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On quaternion Cauchy-Szegö kernel and related boundary value problem

Irina Markina, Der Chen Chang, Wei Wang University of Bergen, Norway Georgetown University, Washington D.C, USA Zhejiang University, PR China

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 1/39

Upper half plain and Cauchy kernel

x = Re z y = Im z U = R × R+ ⊂ C

Any holomorphic function in U, continuous up to the boundary bU, is defined by the values at the boundary

bU. f(z) = 1 2πi

+∞

−∞

fb(t) t − zdt.

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 2/39

Siegel u.h.sp and Cauchy-Szegö kernel

Im z1

R3 ∼

= Re z1 × C U ⊂ C2 i

U = {(z1, z2) ∈ C2| Im z1 > |z2|2} F ∈ H2(U) : holomorphic, F 2H2 = sup

ε>0

• bU

|F(z+iǫ)|2 dµ(z)

• lim

ε→0 F(z + iǫ) = F b exists in L2(bU) for z ∈ bU;

• Set B(bU) of limits F b’s is a closed subset of L2(bU);
• F 2H2(U) = F bL2(bU).

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 3/39

Siegel u.h.sp and Cauchy-Szegö kernel

U = {(z1, z2) ∈ C2| Im z1 > |z2|2} F ∈ H2(U) : holomorphic, F 2H2 = sup

ε>0

• bU

|F(z+iǫ)|2 dµ(z)

Function F from the Hardy space H2(U) is completely defined by its boundary values F b.

F(z) =

• bU

F b(w)S(z, w) dw, z ∈ U

• z → S(z, w) is holomorphic in U for all w ∈ U
• S(z, w) = S(w, z)
• S(z, w) =

1 4π2

• i

2( ¯

w2 − z2) − z1 ¯ w1

−2

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 4/39

Projection operator

The projection operator C : L2(bU) → B(bU) assoiates

L2(bU) ∋ f → Cf = F b

for some F ∈ H2(U) by making use of the Cauchy-Szegö kernel we obtain

Cf(z) = lim

ε→0

• bU

S(z + εi, w)f(w) dβ(w)

by limit in L2(bU) for any z ∈ bU.

STEIN, E. M., Harmonic Analysis: Real-Variable Methods, Orthogonality, and Oscillatory Integrals,1993.

Goal: to construct analogues of S(z, w) for the quaternion Siegel u.h.sp U

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 5/39

Space of quatenions Q

q = (x1 + ix2 + jx3 + kx4) = (x1, − → u ) ∈ R4 σ = (y1 + iy2 + jy3 + ky4) = (y1, − → v ) ∈ R4 q + σ = (x1 + y1, − → u + − → v ), qσ = (x1y1 − − → u · − → v , x1− → v + y1− → u + − → u × − → v ) ¯ q = (x1, −− → u ), |q|2 = q¯ q, qσ = ¯ σ¯ q q−1 = ¯ q |q|2

Space Q is a normed division algebra

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 6/39

Right quaternion space

Let V be a right vector space over Q

V × Q → V, where (v, σ) → vσ.

Let v1, v2 be an Hermitian product on V

v1σ, v2 = ¯ σv1, v2, v1, v2σ = v1, v2σ, σ ∈ Q (V, v) is a Hilbert space w.r.t. v = v, v1/2 L(v) = vL, v,

for any

v ∈ V

bounded right linear functional: L(vσ) = L(v)σ

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 7/39

Siegel upper half space in Qn

Let

V = Qn = (q1, . . . , qn) = q, p, q =

n

• l=1

¯ plql U = {(q1, q2, . . . , qn) = (q1, q′) ∈ Qn| Re q1 > |q′|2}

The space U is invariant under transformations.

• Translation:

τp : (q1, q′) − → q1 + p1 + 2p′, q′, q′ + p′ ,

for p = (p1, p′) ∈ bU.

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 8/39

Siegel upper half space in Qn

U = {(q1, q′) ∈ Qn| Re q1 > |q′|2}

• Rotations:

Ra : (q1, q′) − → (q1, aq′)

for a ∈ Sp(n − 1)

Rσ : (q1, q′) − → (σq1σ, q′σ)

for σ ∈ Q, |σ| = 1

• Dilations:

δr : (q1, q′) − → (r2q1, rq′), r > 0.

