Talk 3: Examples: Star exponential

Transcription

1 Akira Yoshioka 2 7 February 2013 Shinshu Akira Yoshioka 3 2 Dept. Math. Tokyo University of Science

2 1. Star exponential 1 We recall the definition of star exponentials. 2 We show some construction of star exponentials. Based on the joint works with H. Omori, Y. Maeda, N. Miyazaki, Akira Yoshioka 4

3 1.1. Definition 1.1. Definition The space of q-number functions F ρ is a complete topological algebra for 0 < ρ 2 (even ρ > 2 for future under some regularization). We can consider exponential element ( H ) exp t = n=0 t n H n! H } {{ } n in this algebra. For a q-number polynomial H P(P), we define the star exponenial exp t(h/) by the differential equation d ( H ) dt exp t = H ( H ) ( H ) exp t, exp t t=0 = 1 Akira Yoshioka 5

4 1.1. Definition Remark We set the Fréchet space E ρ+ (C 2 ) = λ>ρ E λ (C 2 ) (1) and we donote by E ρ+ the correponding bundle and by F ρ+ the space of the flat sections of this bundle. When H P(P) is a linear element, then exp t ( H ) belongs to the good space F 1+ ( F 2 ). On the other hand, the most interesting case is given by quadratic form H P(P). In this case we can solve the differential equation explicitly, but the star exponential belongs to the space F 2+, which is difficult to treat at present. Although general theory related to the space F 2+ is not yet established, we illustrate the concrete example of the star expoenential of the quadratic forms and its application. Akira Yoshioka 6

5 1.2. Examples 1.2. Examples We very the parameter K S C (2) and at some K we can obtain interesting identities in the algebra of K product. Linear case We consider a liner q-number polynomial. It is written in general form as H = a 1 u + a 2 v = a, u, a 1, a 2 C. Star exponential of H belongs to the space of q-number function F 1+. The star exponetial exp t ( ) H at K, which is denoted by exp K t ( H ), is explicitly given as ( H ) exp K t = exp t2 t 4 ak, a exp a, u. Akira Yoshioka 7

6 1.2. Examples Hence if the real part satisfies an inequality such as R 1 4 ak, a < 0 (A) the term exp t2 4 ak, a is rapidlly decreasing with respect to t and then we can consider an integral ( e et exp K t z + H ) dt Then we define star gamma function by Γ (z) = e et exp t ( z + H ) dt (2) This is evaluated at every K and the value Γ K (z) of the star gamma function at K is given by the integral, where K satisfies the condition (A). We have the identity for K satisfying (A) Γ K (z + 1) = ( z + H ) K Γ K (z). Akira Yoshioka 8

7 2. Examples In this section, we show some examples of star exponetials. Akira Yoshioka 9

8 2.1. Path integral (iterated integral) construction of quadratic case 2.1. Path integral (iterated integral) construction of quadratic case [YM]: Yoshioka A., T. Matsumoto, Path integral for Star Exponential functons of quadratic forms, In: Geometry, Integrability and Quantization. I. M. Mladenov, G. L. Naber (Eds), 4, Coral Press Scientific Publishing, Sofia 2003, pp , Notation We first give the Moyal product on C 2. Let x, y be coordinate functions of C 2 and let be a positive real parameter. The canoncial Poisson bracket is given by ( { f, g} = f g x y f g y x = f x ) y g (3) Here we use a notation such as ( { f, g} = f x ) ( y g = f x y ) y x g (4) Akira Yoshioka 10

9 2.1. Path integral (iterated integral) construction of quadratic case Using the binomial theorem formally, we set bidifferential operators ( x ) n n! ( ) l ( ) m y = l!m! ( 1)m x y y x and ( exp 2 l+m=n x ) y = The Moyal product is then given by n=0 ( ) 1 n ( n! 2 x ) n y Definition ( f 0 g = f exp 2 x ) y g We remark here that the product f 0 g is not necessarily convergent for arbitrary smooth functions, however it is defined when at least one of f, g is a polynomial function. Akira Yoshioka 11

