Cauchy-Riemann (CR) Manifolds, Szeg o kernel asymptotics and Morse - - PowerPoint PPT Presentation

Cauchy-Riemann (CR) Manifolds, Szeg¨

• kernel

asymptotics and Morse inequalities on CR-manifolds

This presentation will about the following paper which constitutes my m´ emoire for M2: C.Y. Hsiao, G. Marinescu. Szeg¨

• kernel asymptotics and Morse

inequalities on CR manifolds

Outline

• 1. Overview of the theory of Cauchy Riemann manifolds
• 2. Sub-elliptic estimates for the Kohn Laplacian
• 3. Szeg¨
• kernel, Maximal function, and Weak Morse Inequalities
• 4. Strong Morse Inequalities

Overview of the theory of Cauchy-Riemann manifolds

Let X be a manifold of real dimension 2n − 1 for n 2. We consider its tangent bundle TX and its complexification: CTX := TX ⊗R C. We suppose that T 1,0X is a complex sub-bundle of the complexified tangent bundle CTX. We say that X is Cauchy-Riemann if it satisfies the following properties

• 1. The complex rank of the tangent bundle T 1,0X is n − 1.
• 2. T 1,0X ∩ T 0,1X = {0} where T 0,1X = T 1,0X.
• 3. Let U be an open set. For any V1, V2 in C ∞(U, T 1,0X), we

have [V1, V2] ∈ C ∞(U, T 1,0X).

Similarities and Differences?

What are the similarities and differences between the things that are done in Cauchy-Riemann manifolds, and the things that are done in complex manifolds? Both have the differential operator d defined locally on smooth (p, q)-forms as follows: d  

• |I|=p, |J|=q

fIJdzI ∧ dzJ   =

• |I|=p, |J|=q
• i

∂fIJ ∂zi dzi ∧ dzI ∧ dzJ + ∂fIJ ∂zi dzi ∧ dzI ∧ dzJ

• .

Similarities and Differences?

For complex manifolds: ∂ maps df to ∂fIJ

∂zi dzi ∧ dzI ∧ dzJ.

For Cauchy-Riemann manifolds, ∂b is defined by the following composition: ∧p,qT ∗X

d

∧p+1,qT ∗X ⊕ ∧p,q+1T ∗X πp,q+1 ∧p,q+1T ∗X .

It is not clear that for Cauchy-Riemann manifolds, ∂b ◦ ∂b = 0. Turns out to be true thanks to Cartan-Lie formula.

Is ∂b ◦ ∂b = 0?

Cartan-Lie Formula gives the following lemma: Lemma Let q 1 and ω be a smooth (0, q)-form. For all x ∈ X, if ω(x) annihilates ∧0,qTxX, then dω(x) annihilates ∧0,q+1TxX. Then for v ∈ ∧0,qTxX, we have df (v) = π0,q+1df (v) so that (1 − π0,q+1)df = 0. Hence by previous lemma, we have d(1 − π0,q+1)df = 0. Therefore, ∂b ◦ ∂bf = π0,q+2d ◦ π0,q+1df = π0,q+2d ◦ π0,q+1df + π0,q+2 d(1 − π0,q+1)df

• =0

= π0,q+2d2f = 0.

Complexes

Let Ωp,q(X) be a space of smooth (p, q)-forms. And thus we have a complex.

Ωp,0(X)

∂b

...

∂b

Ωp,n−1(X)

Let Hp,k

b

(X) = ker ∂b : Ωp,k(X) → Ωp,k+1(X) im∂b : Ωp,k−1(X) → Ωp,k(X) . Now given an inner product (−, −) acting on CTX, we may also have ∂

∗ b.

Let b = (∂b + ∂

∗ b)2

be the Kohn Laplacian.

Hodge I

The second similarity and difference is that Hodge theory applies for both cases, except the ways to prove them are different. For Dolbeault complexes on complex manifolds, the first part of the Hodge theory says that Hp,k

∂

(X) is finite-dimensional. But we see that the cohomology group has a finite-dimension because the kernel of the Laplacian is finite-dimensional. This is due to the Hodge isomorphism. To prove this, we ask a simple question: when is a vector subspace

• f a Hilbert space finite-dimensional?

Sufficient condition: when every bounded sequence has a convergent subsequence. (Also known as Riesz’s theorem.)

