Korman cliffford_thesis

Cliord Algebras and Bilinear Forms on Spinors Honors Thesis Department of Mathematics University of Pittsburgh Eric O. Korman Advisor: Dr. George Sparling Abstract Associated with the vector space Rp+qwith metric gof signature (p; q) is its Cliord algebra, denoted Clp;q. Inside Clp;qlie the groups Pin(p; q) and Spin (p; q), which double cover O(p; q) and SO(p; q), re- spectively. We focus on two issues which seem to be neglected in the standard literature. The rst is when Clp;q,Pin(p; q), and Spin (p; q) are isomorphic to Clq;p,Pin(q; p), and Spin (q; p). While a partial answer can be given implicitly by the representations of the various algebras, our arguments are based purely on the Cliord algebra struc- ture. In the second section we construct natural bilinear forms on the space of spinors such that vectors are self-adjoint (up to sign). These forms are preserved (up to sign) by the Pin and Spin groups. With the Cliord action of k-forms, 0kp+q, on spinors, the bilinear forms allow us to relate spinors with elements of the exterior algebra. We then nd some curious identities involving the norms of various forms. 1 Contents 1 Introduction 3 1.1 Constructing the Cliord algebra . . . . . . . . . . . . . . . . 3 1.2 The Cliord Group . . . . . . . . . . . . . . . . . . . . . . . . 5 1.3Pin(p;q) andSpin (p;q) . . . . . . . . . . . . . . . . . . . . . 6 1.4 Representations of Cliord Algebras . . . . . . . . . . . . . . . 6 2 Space and Time Symmetry on the Cliord Algebra Level 9 2.1Cl0 p;qandCl0 q;pandSpin (p;q) andSpin (q;p) . . . . . . . . . . 9 2.2Clp;qandClq;pandPin(p;q) andPin(q;p) . . . . . . . . . . . 11 3 Bilinear Forms on Spinors 15 3.1 Relations Between Spinors and Forms . . . . . . . . . . . . . . 17 3.2 Identities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17 3.2.1 Real Cliord algebras . . . . . . . . . . . . . . . . . . . 18 3.2.2 Quaternionic Cliord algebras . . . . . . . . . . . . . . 26 4 Conclusion 28 5 Acknowledgements 29 2 1 Introduction Cliord algebras are geometric algebras and can be seen as generalizations of the real numbers, complex numbers, and quaternions (the free real algebra on three variables i;j;k modulo the relations i2=j2=k2=ijk=1). As such, they have been very in uential in the formulation of modern physical theories. For example the Dirac equation , which was the rst successful de- scription of the electron compatible with both special relativity and quantum mechanics, is a dierential equation involving the elements of the Cliord al- gebra associated with the metric signature (+ ). 1.1 Constructing the Cliord algebra We wish to extend a real vector space with a bilinear form into an algebra by dening a notion of multiplication in a suitable way. Here, an algebra is vector space and a ring with identity. Additionally, we want multiplication of vectors to relate in some way to the geometric structure of the space given by the bilinear form. This motivates the following denition: Denition 1.1. Given a vector space Vover the eld Fwith a bilinear form g, its Cliord algebra, Cl(V), is the free algebra on Vmodulo v2=g(v;v): (1) More formally, we can construct Cl(V) by quotienting out from the tensor algebra,T(V), the (two-sided) ideal generated by all elements of the form v vg(v;v), forv2V. Replacingvbyv+win (1) and expanding yields vw+wv= 2g(v;w); (2) from which we see that two vectors anti-commute if and only if they are orthogonal. We denote the vector space Rp+qwith metric signature (+ + :::+|{z} ptimes:::|{z} qtimes) byRp;q. This space will be the sole focus of our work. Further, we abbreviate 3 Cl(Rp;q) byClp;q. We identify Rp;qwith the image of the natural inclusion mapRp+q,!Clp;q. Iffe1;:::;e ngis an orthonormal basis for Rp;qthen by (1) and (2) these elements generate Clp;qwith the rules: e2 i=( 1 if 1ip 1 ifp+ 1ip+q and eiej=ejeiifi6=j: It turns out thatt that any algebra generated by Rp;qwhich satises (1) is unique and is theCliord algebra Clp;q, as long as pq61 (mod 4). These algebras have dimension 2p+q, a basis beingfei1 1ei2 2:::eip+q p+q:ij= 0 or 1g. Ifpq1 (mod 4), then it is possible to have