Tensor product

In mathematics, the tensor product, denoted by <math>\otimes<math>, may be applied in different contexts to vectors, matrices, tensors and vector spaces. In each case the significance of the symbol is the same: the most general bilinear operation.

A representative case is the Kronecker multiplication of any two rectangular arrays, considered as matrices.

Example:

<math>\begin{bmatrix}a_1 & a_2 & a_3\end{bmatrix} \otimes \begin{bmatrix}b_1 & b_2 & b_3 & b_4\end{bmatrix} = \begin{bmatrix}a_1b_1 & a_2b_1 & a_3b_1 \\ a_1b_2 & a_2b_2 & a_3b_2 \\ a_1b_3 & a_2b_3 & a_3b_3 \\ a_1b_4 & a_2b_4 & a_3b_4\end{bmatrix}<math>

Resultant rank = 2, resultant dimension = 12.

Here rank denotes the number of requisite indices, while dimension counts the number of degrees of freedom in the resulting array.

Tensor product of two tensors

There is a general formula for the product of two (or more) tensors

<math>V \otimes U = V_{\left[ i_1,i_2,i_3,...i_n\right] }U_{\left[ j_1,j_2,j_3,...j_m\right] }<math>.

We are assuming here orthogonal tensors, with no distinction of covariant and contravariant indices, for simplicity.

The parameters introduced above work out like this:

<math>rank( U \otimes V )=rank(U)+rank(V)<math>

<math>dim( U \otimes V )=dim(U) \cdot dim(V)<math>

See also: Tensor-classical

Tensor product of multilinear maps

Given multilinear maps <math>f(x_1,...x_k)<math> and <math>g(x_1,... x_m)<math> their tensor product is the multilinear function <math> (f \otimes g) (x_1,...x_{k+m})=f(x_1,...x_k)g(x_{k+1},... x_{k+m})<math>

Tensor product of vector spaces

The tensor product <math>V \otimes W<math> of two vector spaces V and W has a formal definition by the method of generators and relations. The equivalence class under these relations (given below) of <math>(v,w)<math> is called a tensor and is denoted by <math>v \otimes w<math>. By construction, one can prove several identities between tensors and form an algebra of tensors.

To construct <math>V \otimes W<math>, take the vector space generated by <math>V \times W<math> and apply (factor out the subspace generated by) the following multilinear relations:

• <math>(v_1+v_2)\otimes w=v_1\otimes w<math><math>+v_2\otimes w<math>

• <math>v\otimes (w_1+w_2)=v\otimes w_1+v\otimes w_2<math>

• <math>cv\otimes w=v\otimes cw=c(v\otimes w)<math>

where <math>v,v_i,w,w_i<math> are vectors from the appropriate spaces, and <math>c<math> is from the underlying field.

We can then derive the identity <math>0v\otimes w=v\otimes 0w=0(v\otimes w)=0<math>, the zero in <math>V \otimes W<math>.

The resulting tensor product <math>V \otimes W<math> is itself a vector space, which can be verified by directly checking the vector space axioms. Given bases <math>\{v_i\}<math> and <math>\{w_i\}<math> for V and W respectively, the tensors of the form <math>v_i \otimes w_j<math> forms a basis for <math>V \otimes W<math>. The dimension of the tensor product therefore is the product of dimensions of the original spaces; for instance <math>\mathbb{R}^m \otimes \mathbb{R}^n<math> will have dimension <math>mn<math>.

Tensor product for computer programmers

If a, b, and c, are rank-one tensors (i.e. one-dimensional arrays), with indices i,j,k, respectively, then the tensor product of them is a rank-three tensor(i.e. three-dimensional array):

If a is a rank-two tensor and b is a rank-one tensor, with indices i & j, and k, respectively, then the tensor product of them is a rank-three tensor:

Universal property of tensor product

The space of all bilinear maps from V × W to <math>\mathbb R<math> is naturally isomorphic to the space of all linear maps from <math>V \otimes W<math> to <math>\mathbb R<math>. This is built into the construction; <math>V\otimes W<math> has only the relations that are necessary to ensure that a homomorphism from <math>V\otimes W<math> to <math>\mathbb R<math> will be bilinear.

The tensor product in fact satifies the universal property of being a fibered coproduct.

Tensor product of Hilbert spaces

The tensor product of two Hilbert spaces is another Hilbert space, which is defined as described below.

