Tensors are an odd concept, mostly because their definitions look so elusive. In our view, the easiest way to understand them is to treat them as atomic objects, taken tantum et tale, and to concentrate on the rules by which they combine. The basic construction takes two vector spaces and over a field and produces a new vector space , their tensor product, built so that bilinear maps out of become linear maps out of . We give three definitions, from the most concrete to the most abstract, and then prove that they agree.
Definition 9.8.1 (Bilinear map). Let , and be vector spaces over . A map is bilinear if it is linear in each variable when the other is held fixed: for all , and . A bilinear form on (Definition 9.1.1) is the case ,.
9.8.1 Three Definitions
Definition 9.8.2 (The tensor product via bases). Let and be finite-dimensional, with bases of and of . Then is the vector space with basis the symbols (,); in particular . For and we set The map ,, is bilinear.
Definition 9.8.3 (Pure tensors and rank). Elements of of the form are called pure tensors. Every element of is a sum of pure tensors, since it is a linear combination of the pure tensors , and is again pure. The rank of is the smallest number such that is a sum of pure tensors. Thus has rank , and the nonzero pure tensors are exactly the tensors of rank .
Most tensors are not pure. For instance, if , then is not pure: if it were equal to , comparing coefficients would give and , which is impossible. We return to tensor rank in Proposition 9.8.
Definition 9.8.2 is concrete, and it makes the dimension of evident. It is also uncomfortable (Harris and Auroux, in their notes on tensor products, call it ugly): it depends on a choice of bases, and it is not clear that other bases give the same space. Our second definition chooses no bases at all.
Let be the vector space with one basis vector for each pair : its elements are the finite linear combinations of these symbols, with no relations among them. This space is enormous (when and , its basis is uncountable), and in it the symbols are not bilinear in any sense. We force bilinearity by dividing out exactly the relations it requires. Let be the subspace spanned by all elements for , and .
Definition 9.8.4 (The tensor product as a quotient). is the quotient space , and denotes the class of . The map ,, is bilinear, because that is precisely what the relations in say.
Note. Nothing in this construction uses bases or finite dimension. With the same relations it also applies to modules over a commutative ring , which are defined in Chapter 10 (Definition 10.1.1), where the result is written .
The most abstract characterization says what does, without saying how it is built.
Definition 9.8.5 (The tensor product by its universal property). A tensor product of and is a vector space together with a bilinear map with the following property: for every vector space and every bilinear map , there is a unique linear map with . We write .
9.8.2 The Universal Property
Theorem 9.8.6 (Existence and uniqueness of the tensor product). Let and be finite-dimensional vector spaces over .
The space of Definition 9.8.2, with , satisfies the universal property of Definition 9.8.5.
The space of Definition 9.8.4, with the class of , satisfies it as well.
If and both satisfy the universal property, there is a unique linear isomorphism with .
Consequently the three definitions produce the same space up to a unique isomorphism matching the pure tensors ; in particular, the spaces that Definition 9.8.2 builds from different bases are canonically isomorphic. For any bases of and of , the pure tensors form a basis of , and .
Proof. (i) Let be bilinear, and let be the linear map with (a linear map may take arbitrary values on a basis). For and , bilinearity gives Any linear map with satisfies , so , because a linear map is determined by its values on a basis.
(ii) Given , let be the linear map with . Each spanning element of maps to because is bilinear; for example, . So , and by the universal property of the quotient space (Theorem 7.1.7), factors through a linear map with . It is unique because the classes span .
(iii) The universal property of , applied to the bilinear map , gives a linear map with , and it is the only such map. Symmetrically there is with . Then is linear and ; the identity of has the same property, so the uniqueness in the universal property of , applied to , gives . Likewise .
For the last statements: by (i)–(iii), any model of the tensor product is isomorphic to the space built in Definition 9.8.2 from the bases and , by an isomorphism matching pure tensors. That isomorphism carries the basis of the latter to the elements of , which therefore form a basis of .∎
Note. The argument in (iii) applies to every object defined by a universal property, as for products and coproducts (Section 8.3): the universal property determines the object up to unique isomorphism, and a separate construction shows that it exists. From now on means any of these spaces, and we use whichever description is convenient. In practice, two facts get used constantly: to define a linear map out of , it suffices to give a bilinear map out of ; and two linear maps out of that agree on all pure tensors are equal, because pure tensors span.
Note. The assignment is a bijection from the bilinear maps to the linear maps , with inverse . It is linear, so it identifies the vector space of bilinear maps with . In this sense the tensor product converts bilinear problems into linear ones.
9.8.3 Functoriality and Basic Identities
Proposition 9.8.7 (Tensor product of linear maps). Let and be linear maps between finite-dimensional vector spaces. There is a unique linear map with Moreover:
for linear maps ,, and ;
if and , then .