10 2.1. Path integral (iterated integral) construction of quadratic case With the Moyal product, we can define the -exponential function of a quadratic form in the following way. Let us consider a quadratic form on C 2 given by H = ax 2 + 2bxy + cy 2, a, b, c C. (5) The -exponential function is formally given by e t H f = n=0 ( t n H n n! ) f (6) for every polynomial function f (x, y) ( ) H n where = H H. In this } {{ } n note, we study the explicit form of e t H method. by means of the path-integral Akira Yoshioka 12

11 2.1. Path integral (iterated integral) construction of quadratic case Normal forms and the invariance of 0 Given a quadratic function H = ax 2 + 2bxy + cy 2 on C 2, we consider a descriminant D = b 2 ac. (7) We will show that H with nonvanishing descriminant is transformed into x 2 y 2 via linear transformations by S L(2, C). Akira Yoshioka 13

12 2.1. Path integral (iterated integral) construction of quadratic case First we consider the case where H has the descriminat D = 1 4. We will prove Proposition There exists ( p q r s ) S L(2, C) such that where ( w z H = 1 2 w z2, (8) ) ( ) ( p q x = r s y Such matrices ( p q r s ) are not unique, parametrized by C. ). (9) Akira Yoshioka 14

13 2.1. Path integral (iterated integral) construction of quadratic case First we consider the case where H has the descriminat D = 1 4. We will prove Proposition There exists ( p q r s ) S L(2, C) such that where ( w z H = 1 2 w z2, (10) ) ( ) ( p q x = r s y Such matrices ( p q r s ) are not unique, parametrized by C. ). (11) Akira Yoshioka 15

14 2.1. Path integral (iterated integral) construction of quadratic case Furthermore, by a direct calculation we have the following invariance property of the Moyal product: Proposition The Moyal product is expressed in terms of the coordinate functions w, z such as ( f 0 g = f exp w ) z g 2 Akira Yoshioka 16

15 2.1. Path integral (iterated integral) construction of quadratic case Path Integral representation In what follows, we give a path-integral representation of the -exponential function of quadratic forms. 1 Fist we consider H = 1 2 x y2. 2 Then we obtain the -exponential functions for general H with D 0. Akira Yoshioka 17

16 2.1. Path integral (iterated integral) construction of quadratic case -exponential of 1 2 x y2 First, we give the -exponential function of 1 2 x y2. The basic tool is the following Mehler s formula: Lemma Let H n (t) be an Hermite polynomial of degree n. Then it holds e x2 y 2 n=0 z n 2 n n! H n(x)h n (y) = 1 1 z 2 exp ( 1 1 z 2 (x 2 + y 2 2zxy) ) Akira Yoshioka 18

17 2.1. Path integral (iterated integral) construction of quadratic case For the first step of path integral, we consider the product exp ( t H ) 0 exp ( s H ) for H = ( x 2 + y 2 )/2. Mehler s formula gives the formula: Proposition ( exp t H ) ( O exp s H ) = ( ts/4 exp t + s 1 + ts/4 ) H Akira Yoshioka 19

18 2.1. Path integral (iterated integral) construction of quadratic case Direct calculation of star Moyal product gives exp ( ( ) t ) H exp s H = exp ( ) t H 1 ( ) l ( ) ( ) t l t l! 2 2 H l 2 x is l is 2 H l 2 y l=0 ( 1) k ( ) k ( ) ( ) it k it k! 2 2 H k 2 y s k s 2 H k 2 x exp ( ) s H k=0 and then it holds exp ( ( ) t ) H exp s H = exp ( t+s H) k=0 l=0 ( 1 i ts 2 k k! 2 ( 1 2 l l! ) k Hk ( i ) l ( ts 2 Hl ) ( it 2 y H k ) t 2 x ) s 2 x ( ) is H l 2 y (12) Akira Yoshioka 20

