Tensor product
In mathematics, the tensor product V ⊗ W of two vector spaces V and W (over the same field) is itself a vector space, together with an operation of bilinear composition, denoted by ⊗, from ordered pairs in the Cartesian product V × W into V ⊗ W, in a way that generalizes the outer product. The tensor product of V and W is the vector space generated by the symbols v ⊗ w, with v ∈ V and w ∈ W, in which the relations of bilinearity are imposed for the product operation ⊗, and no other relations are assumed to hold. The tensor product space is thus the "freest" (or most general) such vector space, in the sense of having the fewest constraints.
The tensor product of (finite dimensional) vector spaces has dimension equal to the product of the dimensions of the two factors:
In particular, this distinguishes the tensor product from the direct sum vector space, whose dimension is the sum of the dimensions of the two summands:
More generally, the tensor product can be extended to other categories of mathematical objects in addition to vector spaces, such as to matrices, tensors, algebras, topological vector spaces, and modules. In each such case the tensor product is characterized by a similar universal property: it is the freest bilinear operation. The general concept of a "tensor product" is captured by monoidal categories; that is, the class of all things that have a tensor product is a monoidal category.
Intuitive motivation and the concrete tensor product
The intuitive motivation for the tensor product relies on the concept of tensors more generally. In particular, a tensor is an object which can be considered a special type of multilinear map, which takes in a certain number of vectors (its order) and outputs a scalar. Such objects are useful in a number of areas of application, such as Riemannian geometry, famous for its use in Albert Einstein's general theory of relativity in modern physics, where the metric tensor is a fundamental concept: in particular, the metric tensor takes in two vectors, conceived of roughly as small arrows emanating from a specific point within a curved space, or manifold, and returns a local dot product of them relative to that particular point - an operation which encodes roughly the vectors' lengths as well as the angle between them. As the dot product is a scalar, the metric tensor is thus seen to deserve its name. There is one metric tensor at each point of the manifold, and variation in the metric tensor thus encodes how that distance and angle concepts, and so the laws of analytic geometry, vary throughout the manifold.
One can think of the tensor product of two vector spaces, and as representing the set of all tensors that take a vector from and and output a scalar within their common base field (and thus can only be defined if they have such a base field). The two spaces may be the same - in the above they are vectors in the tangent space at a point: roughly the flat space a tiny piece of the manifold most "looks like" when you zoom very, very close into a particular point therein, and thus the metric tensor lives in the tensor product of that space with itself. But they may also be different.
If we have a basis for the vector spaces, and the vector space is finite-dimensional, we can represent the vectors in terms of components under those basis vectors:
- .
where each notation stands for the sum .
A tensor is then a map that works as above, returning a scalar and is linear in both of its arguments. Such a tensor can be represented using a matrix multiplication:
where the superscripted T denotes the matrix transpose which sends the vector to its dual vector.
Given two vectors, we can form a tensor of their own from them rather naturally using the outer product, which is denoted and equals . This tensor comes out as the matrix
and this matrix corresponds to the tensor by the prior construction, which is reminiscent of how it corresponds to a linear map (by multiplying on one side only). These tensors themselves generate a vector space by adding them together and multiplying them by scalars in the usual ways that we do for matrices and functions, and the collection of all such tensors so formed is the tensor product of the two vector spaces themselves. In fact, this space is equivalent to the space of maps represented by every possible matrix of the above size, as can be seen by noting that the simple tensor products (here is the basis of the other vector space, ) have a "1" in the -th position and "0"s everywhere else, which allows them to be multiplied by any number and then added up to get a matrix with arbitrary entries.
The purpose of the succeeding sections is to find a definition that is equivalent to this where it is applicable but which does not require a specific choice of basis and which can also more easily be applied to infinite-dimensional settings where the usual basis concepts (Hamel basis) may be ill-behaved. Not requiring a specific basis is useful from a theoretical point of view since while every vector space has a basis, not all bases are necessarily constructible, and moreover that result itself depends on the acceptance of the axiom of choice which may be rejected in some systems of mathematics. Also, it is useful to find an abstract construction for analysis from the point of view of category theory, the theory of the very zoomed-out "big picture of maths" and how all mathematical objects relate to each other in a very general sense. A very important real-life use for having such a definition can be found in another field of modern physics called quantum mechanics: the tensor product in this form allows us to talk of the wave function of a system of two particles as an abstract Hilbert space vector without having to specify a specific basis of observables.
