Representation of Spin Group Spin(p, q)

Commun. Theor. Phys. Beijing China 48 2007 pp. 415 424 c nternational Academic Publishers Vol. 48 No. 3 September 15 2007 Representation of Spin Group Spinp q WANG Na 1 WU Ke 12 ZHANG Bo 1 1 School of Mathematical Science Capital Normal University Beijing 100037 China 2 Key Laboratory of Mathematics Mechanization the Chinese Academy of Sciences Beijing 100080 China Received December 28 2006 Abstract The representation Yp q of spin group Spinp q in any dimensional space is given by induction the relation between two representations which are obtained in two kinds of inductions from Spinp q to Spinp + 1 q + 1 are studied. PACS numbers: 02.20.-a Key words: spin group group representation 1 ntroduction The theories of spinors Clifford algebras play a major role in contemporary physics mathematics. The spin groups were introduced by Clifford in 1876 1 are well chosen subgroups of units of Clifford algebras. 23 The spinors were discovered by E. Cartan in 1913 in mathematical form who showed that spinors furnish a linear representation of the groups of rotations of a space of arbitrary dimensions. 4 On the other side physicists Pauli 5 Dirac 6 introduced spinors for the representations of the wave functions in quantum mechanics quantum field theory they were also used in general relativity. 7 Recently the spinors in 10 or 11 dimensions are used especially in string theory M-theory. 8 So it is helpful to investigate them in order to get a firm understing. t is known that SL2 R SL2 C are the representations of spin groups Spin1 2 Spin1 3 in 3-4-dimensional Minkowski spaces respectively. n 5-dimensional 6-dimensional spaces the spin groups Spin1 4 Spin1 5 also have classical group representations which are SP1 1 SL2 H respectively. But for any integral number q q 6 there are no classical Lie groups that are the representations of spin groups Spin1 q. Fortunately all the representations Y1 q have been given by Lu et al. 9 very recently. n this article we give the representations of all high dimensional spin groups Spinp q in terms of the method in Ref. 9. n Sec. 2 we recall many notations used in this note give the main results. n Sec. 3 we analyze the relation between the two representations which are obtained in two ways by induction from Spinp q to Spinp + 1 q + 1. 2 Representation Yp q of Spin Group Spinp q n this section we give the representations of spin groups Spinp q. We begin by describing some conventions. Take A to be any 2 2 complex matrix the symbol A is used to denote ɛāɛ i.e. 0 1 A = ɛāɛ ɛ = 1 1 0 ɛ is the transpose matrix of ɛ Ā the complex conjugate matrix of A. The 2 2 Pauli matrices 1 0 0 1 σ 0 = σ 1 = 0 1 1 0 0 i 1 0 σ 2 = σ 3 = 2 i 0 0 1 are all Hermitian matrices satisfy σ j σk + σ k σj = σj σ k + σkσ j = 2η jk 1 3 1 j k = 0 1 2 3. 