Tampereen teknillinen yliopisto. Matematiikan laitos. Tutkimusraportti 9 Tampere University of Technology. Department of Mathematics. Research Report 9 Sirkka-Liisa Eriksson & Heikki Orelma Topics on Hyperbolic Function Theory in Cl_{n+,} Tampereen teknillinen yliopisto. Matematiikan laitos Tampere 29
Topics on Hyperbolic Function Theory in Cl n+, Sirkka-Liisa Eriksson, Heikki Orelma, Department of Mathematics, Tampere University of Technology, P.O. Box 527, FI-33 Tampere, Finland. Abstract Left hypergenic functions are Cliord algebra Cl n+,-valued functions satisfying the equation Df k x Q f = where D is the Dirac operator and Q is the projection-type mapping Cl n+, Cl n, given by composition. Similarly we dene right hypergenic functions taking right actions in the previous denition. We consider some fundamental and local properties of hypergenic functions. We prove our version of the Borel- Pompeiu formula and a new version of Cauchy's theorem. We shall consider hypergenic Cliord algebra-values multivector functions and prove that in that case left- and right hypergenicity coincide. Preliminaries Let {e,..., e n } be the standard basis in R n+. The Cliord algebra Cl n+ is the free associative algebra with unit generated by standard basis vectors together with the dening relations e i e j + e j e i = 2δ ij for each i, j =,..., n. The Cliord algebra Cl n+ has the dimension 2 n+ and the canonical basis is given by e A = e a e ak where A = {a,..., a k } N = {,..., n} and a < < a k. Especially e = and e {j} = e j. The space of k-vectors is dened by Cl k n = span{e A : A = k}. Any a Cl n+ admit the multivector decomposition: a = [a] + [a] + + [a] n+ with [a] k Cl k n+. The space of -vectors is idened with R and the set of -vectors is identied with R n+. Assume that Ω is an open subset of R n+. In the canonical basis every function f : Ω Cl n admit the representation f = A e A f A. A function f is called dierentiable in Ω if f A is dierentiable in Ω for each A. If f is dierentiable we dene the left Dirac operator by n f D l f = e k x k and the right Dirac operator by D r f = k= n k= f x k e k
where derivatives operate componentwise. Denoting Cl n the Cliord algebra generated by {e,..., e n }. We may represent the Cliord algebra Cl n as the direct sum Cl n+ = Cl n e Cl n. Let π and π 2 be the corrensponding projections i.e., π a + e b = a and π 2 a + e b = e b and let µ : Cl n+ Cl n+ be the involution µa = e a. Using the previous mappings we dene P := π and Q := µ π 2. Let Ω be an open subset of R n+ contained in the upper half-space R n+ + := R n+ {x > }. We dene the left- and right-modied Dirac operator on the open set Ω using the previous mappings by H l kf = D l f k x Q f, H r kf = D r f k x Q f where k is an arbitrary real number. We shall also use abbreviated notations H l := Hn l and Hr := Hn r for index k = n. Null solutions of the previous operators are called left- and right- hypergenic functions. As a technical tool we will need P - and Q -parts of the operators Hk l and, represented in the next lemma. H r k Lemma. Let f : Ω Cl n+ be a smooth function. Then a P H l k f = D P f + Qf k x Q f, b Q Hk l Pf f = D Q f, c P H l k 2 f = P f k x P f, d Q H l k 2 f = Q f k x Q f + k x 2 Q f, where D = e x + + e n x n. Proof. We will prove only a and b since the proof for c and d can be found in the proof of the Theorem.8 of [5]. Assume f = P f + e Q f is a smooth function. Since D = e x + + e n we obtain x n P f DP f = e + D P f and De Q f = Q f e D Q f. Previous observations implies that H l kf = H l kp f + e Q f = DP f + De Q f k x Q f { = D P f + Q f k } { P f } Q f + e D Q f. x Applying P and Q we obtain the result. 2