Baby step toward the abstract tensor product: the free vector space
The first step we will consider involves introducing something which is called a "free vector space" over a given set. The thrust behind this idea basically consists of what we said in the last point: since a tensor can be written by the double sum
the most natural way to approach this problem is somehow to figure out how we can "forget" about the specific choice of bases and that are used here. In mathematics, the way we "forget" about representational details of something is to establish an identification that tells us that two different things which are to be considered representations of the same thing are in fact such, i.e. which, given those says either "yes, they are" or "no, they aren't", and then "lump together" all representations as constituting the "thing represented" without reference to any one in particular by packaging them all together into a single set. In formal terms, we first build an equivalence relation, and then take the quotient set by that relation.
But before we can do that, we first need to develop what we are going to take the equivalence relation over. The way we do that is to approach this the other way around, from the "bottom up": since we are not guaranteed a, at least constructible, basis when starting from arbitrary vector spaces, we might instead try to start by guaranteeing we have one - that is, we will start first by considering a "basis", on its own, as given, and then building the vector space on top. To that end, we accomplish the following: suppose that is some set, which we could call an abstract basis set. Now consider all formal expressions of the form
of arbitrary, but finite, length and for which are scalars and are members of . Intuitively, this is a linear combination of the basis vectors in the usual sense of expanding an element of a vector space. We call this a "formal expression" because technically it is illegal to multiply since there is no defined multiplication operation by default on a random set and random field of scalars. Instead, we will "pretend" (similar to defining the imaginary numbers) that this refers to something, and then will go about manipulating it according to the rules we expect for a vector space, e.g. the sum of two such strings of the same length is
where we have used the associative, commutative, and distributive laws to rearrange the first sum into the second. Continuing this way for scalar multiples and all different-length combinations of vectors allows us to build up a vector addition and scalar multiplication on this set of formal expressions, and we call it the free vector space over , writing . Note that the elements of , considered as length-one formal expressions with coefficient 1 out front, form a Hamel basis for this space.
The tensor product expression is then abstracted by considering that if and represent "abstract basis vectors" from two sets and , i.e. that " " and " ", then pairs of these in the Cartesian product , i.e. are taken as standing for the tensor products . (Note that the tensor products in the expression are in some sense "atomic", i.e. additions and scalar multiplications do not split them up into anything else, so we can replace them with something different without altering the mathematical structure.) With such an identification, we can thus define the tensor product of two free vector spaces and as being something (yet to be decided) which is isomorphic to .
Using the free vector space to "forget" about the basis
The above definition will actually work for any vector space in which we can specify a basis, since we can just rebuild it as the free vector space over that basis: the above construction exactly mirrors how you represent vectors via the Hamel basis construction by design. In effect, we haven't gained anything ... until we do this.
Since we are assuming we don't actually have access to a basis for each vector space and that we in general want to form the tensor product of, we will instead do the next best thing and in some sense the one thing we are guaranteed able to do, regardless of any concerns or problematics in finding a specific basis: take all of and as "basis" to build up the tensors - which corresponds to actually what we did in the last part of the "Intuitive motivation" section where we considered adding together arbitrary outer products of arbitrary vectors taken from the two spaces. The only difference here is that if we use the free vector space construction and form the obvious it will have many redundant versions of what should be the same tensor, i.e. going back to our basisful case if we consider the very specific example where in the standard basis, which is manageably small yet nontrivial, we may consider that the tensor formed by the vectors and , i.e.
could also be represented by other sums, such as the sum using individual basic tensors , e.g.
