Some Rarita-Schwinger Type Operators

Transcription

1 Some Rarita-Schwinger Type Operators Junxia Li University of Arkansas April 7-9, th University of Arkansas Spring Lecture Series joint work with John Ryan, Charles Dunkl and Peter Van Lancker

2 A reference J. Bureš, F. Sommen, V. Souček, P. Van Lancker, Rarita-Schwinger Type Operators in Clifford Analysis, J. Funct. Annl. 185 (2001), No.2,

3 Introduction In representation theory for the orthogonal group,o(n), one can consider spaces of homogeneous harmonic polynomials. If one refines to solutions to the Dirac equation, namely to spaces of homogeneous monogenic polynomials one refines the representation theory to the covering group of O(n). Then the Rarita-Schwinger operators arise. The Rarita-Schwinger operators are generalizations of the Dirac operator. They are also known as Stein-Weiss operators. We denote a Rarita-Schwinger operator by R k, where k = 0, 1,, m,. When k = 0 it is the Dirac operator.

4 Goals of my talk 1. Construct the Rarita-Schwinger operators based on the Dirac operator. 2. Construct fundamental solutions to the Rarita-Schwinger equations. 3. Give some basic integral formulas related to the Rarita-Schwinger operators.

5 Clifford algebras We need Clifford algebras: A Clifford algebra, Cl n, can be generated from R n by considering the relationship x 2 = x 2 for each x R n. We have R n Cl n. If e 1,..., e n is an orthonormal basis for R n, then x 2 = x 2 tells us that e i e j + e j e i = 2δ ij, where δ ij is the Kronecker delta function. An arbitrary element of the basis of the Clifford algebra can be written as e A = e j1 e jr, where A = {j 1,, j r } {1, 2,, n} and 1 j 1 < j 2 < < j r n. Hence for any element a Cl n, we have a = A a Ae A, where a A R.

6 For a Cl n, we will need the following anti-involutions: Reversion: ã = ( 1) A ( A 1)/2 a A e A, A where A is the cardinality of A. In particular, e j1 e jr = e jr e j1. Also ãb = bã for a, b Cl n. Clifford conjugation: ā = A ( 1) A ( A +1)/2 a A e A. Further, we have e j1 e jr = ( 1) r e jr e j1 and ab = bā for a, b Cl n.

7 Pin(n) and Spin(n) Pin(n) := {a Cl n : a = y 1... y p : y 1,..., y p S n 1, p N} is a group under multiplication in Cl n. Note for y S n 1, x R n then yxy is a reflection in the y direction. By the Cartan Dieudonné Theorem each O O(n) is the composition of a finite number of reflections. If we choose y 1,..., y p arbitrarily in S n 1 the group homomorphism θ : Pin(n) O(n) : a O a with a = y 1... y p and O a (x) = axã is surjective. Further ax( ã) = axã, so 1, 1 ker(θ). In fact ker(θ) = {±1}. Let Spin(n) := {a Pin(n) : a = y 1... y p and p is even}. Then Spin(n) is a subgroup of Pin(n) and is surjective with kernel {1, 1}. θ : Spin(n) SO(n)

8 Dirac Operator The Dirac Operator in R n is defined to be D := n j=1 e j. x j Note D 2 = n, where n is the Laplacian in R n. The solution of the equation Df = 0 is called a left monogenic function.

9 Almansi-Fischer decomposition Let P k denote the space of Cl n valued monogenic polynomials, homogeneous of degree k and H k be the space of Cl n valued harmonic polynomials homogeneous of degree k. If h k H k then Dh k P k 1. But Dup k 1 (u) = ( n 2k + 2)p k 1 (u), so we have an Almansi-Fischer decomposition H k = P k upk 1. Hence for any h k H k, h k (u) = p k (u) + up k 1 (u). Note that if Df (u) = 0 then f (u) D = f (u)d = 0. So we can talk of right k monogenic polynomials and we have a right Almansi-Fisher decomposition, H k = P k P k 1 ū.

10 Rarita-Schwiger operators R k Suppose U is a domain in R n. Consider a function f : U R n Cl n such that for each x U, f (x, u) is a left monogenic polynomial homogeneous of degree k in u.

11 Rarita-Schwiger operators R k Suppose U is a domain in R n. Consider a function f : U R n Cl n such that for each x U, f (x, u) is a left monogenic polynomial homogeneous of degree k in u. Consider D x f (x, u)

12 Rarita-Schwiger operators R k Suppose U is a domain in R n. Consider a function f : U R n Cl n such that for each x U, f (x, u) is a left monogenic polynomial homogeneous of degree k in u. Consider D x f (x, u) = f 1,k (x, u) + uf 2,k 1 (x, u), where f 1,k (x, u) and f 2,k 1 (x, u) are left monogenic polynomials homogeneous of degree k and k 1 in u respectively.

13 Rarita-Schwiger operators R k Suppose U is a domain in R n. Consider a function f : U R n Cl n such that for each x U, f (x, u) is a left monogenic polynomial homogeneous of degree k in u. Consider D x f (x, u) = f 1,k (x, u) + uf 2,k 1 (x, u), where f 1,k (x, u) and f 2,k 1 (x, u) are left monogenic polynomials homogeneous of degree k and k 1 in u respectively. Let P k be the left projection map P k : H k (= P k upk 1 ) P k, then R k f (x, u) is defined to be P k D x f (x, u). The left Rarita-Schwinger equation is defined to be R k f (x, u) = 0. We also have a right projection P k,r : H k P k, and a right Rarita-Schwinger equation f (x, u)d x P k,r = f (x, u)r k = 0.

14 Are there any non-trivial solutions to the Rarita-Schwinger equation? First for any k-monogenic polynomial p k (u) we have trivially R k p k (u) = 0. Now consider the fundamental solution G(u) = 1 u ω n u n to the Dirac operator D, where ω n is the surface area of the unit sphere, S n 1. Consider the Taylor series expansion of G(v u) and restrict to the kth order terms in u 1,..., u n (u = u 1 e u n e n ). These terms have as vector valued coefficients k v 1 k 1... vn k n G(v) (k k n = k).

15 As DG(v) = n i=1 e G(v) j v j = 0, we can replace by v 1 occurs and collecting v 1 n j=2 e 1 1 e j. Doing this each time v j like terms we obtain a finite series of polynomials homogeneous of degree k in u P σ (u)v σ (v) σ where the summation is taken over all permutations of monogenic polynomials (u 2 u 1 e1 1 e 2),, (u n u 1 e1 1 e n), P σ (u) = 1 k! Σ(u i 1 u 1 e 1 1 e i 1 )... (u ik u 1 e 1 1 e i k ) and V σ (v) = v j 2 2 k G(v)... v jn n, j j n = k.

16 Further V σ (u)up µ (u)ds(u) = δ σ,µ where δ σ,µ is the S Kronecker n 1 delta and µ is a set of n 1 non-negative integers summing to k. Consequently, the expression Z k (u, v) := σ P σ (u)v σ (v)v is the reproducing kernel of P k with respect to integration over S n 1. See [BDS]. Further as Z k (u, v) does not depend on x, R k Z k = 0.

17 Are there any solutions to R k f (x, u) = 0 that depends on x? To answer this we need to look at the links to conformal transformations. Any conformal transformation on R n { } can be expressed as y = φ(x) = (ax + b)(cx + d) 1 where a, b, c, d Cl n and satisfy the following conditions: 1. a, b, c, d are all products of vectors in R n. 2. a b, c d, bc, da R n. 3. a d b c = ±1. When c = 0, φ(x) = (ax + b)(cx + d) 1 = ±axã + bd 1. Now assume c 0, then φ(x) = (ax + b)(cx + d) 1 = ac 1 ± (cx c + d c) 1. These are called Iwasawa decompositions.

18 Conformal invariance of P k Let P k,w and P k,u be the projections with respect to w and u respectively. 1. Orthogonal transformation: Let x = ayã, and u = awã, where a Pin(n). Lemma 1 P k,w ãf (ayã, awã) = ãp k,u f (x, u). 2. Inversion: Let x = y 1, u = ywy y 2. y Lemma 2 P k,w y n f (y 1, ywy y 2 ) = y y n P k,uf (x, u). 3. Translation: x = y + a, a R n. In order to keep the homogeneity of f (x, u) in u, u does not change under translation. Lemma 3 P k f (x, u) = P k f (y + a, u). 4. Dilation: x = λy, where λ R +. Lemma 4 P k f (x, u) = P k f (λy, u).

19 Using the Iwasawa decomposition and combining the previous lemmas, we get the following result: Theorem 1 (cx + d)w(cx + d) P k,w J(φ, x)f (φ(x), cx + d 2 ) = J(φ, x)p k,u f (φ(x), u), (cx where u = + d)w(cx + d) cx cx + d 2 and J(φ, x) = + d cx + d n.