19 2.1. Path integral (iterated integral) construction of quadratic case For the next step, we consider the iterated product of exponential functions e t 1 H e t n H, where we put H = H = ( x2 + y 2 )/2. In what follows, we will show the formula: Proposition where and c n (t) = 1 + s n (t) = 2 e t 1 H e t n H = 1 c n (t) exp ( s n (t) c n (t) H ) k;2 2k n 1 i 1 <i 2 < <i 2k n k;1 2k+1 n 1 i 1 <i 2 < <i 2k+1 n (a) (t i1 /2) (t i2k /2) (b) (t i1 /2) (t i2k+1 /2) (c) Akira Yoshioka 21

20 2.1. Path integral (iterated integral) construction of quadratic case For t > 0, we devide the interval [0, t] into N equal segments for every positive integer N. We put t = t/n. Then we have c N ( t) = a k (t/2) 2k, s N ( t) = 2 b k (t/2) 2k+1, k;0 2k N where the coefficients are given by a k = 1 (2k)! (1 1 N ) ( 1 2 N k;1 2k+1 N ) ( ) 1 2k 1 N and b k = 1 (2k + 1)! (1 1 N ) ( 1 2 N ) ( ) 1 2k N Akira Yoshioka 22

21 2.1. Path integral (iterated integral) construction of quadratic case It is easy to see the coefficients of c N ( t) and s N ( t) converges as N and the limits are lim c N( t) = cosh t N 2, lim N s N( t) = 2 sinh t 2 Thus, we have Theorem The N-itereted product converges lim N e(t/n) H e (t/n) H = 1 cosh t 2 e 2 H tanh t 2 where H = H/() and H = x2 2 + y2 2. Akira Yoshioka 23

22 2.1. Path integral (iterated integral) construction of quadratic case -exponential for general case We consider the -exponential function of H = ax 2 + 2bxy + cy 2, a, b, c C with D = b 2 ac 0. For this case we can show H = 2 D( 1 2 w z2 ), where w = px + qy, z = rx + sy and ( p r q s ) S L(2, C). Notice the commutator of w, z satisfies [w, z] =. Then the trasformation fomula yields Theorem The -exponential function of H is e t H ( 1 = cosh exp Dt H D tanh ) Dt Akira Yoshioka 24

23 2.2. Star Hermite polynomials 2.2. Star Hermite polynomials We recall the following. For any 2 2 complex matrix Λ = ( ) λ 11 λ 12 λ 21 λ 22 M2 (C), we have a biderivation on polynomials p 1 Λ p2 = p 1 λ αβ α β p 2 = λ αβ α p 1 β p 2, p 1, p 2 P(C 2 ). We have Proposition αβ For any Λ M 2 (C), the product Λ is well-defined and associative on P(C 2 ). αβ Akira Yoshioka 25

24 2.2. Star Hermite polynomials In this section, we consider the star product for the simple case where ( ) ρ 0 Λ = 0 0 Then we see easily that the star product is commutative and explicitly ) given by p 1 Λ p 2 = p 1 exp u1 u1 p 2. This means that the algebra is ( ρ 2 essentially reduced to space of functions of one varible u 1. Thus, we consider functions f (w), g(w) of one variable w C and we consider a commutative star product τ with complex parameter τ such that f (w) τ g(w) = f (w)e τ 2 w w g(w) A direct calculation gives exp τ itw = exp(itw (τ/4)t 2 ) Akira Yoshioka 26

25 2.2. Star Hermite polynomials We see the generating function of Hermite polynomilas gives e 2tw 1 2 t2 = n=0 H n (w) tn n! The left hand side is a star exponential function at τ = 1, exp 1 ( 2tw) = e 2tw 1 2 t2. Thus the expansion exp 1 ( 2tw) = n=0 1 n! ( 2tw) n 1 yields ( 2w) n 1 = H n (w). Akira Yoshioka 27