Hodge I

The tool used here is the inclusion of Sobolev spaces: W ∞

(p,q)(X) = Ωp,q(X) ⊆ ... ⊆ W 2 (p,q)(X) ⊆ W 1 (p,q)(X) ⊆ L2 (p,q)(X).

Observation: G˚ arding’s inequality + Sobolev’s theorem implies that solutions to ∆u = 0 are always smooth, following from the fact that there exists k > 0 such that u2

m+k C(u2 L2 + ∆u2 m) = Cu2 L2

for each m (in the weak sense). By Sobolev’s theorem, u is a smooth form. Let {uk}k be a bounded sequence in ker ∆, which is seen as a vector subspace of L2(X). For each l, using G˚ arding’s inequality, we have ukl CukL2.

Hodge I

Therefore, the sequence {uk} is uniformly bounded by some constant M in W l

(p,q)(X) for each l.

This means that {uk} is uniformly bounded in W 1

(p,q)(X). By

Rellich’s lemma, it has a convergent subsequence in L2

(p,q)(X), say

{uki1} Using this subsequence, it is uniformly bounded in W 2

(p,q)(X),

hence we may extract a subsequence of this subsequence which converges in W 1

(p,q)(X).

Continuing this way, we extract the diagonal elements. We then

• btain a subsequence of {uk} that converges in W l

(p,q)(X) for all l.

By Sobolev’s theorem, the limit of this converging subsequence is smooth.

Hodge I

Let the limit be u = lim uk. For all v ∈ L2

(p,q)(X), we have

0 = ∆uk, v = uk, ∆v, where ∆v is defined in the distributional sense. We may in fact avoid this by choosing v = ∆u straightaway. Since strong convergence implies weak convergence, we have ∆u, ∆u = u, ∆∆u = lim

k→∞uk, ∆∆u = lim k→∞∆uk, ∆u = 0.

Hence ∆u = 0, and ker ∆ is finite dimensional. Hence G˚ arding’s inequality is the essential ingredient for Hodge theory.

Hodge II

The second part of the Hodge theory says that Hp,q

∂ (X) is

isomorphic to the kernel of the laplacian restricted to the smooth (p, q) forms. The following ingredients are used for Hodge theory:

• 1. The Laplacian ∆ = (∂ + ∂

∗)2,

• 2. The existence of the Green’s function (which is well-defined by

Hodge I) G on the eigenspaces of ∆ with eigenvalues = 0 such that G∆ = ∆G = id. Extend G by zero on the kernel of ∆. This implies that ∂ ◦ G = G ◦ ∂. Then we have ∆ ◦ G = id − projker ∆. This implies that at the level of Dolbeault cohomology, id = projker ∆.

Hodge II

Why? For α ∈ Hp,q

∂ (X), we have

(∂∂

∗G + ∂ ∗∂G)α = 0 (In cohomology) + ∂ ∗G∂α = 0.

Hence id − projker ∆ = 0 at the level of cohomology. This means that for each α ∈ ker ∂, α − projker ∆α = ∂β for some β ∈ Hp,q−1

∂

(X). We have therefore an important decomposition: ker ∂ = im∂ ⊕ ker ∆. Taking quotient by the image, we get the second important Hodge result: Hp,k

∂

(X) ∼ = ker ∆.

Hodge (Summary)

Therefore, G˚ arding’s inequality is essential. G˚ arding’s inequality + Sobolev’s theorem + Rellich lemma implies the finiteness of dimension of the space of harmonic functions. This implies well-definedness of the Green’s function, and hence the equivalences between the cohomology group and the space of harmonic functions (link between algebra and analysis). Do we have G˚ arding’s inequalities in the case of Cauchy-Riemann manifolds?

Sub-elliptic Estimates of the Kohn Laplacian

Sub-elliptic Estimates

Sub-elliptic Estimates of the Kohn Laplacian

We may not have the G˚ arding’s inequality for the Cauchy-Riemann manifolds, but we may obtain something weaker for the Kohn Laplacian, which still implies the regularity theorem. The ingredients for this to work are:

• 1. The Levi form.
• 2. The eigenvalues of the Levi form (or we call it later the Y (q)

condition).

• 3. Elliptic regularisation techniques.