20 Conformal invariance of the equation R k f = 0 The following is known for the Dirac operator cx + d J 1 (φ, x)d y = D x J(φ, x), where J 1 (φ, x) = cx + d n+2. Hence we have Df (y) = 0 iff DJ(φ, x)f (φ(x)) = 0. Now using the Iwasawa decomposition of (ax + b)(cx + d) 1 and the previous theorem, we obtain that (cx + d)w(cx + d) J 1 (φ, x)r k,u f (x, u) = R k,w J(φ, x)f (φ(x), cx + d 2 ), (cx where u = + d)w(cx + d) cx + d 2, and R k,w, R k,u are the Rarita-Schwinger operators with respect to w and u respectively. So R k,u f (x, u) = 0 implies (cx + d)w(cx + d) R k,w J(φ, x)f (φ(x), cx + d 2 ) = 0.

21 The Fundamental Solution of R k Applying inversion to Z k (u, v) from the right we obtain E k (y, u, v) := c k Z k (u, yvy y 2 ) y y n is a non-trivial solution to f (y, v)r k = 0 on R n \ {0}, where c k = n 2 + 2k. n 2 Similarly, applying inversion to Z k (u, v) from the left we can obtain that y c k y n Z k( yuy y 2, v) is a non-trivial solution to R k f (y, u) = 0 on R n \ {0}. In fact, this function is E k (y, u, v), and E k (y, u, v) is the fundamental solution of R k.

22 Basic Integral Formulas Definition 1 For any Cl n valued polynomials P(u), Q(u), the inner product (P(u), Q(u)) u with respect to u is given by (P(u), Q(u)) u = P(u)Q(u)dS(u). S n 1 For any p k P k, one obtains p k (u) = (Z k (u, v), p k (v)) v = Z k (u, v)p k (v)ds(v). S n 1 See [BDS].

23 Basic Integral Formulas Theorem 2 [BSSV] (Stokes Theorem for R k ) Let Ω and Ω be domains in R n and suppose the closure of Ω lies in Ω. Further suppose the closure of Ω is compact and Ω is piecewise smooth. Then for f, g C 1 (Ω,P k ), we have [(g(x, u)r k, f (x, u)) u + (g(x, u), R k f (x, u)) u ]dx n Ω = Ω where dσ x = (g(x, u), P k dσ x f (x, u)) u = n ( 1) j 1 e j d ˆx j and j=1 Ω d ˆx j = dx 1 dx j 1 dx j+1 dx n. (g(x, u)dσ x P k,r, f (x, u)) u,

24 Basic Integral Formulas Theorem 3 [BSSV](Borel-Pompeiu Theorem ) Let Ω and Ω be as in Theorem 2 and y Ω. Then for f C 1 (Ω, P k ) f (y, u) = (E k (x y, u, v), P k dσ x f (x, v)) v Ω (E k (x y, u, v), R k f (x, v)) v dx n. Ω Here we will use the representation (x y)v(x y) x y E k (x y, u, v) = c k Z k (u, x y 2 ) x y n.

25 Basic Integral Formulas Theorem 4 (Cauchy Integral Formula) If R k f (x, v) = 0, then for y Ω, f (y, v) = (E k (x y, u, v), P k dσ x f (x, v)) v Ω = (E k (x y, u, v)dσ x P k,r, f (x, v)) v. Ω Theorem 5 (E k (x y, u, v), R k ψ(x, v)) v dx n = ψ(y, u) R n for each ψ C0 (Rn ).

26 Basic Integral Formulas Definition 2 For a domain Ω R n and a function f : Ω R n Cl n, which is monogenic in u, the Cauchy, or T k -transform, of f is formally defined to be (T k f )(y, v) = (E k (x y, u, v), f (x, u)) u dx n, y Ω. Ω Theorem 6 R k T k ψ = ψ for ψ C 0 (Rn ). i.e R k R n (E k (x y, u, v), ψ(x, u)) u dx n = ψ(y, v).

27 References L. V. Ahlfors, Old and new in Möbius groups, Ann. Acad. Sci. Fenn. Ser. A I Math., 9 (1984) F. Brackx, R. Delanghe, and F. Sommen, Clifford Analysis, Pitman, London, J. Bureš, F. Sommen, V. Souček, P. Van Lancker, Rarita-Schwinger Type Operators in Clifford Analysis, J. Funct. Annl. 185 (2001), No.2, C. Dunkl and Y. Xu, Orthogonal Polynomials of Several Variables, Cambridge University Press, Cambridge, A. Sudbery, Quaternionic Analysis, Mathematical Procedings of the Cambridge Phlosophical Society, (1979), 85, K. Th. Vahlen, Über Bewegungen und complexe zahlen, (German) Math. Ann., 55(1902), No