26 2.2. Star Hermite polynomials We define a star Hermite function by exp τ ( 2tw) = n=0 H n (w, τ) tn n! The identity d dt exp τ 2tw = 2w τ exp τ 2tw yields τ 2 H n(w, τ) + 2w H n (w, τ) = H n+1 (w, τ). The exponetial law gives k+l=n n! k!l! H k(w, τ) τ H l (w, τ) = H n (w, τ) Akira Yoshioka 28

27 2.3. Star theta functions 2.3. Star theta functions For Rτ > 0 star exponential exp τ niw = exp(niw (τ/4)n 2 ) is rapidly decreasing with respect to n. Then we can consider summations for τ satisfying Rτ > 0 exp τ 2niw n= = exp ( 2niw τ n 2) = n= This is Jacobi s theta function θ 3 (w, τ). q n2 e 2niw, (q = e τ ). n= Akira Yoshioka 29

28 2.3. Star theta functions Then we set star theta functions as θ 1 τ (w) = 1 i ( 1) n exp τ (2n + 1)iw, n= θ 2 τ (w) = exp τ (2n + 1)iw, n= θ 3 τ (w) = exp τ 2niw, n= θ 4 τ (w) = ( 1) n exp τ 2niw n= Akira Yoshioka 30

29 2.3. Star theta functions We see exp τ 2iw τ θ k τ (w) = θ k τ (w), (k = 2, 3) exp τ 2iw τ θ k τ (w) = θ k τ (w), (k = 1, 4) Then using exp τ 2iw = e τ e 2iw and the product formula directly we have the quasi-periodicity e 2iw τ θ k τ (w + iτ) = θ k τ (w), (k = 2, 3) e 2iw τ θ k τ (w + iτ) = θ k τ (w), (k = 1, 4) Akira Yoshioka 31

30 2.3. Star theta functions Star delta function For τ with Rτ > 0, we consider δ τ (w a) = e it(w a) τ dt, a C We see easily for any star polynomial p (w) = a k w k, k a k C it holds p (w) τ δ τ (w a) = p(a) Akira Yoshioka 32

31 2.3. Star theta functions Using star delta functions we have θ 1 τ = 1 2 θ 2 τ = 1 2 ( 1) n δ τ (w + π 2 + nπ) n= θ 3 τ = 1 2 θ 4 τ = 1 2 ( 1) n δ τ (w + nπ) n= δ τ (w + nπ) n= δ τ (w + π 2 + nπ) n= Akira Yoshioka 33

32 2.3. Star theta functions By a direct caluculation we have δ τ (w a) = 2 π exp ( 1 τ τ (w a)2) We see θ 3 τ (w) = θ 3 (w, τ) Then we have θ 3 (w, τ) = δ τ (w + nπ) n = π exp ( 1 τ τ (w + nπ)2) n = π τ e (1/τ)w2 θ 3 (2πw/iτ, π 2 /τ) Akira Yoshioka 34

33 2.3. Clifford algebra 2.3. Clifford algebra In this subsection, we construct a Clifford algebra by means of the star exponential exp t( 2u v ) for certain K. In what follows, we decsribe a rough sketch of the construction. We recall the star exponential for quadratic case. Quadratic case First we consider a generic point in S C (2) ( ) τ κ K = S κ τ C (2) In the star product K algebra, we write the generator u = u 1, v = u 2 satisfying [u, v] K = Akira Yoshioka 35

34 2.3. Clifford algebra Then the star exponential of H = 2u v is explicitly given at a general point K as ( ) 2u v exp K t = 2e t D exp [ e t e t ( (e t e t )τu uv + (e t e t )τ v 2)] D where D = 2 (e t e t )τ τ, = e t + e t κ(e t e t ) Akira Yoshioka 36