3 We recall the definitions of gamma matrices γ j 1 q which have been known in Ref. 9 the results in that article are based on Refs. 10 11. We know that γ j 1 3 are defined by 0 σj γ j 1 3 = σj j = 0 1 2 3. 4 0 n the case of q = 2n 1 n is an integer n > 2 γ j 1 q are defined by 0 γ j 1 2n 2 γ j 1 2n 1 = γ j 1 2n 2 0 j = 0 1... 2n 2 ; 0 n 1 γ 2n 1 1 2n 1. 5 n 1 0 n the case of q = 2n n is an integer n > 1 γ µ 1 q are defined by γ µ 1 2n = γ µ 1 2n 1 µ = 0 1... 2n 1 ; in 1 0 γ 2n 1 2n =. 6 0 i n 1 n this article we use the symbol n to st for the identity matrix of rank 2 n 2 n then we give the definitions of gamma matrices γ j p q by induction which is similar to that of γ j 1 q. n the 2n-dimensional case i.e. p + q = 2n p > 1 q 3 one has 0 γ j p 1 q γ j p q = γ j p 1 q 0 The project partially supported by National Key Basic Research Program of China under Grant No. 2004CB318000 National natural Science Foundation of China under Grant Nos. 10375038 90403018. The authors would like to express their thanks to Moningside Center CAS part of the work was done when we were joining the Workshop on Mathematical Physics there E-mail: wuke@mail.cnu.edu.cn

416 WANG Na WU Ke ZHANG Bo Vol. 48 j = p + 2... 1 0 1... q ; 0 in 1 γ p+1 p q =. 7 i n 1 0 n the 2n + 1-dimensional case one has γ µ p + 1 q = γ µ p q µ = p + 1... 1 0 1... q ; n 1 0 γ p p + 1 q = 8 0 n 1 minus index is used which is changing with the index p. Then by induction we can prove that the gamma matrices γ j p q p 2 q 3 defined above are all the matrices of rank 2 p+q/2 2 p+q/2 satisfy γ j p qγ k p q + γ k p qγ j p q = 2η jk p q p+q/2 j k = p + 1... 1 0 1... q 9 in which p + q p + q matrix η jk p q p+1 jk q is a diagonal matrix for any integral numbers p q p > 1 q > 3 the p elements ahead on diagonal line are 1 the others are 1 we use the symbol p + q/2 to st for the integral part of p + q/2. Lemma 1 Let Yp q be a connected Lie group of 2 p+q/2 2 p+q/2 matrix if any T Yp q satisfies T γ ν p qt 1 = l ν µ p qγ µ p q µ=0 ν = p + 1... 0 1... q ; Lp q = l ν µ p q p+1 µν q SOp q 10 γ µ γ ν p + 1 µ < ν q is a basis of Lie algebra hp q of Yp q then the map T Lp q induces a 2 to 1 homomorphism: Yp q SOp q which is a local isomorphism we also have the following relation η αβ p qγ α p qγ µ p q γ ν p qγ β p q = p + q 4γ µ p q γ ν p q. 11 Proof Since δt = T 1 dt T Yp q is a left invariant matrix it can be expressed as δt = ω µν γ µ p q γ ν p q 12 p+1 µ<ν q ω µν = ω νµ are left invariant 1-forms of the parameter t A A = 1... p + qp + q 1/2 of Yp q. Differentiating the given equation 10 multiplying both sides with T 1 from left T from right δt γ ν γ ν δt = dl ν µ T 1 γ µ T = dl ν µ L α µγ α 13 γ µ = γ µ p q lµ ν = lµp ν q L α µ p+1 µα q = l α µ 1 p+1 µα q. Multiplying both sides of Eq. 13 by η βν γ β from the left summing up the index ν we have η βν γ β δt γ ν p + qδt = η βν dl µ ν L α µγ β γ α = ϕ αβ γ β γ α = α<β ϕ αβ γ α γ β 14 ϕ αβ = L α µ dl µ ν η νβ = ϕ βα are the left invariant 1-form of SOp q. Substituting Eq. 12 into Eq. 14 we have ω µν {η αβ γ α γ µ γ ν γ β p + qγ µ γ ν } = ϕ αβ γ α γ β. 15 Since η αβ γ α γ µ γ ν γ β = p + q 4γ µ γ ν then we have ω µν = 1 4 ϕµν. 16 Let t 1... t p+qp+q 1/2 s 1... s p+qp+q 1/2 be the local coordinates of Yp q SOp q in the neighborhood of the identity respectively then equation 16 can be expressed into the form M ω µν = u µν A dta = 1 M v µν A ds A 4 A=1 A=1 p + 1 µ < ν q 17 M = p + qp + q 1/2 the M M matrices u µν A u µν A are non-singular. Therefore the Jacobian t 1... t p+qp+q 1/2 s 1... s p+qp+q 1/2 deduced from Eq. 17 is non-singular Yp q SOp q is a local isomorphism. f we change T to be T then the identity 10 is unchanged. Globally Yp q SOp q is a homomorphism. The lemma is proved. Now we give the explicit form of group Yp q in which p q is any integer pair which satisfies p > 1 q > 2 we suppose every ξ which is used in the following is not equal to zero. i q is odd a p = 2 q = 3 Y2 3 is defined by Y2 3 = {T T = JU U Y1 3} 18 J = 1 3 aj σ j ξ 3 aj σ j 3 ξ = 1 + η ij a i a j ; i b p = 2 q = 2n 1 n > 2 Y2 2n 1 is defined by Y2 2n 1 = {T T = JU U Y1 2n 1} 19 2n 2 J = 1 a j γ j + a 2n 1 ξ 2n 2 a j γ j + a 2n 1 2n 1 γ j = γ j 1 2n 1 ξ = 1 + η ij a i a j ; i c p = 2m q = 2n 1 m > 1 n > 2 Y2m 2n 1 is defined by Y2m 2n 1 = {T T = JU U Y2m 1 2n 1} 20

No. 3 Representation of Spin Group Spinp q 417 d ii a J = 1 ξ 2n 1 j= 2m+3 aj γ j + ia 2m+2 2n 1 γ j = γ j 2m 2 2n 1 ξ = 1 + j= 2m+3 aj γ j + ia 2m+2 p+2 p = 2m + 1 q = 2n 1 m > 1 n > 2 Y2m + 1 2n 1 is defined by { Y2m + 1 2n 1 = T T = J J = U 0 η ij a i a j ; U Y2m 2n 1 1 2n 1 + i j= 2m+1 aj γ j 0 ξ 0 i 2n 1 j= 2m+1 aj γ j γ j = γ j 2m 2n 1 ξ = 1 + q is even p = 2m q = 2n m 1 n 2 Y2m 2n is defined by { Y2m 2n = T T = J J = U 0 p+2 η ij a i a j. U Y2m 1 2n 1 2n + i j= 2m+2 aj γ j 0 ξ 0 i 2n j= 2m+2 aj γ j γ j = γ j 2m 1 2n ξ = 1 + p+2 η ij a i a j ; } 21 } 22 b p = 2m + 1 q = 2n m 1 n 2 Y2m + 1 2n is defined by Y2m + 1 2n = {T T = JU U Y2m 2n} 23 J = 1 2n j= 2m+2 aj γ j + ia 2m+1 2n ξ j= 2m+2 aj γ j ia 2m+1 γ j = γ j 2m 1 2n ξ = 1 + p+2 η ij a i a j. Theorem 1 For any integer pair p q p 2 q 3 any matrix T in the p + qp + q 1/2-dimensional Lie group Yp q of rank 2 p+q/2 2 p+q/2 satisfies T γ ν p qt 1 = l ν µ p qγ µ p q µ ν = p + 1... 