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 9/39

Holomorphic functions in C

f : Ω ⊂ C → C, f = u + iv, z = x + iy is holomorphic if f′(z0) = lim

h→0

f(z0 + h) − f(z0) h

• r

¯ ∂zf = 0 ← → ∂xu = ∂yv, ∂yu = −∂xv

• r

f(z) =

∞

• n=0

an(z − z0)n

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 10/39

Regular functions in Q

Let f : Ω ⊂ Q → Q, if we require that

lim

h→0 h−1

f(q0 + h) − f(q0)

exists for all

q ∈ Q then f(q) = aq + b

is an affine function

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 11/39

Regular functions in Q

Let f : Ω ⊂ Q → Q, if we require that

f(q) =

∞

• n=0

Pn(q − q0)n

exists for all q0 ∈ Q with some polynomials Pn then it equivalent to requirement that

f(q) is analytic with respect of 4 real variables, defined

by q.

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 12/39

Regular functions in Q

Let f : Ω ⊂ Q → Q. A C1(Ω) function

f = f1 + if2 + jf3 + kf4

is (left) regular if it satisfies Cauchy-Riemann-Fueter equations

¯ ∂qf(q) = 0,

for all

q ∈ Ω,

where

q = x0 + ix1 + jx2 + kx3 = z0 + z1j, ¯ ∂q = ∂x0 + i∂x1 + j∂x2 + k∂x3 = ¯ ∂z0 + ¯ ∂z1j

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 13/39

Regular functions in U

The space of regular functions on U is invariant under mentioned above transformations of U. Namely, if f is regular on U, then

f(τp(·)), p ∈ bU; f(Ra(·)), a ∈ Sp(n − 1); σf(Rσ(·)), σ ∈ Q, |σ| = 1,

and

f(δr(·))

are all regular on U.

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 14/39

Hardy space H2(U)

Re q1

Qn−1 × Im Q1

U ⊂ Qn e bU + εe

L2(BU), f, gL2 =

• bU

f(q)g(q)dβ(q).

Hardy space H2(U) is all regular functions in U

F2

H2(U) = sup ε>0

• bU

|F(q + εe)|2 dβ(q) < ∞

It is right quaternion Hilbert space, invariant under the above mentioned transformations

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 15/39

Boundary values

THEOREM 1. Suppose that F ∈ H(bU). Then

• There exists lim

ε→0 F(q + εe) = F b(q) in L2(bU)

• F bL2(bU) = FH2(U),
• The space B(bU) of all boundary values is a closed

subspace of L2(bU).

• C(f)(q) =

QH S(h−1 ∗ q)f(h)d(h),

q, h ∈ QH

where we used the identification bU ∼

= QH.

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 16/39

Siegel upper half plain in C

y = Im z = bU x = Re z U = R2

+ ⊂ C

(R, +) : U → U (a, b) → (a, b + r)

Identification (R, +) ∼

= Im z = (bU) by the action at (0, 0): (R, +) ∋ r → r ∈ bU

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 17/39

Siegel upper half space in C2

Re z1

R3 ∼

= Im z1 × C U ⊂ C2

U = {(z1, z2) ∈ C2| Re z1 > |z2|2}

H1 = {(t, ω) ∈ R × C with ∗}

(H1, ∗) : U → U (z1, z2) → (z1 + |ω|2 + it + 2ω, z2, z2 + ω)

Identification (H1, ∗) ∼

= (bU):

H1 ∋ (t, ω) → (|ω|2 + it, ω) ∈ bU

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 18/39

Quaternion Heisenberg group

QH = (R3 × R4 ∼

= Im Q × Q, ∗) (x, v) ∗ (y, w) = (x + y + 2 Imv, w, v + w) bU ∋ (q1, q′)

(x,v)

→ (q1 + |v|2 + ix1 + jx2 + kx3 + 2v, q′, q′ + v) ∈ bU bU ∼ = QH ∼ = R3 × R4

via action on (0, 0) ∈ bU

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 19/39

The Cauchy-Szegö kernel

S(q, p): U × U → Q is unique function satisfying

• 1. S(·, p) ∈ H2(U) for each p ∈ U. Sb(q, p) is defined for

each p ∈ U and for a.a. q ∈ bU.

• 2. The kernel S is symmetric: S(q, p) = S(p, q). Sb(q, p)

is defined for each q ∈ U and a.a. p ∈ bU.

• 3. The kernel S satisfies the reproducing property for

F ∈ H2(U) F(q) =

• bU

Sb(q, Q)F b(Q)dβ(Q), q ∈ U

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 20/39

Symmetries of the kernel

S(τQ(q), τQ(p)) = S(q, p), S(Ra(q), Ra(p)) = S(q, p), σS(Rσ(q), Rσ(p))σ = S(q, p), S(δr(q), δr(p))r4n+6 = S(q, p).

for q, p ∈ U,

Q ∈ bU, a ∈ Sp(n − 1), σ ∈ Q, |σ| = 1, and r > 0.