These, while equal expressions in the concrete case, would correspond to distinct elements of the free vector space , namely
in the first case and
in the second case. Thus we must condense them - this is where the equivalence relation comes into play. The trick to building it is to note that given any vector in a vector space, it is always possible to represent it as the sum of two other vectors and not equal to the original. If nothing else, let be any vector and then take - which also shows that if we are given one vector and then a second vector, we can write the first vector in terms of the second together with a suitable third vector (indeed in many ways - just consider scalar multiples of the second vector in the same subtraction.).
This is useful to us because the outer product satisfies the following linearity properties, which can be proven by simple algebra on the corresponding matrix expressions (the vectors below are generic, not the example ones above):
If we want to relate the outer product to, say, , we can use the first relation above together with a suitable expression of as a sum of some vector and some scalar multiple of .
Equality between two concrete tensors is then obtained if using the above rules will permit us to rearrange one sum of outer products into the other by suitably decomposing vectors - regardless of if we have a set of actual basis vectors. Applying that to our example above, we see that of course we have
for which substitution in
gives us
and judicious use of the distributivity properties lets us rearrange to the desired form. Likewise, there is a corresponding "mirror" manipulation in terms of the free vector space elements and , , etc, and this finally leads us to the formal definition of the tensor product.
The definition of the abstract tensor product
The abstract tensor product of two vector spaces and over a common base field is the quotient vector space
where is the equivalence relation of formal equality which is generated by assuming that for each and taken as formal expressions in the free vector space the following hold:
- Identity. .
- Distributivity. and .
- Scalar multiples. and
and then testing equivalence of generic formal expressions through suitable manipulations based thereupon. Arithmetic is defined on the tensor product by choosing representative elements, applying the arithmetical rules, and finally taking the equivalence class. Moreover, given any two vectors and , the equivalence class is denoted .
Properties
Notation
Elements of V ⊗ W are often referred to as tensors, although this term refers to many other related concepts as well.^{[1]} If v belongs to V and w belongs to W, then the equivalence class of (v, w) is denoted by v ⊗ w, which is called the tensor product of v with w. In physics and engineering, this use of the "⊗" symbol refers specifically to the outer product operation; the result of the outer product v ⊗ w is one of the standard ways of representing the equivalence class v ⊗ w.^{[2]} An element of V ⊗ W that can be written in the form v ⊗ w is called a pure or simple tensor. In general, an element of the tensor product space is not a pure tensor, but rather a finite linear combination of pure tensors. For example, if v_{1} and v_{2} are linearly independent, and w_{1} and w_{2} are also linearly independent, then v_{1} ⊗ w_{1} + v_{2} ⊗ w_{2} cannot be written as a pure tensor. The number of simple tensors required to express an element of a tensor product is called the tensor rank (not to be confused with tensor order, which is the number of spaces one has taken the product of, in this case 2; in notation, the number of indices), and for linear operators or matrices, thought of as (1, 1) tensors (elements of the space V ⊗ V^{∗}), it agrees with matrix rank.
Dimension
Given bases {v_{i}} and {w_{j}} for V and W respectively, the tensors {v_{i} ⊗ w_{j}} form a basis for V ⊗ W. Therefore, if V and W are finite-dimensional, the dimension of the tensor product is the product of dimensions of the original spaces; for instance R^{m} ⊗ R^{n} is isomorphic to R^{mn}.
Tensor product of linear maps
The tensor product also operates on linear maps between vector spaces. Specifically, given two linear maps S : V → X and T : W → Y between vector spaces, the tensor product of the two linear maps S and T is a linear map
defined by
In this way, the tensor product becomes a bifunctor from the category of vector spaces to itself, covariant in both arguments.^{[3]}
If S and T are both injective, surjective, or continuous then S ⊗ T is, respectively, injective, surjective, continuous.
By choosing bases of all vector spaces involved, the linear maps S and T can be represented by matrices. Then, the matrix describing the tensor product S ⊗ T is the Kronecker product of the two matrices. For example, if V, X, W, and Y above are all two-dimensional and bases have been fixed for all of them, and S and T are given by the matrices
respectively, then the tensor product of these two matrices is
The resultant rank is at most 4, and thus the resultant dimension is 4. Here rank denotes the tensor rank (number of requisite indices), while the matrix rank counts the number of degrees of freedom in the resulting array.