Definition

Let H₁ and H₂ be two Hilbert spaces with inner products <·,·>₁ and <·,·>₂, respectively. Construct the tensor product of H₁ and H₂ as vector spaces as explained above. We can turn this vector space tensor product into an inner product space by defining

<math> \langle\phi_1\otimes\phi_2,\psi_1\otimes\psi_2\rangle = \langle\phi_1,\psi_1\rangle_1 \, \langle\phi_2,\psi_2\rangle_2 \quad \mbox{for all } \phi_1,\psi_1 \in H_1 \mbox{ and } \phi_2,\psi_2 \in H_2 <math>

and extending by linearity. Finally, take the completion under this inner product. The result is the tensor product of H₁ and H₂ as Hilbert spaces.

Properties

If H₁ and H₂ have orthonormal bases {φ_k} and {ψ_l}, respectively, then {φ_k ⊗ ψ_l} is an orthonormal basis for H₁ ⊗ H₂.

Examples and applications

The following examples show how tensor products arise naturally.

Given two measure spaces X and Y, with measures μ and ν respectively, one may look at L²(X × Y), the space of functions on X × Y that are square integrable with respect to the product measure μ × ν. If f is a square integrable function on X, and g is a square integrable function on Y, then we can define a function h on X × Y by h(x,y) = f(x) g(y). The definition of the product measure ensures that all functions of this form are square integrable, so this defines a bilinear mapping L²(X) × L²(Y) → L²(X × Y). Linear combinations of functions of the form f(x) g(y) are also in L²(X × Y). It turns out that the set of linear combinations is in fact dense in L²(X × Y), if L²(X) and L²(Y) are separable. This shows that L²(X) ⊗ L²(Y) is isomorphic to L²(X × Y), and it also explains why we need to take the completion in the construction of the Hilbert space tensor product.

Similarly, we can show that L²(X; H), denoting the space of square integrable functions X → H, is isomorphic to L²(X) ⊗ H if this space is separable. The isomorphism maps f(x) ⊗ φ ∈ L²(X) ⊗ H to f(x)φ ∈ L²(X; H). We can combine this with the previous example and conclude that L²(X) ⊗ L²(Y) and L²(X × Y) are both isomorphic to L²(X; L²(Y)).

Tensor products of Hilbert spaces arise often in quantum mechanics. If some particle is described by the Hilbert space H₁, and another particle is described by H₂, then the system consisting of both particles is described by the tensor product of H₁ and H₂. For example, the state space of a quantum harmonic oscillator is L²(R), so the state space of two oscillators is L²(R) ⊗ L²(R), which is isomorphic to L²(R²). Therefore, the two-particle system is described by wave functions of the form φ(x₁, x₂). A more intricate example is provided by the Fock spaces, which describe a variable number of particles.

Intrinsic Description -- (please make more accessible)

In abstract algebra, the subject of linear algebra is upgraded to multilinear algebra by introducing the tensor product of two vector spaces.

It is introduced to reduce the study of bilinear operators to that of linear operators. This is sufficient to do the same to all multilinear maps.

The general construction of tensor products over an arbitrary ring comes readily using the language of category theory. Let R be a ring (not necessarily commutative). The product of two R-modules, M and N, viewed as sets, will be denoted M × N. For any R ^op-module, M, any R-module, N, and any abelian group, Z, we set

Bilin_R(M,N;Z) = the set of φ : M × N → Z such that

1. for all m, m′ in M, for all n in N, φ(m + m′,n) = φ(m,n) + φ(m′,n),
2. for all m in M, for all n, n′ in N, φ(m,n + n′) = φ(m,n) + φ(m,n′),
3. for all m in M, for all n in N, for all r in R, φ(m r,n) = φ(m,r n)}

Observe that

• The set Bilin_R(M,N;Z) is an abelian group under addition; i.e., if φ, ψ are in Bilin_R(M,N;Z), then φ + ψ is in Bilin_R(M,N;Z).
• The map Z → Bilin_R(M,N;Z) is a functor from the category of abelian groups to the category of sets. This functor is representable and the representing abelian group is the tensor product. From this comes, for example, the following universal construction of tensor product.

Consider two vector spaces V and W over the same base field F. Their tensor product satisfies the following universal property: it is a vector space T over F, together with a bilinear operator <math> \otimes : V \times W \rarr T<math>, such that for every bilinear operator <math>B : V \times W \rarr X<math> there exists a unique linear operator L: T → X with <math>B = L \circ \otimes<math>, i.e. <math> B(x,y) = L(x \otimes y)<math> for all x in V and y in W.