0 1... q ; Lp q = l ν µ p q p+1 µν q SOp q the map Yp q SOp q induced by T Lp q is a two to one group homomorphism a local isomorphism. Proof Here we only give the proof of the cases p = 2m q = 2n 1 p = 2m + 1 q = 2n 1 for the proof of any other case is similar to one of the cases. c Any T Y2m 2n 1 it is of T = JU U Y2m 1 2n 1 by induction we know Uγ µ U 1 = l ν 1µγ ν µ ν = 2m + 2... 0... 2n 1 l1µ ν 2m+2 µν 2n 1 SO2m 1 2n 1. 24 We also know by induction that any element in group Y2m 1 2n 1 is a block diagonal matrix so we can write U into U1 0 then Uγ 2m+1 U 1 = so we have Uγ µ U 1 = lµγ ν ν 2 0 = γ 2m+1 25 0 µ ν = 2m + 1... 0... 2n 1 26 1 0 lµ ν = 0 l1µ ν SO2m 2n 1 The inverse matrix of J is

418 WANG Na WU Ke ZHANG Bo Vol. 48 then we have Jγ µ J 1 = J 1 1 = 1 + ηjk a j a k ia 2m+2 + a j γ j ia 2m+2 a j γ j 27 1 a j γ j γ µ + a k γ µ γ k A 12 1 + η ij a i a j A 21 a k γ µ γ k a j µ = 2m + 3... 0... 2n 1 28 γ j γ µ A 12 = 1 + η jk a j a k γ µ 2a 2m+2 a µ i 2a j a µ γ j 29 A 21 = 1 + η jk a j a k γ µ + 2a 2m+2 a µ i 2a j a µ γ j 30 i.e. we can write Jγ µ J 1 into Jγ µ J 1 = lµγ ν ν µ = 2m + 3... 0... 2n 1 31 Jγ 2m+2 J 1 1 = 1 + η ij a i a j 2a 2m+2 B 12 32 B 21 2a 2m+2 B 12 = 1 a 2m+2 2 i 2a 2m+2 a k γ k + η jk a j a k i 33 B 21 = 1 a 2m+2 2 i 2a 2m+2 a k γ k η jk a j a k i 34 we can also get the form Jγ 2m+2 J 1 = l ν 2m+2γ ν 35 similarly Jγ 2m+1 J 1 = l ν 2m+1γ ν 36 in which the symbols l ν µ l ν 2m+2 l ν 2m+1 used in Eqs. 31 35 36 respectively st for the coefficients of γ ν so we have Jγ µ J 1 = l ν µγ ν µ ν = 2m+1... 0... 2n 1. 37 Multiplying both sides of Eq. 37 by x µ summing up the index µ we get Jx µ γ µ J 1 = x µ l ν µγ ν µ ν = 2m + 1... 0... 2n 1 38 then η µα x µ x α = Jx µ γ µ J 1 2 = x µ lµγ ν ν 2 = x µ x α lµl ν αη β νβ i.e. η µα = lµl ν αη β νβ so we have l ν µ 2m+1 µν 2n 1 SO2m 2n 1 39 t is easy to check that {γ µ γ ν 2m + 1 µ < ν 2n 1} is a Lie algebra since dt = JdU + dju T 1 = U 1 J 1 40 we have δt = T 1 dt = δu + U 1 δju. 41 By induction we know δj = J 1 dj = η jka j da k 1 + η ij a i a j + 1 1 + η jk a j a k δu = 0 c12 = da 2m+2 γ 2m+1 2m 2n 1 γ 2m+2 2m 2n 1 c 21 0 2m+2 µ<ν 2n 1 ω µν γ µ γ ν ω µν µ < ν are left invariant 1-forms of Lie algebra h2m 1 2n 1 c11 c 12 42 c 21 c 22 + da j γ 2m+1 2m 2n 1 γ j 2m 2n 1 j = 2m + 3... 0... 2n 1 c11 0 = η ij a i da j + a 2m+2 da j a j da 2m + 2γ 2m+2 γ j + 1 0 c 22 2 aj da k γ j γ k. Hence from Eq. 41 we can derive that δt can be written into the following form: δt = ϕ µν γ µ 2m 2n 1 γ ν 2m 2n 1 2m+1 µ<ν 2n 1 ϕ µν are left invariant 1-forms from the Lemma we can get the conclusion. d For Y2m + 1 2n 1 any element T Y2m + 1 2n 1 we know it is of U 0 T = J U Y2m 2n 1 by induction we have Uγ µ 2m 2n 1U 1 = l1µγ ν ν l1µ ν SO2m 2n 1 43 so 1 U 0 U 0 γ µ 2m + 1 2n 1 = l ν 1µγ ν 2m + 1 2n 1