The Levi Forms and the Y (q) condition

We recall that the complexified tangent bundle of X may be written as CTX = T 1,0X ⊕ T 0,1X ⊕ CT. Let w0 be a 1-form dual to T. Here T is the vector field which is transversal to T 1,0X and T 0,1X. Definition of Levi Forms: Let p ∈ X be given. Let Z and W be elements in ∧1,0TpX, and ˜ Z and ˜ W be smooth vector fields C ∞(X, ∧1,0TX) such that ˜ Z(p) = Z and ˜ W (p) = W . The Levi form Lp is the Hermitian quadratic form defined as Lp(Z, W ) := 1 2√−1[˜ Z, ˜ W ](p), w0(p). By Cartan-Lie formula, this is independent of the choice of ˜ Z and ˜ W whose evaluation at p are respectively Z and W . At the point p, the Levi forms may be written as a matrix. Therefore, it makes sense to talk of its eigenvalues.

The Levi Forms and the Y (q) condition

Definition of Y (q) condition: The Levi form acting on (p, q)-forms is said to satisfy the Y (q) condition if there are max(n − q, q + 1) eigenvalues of the same sign, or min(n − q, q + 1) pairs of eigenvalues of the different signs. This leads to an important result: let {Lk} be a set of local

• rthonormal frames of T 1,0X with its corresponding dual ωk.

Proposition: Let X be a compact oriented Cauchy-Riemann manifold, and φ be a (p, q)-form with compact support. Let φIJ be ωI ∧ ωJ component of φ, then φ2

LK + φ2 Lk + |Re(Y (φIJ), φIJ)| bφ2 + φ2 := Qb(φ, φ),

where φ2

Lk = n−1

• k=1

Lkφ2.

The Y (q) conditions implies subelliptic estimates.

The results in the preceding slide implies that Theorem (Kohn): Suppose U is a coordinate neighbourhood and if φ2

Lk + φ2 Lk +

• IJ

|Re(Y φIJ, φIJ)| Qb(φ, φ) holds, then φ2

1 2 Qb(φ, φ).

This in turn implies that Proposition: The Kohn Laplacian b is hypoelliptic and if bφ = α with α ∈ W s

(p,q)(M), with s 0, then

φ2

s+1 bφ2 s + φ2.

Sub-elliptic Estimates implies Regularity Property

Therefore, the subelliptic estimates implies regularity property of the Kohn Laplacian. Hence, by emulating the proof of Hodge theorems, we get the Hodge theorem for the case of Cauchy-Riemann manifolds. To prove the regularity property for manifolds, the regularity property is first proved for coordinate neighbourhoods. Let {ζk}∞

k=1 be a sequence of smooth functions on U (a coordinate

neighbourhood) such that ζk = 1 on the support of ζk+1.

Regularity property

If φ is a smooth solution to (b + I)φ = α, we have for each k, ζkφ2

k 2 ζ1α2 k 2 −1 + α2.

(1) This condition can be weakened as follows: assuming that α|U is

• smooth. Let ζ and ζ1 be cut-off functions with ζ1 = 1 on the

compact support of ζ. Then for φ ∈ Dom(b) which is a solution to (b + I)φ = α, we have ζφ2

s+1 ζ1α2 s + α2.

(2) To show (2), it suffices to show that φ is smooth so that (1) applies.

Elliptic Regularisation

So the elliptic regularisation is as follows: instead of the completion of smooth forms by the metric Qb(φ, ψ) = ∂bφ, ∂bψ + ∂

∗ bφ, ∂ ∗ bψ + φ, ψ,

we complete it by Qε

b(φ, ψ) = Qb(φ, ψ) + ε

• i,j

Djηiφ, Djηiψ where {η}i is a partition of unity. We find the weak solutions for each ε to ( + I)φε = α, i.e. Qε

b(φε, ψ) = α, ψ,

for all ψ in the space completed by Qε

b(−, −).

Elliptic Regularisation

There are some advantages:

• 1. By Riesz representation theorem, φε2

s+1 is uniformly

bounded by α|U2

s+1 for each s. Hence, by Rellich’s lemma,

we may extract a subsequence of {φε} that converges for all

• s. Then Sobolev’s embedding theorem says that this limit

must be smooth, and we let β be the limit.

• 2. Turns out that β is φ as required.

Therefore, it turns out that φ is smooth and the regularity property

• f the Kohn laplacian holds.