2 Local Properties of Hypergenic Functions In this section we consider local properties of hypergenic functions. If Ω is a domain in R n+ + the operator LB g = x 2 g k g x is the Laplace-Beltrami operator for g C 2 Ω, Cl n+ with respect to the Riemannian metric In [5] we obtain the following theorem: ds 2 = dx2 + + dx 2 n. x 2k n Theorem 2. Let Ω R n+ + be a domain and f C 2 Ω, Cl n+ be a k- hypergenic. The function P f is a solution of the Laplace-Beltrami equation i.e. LB P f = and Q f is a solution of the eigenvalue problem LB Q f = kq f. We see, that the Q -part of a hypergenic function is an eigenfunction of the Laplace-Beltrami operator. Theorem 2.2 Let Ω R n+ + be a domain. A smooth function f : Ω Cl n+ is hypergenic if and only if for any a Ω there exist r > satisfying B r a Ω and a map g : B r a Cl n, satisfying f = Dg and LB g =. Proof. Assume rst that the radius and the function g exist. Then in the neighborhood of a H l kf = g k x Q Dg = g k x g = x 2 LB g =. Assume that f : Ω Cl n+ is hypergenic. Let us denote that x = x, x R R n and := D D. Assume that s A is a solution of the Poisson problem s A x = P f A a, x in B r a R n and let s = A {,...,n} s A. The solution of the Poisson probem exits, for example see []. Then we dene and obtain gx = x a Q ft, xsdt + D s x x Dgx = e Q fx + D Q ft, xdt + s x a 3
Using Lemma. we obtain that D Q f = Pf and x P f Dgx = e Q fx + t, xdt + P fa, x = fx. a Similarly we see that which completes the proof. = Df k x Qf = g k x g, 3 On Euler Operator in the Class of Hypergenic Functions The Euler operator is dened by E = n i= x i. x i We see that it is a scalar operator and measures the degree of homogenicity of a function. Lemma 3. If D is the Dirac operator and E is the Euler operator we have: DE = D + ED. Proof. Let f : Ω Cl n+ be a dierentiable function. Then DEf = j,i and we obtain the result. = j,i = j e j f x i x j x i e j δ i,j f x i + j,i e j f x j + j,i = Df + EDf, x i f e j x i x i x j f e j x i x j The similar result holds in the class of hypergenic functions: Theorem 3.2 Let f : Ω Cl n+ be a smooth function. If the function f is k-hypergenic then Ef is k hypergenic. proof. We assume that Df k x Q f =. Applying E to k x Q f we obtain k Q Ef = k k Q f + E Q f. x x x 4
Using previous lemma we obtain The proof is complete. H k Ef = DEf k x Q Ef = Df + EDf k k Q f E Q f x x = Df k Q f + E Df k Q f =. x x If f is a hypergenic function the previous theorem gives us a method how to construct more hypergenic functions: The above theorem implies that for every k-hypergenic function f : Ω Cl n+, there exists the sequnce of hypergenic functions dened by E m f := EE }{{ E } f m copies for each m N and E f = f. This sequence {E m f : m =,,...} is called the homogenized sequence of f. As an example we consider monomials n X i = δij x j e j j= where i =,..., n. Since H l X i = the monomials X i are hypergenic. Moreover EX i = X i for each i =,..., n and hence the homogenized sequence of X i is just {X i }. 4 Further Representation Results In this section we will consider the improved version of the Cauchy's formula represented in [5]. The second aim is to prove the Borel-Pompeiu formula in the language of H-operators. First we recall briey some preliminaries from integration theory. Let M be a be a k-dimensional manifold-with-boundary in R n+ +, see e.g. [2]. The boundary of M is denoted by. If moreover n+ Λ M = Λ p M p= is the exterior algebra over R n+ with basis {dx, dx,..., dx n } we then construct the bundle Cl n+ R Λ k M. If ωx is a section of the previous bundle over x M it is of the form ωx = A,B ω A,B xe A dx B 5
where B = {b,..., b k } N = {,..., n} and dx B = dx b dx bk. The meaning of the symbol e A dx B is clear. Let furthermore M be an oriented k- dimensional manifold-with-boundary in R n+, then we dene ωx = e A ω A,B xdx B. M A,B M In this paper M will be n + dimensional or n-dimensional i.e. the boundary of M. The n + -form dv = dx dx dx n on M is called the Riemannian volume element. In surface integrals we shall often use the n-form n dσ = i e i dˇx i where i= dˇx i = dx dx dx i dx i+ dx n for each i =,,..., n. The exterior derivative d for Cliord algebra valued dierential forms is dened componentwise, i.e., if ω = A,B ω A e B is a k-form then dω = A,B dω A e B. Applying the classical Stokes theorem see [2] it is easy to prove that: Theorem 4. Stokes Let ω be a Cl n+ -valued k-form in the oriented k- dimensional manifold-with-boundary M. Then dω = ω. M Let us denote the interior of the subset U R n by U We recall the Cauchy's formula, represented in [5]. Theorem 4.2 Assume that Ω is an open subset of R n+ + and f : Ω Cl n+ is hypergenic. Let M Ω be an oriented n + -dimensional manifold-withboundary. Then fy = 2n y n ω n+ for each y M. If we decompose x y dσxfx x y dσx fx x y n x ŷ n dσxfx = P dσxfx + e Q dσxfx 6