Proof. The map ,, is bilinear, because and are linear and is bilinear. By the universal property it induces a unique linear map with the stated values.
(i) On pure tensors, and two linear maps that agree on pure tensors are equal. The same argument gives .
(ii) Let and be the matrices of and in bases of and of . Then so the diagonal entry of the matrix of at is . Hence .∎
In the language of Definition 8.2.1, part (i) says that is a functor in two variables: it is a functor in each variable when the other is held fixed, and the two variables do not interfere. In particular, if and are isomorphisms, so is , with inverse .
Proposition 9.8.8 (Basic identities). Let ,, be finite-dimensional vector spaces over . There are isomorphisms
, with ;
, with ;
, with .
Proof. In each case the formula comes from a bilinear map on the product of the two factors, namely , and . By the universal property each induces a linear map with the stated values on pure tensors. Each of these maps sends a basis to a basis, so it is an isomorphism. Indeed, let ,, be bases of ,,. In (i), . In (ii), the vectors and form a basis of , and ,, which together form a basis of . In (iii), .∎
Example 9.8.9 (Complexification as a tensor product). Let be a finite-dimensional real vector space, and regard as a real vector space with basis . Write for the tensor product of these two real vector spaces. For , multiplication by is an -linear map , and we set , so that . Proposition 9.8.7(i) gives and , and checking on pure tensors gives the distributive laws; so is a complex vector space.
If is a basis of , the vectors , form a real basis of , and , so the form a complex basis. The map from the complexification of Definition 7.10.1, is -linear and sends a real basis to a real basis. It also commutes with multiplication by : the vector goes to . So it is an isomorphism of complex vector spaces, and under it corresponds to .
9.8.4 Hom as a Tensor Product
The most useful identity involving tensor products is the following. For a basis of we write for the dual basis of (Definition 6.2.5), so .
Proposition 9.8.10 (). Let and be finite-dimensional. There is an isomorphism defined without any choice of bases. It is compatible with composition: for linear maps and between finite-dimensional spaces, where is the dual map (Definition 6.2.9).
Proof. The map ,, is bilinear: , and similarly in . By the universal property it induces a linear map with the stated values.
Choose bases of and of . The map sends to and every other to , so its matrix is the matrix with a in row , column , and zeros elsewhere. These matrices form a basis of the space of matrices, which is isomorphic to by taking matrices in the chosen bases (Theorem 5.2.7). So sends the basis of to a basis of , and it is an isomorphism.
Both sides of the compatibility formula are linear in , so it suffices to check it for . For ,∎
Under , a pure tensor with and is the map , whose image is the line spanned by : a map of rank one. This is the first case of a general fact, which explains the name “rank” for tensors.
Proposition 9.8.11 (Tensor rank equals rank). Let and be finite-dimensional and . The rank of as a tensor equals the rank of the linear map .
Proof. Let be the tensor rank of and . If , then lies in for every , so .
Conversely, let and . By the Rank–Nullity Theorem (Theorem 6.1.12), . Choose a basis of , extend it to a basis of , and let be the dual basis. The map sends to for and to for , so it equals . Since is injective, , a sum of pure tensors, and .∎
Example 9.8.12 (A matrix as a tensor). Let and have bases and . The linear map with matrix sends and , so it corresponds under to the tensor By Proposition 9.8.11, this tensor is pure exactly when the matrix has rank at most , that is, when its columns are proportional, that is, when . Otherwise it has rank .
Proposition 9.8.13 (Duals and bilinear forms). Let and be finite-dimensional.
There is an isomorphism sending to the linear functional with .
The space of bilinear forms on is isomorphic to ; the tensor corresponds to the form .
Proof. (i) For fixed , the map is bilinear, so it induces a linear functional on . Its values on pure tensors depend bilinearly on , and pure tensors span, so the functional itself depends bilinearly on ; the universal property gives a linear map . With bases of and of , it sends to the functional taking to , which is the dual basis of the basis of . So it sends a basis to a basis.
(ii) By the universal property, bilinear forms on correspond to linear functionals on , and this correspondence is a linear isomorphism . Combine it with (i) for .∎
Note. Part (ii) agrees with the isomorphism ,, of Theorem 9.1.8. Proposition 9.8.10 with gives , and the form has , which corresponds to . So both routes from to give the same isomorphism.
Note. The dual map also has a description in terms of tensors. Chaining isomorphisms, using Proposition 9.8.10, the swap of Proposition 9.8.8(i), the isomorphism of Theorem 6.2.19 on the first factor, and Proposition 9.8.10 again. Follow a rank-one map through the chain: it goes to , then to , then to (where ), and finally to the map . That map is , because . Both ends of the chain are linear in , and rank-one maps span , so the composite sends every to its dual map . This is a construction of the transpose that uses no bases.