Summary

The Y (q) condition is required to prove the elliptic half-estimate, which is then used to prove a special kind of regularity property of the Laplacian. But this special kind of regularity property requires φ to be smooth. Hence, the smoothness of φ is proved by other means known as the elliptic regularisation techniques. It is about selectively finding a sequence φε uniformly bounded in all Sobolev spaces that converges to φ. Once the regularity property is satisfied, the Hodge structure follows.

Morse inequalities

Weak Morse inequalities

Weak Morse inequalities

Let L → X be a line bundle over a compact Cauchy-Riemann

• manifold. Hodge theory shows that

dim Hq

b(X, Lk) = N < ∞,

for some N so that we can define the Szeg¨

• function Π(q)

k (x) by

Π(q)

k (x) = N

• j=1

|fj(x)|2, where fj is an orthonormal frame for Hq

b(X, Lk) Therefore, by

integrating the kernel over X, we have

• X

Π(p)

k (x)dvX(x) = N

• k=1
• X

|fj(x)|2dvX(x) =

N

• k=1

1 = N.

Weak Morse inequalities

This seems a natural choice for proving weak Morse inequalities, if we recall Demailly’s proof of weak Morse inequalities which involves evaluating the heat kernel K(x, y) in the form of Kt(x, y) =

∞

• p=0

e−λptΦp(x)Φp(y) where Φp is the eigenvector of the Laplacian corresponding to the eigenvalue λj. By taking the trace, this gives us Kt(x, x) =

∞

• p=0

e−λptΦp(x)2, and the Demailly’s Morse inequalities depends on evaluating it.

Weak Morse inequalities

The only difference is that here in the case of Cauchy-Riemann manifolds, the weak Morse inequality is evaluated via a way different. In this paper, the weak Morse inequalities are proved by the extremal function: the extremal function S(q)

k,J along the direction

eJ is given by S(q)

k,J : y →

sup

α∈Hq

b(X,Lk),α=1

|αJ(y)|2. Then we have Π(q)

k (y) =

• |J|=q

′S(q) k,J (y).

Proving weak Morse inequalities for LHS = proving weak Morse inequalities for RHS.

Weak Morse inequalities

Let L → X be a line bundle with a trivialising section s such that |s|2 = e−φ, where φ = βθ +

• j,t

µj,tzjzt + O(|z||θ|) + O(|θ|2) + O(|(z, θ)|3)

• φ0

+

n−1

• j=1

αjzj +

n−1

• j,t=1

αj,tzjzt

• R

+

n−1

• j=1

αjzj +

n−1

• j,t=1

αj,tzjzt. Setting: Let X be a compact oriented Cauchy-Riemann manifold, with L⊗k → X a line bundle, with the metric kφ as above and its respective Kohn laplacian.

Weak Morse inequalities

By the correspondence between the L2 spaces L2

(0,q)(D, kφ0) ↔ L2 (0,q)(D, kφ),

we may remove the terms R and R. Moreover, by scaling (z, θ) → (z/ √ k, θ/k), the terms O(|z||θ|), O(|θ|2) and O(|(z, θ)|3) disappear as k → ∞. Hence we are only left to consider the Heisenberg group Cn−1 × R with the metric ψ = βθ +

• µj,tzjzt.

Weak Morse inequalities

We note that while the definition Π(q)

k (0) is global, the maximal

function for the Heisenberg group

|J|=q S(q) J,Hn(0) is local, and

there is a relationship between these two, particularly, lim sup

k→∞

k−nΠ(q)

k (0)

• |J|=q

′S(q) J,Hn(0).

Then the weak Morse inequalities are proved by evaluating the expression on the right.

The MΦη and Bergman projection

Let Φη = − √ 2η

n−1

• j=1

λj|zj|2 +

n−1

• j,t=1

µj,tzjzt. Then we define MΦη : T 1,0

z

(Cn−1) → T 1,0

z

(Cn−1) be a linear map by MΦηU, V = ∂∂Φη, U ∧ V , and Rq be a subset of R consisting of η such that MΦη has q negative eigenvalues and n − 1 − q positive eigenvalues. We also define the Bergman projection: B(q)

Φη : L2 (0,q)(Cn−1, Φη) → ker (q) Φη

to be the orthogonal projection onto the kernel space of the Laplacian defined by the norm · 2

Φη :=

• ·2e−Φη.

