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(Multi)Linear Algebra 1 1.1. Tensor Products 1 1.2. the Grassmann (Exterior) Algebra and Alternating Maps 7 1.3

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(Multi)Linear Algebra 1 1.1. Tensor Products 1 1.2. the Grassmann (Exterior) Algebra and Alternating Maps 7 1.3 MULTILINEAR ALGEBRA NOTES EUGENE LERMAN Contents 1. (Multi)linear algebra 1 1.1. Tensor products 1 1.2. The Grassmann (exterior) algebra and alternating maps 7 1.3. Pairings 9 1. (Multi)linear algebra The goal of this note is to define tensors, tensor algebra and Grassmann (exterior) algebra. Unless noted otherwise all vector spaces are over the real number and are finite dimensional. There are two ways to think about tensors: (1) tensors are multi-linear maps; (2) tensors are elements of a \tensor product" of two or more vector spaces. The first way is more concrete. The second is more abstract but also more powerful. 1.1. Tensor products. We start by reviewing multi-linear maps. Definition 1.1. Let V1;:::;Vn and U be vector spaces. A map n factors z }| { f :V1 × · · · × Vn! U; (v1; : : : ; vn) 7! f(v1; : : : ; vn) is multi-linear if for each fixed index i and a fixed (n − 1)-tuple of vectors v1; : : : ; vi−1; vi+1; : : : ; vn the map Vi ! U; w 7! f(v1; : : : ; vi−1; w; vi+1; : : : ; vn) is linear. When the number of factors is n, as above, we will also say that f is n-linear. n factors 2 z n }| n{ For example, if we identify Rn ' R × · · · × R by thinking of an n × n matrix as an n-tuple of column vectors, then the determinant n factors z n }| n{ det :R × · · · × R ! R; (v1; : : : ; vn) 7! det(v1j ::: jvn) is an n-linear map. Here is an example of a bilinear map. Any inner product on a vector space V : V × V 3 (v; w) 7! v · w 2 R is bilinear. There is no standard notation for the space of n-linear maps from V1 × · · · × Vn to U. We will denote it by Mult(V1 × · · · × Vn;U) = Multn(V1 × · · · × Vn;U) (n is to indicate that these are n-linear maps). This space, Mult(V1 × · · · × Vn;U), is a vector space: any linear combination of two n-linear maps is n-linear. We now take a closer look at the space of bilinear maps Mult2(V × W; U). This case is complicated enough to understand what happens with multi-linear maps in general, but simple enough not to bog down in notation. typeset February 21, 2011. 1 ∗ ∗ Lemma 1.2. Let fvig, fwjg and fukg denote the bases of V , W and U respectively and fvi g, fwj g and ∗ fukg the corresponding dual bases. Then the maps k k ∗ ∗ φij : V × W ! U; φij(v; w) = vi (v)wj (w)uk are bilinear and form a basis of Mult2(V × W; U). Hence dim Mult2(V × W; U) = dim V dim W dim U: k Proof. It is easy to see that φij are bilinear. Next, for any b 2 Mult2(V × W; U), any w 2 W and any v 2 V , X ∗ X ∗ b(v; w) = b( vi (v)vi; wj (w)wj) X ∗ ∗ = vi (v)wj (w)b(vi; wj) i;j X ∗ ∗ ∗ = vi (v)wj (w)uk(b(vi; wj))uk i;j;k X ∗ k = uk(b(vi; wj))φij(v; w): i;j;k k ∗ Hence the maps φij span Mult2(V × W; U). Also, the collection of numbers uk(b(vi; wj)) uniquely determine k the bilinear form b. Hence φij's are linearly independent. We now turn to the definition of the tensor product V ⊗ W [pronounced \V tensor W "] of two vector spaces V and W . Informally it consists of finite linear combinations of symbols v ⊗ w, where v 2 V and w 2 W . Additionally, these symbols are subject to the following identities: (v1 + v2) ⊗ w − v1 ⊗ w − v2 ⊗ w = 0 v ⊗ (w1 + w2) − v ⊗ w1 − v ⊗ w2 = 0 α (v ⊗ w) − (αv) ⊗ w = 0 α (v ⊗ w) − v ⊗ (αw) = 0; for all v; v1; v2 2 V , w; w1; w2 2 W and α 2 R. These identities simply say that the map ⊗ : V ×W ! V ⊗W , (v; w) 7! v ⊗ w, is a bilinear map. The fact that everything in V ⊗ W is a linear combination of symbols v ⊗ w means that the image of the map ⊗ : V × W ! V ⊗ W spans V ⊗ W .1 Here is the formal definition of the tensor product of two vector spaces. Definition 1.3. A tensor product of two finite dimensional vector spaces V and W is a vector space V ⊗ W together with a bilinear map ⊗ : V × W ! V ⊗ W ,(v; w) 7! v ⊗ w2 such that for any bilinear map b : V × W ! U there is a unique linear map ¯b : V ⊗ W ! U with ¯b(v ⊗ w) = b(v; w). That is, the diagram b V × W / U ; ⊗ ¯b V ⊗ W commutes. The existence of the map ¯b satisfying the above conditions is called the universal property of the tensor product. This definition is quite abstract. It is not clear that such objects exist and, if they exist, that they are unique. Setting the question of existence and uniqueness of tensor products aside, let's us sort out the relationship between V ⊗ W and bilinear maps Mult(V × W; U). Recall that Hom(X; Y ) is the space of all linear maps from a vector space X to a vector space Y and is itself a vector space. Lemma 1.4. Assume that V ⊗ W exists. Then there is a canonical isomorphism Hom(V ⊗ W; U) −!' Mult(V × W; U): 1But the image of ⊗ is not all of V ⊗ W . The elements in the image are called decomposable tensors. 2The symbol v ⊗ w stands for the value of the map ⊗ on the pair (v; w) 2 Proof. The isomorphism in question is built into the definition of the tensor product. Given a linear map A : V ⊗ W ! U the composition A ◦ ⊗ : V × W ! U is bilinear. And conversely, given a bilinear map b 2 Mult(V × W; U) there is a unique linear map ¯b : V ⊗ W ! U so that (¯b ◦ ⊗)(v; w) = b(v; w) for all (v; w) 2 V × W . In other words the maps Hom(V ⊗ W; U) 3 A 7! A ◦ ⊗ 2 Mult(V × W; U) and Mult(V × W; U) 3 b 7! ¯b 2 Hom(V ⊗ W; U) are inverses of each other. Next we observed that the uniqueness of the tensor product is also built into the definition of the tensor product. Proposition 1.5. If tensor products exist, they are unique up to isomorphism. Proof. The proof is quite formal and uses nothing but the universal property. Suppose there are two vector spaces V ⊗1W and V ⊗2W with corresponding bilinear maps ⊗1 : V ×W ! V ⊗1W and ⊗2 : V ×W ! V ⊗2W which satisfy the conditions of the Definition 1.3. We will argue that these vector spaces are isomorphic. By the universal property there exist a unique linear map ⊗1 : V ⊗2 W ! V ⊗1 W so that the diagram ⊗1 V × W / V ⊗1 W 8 ⊗2 ⊗ 1 V ⊗2 W commutes. By the same argument, switching the roles of ⊗1 and ⊗2, there is a unique linear map ⊗2 : V ⊗1 W ! V ⊗2 W making the diagram ⊗2 V × W / V ⊗2 W 8 ⊗1 ⊗ 2 V ⊗1 W commute. Define T1 = ⊗1 ◦ ⊗2 : V ⊗1 W ! V ⊗1 W T2 = ⊗2 ◦ ⊗1 : V ⊗2 W ! V ⊗2 W: These are linear maps making the diagrams ⊗1 ⊗2 V × W / V ⊗1 W and V × W / V ⊗2 W 8 8 ⊗1 ⊗2 T1 T2 V ⊗1 W V ⊗2 W commute. But the identity maps idi : V ⊗i W ! V ⊗i W , i = 1; 2, are linear and also make the respective diagrams commute. By uniqueness Ti = idi. Hence ⊗1 and ⊗2 are inverses of each other and provide the desired isomorphisms. Now we construct the tensor product as a quotient of an infinite dimensional vector space by an infinite dimensional subspace thereby proving its existence. Proposition 1.6. Tensor products exist. Proof. Let V and W be two finite dimensional vector spaces. We want to construct a new vector space V ⊗ W and a bilinear map ⊗ : V × W ! V ⊗ W satisfying the conditions of Definition 1.3. We start with a vector space F (V × W ) made of formal finite linear combinations of ordered pairs (v; w), v 2 V , w 2 W . Its basis is the set f(v; w) j v 2 V; w 2 W g = V × W . If you prefer you can think of F (V × W ) as the set of functions ff : V × W ! R j f(v; w) 6= 0 for only finitely many pairs (v; w)g: 3 This set of functions is an infinite dimensional vector space. Its basis consists of functions that take value 1 on a given pair (v0; w0) and 0 on all other pairs. It's tempting to call this function (v0; w0). The vector space F (V × W ) is called the free vector space generated by the set V × W . Note that we have an inclusion map ι : V × W ! F (V × W ), ι(v; w) = (v; w). It is not bilinear since (v1 + v2; w) 6= (v1; w) + (v2; w) in F (V; W ). Consider the smallest subspace K of F (V; W ) containing the following collection of vectors: 8 9 (v1 + v2; w) − (v1; w) − (v2; w) > > < (v; w1 + w2) − (v; w1) − (v; w2) = S = v; v1; v2 2 V; w; w1; w2 2 W and c 2 R c(v; w) − (cv; w) > > : c(v; w) − (v; cw); ; In other words, consider the subspace K of F (V ×W ) spanned by the set S.
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