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 21/39

Main theorem

The Cauchy-Szegö kernel is given by

S(q, p) = s

• q1 + p1 − 2

n

• k=2

¯ p′

kq′ k

• , p, q ∈ U

where

s(φ) = cn ∂2n ∂x2n

1

φ |φ|4, φ = x1 + x2i + x3j + x4k ∈ Q.

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 22/39

Normalization constant

cn = 1 22n+5π2n+1 (2n)!2K(n) 4n − 1 (n + 2)(2n + 3), K(n) =

2n

• k=0

αk

k

• l=0

C2l

k l

• m=0

(−1)k+mCm

l k−2m

• s=0

Cs

k−2m

2k−2m−s+1 (−1)s 2(k − 2m − s + 1) ! (k − 2m − s + 1)!(4n + 5 + k − 2m − s)! αk = (2n+1−k)(2n+2−k)(4n+3+k)

6

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 23/39

Sketch of the proof

Calculations are reduced to s(q) = Sb(q, 0), by translation invariance. By using of rotation symmetry and homogeneity, we end up by solving a system of ODE’s

s′

1(θ) = (2n + 1)s2

sin θs′

2(θ) = −2s2 cos θ − (2n + 3)s1 sin θ,

sin θs′

3(θ) = −s3 cos θ + 2(n + 1)s4 sin θ,

sin θs′

4(θ) = −s4 cos θ − 2(n + 1)s3 sin θ.

with initial value

s1(0) ∈ R1, s2(0) = s3(0) = s4(0) = 0.

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 24/39

Inhomogeneous equation in Qn

We want to solve

¯ ∂qu(z) = f(z), z ∈ Ω ⊂ Qn.

Compatibility condition? Auxiliary tools?

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 25/39

Inhomogeneous equation in Cn

We want to solve

¯ ∂zu(z) = f(z), z ∈ Ω ⊂ Cn. Then in

the case Ω ⊂ C we have

(1) ¯ ∂zu = f ∼ = (2)

• ∆U(z) = f(z),

z ∈ Ω U(z) = 0, z ∈ bΩ

where u = 4∂zU. Since

f = ∆U = 4¯ ∂z∂zU = ¯ ∂z(4∂zU) = ¯ ∂zu

The boundary condition is needed to determine a unique solution: u ∈ [u + h], h is holomorphic

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 26/39

Inhomogeneous equation in Cn

In the case Ω ⊂ Cn we have ¯

∂z = (¯ ∂z1, . . . , ¯ ∂zn) (1)

• ¯

∂zu = f

∂fj ∂¯ zk = ∂fk ∂¯ zj

∼ = (2)

   U(z) = f(z), z ∈ Ω

U(z), ¯ ∂U ∈ Dom() ¯ ∂f = 0

= ¯

∂∗ ¯ ∂ + ¯ ∂ ¯ ∂∗ : Eq → Eq

Is associated with Dolbeault complex or ¯

∂ complex 0 − → E0(Ω)

¯ ∂

− − → E1(Ω)

¯ ∂

− − → . . .

¯ ∂

− − → En(Ω) − → 0.

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 27/39

Inhomogeneous equation in Cn

(1)

• ¯

∂zu = f

∂fj ∂¯ zk = ∂fk ∂¯ zj

∼ = (2)

   U(z) = f(z), z ∈ Ω

U(z), ¯ ∂U ∈ Dom() ¯ ∂f = 0

= ¯

∂∗ ¯ ∂ + ¯ ∂ ¯ ∂∗ : Eq → Eq, (¯ ∂φ, ψ)L2 = (φ, ¯ ∂∗ψ)L2 (f, h)L2 = (U, h)L2 = (¯ ∂U, ¯ ∂h

• =0

)L2 + (¯ ∂( ¯ ∂∗U

• =u

), h)L2,

where h is harmonic. Thus f = ¯

∂u.

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 28/39

Inhomogeneous equation in Cn

Inverting the operator one gets the result: There is an operator N : Eq(Ω) → Eq(Ω) with the following properties:

• 1. If f ∈ Eq(Ω), then f = Nf + Pf,
• 2. ¯

∂N = N ¯ ∂, ¯ ∂∗N=N ¯ ∂∗,

N = N,

PN = NP.

• 3. The operator N : Eq(Ω) → Eq(Ω) is completely

continuous and for each s ≥ 0 we have

NfHs+1/2 ≤ CfHs−1/2.