A dyadic product is the special case of the tensor product between two vectors of the same dimension.
Universal property
In the context of vector spaces, the tensor product and the associated bilinear map are characterized up to isomorphism by a universal property regarding bilinear maps. (Recall that a bilinear map is a function that is separately linear in each of its arguments.) Informally, is the most general bilinear map out of .
The vector space and the associated bilinear map have the property that any bilinear map from to any vector space factors through uniquely. By saying " factors through uniquely," we mean that there is a unique linear map such that .
This characterization can simplify proofs about the tensor product. For example, the tensor product is symmetric, meaning there is a canonical isomorphism:
To construct, say, a map from to , it suffices to give a bilinear map that maps to . Then the universal property of means factors into a map . A map in the opposite direction is similarly defined, and one checks that the two linear maps and are inverse to one another by again using their universal properties.
Similar reasoning can be used to show that the tensor product is associative, that is, there are natural isomorphisms
Therefore, it is customary to omit the parentheses and write .
The category of vector spaces with tensor product is an example of a symmetric monoidal category.
The universal-property definition of a tensor product is valid in more categories than just the category of vector spaces. Instead of using multilinear (bilinear) maps, the general tensor product definition uses multimorphisms.^{[4]}
Tensor powers and braiding
Let n be a non-negative integer. The nth tensor power of the vector space V is the n-fold tensor product of V with itself. That is
A permutation σ of the set {1, 2, ..., n} determines a mapping of the nth Cartesian power of V as follows:
Let
be the natural multilinear embedding of the Cartesian power of V into the tensor power of V. Then, by the universal property, there is a unique isomorphism
such that
The isomorphism τ_{σ} is called the braiding map associated to the permutation σ.
Product of tensors
For non-negative integers r and s a type (r,s) tensor on a vector space V is an element of
Here V^{∗} is the dual vector space (which consists of all linear maps f from V to the ground field K).
There is a product map, called the (tensor) product of tensors^{[5]}
It is defined by grouping all occurring "factors" V together: writing v_{i} for an element of V and f_{i} for elements of the dual space,
Picking a basis of V and the corresponding dual basis of V^{∗} naturally induces a basis for T^{r}
_{s}(V) (this basis is described in the article on Kronecker products). In terms of these bases, the components of a (tensor) product of two (or more) tensors can be computed. For example, if F and G are two covariant tensors of rank m and n respectively (i.e. F ∈ T^{ 0}
_{m}, and G ∈ T^{ 0}
_{n}), then the components of their tensor product are given by
^{[6]} Thus, the components of the tensor product of two tensors are the ordinary product of the components of each tensor. Another 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
and
Relation to dual space
A particular example is the tensor product of some vector space V with its dual vector space V^{∗} (which consists of all linear maps f from V to the ground field K). In this case, there is a canonical evaluation map
which on elementary tensors is defined by
The resulting map
is called tensor contraction (for r, s > 0).
On the other hand, if V is finite-dimensional, there is a canonical map in the other direction (called the coevaluation map)
where v_{1}, ..., v_{n} is any basis of V, and v_{i}^{∗} is its dual basis. Surprisingly, this map does not depend on our choice of basis.^{[7]}
The interplay of evaluation and coevaluation map can be used to characterize finite-dimensional vector spaces without referring to bases.^{[8]}
Tensor product vs. Hom
Given two finite dimensional vector spaces U, V, denote the dual space of U as U*, we have the following relation:
an isomorphism can be defined by , when acting on pure tensors
its "inverse" can be defined in a similar manner as above (Relation to dual space) using dual basis ,
This result implies
which automatically gives the important fact that forms a basis for where are bases of U and V.
Furthermore, given three vector spaces U, V, W the tensor product is linked to the vector space of all linear maps, as follows:
Here Hom(-,-) denotes the K-vector space of all linear maps. This is an example of adjoint functors: the tensor product is "left adjoint" to Hom.
Adjoint representation
The tensor may be naturally viewed as a module for the Lie algebra End(V) by means of the diagonal action: for simplicity let us assume r = s = 1, then, for each u ∈ End(V),
where u^{∗} in End(V^{∗}) is the transpose of u, that is, in terms of the obvious pairing on V ⊗ V^{∗},
- .
