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TENSOR ALGEBRA in These Notes We Will Be Working in a Field F with Charf = 2. the Goal of These Notes Is to Introduce Tensor

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TENSOR ALGEBRA in These Notes We Will Be Working in a Field F with Charf = 2. the Goal of These Notes Is to Introduce Tensor TENSOR ALGEBRA INNA ZAKHAREVICH In these notes we will be working in a field F with char F 6= 2. The goal of these notes is to introduce tensor products and skew-symmetric tensor products of vector spaces, with the goal of introducing determinants as an invariant of a linear transformation. Hoffman & Kunze has a discussion of determinants from this perspective as well, in Chapter 5, Sections 6 and 7; however, they discuss it from the dual perspective, in terms of multilinear functionals. Let V be a vector space over F . Recall that V ∗ = L(V; F ), and that we discussed that for finite-dimensional V there exists a (non-canonical!) isomorphism V ! V ∗. However, for infinite- dimensional V this is not necessarily the case. We do have a linear transformation V ! V ∗∗, ∗ however, defined by sending the vector v to the map fv : V ! F given by fv(α) = α(v). Equiva- lently,1 there is a map ev : V × V ∗ −! F given by (v; α) 7! α(v). Note that this map is not linear: if it were then α(v) + β(w) = ev(v; α) + ev(w; β) = ev((v; α) + (w; β)) = ev(v + w; α + β) = (α + β)(v + w): However, note that directly from the definition we can see that ev is linear in each variable sepa- rately. Thus (α + β)(v + w) = (α + β)(v) + (α + β)(w) = α(v) + β(v) + α(w) + β(w): Thus we need to either develop a separate theory of bilinear maps, or to construct a representing object: a vector space W such that bilinear maps L(W; F ) =∼ : V × V ∗ ! F This is where we diverge from Hoffman & Kunze. They proceed to develop the theory of bilinear maps, and we will construct the representing object. In addition, we solve a slightly more general problem which will in fact be easier to work with: for any two vector spaces V and W , we will construct a vector space V ⊗ W which will have the property that bilinear maps L(V ⊗ W; Z) =∼ V × W ! Z for any vector space Z. We will do the construction in two ways. First, we will do a completely general construction, which will work for both vector spaces over fields and modules over rings. The advantage of this construction is that it will involve no choices, and thus showing that various nice properties hold will be easy with this construction. After this we will do a much more computational construction using bases; this will not be a natural construction, but it will make some later computations easier. Let A be the free vector space on pairs v ⊗ w, with v 2 V and w 2 W ; by this we mean that as a set, A is the set of formal linear combinations of vectors v ⊗ w, for v 2 V and w 2 W . A is very large, and if F is infinite it will always be infinite-dimensional. From this definition we have that set maps L(A; Z) = : V × W ! Z 1You are asked to show that this equivalence holds on the next homework 1 2 INNA ZAKHAREVICH Now obviously the set of set maps V ×W ! Z is much larger than the set of linear maps, so we need to make A smaller. We do this by enforcing the bilinearity: for any f 2 L(V ⊗ W; Z) we want to have f((v+v0)⊗w) = f(v⊗w)+f(v0 ⊗w); we can enforce this by setting (v+v0)⊗w = v⊗w+v0 ⊗w in V ⊗ W . We define 8 9 (v + v0) ⊗ w − v ⊗ w − v0 ⊗ w > > < v ⊗ (w + w0) − v ⊗ w − v ⊗ w0 = A0 = span a 2 F; v 2 V; w 2 W : (av) ⊗ w − a(v ⊗ w) > > : v ⊗ (aw) − a(v ⊗ w) ; We then define V ⊗ W = A=A0; since A and A0 are abelian groups this definition is well-defined as an abelian group. All that needs to be checked is that it inherits a scalar multiplication from A.2 To get a bilinear map from a linear map V ⊗ W ! Z it suffices to check that (1) the map V × W ! V ⊗ W given by (v; w) 7! v ⊗ w is bilinear, and (2) the composition of a bilinear map and a linear map is bilinear. Thus to check that V ⊗ W has the right bijection it remains to