Tensor product

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

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

Example:

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

Resultant rank = 2, resultant dimension = 4×3 = 12.

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

1 Tensor product of two tensors
- 1.1 Example
2 Kronecker product of two matrices
3 Tensor product of multilinear maps
4 Tensor product of vector spaces
5 Universal property of tensor product
6 Tensor product of Hilbert spaces
7 Relation with the dual space
8 Types of tensors, e.g., alternating
9 Over more general rings
10 Tensor product for computer programmers
- 10.1 Array programming languages
11 See also

[edit] Tensor product of two tensors

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

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

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

What does the general formula mean? It means that if a pair of tensors are juxtaposed (placed side-by-side) then they combine by mere aggregation to form a new tensor which can be subsequently called the tensor product of the pair of juxtaposed tensors. The number of independent components multiplies

The parameters introduced above work out like this:

$\mathrm{rank}( U \otimes V )=\mathrm{rank}(U)+\mathrm{rank}(V)$

$\mathrm{dim}( U \otimes V )=\mathrm{dim}(U) \mathrm{dim}(V)$

[edit] Example

Let U be a tensor of type (1,1) with components U^α_β, and let V be a tensor of type (1,0) with components V^γ. Then

$U^\alpha {}_\beta V^\gamma = (U \otimes V)^\alpha {}_\beta {}^\gamma$

and

$V^\mu U^\nu {}_\sigma = (V \otimes U)^{\mu \nu} {}_\sigma$ .

The tensor product inherits all the indices of its factors.

[edit] Kronecker product of two matrices

Main article: Kronecker product.

With matrices this operation is usually called the Kronecker product, a term used to make clear that the result has a particular block structure imposed upon it, in which each element of the first matrix is replaced by the second matrix, scaled by that element. For matrices $U$ and $V$ this is:

$U \otimes V = \begin{bmatrix} u_{11}V & u_{12}V & \cdots \\ u_{21}V & u_{22}V \\ \vdots & & \ddots \end{bmatrix} = \begin{bmatrix} u_{11}v_{11} & u_{11}v_{12} & \cdots & u_{12}v_{11} & u_{12}v_{12} & \cdots \\ u_{11}v_{21} & u_{11}v_{22} & & u_{12}v_{21} & u_{12}v_{22} \\ \vdots & & \ddots \\ u_{21}v_{11} & u_{21}v_{12} \\ u_{21}v_{21} & u_{21}v_{22} \\ \vdots \end{bmatrix}$ .

[edit] Tensor product of multilinear maps

Given multilinear maps $f (x 1,... x k)$ and $g (x 1,... x m)$ their tensor product is the multilinear function

$(f \otimes g) (x_1,...,x_{k+m})=f(x_1,...,x_k)g(x_{k+1},...,x_{k+m})$

[edit] Tensor product of vector spaces

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

To construct $V \otimes W$ , take a vector space over $K$ with basis $V \times W$ and apply (factor out the subspace generated by) the following multilinear relations:

$(v_1+v_2)\otimes w=v_1\otimes w$ $+v_2\otimes w$
$v\otimes (w_1+w_2)=v\otimes w_1+v\otimes w_2$
$cv\otimes w=v\otimes cw=c(v\otimes w)$

where $v, v i, w, w i$ are vectors from the appropriate spaces, and $c$ is from the underlying field $K$ .

We can then derive the identity

$0v\otimes w=v\otimes 0w=0(v\otimes w)=0$ ,

the zero in $V \otimes W$ .

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

[edit] Universal property of tensor product

The space of all bilinear maps from $V\times W$ to another vector space $K$ is naturally isomorphic to the space of all linear maps from $V \otimes W$ to $K$ . This is built into the construction; $V\otimes W$ has all and only the relations that are necessary to ensure that a homomorphism from $V\otimes W$ to $K$ will be linear.

[edit] Tensor product of Hilbert spaces

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

[edit] Definition

Let H₁ and H₂ be two Hilbert spaces with inner products $\langle \cdot,\cdot\rangle_1$ and $\langle \cdot,\cdot\rangle_2$ , 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

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

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.

[edit] 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.

[edit] 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.

[edit] Relation with the dual space

Note that the space $(V \otimes W)^\star$ (the dual space of $V \otimes W$ , containing all linear functionals on that space) corresponds naturally to the space of all bilinear functionals on $V \times W$ . In other words, every bilinear functional is a functional on the tensor product, and vice versa.

Whenever $V$ and $W$ are finite dimensional, there is a natural isomorphism between $V^\star \otimes W^\star$ and $(V \otimes W)^\star$ , whereas for vector spaces of arbitrary dimension we only have an inclusion $V^\star \otimes W^\star\subset (V \otimes W)^\star$ . 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.

[edit] 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.

[edit] Over more general rings

See tensor product of modules over a ring

[edit] Tensor product for computer programmers

[edit] Array programming languages

Array programming languages may have this pattern built in. For example, in APL the tensor product is expressed as $\circ . \times$ (for example $A \circ . \times B$ or $A \circ . \times B \circ . \times C$ ). In J the tensor product is the dyadic form of */ (for example a */ b or a */ b */ c).

Note that J's treatment also allows the representation of some tensor fields (as a and b may be functions instead of constants -- the result is then a derived function, and if a and b are differentiable, then a*/b is differentiable).

However, these kinds of notation are not universally present in array languages. Other array languages may require explicit treatment of indices (for example, Matlab), and/or may not support higher-order functions such as the Jacobian derivative (for example, Fortran/APL).

Categories: Abstract algebra | Multilinear algebra