There are several constructions of new dual pairs from given ones. In particular, as all vector spaces involved, any tensor space such as

We start with a vector space

Here,

Now let ϕ :

Finally, let

The fact

Thus in finite dimensional cases,

One reason to use this dual pair (

But the standard reason is that it is required for the tensor product to satisfy its below universality.

Here,

- It is bilinear : ∀
*x*∈*E*', ∀*T*∈*E*⊠*F*,*T*_{ij}*x*∈_{i}*F*and the same on the other side. - It is universal for this bilinearity : for any dual pair (
*G*,*G*') and any bilinear operation*g*:*E'*×*F'*→*G'*there is a unique linear map*h*:*E'*⊗*F'*→*G'*such*g*=*h*⚬⊗. Indeed defining*h*by restriction of^{t}(*G*∋*z*↦*z*⚬*g*) : (*E*⊠*F*)*→*G**, we get*g*=*h*⚬⊗ (why ?); then Im(*h*⚬⊗)⊂*G'*implies Im*h*⊂*G'*because Im ⊗ generates*E'*⊗*F'*= Dom*h*, which also ensures the uniqueness of*h*.

To justify these natural isomorphisms, note that our construction of a space of the form

Remains the issue of the difference between ⊗ and ⊠, which disappears with finite dimensional spaces.

g⊗f

〈x, y〉

〈

A monomial tensor expression is a tree if between any 2 vertices
there exists one and only one path that does not repeat any vertex
(so as to forget paths that can be obviously shortened by cutting
some edges). In other words it is a graph that :

- is connected : it cannot be split without cutting some edge (existence of a path between any two vertices)
- does not contain any loop (that give different paths between some vertices). In particular it does not contain any edge with its two ends at the same vertex (which is otherwise admissible in other tensor expressions).

Indeed, every vertex, or edge, or free end, can be chosen to be seen as the main symbol of the expression (if a vertex it is an operation with values in ℝ; if an edge it is a scalar product; if a free end it is an operation symbol with vectorial value); every edge is marked by the orientation of the unique path to this "main symbol", and every vertex is interpreted as the operation symbol that follows these orientations.

All these interpretations give the same result because between any two choices of main symbol there is a path, and the result is preserved at every step of this path (from each edge to each next edge), thanks to the identities between the different algebraic interpretations of each symbol (vertex).

Any graph that is not connected is divided into several separate
connected components in a unique way.

This possibility to interpret disconnected graphs can also be expressed as follows:

This rank is also equal to the dimension of the image set of each of both maps that t defines as an element of E⊠F, from E' to F and from F' to E.

Proof: The image in F of the map defined by x

If the y

Note that the proof of this equality between expressions in E⊗F is processed in the classical concept of E⊗F (universal algebraic : as quotient of the set of formal combinations of elements of E'×F' by the relations in each of E' and F') rather than as a dual to E'⊠F'. From this we can deduce that the map from the classical E⊗F to E⊠F is injective, and therefore both definitions of E⊗F coincide. We shall identify E⊗F to its image in E⊠F.

Now let us directly define the rank of an element of E⊠F as the dimension of each of its images in E and in F (and thus cannot exceed the smallest of the dimensions of E and F); it is also the dimension of the dual vector spaces it defines in the role of a scalar product between (the quotiented) E' and F'.

An element t of E⊠F belongs to E⊗F if and only if its rank is finite. In this case, its two images A in E and B in F are in duality to each other by defining for every x ∈A and y∈B, their scalar product by x.y'=y.x' (= x'ty') for any elements y'∈F' and x'∈E' such that ty'=y and x't=x.

The choice of a basis x

Thus E⊗F = E⊠F when E or F is finite dimensional, but generally not otherwise. Anyway both E⊗F and E⊠F are duals to E'⊗F' .

Now if E has finite dimension then we have a basis of n vectors e

Let us apply the trace function to this element I of E⊗E' itself. This transforms each tensor product into a scalar product, thus giving (e

- They are multilinear with respect to each symbol (distributive
over addition, and scalar factors can be put outside):

- If a tensor symbol x satisfies x=y+z then for each occurence of x in a tensor expression the result is the sum of those obtained by replacing this occurence of x with those of y and z.
- For any scalar a, replacing an occurence of x by ax
multiplies the result by a; isolated components of the graph
mean such scalar multiplication.

- Any subgraph of a graph distinguished by taking a subset of the set of vertices, can be replaced by a single symbol equal to the monomial expression defined by that subgraph. Thus when a symbol equals a linear combination of graphs, the whole equals the same linear combination where one occurence of this symbol is replaced by each graph in the combination.
- Any edge can be replaced by a "long edge" trough the identity
symbol.

But let us show that it makes sense in all other cases, that is

- Whatever tensor expression only using finite-dimensional
spaces

- More generally, expressions that may contain infinite-dimensional spaces but that do not form any loop (every loop in the graph goes through some finite-dimensional space or some tensor of finite rank).

- at a vertex representing a tensor of finite rank, or

- at an edge labelled by a finite-dimensional space, replaced by a "long edge" through the identity element that has finite rank.

This result does not depend on the choice of decomposition. Indeed if you have 2 decompositions applied to edges (or at least that does not apply 2 different decompositions to the same vertex), then let us consider making both decompositions together (if an edge is decomposed two ways, let us see it as a long edge through 2 copies of I, and apply the decomposition to a different copy of I in both cases). Then we can verify that the result of the double decomposition equals that of each of both decompositions.

Another way of seeing it, is to consider that an element of E⊠F with finite rank is identified to an element of E⊗F, which is in duality with E'⊠F'. The use of this duality gives meaning to the expression.

This formalism provides the computation : dim(E⊗F)= (dim E) (dim F)

Any family of n vectors in a space E can be formalized as an element of ℝ

In many cases, the trick to avoid any risk of mistake is to introduce vector spaces with special names to label edges in the tensor expressions, such as "ℝ

(to be continued)