There is a canonical isomorphism given by
Under this isomorphism, every u in End(V) may be first viewed as an endomorphism of and then viewed as an endomorphism of End(V). In fact it is the adjoint representation ad(u) of End(V).
Tensor products of modules over a ring
The tensor product of two modules A and B over a commutative ring R is defined in exactly the same way as the tensor product of vector spaces over a field:
where now F(A × B) is the free R-module generated by the cartesian product and G is the R-module generated by the same relations as above.
More generally, the tensor product can be defined even if the ring is non-commutative (ab ≠ ba). In this case A has to be a right-R-module and B is a left-R-module, and instead of the last two relations above, the relation
is imposed. If R is non-commutative, this is no longer an R-module, but just an abelian group.
The universal property also carries over, slightly modified: the map φ : A × B → A ⊗_{R} B defined by (a, b) ↦ a ⊗ b is a middle linear map (referred to as "the canonical middle linear map".^{[9]}); that is,^{[10]} it satisfies:
The first two properties make φ a bilinear map of the abelian group A × B. For any middle linear map ψ of A × B, a unique group homomorphism f of A ⊗_{R} B satisfies ψ = f ∘ φ, and this property determines within group isomorphism. See the main article for details.
Tensor product of modules over a non-commutative ring
Let A be a right R-module and B be a left R-module B. Then the tensor product of A and B is an abelian group defined by
where is a free abelian group over and G is a subgroup of generated by relations
The universal property can be stated as follows. Let G be an abelian group with a map which is bilinear, in the sense that
Then there is a unique map such that for every .
Furthermore, we can give a module structure under some extra conditions.
- 1) If A was a (S,R)-bimodule, then is a left S-module where .
- 2) If B was a (R,S)-bimodule, then is a right S-module where .
- 3) If R was a commutative ring, then A and B are (R,R)-bimodules where and . By 1), is a left R-module. By 2), is a right R-module. So we can conclude is a (R,R)-bimodule.
Computing the tensor product
For vector spaces, the tensor product V ⊗ W is quickly computed since bases of V of W immediately determine a basis of V ⊗ W, as was mentioned above. For modules over a general (commutative) ring, not every module is free. For example, Z/nZ is not a free abelian group (= Z-module). The tensor product with Z/nZ is given by
More generally, given a presentation of some R-module M, that is, a number of generators m_{i} ∈ M, i ∈ I together with relations , with a_{ji} ∈ R, the tensor product can be computed as the following cokernel:
Here N^{J} := ⨁_{j ∈ J} N and the map is determined by sending some n ∈ N in the jth copy of N^{J} to a_{ji}n (in N^{I}). Colloquially, this may be rephrased by saying that a presentation of M gives rise to a presentation of M ⊗_{R} N. This is referred to by saying that the tensor product is a right exact functor. It is not in general left exact, that is, given an injective map of R-modules M_{1} → M_{2}, the tensor product
is not usually injective. For example, tensoring the (injective) map given by multiplication with n, n : Z → Z with Z/nZ yields the zero map 0 : Z/nZ → Z/nZ, which is not injective. Higher Tor functors measure the defect of the tensor product being not left exact. All higher Tor functors are assembled in the derived tensor product.
Tensor product of algebras
Let R be a commutative ring. The tensor product of R-modules applies, in particular, if A and B are R-algebras. In this case, the tensor product A ⊗_{R} B is an R-algebra itself by putting
For example,
A particular example is when A and B are fields containing a common subfield R. The tensor product of fields is closely related to Galois theory: if, say, A = R[x] / f(x), where f is some irreducible polynomial with coefficients in R, the tensor product can be calculated as
where now f is interpreted as the same polynomial, but with its coefficients regarded as elements of B. In the larger field B, the polynomial may become reducible, which brings in Galois theory. For example, if A = B is a Galois extension of R, then
is isomorphic (as an A-algebra) to the A^{deg(f)}.