35 2.3. Clifford algebra In the sequel, we assum τ = 0, that is, we take a point ( ) 0 κ K = κ τ Then D =. Akira Yoshioka 37

36 2.3. Clifford algebra We have a limit ( ) 2u v lim ϖ 00 = exp t K t which we call a vacuum. In fact, we have Lemma i) ϖ 00 K ϖ 00 = ϖ 00 ii) v K ϖ 00 = ϖ 00 K u = 0. = 2 1+κ exp ( 1 (1+κ) (2uv τ 1+κ u2 ) ) (13) Akira Yoshioka 38

37 2.3. Clifford algebra Putting t = πi, we have the identity ( ) 2u v exp K πi = 1 (14) Using v K (u K v) = (v K u) K v = (u K v + ) K u we see that the star exponential satisfies ( ) ( ) 2u v 2v u v K exp K t = exp K t K v ( 2u v + 2 = exp K t ( 2u v = e 2t exp K t ) K v ) K v Akira Yoshioka 39

38 2.3. Clifford algebra Then the integral 1 2 and Then we have Lemma 0 exp t( 2v u K 1 0 exp 2 K t )dt converges and then we define ( 2v u The element v is the right inverse of v satisfying ) dt = (v K u) 1 + (15) v = u K (v K u) 1 +. (16) v K v = 1, v K v = 1 ϖ 00 Akira Yoshioka 40

39 2.3. Clifford algebra Now we fix an integer l. By putting t = t l = πi 2 l we obtain 2 l roots of the unity such that Then we have Lemma These satisfy Ω l = exp K ( ) πi 2u v 2 l, ϖl = exp 2 ( ) πi 2 l Ω 2l l K = Ω l K K Ω } {{ } l = 1, ϖl = 1 2 l 2 l (17) Ω k l K K u m K K ϖ 00 K v m K = ϖ km l u m K K ϖ 00 K v m K Akira Yoshioka 41

40 2.3. Clifford algebra Now we take appropriate complex numbers a 0, a 1,, a 2 l 1 element 2 l 1 E = a k Ω k l K satisfies the identies k=0 so that an E K u m K K ϖ 00 K v m K = { K u m K K ϖ 00 K v m K 0 m 2 l l 1 m 2 l 1 Akira Yoshioka 42

41 2.3. Clifford algebra We see this is equivalent to 2 l 1 { 1 0 m 2 a k ϖ km l = l l 1 m 2 l 1 k=0 The complex numbers a 0, a 1,, a 2 l 1 are uniquely determined by these equations. Then we have Lemma The element E satisfies and the element F = 1 E satisfies E K E = 1 F K F = 1, E K F = F K E = 0 Akira Yoshioka 43

42 2.3. Clifford algebra Further we have Lemma where (v) 2l 1 K E K (v) 2l 1 K = (v) 2l 1 K K F, ( v) 2l 1 K K F = E ( v) 2l 1 K = v K K v and ( v) 2l 1 = v } {{ } K K K v } {{ } 2 l 1 2 l 1 Akira Yoshioka 44

43 2.3. Clifford algebra Now we set Then we have ξ = E K (v) 2l 1, η = ( v) 2l 1 K K K F Theorem The elements ξ and η of the K product algebra satisfies the identities ξ K ξ = η K η = 0 ξ K η + ξ K η = 1 Akira Yoshioka 45

44 [OM] H. Omori, Y. Maeda, Quantum Calculus (in Japanese), Springer, 2004 [TY] Tomihisa T. and Yoshioka A., Star Products and Star Exponentials, J. Geometry and Symmetry in Physics, 19 (2010) [YM] Yoshioka A., T. Matsumoto, Path integral for Star Exponential functons of quadratic forms, In: Geometry, Integrability and Quantization. I. M. Mladenov, G. L. Naber (Eds), 4, Coral Press Scientific Publishing, Sofia 2003, pp , Akira Yoshioka 46