Result of G.Marinescu

The result of G. Marinescu relates the function MΦη with the Bergman projection. If η / ∈ Rq, then B(q)

Φη (z, z) = 0. Otherwise, let {Zj(η)}n−1 j=1 be an

• rthonormal frame of T (1,0)

z

(Cn−1) that diagonalises MΦη with eigenvalues (νj(η))n−1

j=1 . Let Tj(η) be a basis of T ∗0,1 z

(C) dual to Z j(η). Suppose that νj(η) < 0 for j = 1, ..., q and νj(η) > 0 for the rest, then B(q)

Φη (z, z) = eΦη(2π)−n+1|ν1(η)|...|νn−1(η)| q

• j=1

Tj(η) ∧ (Tj(η)∧)∗. Therefore, TrB(q)

Φη (z, z) = eΦη(2π)−n+1| det MΦη|1Rq.

First estimation

The first estimate proven is that

• S(q)

J,Hn(0)

• CJ,

CJ = 1 4π

• R

B(q)

Φη (0, 0)dzJ, dzJdv(η).

Therefore, we have

• S(q)

J,Hn(0) 1

4π

• R

TrB(q)

Φη (0, 0)dv(η).

While noting that Φη(0) = 0, applying Marinescu’s result, we get

q

• |J|=q

S(q)

J,Hn(0)

1 2(2π)n

• R

| det MΦη|dv(η).

First estimation

This alone is in fact sufficent for us to prove the weak Morse

• inequalities. A more precise estimate shows that

q

• |J|=q

S(q)

J,Hn(0) =

1 2(2π)n

• R

| det MΦη|dv(η), by finding a maximal element. It exists and is found to be u(z, θ) = 1 2π

• eiθη+ βθ

2 +

• η

√ 2− iβ 2 √ 2 λ|z|2

α(z, η)dv(η), where α(z, η) = 1 √ 2 C0| det MΦη|1Rqeν1(η)|z1(η)|2+...+νq(η)|zq(η)|2dz1(η) ∧ ... ∧ dzq(η). for some known constant C0.

Weak Morse inequalities

Using the fact that Φη =

• µj,tzjzt −

√ 2η

• λj|zj|2,

and MΦηU, V = ∂b∂bΦη, U ∧ V , we get det MΦη = det(µj,t − √ 2ηλjδj,t). Let Mφ

x is a Hermitian quadratic form given by the following form

Mφ

x (U, V ) = 1

2U ∧ V , d(∂b + ∂b)φ. which can be written as a matrix of the form (µj,t). Let λjs’ are the eigenvalues of the Levi form Lx. Also, let Rφ,q be a subset of R give by Rφ,q = {s ∈ R : the matrix: (µj,t + sδj,tλj)n−1

j,t=1 has q

negative eigenvalues and n − 1 − q positive eigenvalues.}

Weak Morse inequalities

It turns out that

• Rq

| det MΦη|dv(η) =

• Rφ,q

| det(Mφ

x + sLx)|ds.

is independent of the choice of trivialisation. The fact that it is independent of the choice of trivialisations means that this integration is well-defined, and the reason is as follows: If we have two open trivialisation with non-empty intersection, say ˜ s(x) = g(x)s(x), (with respective norm e−˜

φ and e−φ), then it is

true that R˜

φ,q = Rφ,q + i

Tg g − Tg g

• (p).

Moreover, det(M

˜ φ x + sLx) = det

• Mφ

x +

• s − i

Tg g − Tg g

• Lp
• .

Weak Morse inequalities

Hence the integration remains unchanged. i.e.

• R ˜

φ,q

| det(M

˜ φ x + sLx)|ds

=

• Rφ,q+i
• Tg

g − Tg g

• det
• Mφ

x +

• s − i

Tg g − Tg g

• Lp
• ds

=

• Rφ,q

| det(Mφ

x + sLx)|ds.

In summary, we have the weak Morse inequalities lim sup

k→∞

k−nΠ(q)

k (0)

• |J|=q

′S(q) J,Hn(0) =

1 2(2π)n

• Rφ,q

| det(Mφ

p +sLp)|ds.