an algebra generated by Rp;qand satisfying (1) but with the property that e1e2:::e p+q=1. These algebras therefore have dimension 2p+q1. However, we can get the so-called universal Cliord algebra (of dimension 2p+q) by taking the direct sum of these two algebras [2] [3]. As the above discussion hints, the element e1e2:::e p+qis of special interest. It is called the pseudoscalar and is denoted by . Though the e0 isare obviously basis dependent, is canonical in that it remains unchanged (up to sign) under any orthogonal transformation [1]. We see that 2= (1)(p+q1)+(p+q2)+:::+1p+qY i=1e2 i = (1)(p+q1)(p+q) 2+q = (1)(p+q)2+qp 2 (3) and u=( u ip+qis odd oruis even u ip+qis even and uis odd.(4) In constructing isomorphisms and representations of Cliord algebras, we will be implicitly using the following universal property of Cliord algebras: 4 Theorem 1.1. LetAbe a real algebra and j:Rp;q!Abe linear and have the property that j(v)2=g(v;v)1Afor allv2Rp;q, where 1Ais the identity element ofA. Then there exists a unique homomorphism h:Clp;q!Asuch thath(v) =j(v). The function his given by h c0+X iciei1:::e ik! =c0+X icij(ei1):::j(eik) whereci2R. Clp;qhas a graded structure provided by the involution , induced by v7!v forv2Rp;q. We dene Cl0 p;q=fu2Clp;q:(u) =ug. It is not hard to verify thatCl0 p;qis a subalgebra of Clp;q, called the even algebra , of dimension 2p+q1. 1.2 The Cliord Group Supposeuis an invertible element in Clp;qsuch thatu(v) =uv(u1)2Rp;q for allv2Rp;q. Thenuis an orthogonal (i.e. preserves g) automorphism of Rp;q. To see this, note that the inverse of uisu1and g(u(v);u(v)) =u(v)u(v) =(u(v))u(v) =(uv(u1))(uv(u1)) =(u)(v)u1uv(u1) =(u)(v2)(u1) =(u)g(v;v)(u1) =g(v;v): The set of all such u2Clp;qforms a group called the Cliord Group , and is denoted by p;q. Dene 0 p;q= p;q\Cl0 p;q Letu2Rp;qnot be null (i.e. g(u;u)6= 0). Then, by 1, uhas inverse u g(u;u). Further, we see that u(u) =uu(u g(u;u)=uand, ifg(u;v) = 0,u(v) =uv(u1) =uvu1=vuu1=v. Therefore u2p;qand urepresents a re ection in the hyperplane perpendicular to u. Since any 5 orthogonal transformation is a composition of re ections, we see that the map p;q!O(p;q),u7!uis surjective. Furthermore, we claim that the map is actually a homomorphism with kernel R. It can easily be veried that the map is a homomorphism. To see that its kernel is R, suppose that u(v) =vfor allv2Rp;q. Then, for v2Rp;q, we have uv(u1) =v uv=v(u): Putu=u0+u1withu020 p;qandu12p;qn0 p;q. Then since (u0) =u0and (u1) =u1we haveu0v=vu0andu1v=vu1. We need to show that u02R andu1= 0. Assume that u0=2R, then there exists a basis element ei1:::e i2k on whichu0has a non-zero component. Then ( ei1:::e i2k)ei1=e2 i1ei2:::e i2k butei1(ei1:::e i2k) =e2 i1ei2:::e i2ksinceei1must pass an odd amount of el- ements with which it anti-commutes. Contradiction. A similar argument shows that u1= 0. The above result also tells us that p;qis the group generated by non-null vectors: since any orthogonal transformation is a product of re ections, for anyu2p;qthere exist v1;:::;v n2p;qsuch thatu=v1:::vn. But since the mapu7!uis injective up to scale, we have that u=kv1:::v nfor somek2R. Note that the restriction of the homomorphism to 0 p;qgives a surjective homomorphism to SO(p;q). 1.3Pin(p;q)andSpin (p;q) To limit the kernel of the homomorphism u7!ufrom p;q(0 p;q) toO(p;q) (SO(p;q)), we dene the Pin (Spin) group: Pin(p;q) =fv1v2:::v njg(vi;vi) =1 for allig Spin (p;q) =Pin(p;q)\Cl0 p;q: The mapsPin(p;q)!O(p;q) andSpin (p;q)!SO(p;q),u7!u, are now surjective homomorphisms with kernel f1;1g. 1.4 Representations of Cliord Algebras We can always represent Clp;qas the set of all nnmatrices with entries in R;C, orH(the quaternions). We denote the set of all nnmatrices with 6 entries in FbyF[n]. The representation space, i.e. Fnis called the space of spinors . Ifpq6= 1 (mod 4), then the representation is unique. If pq= 1 (mod 4), then there are two inequivalent representations; one has = 1 and the other has =1. Furthermore we can always chose our representations such thatey i=eiie2 i= 1 andey i=ey iie2 i=1, whereyis the conjugate transpose. The full matrix algebra that Clp;qis