No. 3 Representation of Spin Group Spinp q 419 1 U 0 U 0 γ 2m 2m + 1 2n 1 = γ 2m2m + 1 2n 1. Hence we know The inverse matrix of J is then one can get Jγ µ 2m + 1 2n 1J 1 = Uγ µ 2m 2n 1U 1 = lµγ ν ν 44 1 0 lµ ν = 0 l1µ ν SO2m + 1 2n 1. 45 J 1 = 1 ia j γ j 0 1 + ηjk a j a k 0 + ia j γ j 1 0 γ µ + 2ia µ + a j a µ γ j η jk a j a k γ µ 1 + η ij a i a j γ µ 2ia µ + a j a µ γ j η jk a j a k γ µ 0 = lµγ ν ν µ = 2m + 1... 0... 2n 1 ν = 2m... 0... 2n 1 Jγ 2m 2m + 1 2n 1J 1 1 0 i1 η jk a j a k = 1 + η ij a i a j i1 η jk a j a k =l ν 0 2mγ ν ν = 2m... 0... 2n 1 in which the symbols lµ ν l 2m ν in the equations above st for the coefficients of γ ν respectively then we have Jγ µ J 1 = lµγ ν ν µ ν = 2m... 0... 2n 1 47 similarly lµ ν 2m µν 2n 1 SO2m + 1 2n 1. t is easy to check that {γ µ γ ν 2m µ < ν 2n 1} is a Lie algebra since we get By induction we know dt = J du 0 0 du δt = T 1 dt = δu = + dj 2m+1 µ<ν 2n 1 U 0 T 1 = U 1 0 1 46 J 1 48 δu 1 0 + 0 δu 1 δj. 49 ω µν γ µ 2m 2n 1 γ ν 2m 2n 1 ω µν µ < ν are left invariant 1-forms of Lie algebra h2m 1 2n 1 hence we have δu 0 = ω µν γ µ 2m + 1 2n 1 γ ν 2m + 1 2n 1. 0 δu Since 2m+1 µ<ν 2n 1 δj = J 1 dj = Then from Eq. 49 we can derive that δt = 2m µ<ν 2n 1 ϕ µν are left invariant 1-forms then the conclusion is proved. da j 1 + η ij a i a j γ a j da k 2m γ j + 2 + 2η ij a i a j γ j γ k. 50 ϕ µν γ µ 2m 2n 1 γ ν 2m 2n 1 3 Relation of two Representations in Spinp + 1 q + 1 According to Ref. 9 the last section we know that Yp q is matrix representation of Spinp q for p 1 q 3. We get the representation by induction there are two ways from Yp q to Yp + 1 q + 1 i.e. a Yp q Yp q + 1 Y 1 p + 1 q + 1 b Yp q Yp + 1 q Y 2 p + 1 q + 1. The induction from Yp q to Yp+1 q was discussed in Sec. 2. Then we will discuss the relation of the representations which are obtained in above two ways. Firstly we consider Yp q 1 Yp q. i When p+q is odd the element of Yp q is of the form T p q = 1 ξ JT p q 1 51 ξ = 1 q 1 jk= p+1 η jk a j a k