Strong Morse inequalities

Strong Morse inequalities

The following result to be proved is this: assuming that the Y (q) condition holds at each point of X. Then for any sequence νk > 0 with νk → 0 as k → ∞, there exists a constant C ′

0 independent of

k such that 1 kn Π(q)

k,kνk(x) C ′

for all x ∈ X. Moreover, there is a sequence µk > 0, µk → 0 as k → ∞ such that for any sequence νk > 0 with limk→∞

µk νk = 0

and νk → 0 as k → ∞, we have lim

k→∞

1 kn Π(q)

k,kνk(x) =

1 2(2π)n

• Rφ,q
• det(Mφ

x + sLx)

• ds.

Strong Morse inequalities

Previously, we have see that this is true lim sup

k→∞

1 kn Π(q)

k (p)

1 2(2π)n

• Rφ(p),q

| det(Mφ

p + sLp)|ds.

The proof which is used to show the above can also be adapted to show that for any sequence νk → 0 as k → ∞, lim sup

k→∞

1 kn Π(q)

k,kνk(0)

1 2(2π)n

• Rφ(p),q

| det(Mφ

p + sLp)|ds.

It only remains to show that lim inf

k→∞

1 kn Π(q)

k,kνk(0)

1 2(2π)n

• Rφ(p),q

| det(Mφ

p + sLp)|ds.

Strong Morse inequalities

But this inverse inequality requires the existence of µk such that limk→∞

µk νk = 0. Let u(z, θ) be the maximal element as before. Let

χ be a smooth function supported on D1 and which is 1 on D1/2. Let βk = χk( √ kz, kθ)

• |J|=q

uJ( √ kz, kθ)eJ(z, θ). Let αk(z, θ) = skk

n 2 ekRβk(z, θ).

We have both lim

k→∞ αk = 1,

lim

k→∞(1

k (q)

b,kαk|αk)k = 0.

Thus we may find a sequence µk such that (1 k (q)

b,kαk|αk)k µk.

Strong Morse inequalities

To finish the proof, we write Ω0,q(X, Lk) = Hq

b,λ(X, Lk) ⊕ Hq b,>λ(X, Lk).

Hence, αk = α1

k + α2

• k. We note that α2

k goes to zero as k → ∞.

This is true because α2

k2

1 kνk ((q)

b,kα2 k|α2 k)k

1 kνk ((q)

b,kαk|αk) µk

νk → 0. And hence α1

k → 1 as k → ∞. Moreover, we have

lim

k→∞

1 kn |α2

k(0)|2 = 0.

The proof of this uses elliptic estimate.

Strong Morse inequalities

Therefore, lim

k→∞ k−n|αk|2 = lim k→∞ k−n(|α1 k|2 + |α2 k|2) = lim k→∞ k−n|α1 k|2

while the expression on the LHS is none other than lim

k→∞ k−n|αk|2 =

1 2(2π)n

• Rφp,q

| det(Mφ

p + sLp)|ds.

This is because we have the expression of α involving u which is the maximal function as we have seen earlier. By definition of Π(q)

k,kνk, we have

k−nΠ(q)

k,νk(0) k−n |α1 k(0)|2

α2

k

→ 1 2(2π)n

• Rφ(p),q

| det(Mφ

p +sLp)|ds.

We have the following strong Morse inequalities in the next few slides.

Strong Morse inequalities

Therefore, we have the following result: Theorem: Assume that the condition Y (q) holds at each point of X, then for any sequence νk with νk → 0 as k → ∞, there is a constant C ′

0 independent of k such that

k−nΠ(q)

k,kνk(x) C ′ 0.

Moreover, there is a sequence µk > 0 such that µk → ∞ and

µk νk → 0 as k → ∞ such that

lim

k→∞ k−nΠ(q) k,kνk(x) =

1 2(2π)n

• Rφ(p),q

| det(Mφ

x + sLx)|ds.

Integrating over X, we have

• X

lim

k→∞ k−nΠ(q) k,kνk(x)dvX(x)

= 1 2(2π)n

• X
• Rφ(p),q

| det(Mφ

x + sLx)|ds dvX(x) + C.

Strong Morse inequalities

Since X is compact, by the first equation of the previous theorem, we have the expression on the LHS to be lim

k→∞

• X

k−nΠ(q)

k,kνk(x)dvX(x)

= 1 2(2π)n

• X
• Rφ(p),q

| det(Mφ

x + sLx)|ds dvX(x) + C.

