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APPENDIX a Tensors, Filtrations, and Gradings

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APPENDIX A Tensors, Filtrations, and Gradings Abstract. If Eisa vector space, the tensor algebra T(E) of E is the direct sum over n :::: 0 of the n-fold tensor product of E with itself. This is an associative algebra with the following universal mapping property: Any linear mapping of E into an associative algebra A with identity extends to an algebra homomorphism of T(E) into A carrying 1 into 1. The symmetric algebra S(E) is a quotient of E with the following universal mapping property: Any linear mapping of E into a commutative associative algebra A with identity extends to an algebra homomorphism of S(E) into A carrying 1 into 1. The symmetric algebra is commutative. Similarly the exterior algebra 1\(E) is a quotient of E with this universal mapping property: Any linear mapping I of E into an associative algebra A with identity such that l(v)2 = 0 for all v e E extends to an algebra homomorphism of A<E) into A carrying 1 into 1. The tensor algebra, the symmetric algebra, and the exterior algebra are all examples of graded associative algebras. A more general notion than a graded algebra is that of a filtered algebra. A filtered associative algebra has an associated graded algebra. The notions of gradings and filtrations make sense in the context of vector spaces, and a linear map between filtered vector spaces that respects the filtration induces an associated graded map between the associated graded vector spaces. If the associated graded map is an isomorphism, then the original map is an isomorphism. 1. Tensor Algebra Just as polynomial rings are often used in the construction of more general commutative rings, so tensor algebras are often used in the construction of more general rings that may not be commutative. In this section we construct the tensor algebra of a vector space as a direct sum of iterated tensor products of the vector space with itself, and we establish its properties. We shall proceed with care, in order to provide a complete proof of the associativity of the multiplication. Fix a field k. Let E and F be vector spaces over the field k. A tensor product V of E and F is a pair (V, t) consisting of a vector space V over lk: together with a bilinear map t : E x F ~ V, with the following 487 488 A. Tensors, Filtrations, and Gradings universal mapping property: Whenever b is a bilinear mapping of E x F into a vector space U over k, then there exists a unique linear mapping B of V into U such that the diagram V (=tensor product) (A.l) -,_ B ExF-----~u b commutes. We call B the linear extension of b to the tensor product. It is well known that a tensor product of E and F exists and is unique up to canonical isomorphism, and we shall not repeat the proof. One feature of the proof is that it gives an explicit construction of a vector space that has the required property. A tensor product of E and F is denoted E ®k F, and the associated bilinear map tis written (e, f) ~-+> e ®f. The elements e ® f generate E ®k F, as a consequence of a second feature of the proof of existence of a tensor product. There is a canonical isomorphism (A2) E®kF~F®kE given by taking the linear extension of (e, f)~-+> f ® e as the map from left to right. The linear extension of(/, e) ~-+> e ® f gives a two-sided inverse. Another canonical isomorphism of interest is (A.3) Here the map from left to right is the linear extension of (e, c) ~-+> ce, while the map from right to left is e ~-+> e ® 1. In view of (A.2) we have lk®k E ~ E also. Tensor product distributes over direct sums, even infinite direct sums: (A.4) E ®k ($Fa)~ $<E ®k Fa). a a The map from left to right is the linear extension of the bilinear map (e, L fa)~--+ L (e ®fa). To define the inverse, we have only to define it on each E®kFa. where it is thelinearextensionof(e, fa)~-+> e®(ia<fa)); here ia : Fa ___. ffi Fp is the injection corresponding to a. It follows from 1. Tensor Algebra 489 (A.3) and (A.4) that if {x;} is a basis of E and {yj} is a basis ofF, then {x; ® Yj} is a basis of E ®k F. Consequently (A.5) dim(£ ®k F) = (dim E)( dim F). Let Homk (E, F) be the vector space of lk linear maps from E into F. One special case is E = lk, and we have (A.6) The map from left to right sends ffJ into ffJ(l), while the map from right to left sends f into ffJ with ffJ(c) = cf. Another special case of interest occurs when F = lk. Then Hom(£, lk) = E* is just the vector space dual of E. We can use ®k to construct new linear mappings. Let £ 1, F1, £ 2 and F2 be vector spaces, Suppose that L 1 is in Homk(E~o FI) and L 2 is in Homk(£2, F2). Then we can define (A.7) L1 ® L2 in Homk(EI ®k £2, F1 ®k F2) as follows: The map (e~o e2) ~---+ L1 (e1) ® L 2(e2) is bilinear from £ 1 x E2 into F1®k F2, and we let L 1® L2 be its linear extension to E 1 ®k £ 2. The uniqueness in the universal mapping property allows us to conclude that (A.8) when the domains and ranges match in the obvious way. Let A, B, and C be vector spaces over lk. A triple tensor product V = A ®k B ®k C is a vector space over k with a trilinear map t : A x B x C ~ V having the following universal mapping property: Whenever t is a trilinear mapping of A x B x C into a vector space U over k, then there exists a linear linear mapping T of V into U such that the diagram V (=triple tensor product) (A.9) ',, T A X B X c -------+ u t commutes. It is clear that there is at most one triple tensor product up to canonical isomorphism, and one can give an explicit construction just as for ordinary tensor products E ®k F. We shall use triple tensor products to establish an associativity formula for ordinary tensor products. 490 A. Tenso.rs, Filtmtions, and Gradings Proposition A.lO. (a) (A ®k B) ®k C and A ®k (B ®k C) are triple tensor products. (b) There exists a unique isomorphism <I> from left to right in (A.ll) (A ®k B) ®k C ~ A ®k (B ®k C) such that <l>((a ®b)® c)= a® (b ®c) for all a e A, be B, and c e C. PROOF. (a) Consider (A ®k B) ®k C. Lett: Ax B x C ~ U be trilinear. For c e C, define tc : A x B ~ U by tc(a, b) = t(a, b, c). Then tc is bilinear and hence extends to a linear Tc : A ®k B ~ U. Since t is trilinear, tc1+<'2 = fc 1 + tc2 and fxc = xtc for scalar x; thus uniqueness of the linear extension forces Tc 1+<'2 = Tc1 + T<'l and Txc = xTc. Consequently t' : (A ®k B) X c ~ u given by t'(d, c) = Tc(d) is bilinear and hence extends to a linear T : (A ®k B) ®k C ~ U. This T proves existence of the linear extension of the given t. Uniqueness is trivial, since the elements (a ® b) ® c generate (A ®k B) ®k C. So (A ®k B) ®k C is a triple tensor product. In a similar fashion, A ®k (B ®k C) is a triple tensor product. (b) In (A.9) take V = (A ®k B) ®k C, U = A ®k (B ®k C), and t(a, b, c) = a ® (b ® c). We have just seen in (a) that V is a triple tensor product with t(a, b, c) = (a® b)® c. Thus there exists a linear T : V ~ U with Tt(a, b, c) = t(a, b, c). This equation means that T((a ®b)® c) =a® (b ®c). Interchanging the roles of (A ®k B) ®k C and A ®k (B ®k C), we obtain a two-sided inverse for T. Thus T will serve as <I> in (b), and existence is proved. Uniqueness is trivial, since the elements (a® b)® c generate (A ®k B) ®k C. When this proposition is used, it is often necessary to know that the isomorphism <I> is compatible with maps A ~ A', B ~ B', and C ~ C'. This property is called naturality in the variables A, B, and C, and we make it precise in the next proposition. Proposition A.12. Let A, B, C, A', B', and C' be vector spaces over k, and let LA: A~ A', Ls: B ~ B', and Lc: C ~ C' be linear maps. Then the isomorphism <I> of Proposition A.lOb is natural in the sense that the diagram A®k (B ®kC) 4> (A' ®k B') ®k C' ----+ A' ®k (B' ®k C') commutes. 1. Tensor Algebra 491 PRooF. We have ((LA® (Ls ® Le)) o cl>)((a ®b)® c) =(LA® (Ls ® Le))(a ® (b ®c)) = LAa ® (Ls ® Le)(b ®c) = ci>((LAa ® Lsb) ®Lee) = ci>((LA ® Ls)(a ®b)® Lee) =(<I> o ((LA® Ls) ® Le))((a ®b)® c), and the proposition follows.
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