The tensor product is up to a unique isomorphism uniquely specified by this requirement, and we may therefore write <math>V \otimes W<math> instead of T. By direct construction, as suggested in the previous section, one can show that the tensor product for any two vector spaces exists.

The space <math>V \otimes W<math> is generated by the image of <math> \otimes <math>, and even more: if S is a basis of V and T is a basis of W, then <math> { s \otimes t : s \ \mbox{in}\ S \ \mbox{and}\ t \ \mbox{in} \ T} <math> is a basis for <math>V \otimes W<math>. The dimension of the vector space <math>V \otimes W<math> is therefore given by the product of the dimensions of V and W.

It is possible to generalize the definition to a tensor product of any number of spaces. For example, the universal property of <math>V \otimes W \otimes X<math> is that every tri-linear operator on <math>V \times W \times X<math> corresponds to a unique linear operator on <math>V \otimes W \otimes X<math>. The binary tensor product is associative: <math> (V \otimes W) \otimes Z<math> is naturally isomorphic to <math>V \otimes (W \otimes Z)<math>. The tensor product of all three may therefore be identified with either of those: the binary <math> \otimes <math> will suffice. Tensor spaces allow us to use the theory of linear operators to study multi-linear operators, and this says the bilinear case is the main hurdle.

Relation with the dual space

Note that the space <math> (V \otimes W)^\star<math> (the dual space of <math>V \otimes W<math> containing all linear functionals on that space) corresponds naturally to the space of all bilinear functionals on <math>V \times W<math>. In other words, every bilinear functional is a functional on the tensor product, and vice versa. There is a natural isomorphism between <math> V^\star \otimes W^\star <math> and <math>(V \otimes W)^\star<math>. So, the tensors of the linear functionals are bilinear functionals. This gives us a new way to look at the space of bilinear functionals, as a tensor product itself.

Types of tensors, e.g. alternating

Linear subspaces of the bilinear operators (or in general, multilinear operators) determine natural quotient spaces of the tensor space, which are frequently useful. See wedge product for the first major example. Another would be the treatment of algebraic forms as symmetric tensors.

Over more general rings

It is also possible to generalize the definition to tensor products of modules over the same ring. If the ring is non-commutative, we'll need to be careful about distinguishing right modules and left modules. We will write _RM for a left module, and M_R for a right module. If a module M has both a left module structure over a ring R and a right module structure over a ring S, and in addition for every m in M, r in R and s in S we have r(ms) = (rm)s, then we will say M is a bimodule, and will denote it by _RM_S. Note that every left-module is a bimodule with Z acting by mn = m + m + ... + m as the right ring, and vice versa.

When defining the tensor product, we need to be careful about the ring: most modules can be considered as modules over several different rings or over the same ring with a different actions of the ring on the module elements.

The most general form of the tensor product definition is as follows: let M_R and _RN be a right and a left module, respectively. Their tensor product over R is an abelian group P together with an R-bilinear operator T: M × N → P such that for every R-bilinear operator B: M × N → O there is a unique group homomorphism L: P → O such that L o T = B. P need not be a module over R. However, if _S1M_R is an S₁-R-bimodule, then there is a unique left S₁-module structure on P which is compatible with T. Similarly, if _RM_S2 is an R-S₂-bimodule, then there is a unique right S₂-module structure on P which is compatible with T. If M and N are both bimodules, then P is also a bimodule, again in a unique way. (P, T) are unique up to a unique isomorphism, and are called the "tensor product" of M and N.

If R is a ring, _RM is a left R-module, and the commutator rs-sr of any two elements r and s of R is in the annihilator of M, then we can make M into a right R module by setting mr = rm. Note that in this situation the action of R on M factors through an action of the commutative ring R/Z(R), R modulo its center. In this case the tensor product of M with itself over R is again an R-module. If M and N are both R-modules satisfying this condition, then their tensor product is again an R-module. This is a very common technique in commutative algebra.

Example: Consider the rational numbers Q and the integers modulo n Z_n. Both can be considered as modules over the integers, Z. Let B: Q × Z_n → M be a Z-bilinear operator. Then B(q,i) = B(q/n, n*i) = B(q/p, 0) = 0, so every bilinear operator is identically zero. Therefore, if we define P to be the trivial module, and T to be the zero bilinear function, then we see that the properties for the tensor product are satisfied. Therefore, the tensor product of Q and Z_n is {0}.