isomorphic to is determined by the quantity=qp1 mod 8: Clp;q'8 >< >:R[2[p+q]=2] if=5,6, or 7 H[2[p+q]=21] if=1,2, or 3 C[2p+q1=2] if=0, or 4; where [p+q] denotes the integer part of p+q. We say that Clp;qis type Fif it is isomorphic to a full matrix algebra with entries in F(F=R;CorH). Given a representation of Clp;qonCn, we can get a representation on R2nby replacingiwith01 1 0 and 1 with1 0 0 1 . Note that although Clp;q is isomorphic to a proper subalgebra of R[2n], the spin spaces, CnandR2n, are isomorphic as real vector spaces. Similarly, if we have a representation of 7 Clp;qonHn, we can get a representation on R4nby making the replacements i!0 BB@0 1 0 0 1 0 0 0 0 0 01 0 0 1 01 CCA j!0 BB@0 0 1 0 0 0 0 1 1 0 0 0 01 0 01 CCA k!0 BB@0 0 0 1 0 01 0 0 1 0 0 1 0 0 01 CCA 1!0 BB@1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 11 CCA: Representations are built up from lower dimensional representations using identities we will derive in later sections and tensor products of matrices. For example, Clp;q'Clp1;q1 R[2]. To see this let feigbe the standard generators for Clp1;q1. PutA=0 1 1 0 ,B=01 1 0 , andC= 1 0 0 1 . We have that A2=B2=C2= 1,AB=BA,AC=CA, andBC =CB. Therefore,fei Ag[fI C;I BggenerateClp;q. Additionally, a form of Bott periodicity states that Clp;q+8'Clp;q R[16]. Upon seeing that Cl0;8'R[16], we can prove this as follows: let feigand fEiggenerateClp;qandCl0;8, respectively. Let be theCl0;8pseudoscalar and note that 2= 1 (3) and anti-commutes with Ei(4). Then it is easily seen thatfei ;1 EiggenerateClp;q+8. 8 2 Space and Time Symmetry on the Cliord Algebra Level From a purely geometric standpoint, there is no dierence between Rp;qand Rq;p. Indeed,O(p;q) andSO(p;q) are naturally isomorphic to O(q;p) and SO(q;p), respectively. However, the relationship with Clp;qandClq;pis not immediately evident. We will see however, that Clp;q'Clq;pandPin(p;q)' Pin(q;p) if and only if pq0 (mod 4) and that Spin (p;q) andSpin (q;p) are always isomorphic. 2.1Cl0 p;qandCl0 q;pandSpin (p;q)andSpin (q;p) We begin by proving the following proposition. Proposition 2.1. Clp;q1=Cl0 p;q=Clq;p1, assuming for the rst isomor- phism that q1and for the second isomorphism that p1. Proof. LetfeigandfEigbe standard generators for Clp;q1andClp;qrespec- tively. We claim that the map 'fromCl0 p;qtoClp;q1, dened on generators by, wherei<j , EiEj7!( ei ifj=p+q eiejotherwise is an isomorphism. To check that it is a homomorphism, it is sucient, because of the universal property, to verify that the elements fEiEp+qg anti-commute with each other and that ( EiEp+q)2=e2 i. Indeed, if i6=j then we have that ( EiEp+q)(EjEp+q) =EiEjEp+qEp+q=EjEiEp+qEp+q= (EjEp+q)(EiEp+q) and (EiEp+q)2=EiEp+qEiEp+q=E2 iE2 p+q=e2 i. It remains to show that 'is a bijection. Since 'is a linear map of nite di- mensional vector spaces of the same dimension, it is a bijection if and only if it is a surjection. We can easily see that 'is surjective since all of the generators of Clp;q1are inim('). Letfeigbe the standard generators for Clq;p1and letfEigbe anti-commuting generators for Cl0 p;qwhere we break convention by having: E2 i=( 1 if 1iq 1 ifq+ 1ip+q: 9 Thuse2 i=E2 ifor 1iq+p1. By the same argument as before, we see that the map ':Cl0 p;q!Clq;p1dened on generators by EiEj7!( ei ifj=p+q eiejotherwise is an isomorphism. Corollary 2.1. Cl0 p;q=Cl0 q;p. Proof. One ofp;qmust be non-zero ( Cl0;0is not a very interesting algebra). Without loss of generality we can assume that p6= 0. Then from the previous theorem we have that Cl0 p;q=Clq;p1. But we also have from the above theorem that Clq;p1=Cl0 q;p. Thus we have that Cl0 p;q=Cl0 q;p. Though we now know that Clp;q'Clq;p, we construct an isomorphism from Cl0 p;qtoCl0 q;pby composing the isomorphism from Cl0 p;qtoClp;q1with the isomorphism from Clp;q1toCl0 q;p. We do this to then show that it the map restricts to an isomorphism of the Spin groups. Letfeigbe standard generators for Clp;qandfEigbe generators for Clq;p whereE2 i=e2 i. Also letgand gbe the metrics on Clp;qandClq;p, respec- tively. By composing the type of maps constructed in the previous corollary, we get an