420 WANG Na WU Ke ZHANG Bo Vol. 48 in which we let ξ 0 J = iζ + ia q 1 ζ = q 2 a j γ j p q 2. iζ + ia q 1 We transform J into the following form: q 1 J = γ q p qγ q p q + a j γ j p q 52 so T p q is of T p q = 1 ξ γ q p qγ q p q + q 1 a j γ j p qt p q 1. 53 ii When p + q is even the element of Yp q is of the form T p q = 1 T p q 1 0 J ξ 0 T p q 1 ξ = 1 in which we let ξ 0 q 1 jk= p+1 η jk a j a k Similarly J can be written as therefore + J = Here for any matrix A we define q 1 a j γ j p q 1 0 0 q 1. a j γ j p q 1 J = γ q p q γ q p q + T p q = 1 γ q p q γ q p q + ξ q 1 RA A = q 1 a j γ j p q a j γ j p q RT p q 1 T p q 1. A 0. 0 A Secondly we consider the case Yp 1 q Yp q. i When p + q is odd the element of Yp q is of the form T p q = 1 JT p 1 q 54 λ in which we let λ 0 which can be written into then ii λ = 1 + jk= p+2 J = q 1 a j γ j p 1 q 1 a q J = γ p+1 p q γ p+1 p q + T p q = 1 γ p+1 p q γ p+1 p q + λ η jk a j a k q 1 q 1 When p + q is even the element of Yp q is of the form T p q = 1 T p 1 q 0 J λ 0 T p 1 q λ = 1 + jk= p+2 η jk a j a k a j γ j p 1 q 1 + a q a j γ j p q 55 a j γ j p q T p 1 q

No. 3 Representation of Spin Group Spinp q 421 in which we let λ 0 + i J = so then a j γ j p 1 q 0 0 i a j γ j p 1 q J = γ p+1 p qγ p+1 p q + γ p+1 p q T p q = 1 γ p+1 p q γ p+1 p q + λ a j γ j p q 56 a j γ j p q RT p 1 q T p 1 q. Now we can obtain the form of Yp + 1 q + 1 from Yp q by induction. Then we do it in those two ways find the relationship between their results. As we know the situations are different when p + q is odd or even So we discuss them separately we take the cases p = 1 q = 4 p + q = 5 p = 2 q = 4 p + q = 6 as examples to introduce them. Let us begin with p = 1 q = 4 the two ways from Y1 4 to Y2 5 are a Y1 4 Y1 5 Y 1 2 5 T 1 2 5 Y 1 2 5 b Y1 4 Y2 4 Y 2 2 5 T 2 2 5 Y 2 2 5. t is similar with the computation of T p q that we obtain the following four equations: ξ 1 = 1 4 jk=0 T 1 5 = 1 ξ1 4 a j γ j 1 5 γ 5 1 5 γ 5 1 5RT 1 4 T 1 4 57 T 1 2 5 = 1 γ 1 2 5 γ 1 2 5 + λ1 T 2 4 = 1 γ 1 2 4 γ 1 2 4 + λ2 T 2 2 5 = 1 ξ2 η jk a j a k ξ 2 = 1 5 b j γ j 2 5 T 1 5 58 4 c j γ j 2 4 RT 1 4 T 1 4 59 4 d j γ j 2 5 γ 5 2 5 γ 5 2 5T 2 4 60 j= 1 4 jk= 1 in which ξ 1 ξ 2 λ 1 λ 2 are not equal to zero. Substituting Eq. 57 into Eq. 58 we obtain T 1 2 5 = 1 γ 1 2 5 γ 1 2 5 + λ1 ξ 1 η jk d j d k λ 1 = 1 + 5 b j γ j 2 5 5 jk=0 4 η jk b j b k λ 2 = 1 + 4 jk=0 a j γ j 1 5 γ 5 1 5 γ 5 1 5 η jk c j c k RT 1 4 T 1 4 61 γ 1 2 5 γ 1 2 5 + 5 b j γ j 2 5 4 a j γ j 1 5 γ 5 1 5 γ 5 1 5 2 + a m γ m b 5 2 b j a m γ j γ m + b j γ j b 5 a m γ m = b 5 2 b j a m γ j γ m b j γ j + b 5 a m γ m 2 a m γ m in which Einstein convention is used the index m j are summing up from 0 to 4 hence T 1 2 5 is of the form T 1 2 5 = 1 2 + a m γ m A 1 T 1 4 0 ξ1 λ 1 A 2 2 a m 62 γ m 0 T 1 4. A 1 = b 5 2 b j a m γ j γ m + b j γ j b 5 a m γ m A 2 = b 5 2 b j a m γ j γ m b j γ j + b 5 a m γ m.