we obtain If we dene a new kernel we get x y dσxfx x y dσx fx = x y x y P dσxfx + x y + x y e Q dσxfx. Λx, y := x y x y x y n x ŷ n. Lemma 4.3 x y + x y x y n x ŷ n e x = Λx, yy P x. Using the identity and the above lemma we obtain that x y dσxfx x y dσx fx x y n x ŷ n = Λx, yp dσxfx + Λx, y y P x x Q dσxfx. Hence we obtain an other version of the Cauchy's formula: Theorem 4.4 Assume that Ω is an open subset of R n+ + and f : Ω Cl n+ is hypergenic. Let M Ω be an oriented n + -dimensional manifold-withboundary. Then fy = 2n y n ω n+ for each y M. y P x Λx, y P dσxfx + Q dσxfx. x Theorem 4.5 Borel-Pompeiu Formula Assume that Ω is an open set of R n+ + and f : Ω Cl n+ is dierentiable. Let M Ω be an oriented n + - dimensional manifold-with-boundary. Then fy = 2n y n { } ω n+ x n P px, ydσxfx + e Q qx, ydσxfx + 2n y n ω n+ M x n { P px, yh l fx + e Q qx, yh l fx } dv. for each y M where px, y := x n x y x ŷ x y n x ŷ n are hypergenic with respect to x. and qx, y := x y +x ŷ x y n x ŷ n Remark 4.6 In Borel-Pompeiu formula hypergenicity has no role. Assuming that f is hypergenic we obtain original version of Cauchy's formula. In order to prove Borel-Pompeiu's theorem, let us rst state a few basic lemmata. All proofs we refer [5]. 7
Lemma 4.7 The function x px, y is right n -hypergenic and the function x qx, y is right n -hypergenic on R n+ \{y, ŷ} for each y R n+. Lemma 4.8 Assume that Ω is an open set of R n+ + and f, g : Ω Cl n+ are dierentiable functions. Let M Ω be an oriented n + -dimensional manifold-with-boundary. If g is a right hypergenic then x n P gdσf = and if g is a right hypergenic then Q gdσf = M M x n P gh l f dv Q gh l f dv. Proof. This is a corollary of Lemma 2.6 and Lemma 2.9 in [5]. Lemma 4.9 If f : Ω Cl n+, is continuous and B r y Ω R n+ + then 2 n y n P px, ydσxfx P fy ω n+ and as r. 2 n y n ω n+ B ry B ry x n Q qx, ydσxfx Q fy Proof. See the proofs of Theorem 2.4 and 2.7 in [5]. Proof of Theorem 4.5. Since M is an open set for each y M there exists r > such that B r y M. We denote M r y := M\B r y. Then P px, ydσxfx + e Q qx, ydσxfx x n = ry + B ry x n x n P px, ydσxfx + e P px, ydσxfx + e Using Lemma 4.9 we obtain P px, ydσxfx +e B ry x n B ry as r. Using Lemma 4.7 and 4.8 we obtain P px, ydσxfx = and ry x n ry Since M ry M Q qx, ydσxfx = M ry M ry ry B ry Q qx, ydσxfx Q qx, ydσxfx. 2 ω n+ Q qx, ydσxfx 2 n y n fy x n P px, yh l fx dx Q qx, yh l fx dx. as r, we obtain the result just taking r in 2. Remark 4. In the proof there are no problems with integration since the Stokes theorem holds also on the manifold with corners, see [8] 8
5 Multivector Calculus Lastly in [5] we study some connections between Q -operator, H k -operators and left- and right contraction -operators. We recall some notions and properties. Let x Cl n+ be a vector and u Cl n+ be an arbitrary Cliord number. The left contraction is dened by and the right contraction by x u = 2 xu u x, 3 u x = 2 ux xu. Marcel Riesz has introduced in 958 the exterior product in Cliord algebras. The exterior product is dened by x u = 2 xu + u x. 4 Adding 3 and 4 together we obtain the Cliord product Moreover if u is an r-vector we have xu = x u + u x. x u = r u x for each vector x. Historical remarks and more comprehensive introduction to the contraction products and the exterior product, see []. Remark 5. If u, x Cl n+,, then x u = u x = x, u where, is the usual Euclidean inner product. As in previous remark we noticed that for each x Cl n+ the mapping u x u or u u x is a mapping Cl k n+ Cl k n+ for each k. It is easy to notice that this phenomen should be true in generally. To see that, let us denote L x u := x u, R x u := u x and W x u = x u where all mappings are from Cl n+ into Cl n+. Hence it's obvious that for each k n. L x, R x : Cl k n+ Cl k n+ and W x : Cl k n+ Cl k+ n+ Let f : Ω Cl n+ be a C -function where Ω is an open set. As in [5] we deduce that hyperbolic Dirac operators are represented using the previous operators in the form H l kf = D l f k x e f, H r kf = D r f k x f e, 9