isomorphism :Cl0 p;q!Cl0 q;p, eiej7!EiEj (5) Note that (5) holds even when i=jsincee2 i=E2 i. Denote the restriction oftoSpin (p;q) byjs. To see that jsmaps intoSpin (q;p), it is sucient to check that a product of two unit vectors maps to a product of two unit vectors, for then the fact that jsis a group homomorphism (since is an algebra homomorphism) will establish that the image of a product of any even amount of unit vectors in Spin (p;q) is a product of an even amount of unit vectors in Spin (q;p). Thus let u=P iuieiandv=P ivieibe unit 10 vectors in Rp;q. Then (uv) = X iuiei! X jviei!! = X i;juivjeiej! =X i;juivjEiEj = X iuiEi! X jvjEj! is a product of two unit vectors since  g(P iuiEi;P iuiEi) =g(u;u) = 1 and gP jvje0 j;P jvje0 j; =g(v;v) =1. Besides showing that js maps intoSpin (q;p), the above calculation makes it clear that jsis sur- jective. Finally, jsis injective since is injective. This establishes that Spin (p;q)=Spin (q;p). One may think that since Spin (p;q)=Spin (q;p), a similar type of argument can establish that Pin(p;q)=Pin(q;p). However, in the next section we show that this is not generally true. 2.2Clp;qandClq;pandPin(p;q)andPin(q;p) We begin by proving the only armative case for Clp;q'Clq;p. Theorem 2.1. Ifqp0(mod 4) then Clp;q=Clq;p. Proof. Letfeigbe the standard generators for Clp;qand letfEigbe gener- ators forClq;psuch thatE2 i=e2 i. Since 4j(qp),qpis even so that p+q=qp+2pis even. Therefore 4 j(p+q)2so that 4j((p+q)2+qp)) (p+q)2+qp 2is even. Thus, by (3) we see that =Q iEisquares to the identity. We claim that the map :Clp;q!Clq;pgiven on generators by (ei) = Ei extends to an isomorphism. To see that is a bijection, we note that since  is a linear map between two nite dimensional vector spaces with the same 11 dimension, it is a bijection if and only if it is a surjection. Indeed, the map is surjective since for any generator EiofClq;p,im() contains Y j6=i Ej= p+q1Y j6=iEj = p+q2 Y j6=iEj = Y j6=iEj =Ei where the third equality follows from the (crucial) fact that p+qis even. Finally, to show that is an isomorphism it is sucient, by the universal property, to show that the elements f Eigobey the same Cliord relations asfeig, i.e. they anti-commute with each other and pof them square to the identity and qof them square to negative the identity. Since anti- commutes with each Eiby (4), fori6=jwe have that ( Ei)( Ej) = EjEi = ( Ej)( Ei). Lastly, note that ( Ei)2= Ei Ei= 2E2 i=E2 i=e2 i. Corollary 2.2. Ifqp0(mod 4) then Pin(p;q)=Pin(q;p), an isomor- phism being the restriction of (as above) to Pin(p;q). Proof. As before, letfeigandfEigbe generators for Clp;qandClq;prespec- tively. Let gand gbe the metrics on Clp;qandClq;p, respectively, and let jpbe the restriction of toPin(p;q). SincePin(p;q) is made up of unit vectors, we need only check that jpmaps unit vectors to products of unit vectors and that any unit vector in Clq;pis inim(p). Letv=P ivieibe a unit vector in Rp;q, i.e.g(v;v) =1. Then we have that jp(v) =X ivi Ei= X iviEi! : But is obviously a product of unit vectors andP ivieiis a unit vector since g X iviEi;X iviEi! =g(v;v) =1: 12 Thus(v) is a product of unit vectors and so is in Pin(q;p). Finally, to see that any unit vector in Pin(q;p) is inim(p), we rst consider jp( 0) where 0=Qp+q i=1ei. Since andEianti-commute by (4) and 2= 1 by (3) we have jp( 0) =p+qY i=1 Ei = (1)(p+q)=2 p+qp+qY i=1Ei = (1)(p+q)=2p+qY i=1Ei = where the third equality follows from the fact that p+q= 1 sincep+q= pq+ 2qis even. Now let u=P iuiEibe a unit vector in Pin(q;p). Then u0=P iuiei2Pin(p;q) (sinceg=g) and jp( 0u0) =jp( 0)(u0) = X iui Ei = 2X iuiEi =u: Theorem 2.2. Ifpq1or 3 (mod 4) then Clp;q6=Clq;pandPin(p;q)6= Pin(q;p). Proof. Note that we need only prove this when pq1 (mod 4). For say it holds for pq1 (mod 4) and we have that pq3 (mod 4). Then qp1 (mod 4) so that Clq;p6=Clp;q. Letpq1 (mod 4). Then qp3 (mod 4). We also have that p+q=pq+ 2qis odd since pqis odd. Thus p+qis congruent to 13 either 1 or 3 modulo 4. In either case, we have that ( p+q)21 (mod 4) so that (p+q)2+qp0 (mod 4). Thus from (3) we have that 2= 1, where is the pseudoscalar in