Eigenconfigurations of tensors
Square matrices A with entries in a field K represent linear maps of vector spaces, say , and thus linear maps of projective spaces over . If A is nonsingular then is well-defined everywhere, and the eigenvectors of correspond to the fixed points of . The eigenconfiguration of A consists of points in , provided is generic and K is algebraically closed. The fixed points of nonlinear maps are the eigenvectors of tensors. Let be a -dimensional tensor of format with entries lying in an algebraically closed field of characteristic zero. Such a tensor defines polynomial maps and with coordinates
Thus each of the coordinates of is a homogeneous polynomial of degree in . The eigenvectors of are the solutions of the constraint
and the eigenconfiguration is given by the variety of the minors of this matrix.^{[11]}
Other examples of tensor products
Tensor product of sheaves of modules
Tensor product of Hilbert spaces
Topological tensor product
Tensor product of graded vector spaces
Tensor product of multilinear forms
Given two multilinear forms and on a vector space over the field their tensor product is the multilinear form
- ^{[12]}
This is a special case of the product of tensors if they are seen as multilinear maps (see also tensors as multilinear maps). Thus the components of the tensor product of multilinear forms can be computed by the Kronecker product.
Tensor product of graphs
It should be mentioned that, though called "tensor product", this is not a tensor product of graphs in the above sense; actually it is the category-theoretic product in the category of graphs and graph homomorphisms. However it is actually the Kronecker tensor product of the adjacency matrices of the graphs. Compare also the section Tensor product of linear maps above.
Monoidal categories
A general context for tensor product is that of a monoidal category.
Applications
Exterior and symmetric algebra
Two notable constructions in linear algebra can be constructed as quotients of the tensor product: the exterior algebra and the symmetric algebra. For example, given a vector space V, the exterior product
is defined as
Note that when the underlying field of V does not have characteristic 2, then this definition is equivalent to
The image of in the exterior product is usually denoted and satisfies, by construction, . Similar constructions are possible for (n factors), giving rise to , the nth exterior power of V. The latter notion is the basis of differential n-forms.
The symmetric algebra is constructed in a similar manner:
That is, in the symmetric algebra two adjacent vectors (and therefore all of them) can be interchanged. The resulting objects are called symmetric tensors.
Tensor product of line bundles
Tensor product in programming
Array programming languages
Array programming languages may have this pattern built in. For example, in APL the tensor product is expressed as ○.×
(for example A ○.× B
or A ○.× B ○.× 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. This product of two functions is 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).
See also
Look up tensor product in Wiktionary, the free dictionary. |
Notes
- ↑ See Tensor or Tensor (intrinsic definition).
- ↑ This similar to how the engineering use of " " specifically returns the remainder, one of the many elements of the equivalence class.
- ↑ Hazewinkel, Michiel; Gubareni, Nadezhda Mikhaĭlovna; Gubareni, Nadiya; Kirichenko, Vladimir V. (2004). Algebras, rings and modules. Springer. p. 100. ISBN 978-1-4020-2690-4.
- ↑ "Archived copy". Archived from the original on 2017-09-02. Retrieved 2017-09-02.
- ↑ Bourbaki (1989), p. 244 defines the usage "tensor product of x and y", elements of the respective modules.
- ↑ Analogous formulas also hold for contravariant tensors, as well as tensors of mixed variance. Although in many cases such as when there is an inner product defined, the distinction is irrelevant.
- ↑ "The Coevaluation on Vector Spaces". The Unapologetic Mathematician. 2008-11-13. Archived from the original on 2017-02-02. Retrieved 2017-01-26.
- ↑ See Compact closed category.
- ↑ Hungerford, Thomas W. (1974). Algebra. Springer. ISBN 0-387-90518-9.
- ↑ Chen, Jungkai Alfred (Spring 2004), "Tensor product" (PDF), Advanced Algebra II (lecture notes), National Taiwan University, archived (PDF) from the original on 2016-03-04
- ↑ Abo, H.; Seigal, A.; Sturmfels B. arXiv:1505.05729 [math.AG]
- ↑ Tu, L. W. (2010). An Introduction to Manifolds. Universitext. Springer. p. 25. ISBN 978-1-4419-7399-3.
References
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