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 29/39

Inhomogeneous equation in Qn

The inhomogeneous k-Cauchy-Fueter equation

• Dk

0u(q) = f(q),

q ∈ Ω ∈ H, Dk

1f(q) = 0

compatibility condition

k-CF operator Dk

0 and compatibility operator Dk 1 are

given by k-CF complex

0 − → C∞(Ω, ⊙kC2)

D(k)

− − − − → C∞(Ω, ⊙k−1C2 ⊗ C2)

D(k)

1

− − − − → C∞(Ω, ⊙k−2C2 ⊗ Λ2C2) − → 0.

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 30/39

Motivation

k-CF operator is the Euclidean version of k/2-spin

massless free field operator on the Minkowski space:

k = 1 it is the Dirac-Weyl operator k = 2 it is the Maxwell operator k = 3 it is the Rarita-Schwinger operator k = 4 it is the linearised Einstein operator. . .

All of them are the quaternion analogous of Cauchy-Riemann operator in Cn.

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 31/39

Some general properties

The inhomogeneous ¯

∂-equation on Ω ⊂ C is always

solvable. The inhomogeneous 1-CF equation on Ω ⊂ H is always solvable too, since it is the Dirac equation on R4. The k-CF operator for k ≥ 2 is overdetermined and

• nly can be solved under the compatibility condition

given by the k-CF complex. The k-CF complex on Qn, which is a counterpart of Dolbeault complex in Cn is more complicate but known explicitly.

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 32/39

Previous results

W.W. solved the non-homogeneous k-CF equation on

Ω = Qn

We solved the non-homogeneous k-CF equation on smooth domain Ω Q The next step is to find the relation between the geometry of domain ("pseudoconvexity") and the vanishing property of k-CF cohomology.

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 33/39

Some results

• Dk

0u(q) = f(q),

Dk

1f(q) = 0

∼ =

   U = f(q), q ∈ Ω ⊂ H

Dk

0, Dk 1 ∈ Dom()

¯ ∂f = 0

= Dk

0(Dk 0)∗ + (Dk 1)∗Dk 1 associated to the complex

complex

0 − → C∞(Ω, Ck+1)

D(k)

− − − − → C∞(Ω, C2k)

D(k)

1

− − − − → C∞(Ω, Ck−1) − → 0.

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 34/39

Main result

Suppose Ω ⊂ Q is a smooth and bounded domain. If

f ∈ W s,2(Ω, C2k), s = 0, 1, 2, . . ., is orthogonal to the first

cohomology group H1

k = {f∈C∞(Ω;C2k); D(k)

1 f=0}

{D(k)

1 u=0; u∈C∞(Ω;C2k)} then the

b.v. problem has a solution U = N(k)f:

UHs+2(Ω,C2k) ≤ CΩ,k,sfHs(Ω,C2k)

We have the Hodge-type decomposition:

ψ = (k)

1 N(k)ψ + Pψ,

for any

ψ ∈ W s,2(Ω, C2k),

where P is the orthonormal projection to the first cohomology group under the L2 product.

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 35/39

Details of the proof, k = 2

= −     

∆ + ∆1 L ¯ L ∆ + ∆2 ∆ + ∆2 −L −¯ L ∆ + ∆1

    

∆1 = ∂z0 ¯ ∂z0, ∆2 = ∂z1 ¯ ∂z1, L = ¯ ∂z0 ¯ ∂z1, q = z0 + z1j.

         ψ(x) = f(x),

x ∈ Ω ⊂ R4 ψ1 = ψ2 = 0, ψ0 − ψ4 = 0, x ∈ bΩ ∂x0(ψ0 + ψ3) = 0,

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 36/39

Details of the proof, k = 2

We use Shapiro - Lopatinkskii condition to show that the b.v. problem is regular. Then we need to check that the complex is exact

0 − → C∞(Ω, C3)

D0

− − − → C∞(Ω, C4)

D1

− − − → C∞(Ω, C1) − → 0. D0, D1 are matrix operators in = D0D∗

0 + D∗ 1D1

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 37/39

Details of the proof, k = 2

All above ensures that the operator : W 2+s,2

b

(Ω, C4) → W s,2

b

(Ω, C4)

is Fredholm: the kernel and co-kernel is finite

• dimensional. Thus the operator

N : W s,2(Ω, C4) → W 2+s,2(Ω, C4)

exists and gives

ψ = Nψ + Pψ,

for any

ψ ∈ W s,2(Ω, C4), P is the orthogonal projection to ker .

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 38/39

The end

Thanks for the attention

On quaternion Cauchy-Szeg¨

• kernel and related boundary value problem – p. 39/39