Clp;q. However, ( p+q)2+pq2 (mod 4) so that the pseudoscalar 0inClq;psquares to -1. Since p+qis odd, we see from (4) that 2Z(Clp;q) and  2Z(Clq;p). Additionally, it is not hard to see thatZ(Clp;q) =fa+b :a;b2RgandZ(Clq;p) =fa+b 0:a;b2Rg However, these centers are certainly not isomorphic since the latter contains an element ( 0) which squares to -1 but the former does not. Since the cen- ters are not isomorphic, neither are the algebras. The same argument works to show that Pin(p;q)6=Pin(q;p) sinceZ(Pin(p;q)) =f ; ;1;1gand Z(Pin(q;p)) =f 0; 0;1;1g. For the last case, pq2 (mod 4), there seems to be no natural argument and we are forced to appeal to representations of Cliord algebras. By Bott periodicity, we can know the type ( R;C, orH) of a Cliord algebra by know- ing the types of Clp;qfor 0p;q8. Such tables can be found throughout the literature [3] and [2]. There we see that the types for Clp;qandClq;pare indeed dierent if pq2 (mod 4). The question now becomes how do we know, for example, that R[4]6'H[2]? After all, they are isomorphic as (real) vector spaces. To show that they are not isomorphic as algebras, we consider minimal left ideals. Since the product of a matrix with a rank one matrix has rank at most one, any minimal left ideal of F[n] is generated by one rank one matrix. Let I be the ideal generated by the rank one matrix Mand letvbe a non-zero column vector of M. We claim that the map AM7!Avgives a vector space isomorphism from ItoFn. Clearly the map is linear and it is surjective since for any non-zero vector w2Fn, there exists a matrix Asuch thatAv=w. To see that it is injective, suppose that Av= 0. Since Mhas rank one, all column vectors are multiples of v. ThusAmaps each column vector to 0, so thatAM = 0. Since Iis isomorphic to Fn, any minimal left ideal has real dimensionndimRF. Now suppose that F1[n] and F2[m] have the same real dimension. Then n2dimRF1=m2dimRF2. If they are isomorphic then their minimal left ide- als must also be isomorphic and, in particular, must have the same dimension. Therefore we must also have that ndimRF1=mdimRF2. Dividing these two equations gives n=mwhich further implies that dim RF1= dim RF2. Since 14 Fiis either R;C, orH, we must have that F1=F2. This shows that for pq2 (mod 4), Clp;q6'Clq;p, but what about the Pingroups? Since Clp;qis isomorphic to a matrix algebra, Pin(p;q) must be isomorphic to a group that is a subset of the matrix algebra. Since Pin(p;q) contains all of the generators for the algebra, if it were isomorphic to Pin(q;p) then we could represent Clq;pandClp;qon the same space. However, since Clq;pandClp;qhave the same dimensions, it would then follow that they are isomorphic. 3 Bilinear Forms on Spinors Bilinear forms on spinors are discussed in [2] but from a dierent perspective. Our approach is to look for bilinear forms on the space of spinors, S, such that vectors are self-adjoint, up to sign. That is, a bilinear function (
;
) : SS!Rsuch that (;v ) =(v; ) (6) for allv2Rp;q,; 2S. The form can be represented as ( ; )7!yA , whereA2Clp;q. The condition (6) then becomes Av=vyA for allv2Rp;q. If 1iptheney i=eiand ifp+ 1ip+qthen ey i=ei. Therefore we must have that Aei=( eiAif 1ip eiAifp+ 1ip+q: Put A=X If1;2;:::;p+qgAIeI: Sinceeieither commutes or anti-commutes with each eI, ifAI6= 0 then we must have that eIei=( eieIif 1ip eieIifp+ 1ip+q 15 It follows that if AI6= 0 theneImust be either e1e2:::e porep+1ep+2:::e p+q. ThusAis, up to scale, either e1e2:::e porep+1ep+2:::e p+q. We denote the former element by pand the latter by q. We dene two real bilinear forms (; )+=Re(y p ) (; )=Re(y q ): We see that 2 p= (e1e2:::e p)(e1e2:::e p) = (1)(p1)+(p2)+:::+1e2 1e2 2:::e2 p= (1)p(p1)=2 and, similarly, 2 q= (1)q(q1)=2(1)q= (1)q(q+1)=2: Thus (
;
)+is symmetric if p= 0 or 1 (mod 4) and anti-symmetric if p= 2 or 3 (mod 4) and (
;
)is symmetric if q= 0 or 3 (mod 4) and anti-symmetric ifp= 1 or 2 (mod 4). Given any vector v2Rp;q, we can put v=v++v, withv+2spanfe1;e2;:::e pg andv2spanfep+1;ep+2;:::;e p+qg. We then have (;v )+=y p(v++v) =y((1)p+1v++ (1)pv) p =y((1)p+1vy ++ (1)p+1vy ) p = (1)p+1(v; )+: (7) A similar calculation yields (;v )= (1)q(v; ): (8) Recall that the action of the Pin and Spin groups on vectors preserves the metric. We also see that the action of the pin and spin groups on spinors (which is just left multiplication) preserves (