on Ω. We interpert the Dirac operator D l as a vector. As in [7] we also interpret D l f as a product of a vector and a Cliord number i.e. D l f = D l f + D l f. Since D f = L D f and D f = W D f we obtain that D l = L D + W D. That allow us to ask: Is there any corresponding splitting for Hk l -operator? An aswer is simple. Using the above formulas we compute: Hence H l kf = D l f k x e f = D l f + D l f k x e f = D l k x e f + D l f. H l k = L Dl k x e + W D. Similarly as in the previous discussion we obtain that H r k = R Dr k x e + W D. Recall that H l := H l n, Hr := H r n, Hr := R Dr k x e, Hl := R Dl n x e and H r + = H l + = E D. Thus in the class of k-vector valued functions H l = H l + H l + and H r = H r + H r +. Theorem 5.2 Let f C Ω, Cl p n+ be a p-vector, for p =,..., n. Then if and only if Proof. Assume that Since and so we obtain the result. H l f =, H l +f = H l f =. H l f = D l n x e f + D l f =. [H l f] p = D l n x e f = [H l f] p+ = D l f =
If the function f C Ω, Cl p n+ is a product of functions i.e., there exist a vector valued function g and a p-vector h satisfying f = gh. That allow us to split f = f + + f where f + = g h and f = g h. Using previous theorem we obtain: Corollary 5.3 Let f C Ω, Cl n+ be a product f = gh where g C Ω, Cl n+ and h C Ω, Cl p n+. If H l f + =, H l +f + =, H l f =, H l +f =, where f + = g h and f = g h, then f is a hypergenic function. In the case of k-vector valued functions it is comfortable to represent P - and Q -operators using the Riesz's products: Proposition 5.4 If u Cl n+ then. P u = e e u and Q u = e u, 2. P u = e u e and Q u = u e, Proof.. The proof for Qu = e u, see [5]. Using we obtain e u + u e = e P u + Qu + P ue e Q ue = 2e P u since e e u =. P u = e 2 e u + u e = e e u = e e u 2. Recall P u = P u and Q u = Q u. Since e v = 2 e v v e = 2 ve e v = v e for each v Cl n+, we obtain that and Q u = u e. P u = e e u = e u e The Theorem 5.2 is formulated and proved only for the left hypergenic functions. In the next theorem we shall show that this is not any shortcoming of the theory. Theorem 5.5 For any k-vector valued function the notions of left hypergenicity and right hypergenicity coincide.
Proof. Assume that f is a k-vector valued function. Then [H l ] k = n e i f n e f x i i= = k n i= = k [H r ] k. Since [H l ] k = [H r ] k we obtain the result. References x f n e i f e x i x [] S. Axler, P. Bourdon and W. Ramey, Harmonic Function Theory, 2nd. ed., Springer, New York, 2. [2] F. Brackx, R. Delanghe and F. Sommen, Cliord Analysis, Research Notes in Mathematics, 76, Pitman, Boston, MA, 982. [3] S.-L. Eriksson-Bique, and H. Leutwiler, Hypermonogenic functions, In Clifford algebras and their application in mathematical physics. Vol 2. Cliord Analysis, Birkhäuser, Boston, 287-3, 2. [4] S.-L. Eriksson, Integral formulas for hypermonogenic functions, Bull. Belg. Math. Soc. 24, 75-77. [5] S.-L. Eriksson and H. Orelma, Hyperbolic function theory in the Cliord algebra Cl n+,, Advances in Applied Cliord algebras to appear. [6] K. Gürlebeck, K. Habetha and W. Sprössig Holomorphic Functions in the Plane and n-dimensional Space, Birkhäuser, Basel, 28. [7] D. Hestenes and G. Sobczyk, Cliord Algebra to Geometric Calculus, Reidel, Kluwer, Dordrecht, 984. [8] J. Lee, Introduction to smooth manifolds, Graduate Texts in Mathematics, 28. Springer-Verlag, New York, 23. [9] H. Leutwiler, Generalized Function Theory, Proceedings of the Summer School: Cliord Analysis and Applications, Tampere: August 2.-6., -23, 24. [] H. Leutwiler, Modied Cliord analysis, Complex Variables 7 992, 53-7. [] P. Lounesto, Cliord Algebras and Spinors, London Mathematical Society, Lecture Note Series 239, Cambridge University Press, Cambridge, 997. [2] M. Spivak, Calculus on Manifolds, Benjamin, New York, 965. 2