Tensor product
In mathematics, the tensor product, denoted by , may be applied in different contexts to vectorss, 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:
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.
There is a general formula for the product of two (or more) tensors
The parameters introduced above work out like this:
Tensor product of two tensors
We are assuming here orthogonal tensors, with no distinction of covariant and contravariant indices, for simplicity.
See also: Tensor-classical
Given multilinear maps and
their tensor product
is the multilinear function
The tensor product of two vector spaces V and W has a formal definition by the method of generators and relations. The equivalence class of (v,w) is called a tensor and is denoted by . By construction, one can prove just as many identities between tensors, and sums of tensors, as follow from the relations used.
Take the vector space generated by and apply (factor out the subspace generated by) the multilinear relations detailed just below. With this notation the relations take the form:
Given bases for V and W, the set of tensors of basis vectors, one from V and one from W,
forms a basis for . The dimension of the tensor product therefore is the product of dimensions.
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):
The space of all bilinear maps from V × W to is naturally isomorphic to the space of all linear maps from to . This is built into the construction; has only the relations that are necessary to ensure that a homomorphism from to will be bilinear.
The tensor product in fact satifies the universal property of being a fibered coproduct.
The tensor product of two Hilbert spaces is another Hilbert space, which is defined as described below.
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
If H1 and H2 have orthonormal bases {φk} and {ψl}, respectively, then {φk ⊗ ψl} is an orthonormal basis for H1 ⊗ H2.
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.
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.
Formally, the tensor product of the two vector spaces V and W over the same base field F is defined by the following universal property: it is a vector space T over F, together with a bilinear operator , such that for every bilinear operator
The tensor product is up to a unique isomorphism uniquely specified by this requirement, and we may therefore write 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 is generated by the image of , and even more: if S is a basis of V and T is a basis of W, then is a basis for . The dimension of the vector space 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 is that every tri-linear operator on
Note that the space (the dual space of containing all linear functionals on that space)
corresponds naturally to the space of all
bilinear functionals on . In other words, every bilinear functional is a functional
on the tensor product, and vice versa.
There is a natural isomorphism between
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.
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 MR 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
RMS. 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 MR 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 S1MR is an S1-R-bimodule, then there is a unique left S1-module structure on P which is compatible with T. Similarly, if RMS2 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 Zn. Both can be considered as modules over the integers, Z.
Let B: Q × Zn → 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 Zn is {0}.Tensor product of multilinear maps
Tensor product of vector spaces
Each element of the tensor product is a finite sum of tensors: more than one tensor is usually required to do that. It is simply shown how to construct a basis of . Tensor product for computer programmers
for( int i = 0; i < i_dim; i++)
for( int j = 0; j < j_dim; j++)
for( int k = 0; k < k_dim; k++)
result[i][j][k] = a[i]*b[j]*c[k];
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: for( int i = 0; i < i_dim; i++)
for( int j = 0; j < j_dim; j++)
for( int k = 0; k < k_dim; k++)
result[i][j][k] = a[i][j]*b[k];
Universal property of tensor product
Tensor product of Hilbert spaces
Definition
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
Examples and applications
Intrinsic Description -- (please make more accessible)
there exists a unique linear operator
L: T → X with , i.e. for all x in V and y in W. corresponds to a unique linear operator on
. The binary tensor product is associative: is naturally isomorphic to . The tensor product of all three may therefore be identified with either of those: the binary 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
and .
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
Over more general rings