;
)up to sign: (u;u )=(; ); for allu2Pin(p;q);; 2S. This is evident from (7), (8), and the fact that foru2Pin(p;q);u~u=1. We can dene a subgroup Pin +(p;q) of Pin(p;q) by Pin +(p;q) =fu2Pin(p;q) :u~u= 1g: Then we see that the action of Pin +(p;q) on spinors preserves (
;
)+ifpis odd and preserves (
;
)ifqis even. Furthermore, Spin +(p;q) =Pin +(p;q)\ Spin (p;q) always preserves (
;
). 16 3.1 Relations Between Spinors and Forms Denote the exterior algebra on Rp;qby (Rp;q). The metric gonRp;qinduces a metric on ( Rp;q) by g(ei1ei2:::e im;ej1ej2:::e jn) =Y ik=jlg(eik;ejl): We can use the bilinear forms to associate an element vin the dual space of (Rp;q) with spinors ; by v(u) = (;u );u2(Rp;q): Since the metric provides an identication of forms with dual forms, we can associate two spinors with an element of ( Rp;q). Denote the k-form associated to the spinors ; using (
;
)byvk . Under this identication, the component of vk alongei1^ei2^:::^eikis (;ei1ei2:::e ik )and g(vk ;vk ) =X 1i1<:::<i kp+qi1;:::;ik(;ei1:::e ik )2 where i1;:::;ik=g(ei1:::e ik;ei1:::e ik) =e2 i1:::e2 ik: 3.2 Identities Upon playing around with these constructions in certain dimensions, we found some identities relating the norms of various forms associated with spinors. We then used the scripting language Python and Mathematica to search for similar identities in general. The identities resemble some of the Fierz identities, as seen in the context of Cliord algebras, for example, in [2]. We use a Monte-Carlo method whereby we created random spinors and saw if there were any identities which held in those special cases. We then checked those to see if they held in general. We rst give the cases when the Cliord algebra is isomorphic to a real matrix algebra. 17 3.2.1 Real Cliord algebras We rst consider the real corner algebras (non-universal Cliord algebras where =1) up to dimension 13, from which we can derive identities in the subordinate algebras. Recall that for corner algebras p= qso that there is only one bilinear form, which we denote by (
;
). We denote the kform which acts on a kform u as (;u ) asvk. Cl3;2: 2g(v;v) +g(v2;v2) = 0 (; )2+g(v;v) = 0 Cl4;3andCl0;7: 3g(v;v) +g(v2;v2) = 0 7(; )2+ 4g(v;v)g(v3;v3) = 0 7(;)( ; ) + 3g(v;v) +g(v3;v3) = 0 Cl9;0,Cl5;4, andCl1;8: 28g(v;v) + 7g(v2;v2)3g(v3;v3)2g(v4;v4) = 0 6(; )2+ 6g(v;v) +g(v2;v2)g(v3;v3) = 0 24(;)( ; ) + 24g(v;v) + 7g(v2;v2)g(v3;v3) = 0 Cl10;1,Cl6;5, andCl2;9: 5(; )2+ 5g(v;v) +g(v2;v2)g(v3;v3) = 0 75g(v;v) + 21g(v2;v2)16g(v3;v3)5g(v5;v5) = 0 15g(v;v) + 3g(v2;v2)2g(v3;v3)g(v4;v4) = 0 18 Cl11;2,Cl7;6, andCl3;10: 30g(v;v) + 5g(v2;v2)5g(v3;v3)2g(v4;v4) = 0 60g(v;v) + 21g(v2;v2)21g(v3;v3)4g(v5;v5) = 0 32g(v;v) + 9g(v2;v2)7g(v3;v3) +g(v6;v6) = 0 4(; )2+ 4g(v;v) +g(v2;v2)g(v3;v3) = 0 From the identities on corner algebras, we can get identities on the subor- dinate algebras. Consider the corner algebra Clp;q, generated byfeij1 ip+qg. To pass to Clp;q1we use generators fEij1ip+q1g whereEi=ei. Letting be the pseudoscalar in Clp;q1, we have that =ep+q. Denote by gthe metric on Clp;q, gthe metric on Clp;q1,vkthe form (; )2k(Rp;q), andvk the form (;() )2k(Rp;q1). Note that (
;
)+= (
;
) since the pof the two algebras are the same. Dene i1;:::;ik=g(ei1:::e ik;ei1:::e ik) =e2 i1:::e2 ik i1;:::;im= g(Ei1:::E im;Ei1:::E im) =E2 i1:::E2 im: Then we have g(vk;vk) =X 1i1<:::<i kp+qi1;:::;ik(;ei1:::e ik )2 =X 1i1<:::<i kp+q1i1;:::;ik(;E i1:::E ik )2 ++X 1i1<:::<i k1p+q1i1;:::;ik1;p+q(; E i1:::E ik1)2 + =X 1i1<:::<i kp+q1i1;:::;ik(;E i1:::E ik )2 +X 1i1<:::<i k1p+q1i1;:::;ik1(;E i1:::E ik1)2 = g(vk +;vk +)g(vk1 ;vk1 ): (9) A similar argument yields the following equation for passing from Clp;qto Clp1;q g(vk;vk) = g(vk ;vk ) + g(vk1 +;vk1 +): (10) Using (9) and (10) we get the following identities 19 Cl3;1: 2(; )2 + 2g(v+;v+)g(v;v) +g(v2 +;v2 +) = 0 (; )2 +(; )2 +g(v+;v+) = 0: Cl2;2: 2(; )2 ++g(v+;v+) + 2g(v;v) +g(v2 ;v2 ) = 0 (; )2 +(; )2 +g(v;v) = 0: Cl4;2andCl0;6: 3(; )2 + 3g(v+;v+)g(v;v) +g(v2 +;v2 +) = 0 7(; )2 +4(; )2 + 4g(v+;v+) +g(v2 ;v2 )g(v3 +;v3 +) = 0 7(;)+( ; )+3(; )2 + 3g(v+;v+)g(v2 ;v2 ) +g(v3 +;v3 +) = 0: Cl3;3: 3(; )2 ++g(v+;v+) + 3g(v;v) +g(v2 ;v2 ) = 0 4(; )2 +7(; )2 + 4g(v;v)g(v2 +;v2 +)g(v3 ;v3 ) = 0 3(; )2 ++ 7(;)( ; )+ 3g(v;v) +g(v2 +;v2 +) +g(v3 ;v3 ) = 0: 20 Cl5;3andCl1;7: 28(; )2 + 28g(v+;v+)7g(v;v) + 7g(v2 +;v2 +) + 3g(v2 ;v2 ) 3g(v3 +;v3 +) + 2g(v3 ;v3 ) + 2g(v4 +;v4 +) = 0 6(; )2 +6(; )2 + 6g(v+;v+)g(v;v) +g(v2 +;v2 +) +g(v2 ;v2 )g(v3 +;v3 +) = 0 24(;)+( ; )+24(; )2 + 24g(v+;v+)7g(v;v)+ 7g(v2 +;v2 +) +g(v2 ;v2 )g(v3 +;v3 +) = 0: Cl8;0,Cl4;4, andCl0;8: 28(; )2 ++ 7g(v+;v+) + 28g(v;v)3g(v2 +;v2 +) + 7g(v2 ;v2 ) 2g(v3 +;v3 +)3g(v3 ;v3 )2g(v4 ;v4 ) = 0 6(; )2 +6(; )2 +g(v+;v+) + 6g(v;v)g(v2 +;v2 +) +g(v2 ;v2 )g(v3 ;v3 ) = 0 24(;)( ; )+ 24(; )2 ++ 7g(v+;v+) + 24g(v;v) g(v2 +;v2 +) + 7g(v2 ;v2 )g(v3 ;v3 ) = 0: Cl10;0,Cl6;4, andCl2;8: 5(; )2 +5(; )2 + 5g(v+;v+)g(v;v) +g(v2 +;v2 +) +g(v2 ;v2 )g(v3 ;v3 ) = 0 75(; )2 + 75g(v+;v+)21g(v;v) + 21g(v2 +;v2 +) + 16g(v2 ;v2 ) 16g(v3 +;v3 +) + 5g(v4 ;v4 )5g(v5 +;v5 +) = 0 15(; )2 + 15g(v+;v+)3g(v;v) + 3g(v2 +;v2 +) + 2g(v2 ;v2 ) 2g(v3 +;v3 +) +g(v3 ;v3 )g(v4 +;v4 +) = 0: 21 Cl9;1,Cl5;5, andCl1;9: 5(; )2 +5(; )2 +g(v+;v+) + 5g(v;v)g(v2 +;v2 +) +g(v2 ;v2 )g(v3 ;v3 ) = 0 75(; )2 ++ 21g(v+;v+) + 75g(v;v)16g(v2 +;v2 +) + 21g(v2 ;v2 ) 16g(v3 ;v3 )5g(v4 +;v4 +)5g(v5 ;v5 ) = 0 15(; )2 ++ 3g(v+;v+) + 15g(v;v)2g(v2 +;v2 +) + 3g(v2 ;v2 ) g(v3 +;v3 +)2g(v3 ;v3 )g(v4 ;v4 ) = 0: Cl11;1,Cl7;5, andCl3;9: 30(; )2 + 30g(v+;v+)5g(v;v) + 5g(v2 +;v2 +) + 5g(v2 ;v2 ) 5g(v3 +;v3 +) + 2g(v3 ;v3 )2g(v4 +;v4 +) = 0 60(; )2 + 60g(v+;v+)21g(v;v) + 21g(v2 +;v2 +) + 21g(v2 ;v2 ) 21g(v3 +;v3 +) + 4g(v4 ;v4 )4g(v5 +;v+5) = 0 32(; )2 + 32g(v+;v+)9g(v;v) + 9g(v2 +;v2 +) + 7g(v2 ;v2 ) 7g(v3 +;v3 +)g(v5 ;v5 ) +g(v6 +;v6 +) = 0 4(; )2 +4(; )2 + 4g(v+;v+)g(v;v) +g(v2 +;v2 +) +g(v2 ;v2 )g(v3 +;v3 +) = 0: 22 Cl10;2,Cl6;6, andCl2;10: 30(; )2 ++ 5g(v+;v+) + 30g(v;v)5g(v2 +;v2 +) + 5g(v2 ;v2 ) 2g(v3 +;v3 +)5g(v3 ;v3 )2g(v4 ;v4 ) = 0 60(; )2 ++ 21g(v+;v+) + 60g(v;v)21g(v2 +;v2 +) + 21g(v2 ;v2 ) 21g(v3 ;v3 )4g(v4 +;v4 +)4g(v5 ;v5 ) = 0 32(; )2 ++ 9g(v+;v+) + 32g(v;v)7g(v2 +;v2 +) + 9g(v2 ;v2 ) 7g(v3 ;v3 ) +g(v5 +;v5 +) +g(v6 ;v6 ) = 0 4(; )2 +4(; )2 +g(v+;v+) + 4g(v;v)g(v2 +;v2 +) +g(v2 ;v2 )g(v3 ;v3 ) = 0: Since (
;
)+is symmetric if p= 0 or 1 (mod 4) and anti-symmetric otherwise, we have that (; )+= (1)p(p+3)=2( ;)+: Similarly, since (
;
)is symmetric if q= 0 or 3 (mod 4) and anti-symmetric otherwise, we can write (; )= (1)q(q+1)=2( ;): Combining this with (7) and (8), we can nd the conditions under which vk 6= 0 when= (fork0): (;ei1:::e ik)+= (1)k(p+1)(eik:::e i1;)+ = (1)k(p+1)+ k(k1)=2(ei1:::e ik;)+ = (1)k(p+1)+ k(k1)=2+p(p+3)=2(;ei1:::e ik)+ Thus forvk +to be nonzero it is necessary to have k(p+ 1) +k(k1)=2 + p(p+ 3)=20 (mod 2). This is equivalent to (k+p)2p+k0 (mod 4): A similar calculation shows that for vk to be nonzero we need (k+q)2+qk0 (mod 4): Therefore many of the terms in the above identities vanish when we specialize to the case where = . The simplied identities for corner algebras are: 23 Cl3;2: g(v2;v2) = 0 Cl4;3andCl0;7: 7(;)2+g(v3;v3) = 0 Cl9;0,Cl5;4, andCl1;8: 14g(v;v)g(v4;v4) = 0 (; )2+g(v;v) = 0 Cl10;1,Cl6;5, andCl2;9: 5g(v;v) +g(v2;v2) = 0 75g(v;v) + 21g(v2;v2)5g(v5;v5) = 0 Cl11;2,Cl7;6, andCl3;10: g(v2;v2)g(v3;v3) = 0 9g(v2;v2)7g(v3;v3) +g(v6;v6) = 0 and for subordinate algebras are: Cl3;1: g(v;v)g(v2 +;v2 +) = 0 24 Cl2;2: g(v+;v+) +g(v2 ;v2 ) = 0 Cl4;2andCl0;6: 7(;)2 ++g(v2 ;v2 )g(v3 +;v3 +) = 0 Cl3;3: 7(;)2 +g(v2 +;v2 +) +g(v3 ;v3 ) = 0 Cl5;3andCl1;7: 28(;)2 + 14g(v+;v+) +g(v3 ;v3 ) +g(v4 +;v4 +) = 0 (;)2 ++ (;)2 g(v+;v+) = 0 Cl8;0,Cl4;4, andCl0;8: 14(;)2 ++ 14g(v;v)g(v3 +;v3 +)g(v4 ;v4 ) = 0 (;)2 +(;)2 +g(v;v) = 0 Cl10;0,Cl6;4, andCl2;8: 5(;)2 + 5g(v+;v+)g(v;v) +g(v2 +;v2 +) = 0 75(;)2 + 75g(v+;v+)21g(v;v) + 21g(v2 +;v2 +) + 5g(v4 ;v4 )5g(v5 +;v5 +) = 0 Cl9;1,Cl5;5, andCl1;9: 5(;)2 ++g(v+;v+) + 5g(v;v) +g(v2 ;v2 ) = 0 75(;)2 ++ 21g(v+;v+) + 75g(v;v) + 21g(v2 ;v2 )5g(v4 +;v4 +)5g(v5 ;v5 ) = 0 25 Cl11;1,Cl7;5, andCl3;9: g(v;v)g(v2 +;v2 +)g(v2 ;v2 ) +g(v3 +;v3 +) = 0 9g(v;v) + 9g(v2 +;v2 +) + 7g(v2 ;v2 )7g(v3 +;v3 +)g(v5 ;v5 ) +g(v6 +;v6 +) = 0 Cl10;2,Cl6;6, andCl2;10: g(v+;v+)g(v2 +;v2 +) +g(v2 ;v2 )g(v3 ;v3 ) = 0 9g(v+;v+)7g(v2 +;v2 +) + 9g(v2 ;v2 )7g(v3 ;v3 ) +g(v5 +;v5 +) +g(v6 ;v6 ) = 0: 3.2.2 Quaternionic Cliord algebras We consider rst the quaternionic corner algebras, from which we can deduce identities in the subordinate algebras using the same argument as the previ- ous section. Recall that if we have a representation of Clp;qonHn, we can get a representation on Rn. Given a type Hcorner algebra, Clp;q, we can nd generators for it using p+q1 many generators from a type Rcorner algebra in dimensions p+q+2. By taking products of the three unused we generators, we give a quaternionic structure on the spinors of Clp;q, i.e. three maps I;J;K such thatI2=J2=K2=IJK =1 by which we can dene an action of H on the space of spinors by ( q0+q1I+q2J+q3K)=q0+q1I+q2J+q3K. Furthermore, these maps commute with Clp;q. Lastly, this process is such that ifE1;:::;E p+qare generators for Clp;qthen there is one of them, Ei, such thatE1;:::;E p+q;iEi;jE i;kE igenerate the type Ralgebra. In this section eiandEiwill be the generators, (
;
) and<
;
>the scalar products, and gand gthe metrics for type RandHalgebras, respectively. We denote the m-forms corresponding to ;I ,;J , and;K byivm , jvm , andkvm . Cl0;3viaCl3;2:E1=e4;E2=e5;andE3=e4e5. Note that since e1e2e3e4e5= 1,E3is equal, up to sign, to e1e2e3. Quaternionic structure is given by I=e1e2,J=e3e1, andK=e2e3. We see that e1=KE 3;e2=JE3;e3= IE3;e4=E1;e5=E2. We see that (where everything is modulo sign) (; ) =te4e5 =tE3 =<;E 3 >: 26 Therefore g(v;v) =3X i=1(;ei )25X i=4(;ei )2 =<;I >2+<;J >2+<;K >2<;E 1 >2<;E 2 >2: Hence the identity (; )2+g(v;v) = 0 becomes <;I >2+<;J >2+<;K >2+g(v;v) = 0: We have g(v2;v2) =X 1i<j3(;eiej )2+ (;e4e5 )X 1i3X 4j5(;eiej )2 =<;IE 3 >2+<;JE 3 >2+<;KE 3 >2+<; >2 2X i=1(<;IE i >2+<;JE i >2+<;KE i >2) So by theCl3;2identity 2g(v;v) +g(v2;v2) = 0, we have that 2<;I >2+2<;J >2+2<;K >22<;E 1 >22<;E 2 >2 +<;IE 3 >2+<;JE 3 >2+<;KE 3 >2+<; >2 2X i=1(<;IE i >2+<;JE i >2+<;KE i >2) = 0 By symmetry, the above equation must be valid if we permute Ei;Ej;Ek. Thus we get three equations which, when added together, give 3<; >2+6<;I >2+6<;J >2+6<;K >2+ 4g(v;v) + g(iv;iv) + g(jv;jv) + g(kv;kv) = 0 If we go down to Cl0;2then< ; > +=< ; > and< ; >=< ;E 3 >. Hence g(v;v) =<;I >2+<;J >2+<;K >2g(v;v) g(v2;v2) =<;I >2 +<;J >2 +<;K >2 +<; >2 + g(iv;iv)g(jv;jv)g(kv;kv): 27 Cl1;4viaCl4;3:E1=e4;E2=e5;E3=e6;E4=e7;E5=e4e5e6e7;I= e1e2;J=e3e1;K=e2e3. Hencee1=KE 5;e2=JE5;e3=IE5;e4= E1;e5=E2;e6=E3;e7=E4. ForCl1;3we takeE1;E2;E3;E4andE5becomes . Then (; ) =te5e6e7 = tE2E3E4 =<; >. We have g(v;v) =4X i=1(;ei )27X i=5(;ei ) =<;K >2 +<;J >2 +<;I >2 +<;E 1 >4X i=2<;E i > =<;I >2 ++<;J >2 ++<;K >2 ++g(v;v) and g(v2;v2) =X 1i<j4(;eiej )2+X 5i<j7(;eiej )24X i=17X j=5(;eiej )2 =<;I >2 +<;J >2 +<;K >2 +<; IE 1 >2 +<; JE 1 >2 +<; KE 1 >2 +X 2i<j4<;E iEj >2 4X i=2(<; IE i >2 +<; JE i >2 +<; KE i >2 +<;E 1Ei >2 ) =<;I >2 +<;J >2 +<;K >2 +g(iv+;iv+) + g(jv+;jv+) + g(kv+;kv+) + g(v2 ;v2 ) and g(v3;v3) =<; >2 ++g(iv;iv) + g(jv;jv) + g(kv;kv) +g(iv2 +;iv2 +) + g(jv2 +;jv2 +) + g(kv2 +;kv2 +)g(v+;v+) 4 Conclusion In the area of bilinear forms on spinors, there is still much work to be done. This includes identifying a pattern with the identities, studying the rela- tionships between algebras of dierent type, and considering more general expressions, like g(v;w) wherev(u) = (;u )andw(u) = (;u ). 28 5 Acknowledgements I am greatly indebted to my advisor Dr. George Sparling who has spent countless hours with me. Additionally, I am grateful for the support of the University of Pittsburgh Honors College through two fellowships, during which most of this work was done: the Chancellor's Undergraduate Research Fellowship in Spring 2008 and the Brackenridge Fellowship in Summer 2008. References [1] Marie-Louise Michelsohn H. Blaine Lawson. Spin Geometry . Princeton, 1990. [2] Pertti Lounesto. Cliord Algebras and Spinors . Cambridge, 2001. [3] Ian Porteous. Cliord Algebras and the Classical Groups . Cambridge, 1995. 29