The expression on the LHS is in fact k−n dim Hq

b,kνk(X, Lk).

Therefore, we have the semi-classical Weyl’s law dim Hq

b,kνk(X, Lk)

=

• X

kn 2(2π)n

• Rφ(p),q

| det(Mφ

x + sLx)|ds dvX(x) + o(kn).

Strong Morse inequalities

Hence, we get the strong Morse inequalities: Theorem: Let q ∈ {0, ..., n − 1}. If Y (j) holds for all j = 0, ..., q, then as k → ∞,

∞

• j=0

(−1)q−j dim Hj

b(X, Lk)

• kn

2(2π)n

q

• j=0

(−1)q−j

• X
• Rφ(p),q

| det(Mφ

x + sLx)|ds dvX(x) + o(kn).

If Y (j) holds for all j = q, q + 1, ..., n − 1, then as k → ∞,

∞

• j=0

(−1)q−j dim Hj

b(X, Lk)

• kn

2(2π)n

n−1

• j=q

(−1)q−j

• X
• Rφ(p),q

| det(Mφ

x + sLx)|ds dvX(x) + o(kn).

Conclusion

The proof of this is an adaption of the same proof for the case of complex manifolds. We refer to the following article

• R. Berman. Bergman Kernels and Local Holomorphic Morse

Inequalities, which can be found on the website http://arxiv.org/abs/math/0211235 with more details of the proofs on the website http://www.math.chalmers.se/Math/Research/Preprints/2002/34.ps.gz

Final Similarity and Conclusion

We draw the final similarity between CR manifolds and complex manifolds before concluding this talk. To quote from the paper taken from arxiv, The starting point is the formula dimC H0,q

∂ (X, Lk) =

• X
• i

|Ψi(x)|2 (3) where Ψi is any orthonormal basis for the space of ∂-harmonic (0, q)-forms, when X is compact (complex manifold)... Note: The expression Bq,k

X (x) =

• i

|Ψ(x)|2, is called the Bergman kernel function in the paper (versus the Szeg¨

• kernel function for CR-manifolds).

Final Similarity and Conclusion

The main point of the proof (of Demailly’s Morse inequality) is to show the corresponding localisation property for the closely related extremal function Sq,k

X (x) defined as

sup |α(x)|2 α2

X

. where the supremum is taken over all ∂-harmonic (0, q)-forms with values in Lk. Since the estimates are purely local, they hold on any complex manifold, and yield local weak holomorphic Morse inequalities for the corresponding L2-objects.

Final Similarity and Conclusion

One final remark, it is fair to say that the formula (3) is the starting point for the previous writers’ approaches to Demailly’s inequalities as well. The heat kernel approach is based on the

• bservation that the term corresponding to the zero eigenvalue in

the heat kernel on the diagonal eq,k(x, x; t) is precisely the Bergman kernel function Bq,k

X (x) (if X is compact). Moreover,

when t tends to infinity the contribution of the other eigenvalues tends to zero. The main problem, then, is to obtain the asymptotic expression for the heat kernel in k and t and investigate the interchanging or the limits in k and t.

References

The main reference for the m´ emoire is the following C.Y. Hsiao, G. Marinescu. Szeg¨

• kernel asymptotics and Morse

inequalities on CR manifolds. For references to Hodge Theory on CR manifolds and subelliptic estimates, one of the earliest references is J.J Kohn. Boundaries of Complex Manifolds. Proc. Conf. Complex Analysis, Minneapolis 1964, 81-94 (1965). The book which gives more details to the paper above is: S.S. Chen and M.C. Shaw. Partial Differential Equations in Several Complex Variables, Chapters 5 and 8. Providence, RI: American Mathematical Society (AMS). Somerville, MA: International Press, xii, 380 o. (2001. Chapters 5 and 8.

References

For general theory on Cauchy-Riemann manifolds,

• R. Beals and P. Greiner. Calculus on Heisenberg Manifolds. Annals
• f Mathematical Studies, Princeton University Press.

C.Y. Hsiao. Projections in several complex variables, Ph.D thesis at Ecole Polytechnique. (avaliable on his personal webpage) For more information on Bergman kernels and holomorphic Morse inequalities, we refer to

• R. Berman. Bergman Kernels and Local Holomorphic Morse

Inequalities.