422 WANG Na WU Ke ZHANG Bo Vol. 48 With the same method for b : Y1 4 Y2 4 Y 2 2 5 we obtain the form of T 2 2 5 T 2 2 5 = 1 2 + ic m γ m B 1 T 1 4 0 ξ2 λ 2 B 2 2 ic m 63 γ m 0 T 1 4 the index m j are also summing up from 0 to 4 B 1 = d 1 2 + d j c m γ j γ m + id j γ j + id 1 c m γ m then we obtain when B 2 = d 1 2 + d j c m γ j γ m id j γ j id 1 c m γ m T 1 2 5 = T 2 2 5 b 5 = d 1 a m = ic m b j = id j for m j = 0 1 2 3 4. Next we discuss the case p = 2 q = 4 p + q = 6. The two ways from Y2 4 to Y3 5 are a Y2 4 Y2 5 Y 1 3 5 b Y2 4 Y3 4 Y 2 3 5. For a similar to Eq. 58 we can write T 1 3 5 as T 1 3 5 = 1 γ 2 3 5 γ 2 3 5 + λ3 λ 3 = 1 + 5 m= 1 5 mn= 1 we let λ 3 0. Substituting Eq. 60 into Eq. 64 we obtain T 1 3 5 = 1 γ 2 3 5 γ 2 3 5 + λ3 A = 3 + γ 5 2 5 = 1 ξ2 λ 3 A + B 0 0 A B 4 a j γ j 2 4 j= 1 B = ib 5 γ 5 2 5 ib 5 4 a j γ j 2 4 + i j= 1 b m γ m 3 5 RT 2 5 T 2 5 64 η mn b m b n 5 m= 1 b m γ m 3 5 RT 2 5 T 2 5 T 2 4 0 0 T 2 4 4 jm= 1 b m a j γ m 2 4γ 5 2 5γ j 2 4. n the same way we obtain T 2 3 5 = 1 C + D 0 T 2 4 0 65 ξ3 λ 4 0 C D 0 T 2 4 4 4 4 ξ 3 = 1 η mn d m d n λ 4 = 1 + η jk c j c k C = 3 + γ 2 3 4 c j γ j 2 4 D = 4 m= 1 + mn= 2 jk= 1 d m γ m 2 4 + d 2 γ 2 3 4 + d 2 4 mj= 1 d m c j γ m 2 4γ 2 3 4γ j 2 4 in which we let ξ 3 0 λ 4 0 then we have when 4 c j γ j 2 4 j= 1 T 1 3 5 = T 2 3 5 j= 1 b 5 = d 2 a j = ic j b m = id m for j m = 1 0 1 2 3 4. Then we discuss the situation for any integer pair p q in which p 1 q 4. Theorem 2 When p + q = 2n + 1 for p > 1 q > 4 the two ways from Yp q to Yp + 1 q + 1 by induction are a Yp q Yp q + 1 Y 1 p + 1 q + 1

No. 3 Representation of Spin Group Spinp q 423 And we can obtain i The form of element in Y 1 p + 1 q + 1 is we let ξ 0 λ 0 ii b Yp q Yp + 1 q Y 2 p + 1 q + 1. T 1 p + 1 q + 1 = 1 p+q/2 + Ha p+q A + B ξλ A B p+q/2 Ha p+q Ha p+q = ξ = 1 mn= p+1 A = b q+1 p+q/2 B = The form of element in Y 2 p + 1 q + 1 is T 2 p + 1 q + 1 = 1 ξ λ a m γ m p q η mn a m a n λ = 1 + j b j γ j p q + b q+1 q+1 jk= p+1 b j a m γ j p qγ m p q a m γ m p q p+q/2 + ihc p+q C + D C D p+q/2 ihc p+q Hc p+q = ξ = 1 jk= p C = d p p+q/2 + D = i j= P +1 c m γ m p q η jk d j d k λ = 1 + j d j γ j p q + id p T p q 0 66 0 T p q η jk b j b k η mn c m c n d j c m γ j p qγ m p q c m γ m p q. we let ξ 0 λ 0. When b q+1 = d p a m = ic m b j = id j for m j = p + 1 p + 2... q 1 q we obtain Theorem 3 T 1 p + 1 q + 1 = T 2 p + 1 q + 1. T p q 0 67 0 T p q When p + q = 2n for p 2 q 4 i.e. p + q is even the two ways from Yp q to Yp + 1 q + 1 are then we can obtain that i The element of Y 1 p + 1 q + 1 is of the form a Yp q Yp q + 1 Y 1 p + 1 q + 1 b Yp q Yp + 1 q Y 2 p + 1 q + 1 T 1 p + 1 q + 1 = 1 ξλ A + B 0 0 A + B ξ = 1 jk= p+1 η jk a j a k λ = 1 + A = p+q/2 + γ q+1 p q + 1 B = i q+1 mn= p+1 a j γ j p q T p q 0 68 0 T p q η mn b m b n b m γ m p q + ib q+1 γ q+1 p q + 1 ib q+1 a j γ j p q