check that if we are given a bilinear map V × W ! Z it gives us a linear map V ⊗ W ! Z. Given a bilinear map f : V × W ! Z we can define a linear map g : A ! Z by defining g(v ⊗ w) = f(v; w) for all v 2 V and w 2 W . Since these form a basis of A, g is a well-defined linear map. In order to check that we get a linear map V ⊗ W ! Z from g we just need to check that g(A0) = f0g; then g descends to a well-defined linear map on A=A0. In particular, we need to check that g is 0 on each of the four types of generators of A0; that this holds follows directly from the bilinearity of f. This construction of V ⊗W is very clean and formal, but it does not give us a good model for how to work with V ⊗ W explicitly. For example, if dim V = m and dim W = n, what is dim V ⊗ W ? To answer this question, we give a second construction of V ⊗ W , this time using a basis. Let fv1; : : : ; vmg be a basis for V and fw1; : : : ; wng be a basis for W . Then we claim that fvi ⊗ wj j 1 ≤ i ≤ m; 1 ≤ j ≤ ng is a basis for V ⊗ W . Indeed, note that any v ⊗ w 2 V ⊗ W can Pm Pn be written in terms of these: if v = i=1 aivi and w = j=1 bjwj then m ! m m X X X v ⊗ w = aivi ⊗ w = (aivi) ⊗ w = ai(vi ⊗ w) i=1 i=1 i=1 m 0 n 1 m n m n X X X X X X = ai @vi ⊗ bjwjA = ai vi ⊗ (bjwj) = ai bj(vi ⊗ wj) i=1 j=1 i=1 j=1 i=1 j=1 m n X X = aibj(vi ⊗ wj): i=1 j=1 Thus V ⊗ W = span(v ⊗ w j v 2 V; w 2 W ) = span(vi ⊗ wj j 1 ≤ i ≤ m; 1 ≤ j ≤ n): We claim that this is actually a basis for V ⊗W , and thus that we can define V ⊗W to be the vector space with this basis. Checking that these are linearly independent directly from the definition is somewhat difficult, so we do something slightly indirect. Let Z be a vector space on a basis zij, with 1 ≤ i ≤ m and 1 ≤ j ≤ n. We clearly have a linear transformation Z ! V ⊗ W given by zij 7! vi ⊗ wj. We will show that this is an isomorphism by constructing an inverse. To do this, note that the above calculation of v ⊗ w in terms of the vi ⊗ wj defines a linear map A ! Z, so it suffices to check that this is 0 on A0; this follows directly from the definition of A0. Note that since vi ⊗ wj 2 A, the map V ⊗ W ! Z is surjective, and checking that it is in fact the inverse to 2Note that this definition works equally well if V and W are R-modules; this may be helpful on the homework. TENSOR ALGEBRA 3 the map Z ! V ⊗ W is straightforward.3 Thus we see that we could alternately define V ⊗ W as the vector space with basis vi ⊗ wj. From this it follows that dim V ⊗ W = mn. Important: the vector space V ⊗ W is spanned by vectors of the form v ⊗ w, but not every vector in V ⊗ W can be written in this form. For example, if dim V; dim W ≥ 2 and fv1; v2g and fw1; w2g are linearly independent in V and W , respectively, then v1 ⊗ w1 + v2 ⊗ w2 is not a pure tensor. Example 0.1. If W = F then V ⊗ W =∼ V . Example 0.2. Let V = W = F , and let µ : F × F ! F be the usual multiplication in F . This is a bilinear map by the distributivity property, so it gives us a linear map F ⊗ F ! F . Example 0.3. Let V; W; Z be vector spaces. Composition of linear maps is bilinear (check this!) and thus composition gives us a linear map L(W; Z) ⊗ L(V; W ) !L(V; Z). Note that if T : V ! W and S : V 0 ! W 0 are linear transformations, then we get a linear transformation T ⊗ S : V ⊗ V 0 ! W ⊗ W 0 by defining (T ⊗ S)(v ⊗ v0) = (T v) ⊗ (Sv0): To check that it is linear it suffices to check that the map V × V 0 ! W ⊗ W 0 given by (v; v0) 7! (T v) ⊗ (Sv0) is bilinear.4 If all of these vector spaces are finite dimensional, then the matrix of the linear transformation T ⊗ S will be the (dim V )(dim V 0) × (dim W )(dim W 0) matrix whose entries will be the pairwise products of entries in the matrix of T with entries in the matrix of S. There is one more important definition we need before we can introduce the determinant. We will want to be able to look at the skew-symmetric tensor product: we want to look at the tensor product with one extra relation imposed: that v ⊗ v0 = −v0 ⊗ v.
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