T = γ(|y|)

√αx,T

exp(−T |y|²)dy¹∧ dy²∧ · · · ∧ dy^nf

(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 ∈ H⁰(M ), we have:

pTs = X

x∈crit(f )

hρx,T, si_H0(M )ρx,T (39)

and we also have:

DT fhρx,T, si_H0(M )ρx,T ∈ Ω^nf−1(M ) ⊕ Ω^nf+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

Lemma 11.

There exist C₁> 0, T₃> 0 such that for any T ≥ T₃ 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 ∈ H¹(M ),

||(λ − DT fs)||_o≥1

2||λpTs − D_{T ,2}p^⊥_Ts||₀+1

2||λp^⊥_Ts − D_{T ,3}p_Ts − D_{T ,4}p^⊥_Ts||₀

≥1 2((c − 1

T)||p_Ts||₀+ (C√

T − c − 1

T)||p^⊥_Ts||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:

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)⁻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.

proof of proposition 6:

By lemma 11, there is T₅> 0 such that for T > T₅ 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 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,

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

CC4

√

T ||s||0≤ ||DT fp^⊥_Ts||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

Ωⁱ(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 = µ²Qis

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

||Q_nf(x)PT(c)ρx,T − ρx,T|| ≤ C1

T (52)

One can see that when T > 0 large enough, the forms Q_nf(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_iE_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^TM 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^⊥_tW^s(y) to Tγ_tW^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_γ⁰_tW^u(x) to −∇f in Tγ_tW^u(x) can be oriented in such a way that s is an oriented basis of T_γ⁰_tW^u(x) if

(−∇f (γt), s) is an orented basis of Tγ_tW^u(y).

Since W^u(x) and W^s(y) intersect transversally along γ, for any t ∈ (−∞, ∞), T_γ^⊥_tW^s(y) and T_γ⁰_tW^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 ∈ 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_∗(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_Tρ_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})⁻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 v_p> 0 such that as T → ∞,

||D_{T f}ρ_x,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 ∈ Ω^∗M ):

||s||q ≤ C1(||Ds||q−1+ ||s||0)

≤ C₁(||(λ − D_{T f})s||_q−1+ C₂T ||s||_q−1+ ||s||₀)

≤ CT^q(||(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)

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

||(λ − DT f)⁻¹s||q ≤ CT^q(||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)⁻¹D_{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})⁻¹D_{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)^{nf(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(y⁰), with n_f(y⁰) < 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^−c0T) (81)

on ¯W^u(y) for some c⁰> 0. Hence we get:

Z

W (y)¯

eT (f −f (y))e_Ts = 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 (ω⁰) = P (ω ∪ ω⁰) (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, y₁) ∪ M (y₁, y₂) ∪ ... ∪ M (y_q, z) (91) The union being over all sequences y₁, ..., y_q of critical points such that

M (x, y₁), M (y₁, y₂), ..., M (y_q, z) are all non empty. Moreover, for any such sequence (y₁, ..., 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(a₀, ..., 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, y₁) × ... × ˆP (y_q, z)∆^◦_q× {∞} (97) By the third property of G, this integration is on:

P (yˆ ₁, y₂) × ... × 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)

ω₁∧ ω2=X

y

Z

M (x,y)

ω₁ Z

M (y,z)

ω₂ (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 (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.