424 WANG Na WU Ke ZHANG Bo Vol. 48 in which we let ξ 0 λ 0. ii + i j The element of Y 2 p + 1 q + 1 is of the form T 2 p + 1 q + 1 = 1 ξ λ we let ξ 0 λ 0. ξ = 1 mn= p b m a j γ m p qγ q+1 p q + 1γ j p q C + D 0 0 C D η mn d m d n λ = 1 + C = p+q/2 iγ q+1 p q + 1 D = i jk= p+1 T p q 0 69 0 T p q c j γ j p q d m γ m p q id p γ q+1 p q + 1 + d p j d m c j γ m p qγ q+1 γ j p q η jk c j c k c j γ j p q When b q+1 = d p a j = ic j b m = id m for j m = p + 1... q we have T 1 p + 1 q + 1 = T 2 p + 1 q + 1. The results proofs of Theorem 2 Theorem 3 are very similar to those of the situations for p = 1 q = 4 p + q = 5 p = 2 q = 4 p + q = 6 so we omit them. 4 Discussion Clifford algebras spin groups are widely studied since they are fundamental tools of physics mathematics. n this article we analyze the representation of high dimensional spin group obtain the following results. i We give the representation of spin group Spinp q in any high dimensional space. ii We find the relation between the two representations which are obtained in two ways by induction from Spinp q to Spinp + 1 q + 1. However spin group may not have yet revealed all their mysteries 12 we hope that the conclusions obtained in this article can help us to derive more understings of spin groups. Acknowledgments We are deeply grateful to Profs. Qi-Keng Lu Han- Ying Guo Shi-Kun Wang for their valuable discussions which are essentially stimulating to us in writing down this work. References 1 W.K. Clifford A Preliminary Sketch of Biquaternions Proceedings of the London Mathematical Society 4 1873 381. 2 J. Gallier Clifford Algebras Clifford Groups a Generalization of the Quaternions: The Pin Spin Groups lecture note avaluable at www.cis.upenn.edu/ jean/. 3 V.V. Varlamov preprint arxiv: math-ph/0108022. 4 S.. Vacaru N.A. Vicol arxiv: math-ph/0406585. 5 W. Pauli Z. Physik 43 1927 601. 6 P.A.M. Dirac Proc. Roy. Soc. A 117 1982 610. 7 R. Penrose Ann. Phys. N.Y. 10 1960 171. 8 J. Polchinski String Theory Cambridge University Press Cambridge 1998. 9 Q.K. Lu K. Wu Acta Mathematica Sinica 23 2007 577. 10 Q.K. Lu Dirac s Conformal Space de-sitter Sapce MCM-Workshop Series Vol. 1 Morningside Center of Mathematics Chinese Academy of Sciences Beijing 2005. 11 Q.K. Lu Differential Geometry ts Application to Physics Science Press Beijing 1982 in Chinese. 12 R. Coquereaux Cliford Algebras Spinors Fundamental nteractions: Twenty Year After preprint arxiv:math-ph/0509040.