This page under construction is to provide some technical details to
justify more completely the formalism of tensors
There are several constructions of new dual pairs from given ones.
In particular, as all vector spaces involved, any tensor space such
as E⊠F needs to be seen as a component of a dual
pair, namely (E⊠F,E'⊗F'). This and other
constructions follow the same procedure as follows.
We start with a vector space H of ℝ-valued operations, thus
a subspace of ℝX for some set X.
Here, X = E'×F' and H = E⊠F.
Now let ϕ : X → H* defined by ∀x∈X,∀y∈H,
ϕ(x)(y) = y(x).
Finally, let H' be the subspace of H* generated by
Im ϕ. This way, we have an operation ϕ : X → H'.
The fact H' is indeed a possible dual of H is easily
deducible from Im ϕ ⊂H.
Thus in finite dimensional cases, H' = H*.
One reason to use this dual pair (H,H') instead of (H,H*)
is that, when they differ in infinite dimensional cases, H'
is anyway what we need for physics, while other possible elements of
H* are often just "mathematical monsters" whose existence
depends on the axiom of choice.
But the standard reason is that it is required for the tensor
product to satisfy its below universality.
Here, H' = E'⊗F' and ϕ is the tensor product
⊗ : E'×F' → E'⊗F'. It has
The tensor product is more generally defined on any list of dual
pairs: for example the tensor product of 3 dual pairs (E,E'),(F,F'),(G,G')
is a dual pair (E⊗F⊗G, E'⊠F'⊠G')
where essentially E⊗F⊗G=(E⊗F)⊗G=E⊗(F⊗G)
is generated by E×F×G, and E'⊠F'⊠G'
= (E'⊠F')⊠G'= E'⊠(F'⊠G')
is a tensor space with order 3.
- It is bilinear : ∀x∈E', ∀T∈ E⊠F,
Tij xi ∈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 E'⊠F' is essentially unaffected
when replacing E
by one of its generating subsets, and the same for F. In
particular, for any basis B⊂E, the space E'
of all linear forms on E is identifiable as the space ℝB
of all functions on B (without any more linearity
condition), and the space E'⊠F' is essentially
unchanged when replacing E by B.
Remains the issue of the difference between ⊗ and ⊠, which
disappears with finite dimensional spaces.
Meaning of a tree-like tensor expression
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 :
Such an expression can be interpreted as an algebraic expression in
as many ways as there are vertices, edges and free ends in the
graph; and all these ways give the same result in the expected
- 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
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).
Meaning of tensor expressions without loop
Any graph that is not connected is divided into several separate
connected components in a unique way.
The result of such an expression is given by separately making the
computation in each tree in the above way, then multiplying the
results. This can either be seen as a multiplication between
numbers, or between numbers and one vector, depending on the
interpretation of the expression.
This possibility to interpret disconnected graphs can also be
expressed as follows:
The injection from E⊗F to E⊠F
There is a natural map from E⊗F to E⊠F defined by mapping each (x,y)
in E×F to the map sending each (x',y') in E'×F' to the real number
obtained by multiplying in ℝ the scalar products : (x.x')(y.y').
Indeed for every (x,y) ∈ E×F, this map from E'×F' to
ℝ has its currified forms defining x'↦(x.x')y from E' to F, and
y'↦(y.y')x from F' to E, thus indeed belongs to E⊠F.
Every element t of a tensor product E⊗F has finite rank (in the
sense of the rank of a matrix, to not confuse with the traditional
use of this word for tensors, that we call here its arity or degree,
here equal to 2), defined as the minimum number of elements of the
form x⊗y (the rank 1 element) whose sum gives t.
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 x1⊗y1+...+xn⊗yn
from E' to F, is contained in the subspace of F generated by y1,...,yn,
thus its dimension is no larger than n.
If the y1,...,yn were not linearly
independent, they could be decomposed in another basis with a
smaller number of elements (made of some of them), providing another
expression of that element of E⊗F as a sum of a smaller number of
rank 1 elements. The same for the x1,...,xn.
The linear independence of the x1,...,xn
ensures that the image is equal to the subspace of F generated by y1,...,yn,
that has dimension n.
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 x1,...,xn of A and its
dual basis y1,...,yn of B gives a
decomposition of t as t = x1⊗y1+...+xn⊗yn.
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' .
The identity element, the trace
Consider the case of F=E'. Then we have the natural element I in
E⊠E' (also usually noted as δ (small delta) named "Kronecker
symbol") that gives the scalar product itself (as a map from E'×E to
ℝ) and behaves as the identity from E to itself, and the identity
from E' to itself. Seen as a map from E'⊗E to ℝ, this element is
also called the trace. Its rank equal to the common
dimension of E and E', thus belongs to E⊗E' only if this dimension
Now if E has finite dimension then we have a basis of n vectors e1
... en, and its dual basis e'1...e'n,
by which the element I can be written as the image in E⊠F of the
Let us apply the trace function to this element I of E⊗E' itself.
This transforms each tensor product into a scalar product, thus
1+...+1 = n. This cannot be done in an infinite dimensional space,
as the result would be infinite. So we have this rule : for every
vector space, its dimension equals the trace of the identity in this
Meaning of any tensor expression without infinite-dimensional
Tensor expressions have the following properties, that we can verify
in previous cases (without loop), and that will be postulated as
We saw above (with the trace of the identity) that it is not
generally possible to make sense of a tensor expression containing
an infinite-dimensional loop, that is a loop (path in the graph that
comes back to itself) where all edges are labelled with
infinite-dimensional spaces, and vertices have infinite rank. (At
least in an algebraic manner ; let us not mention constructions
where they could be defined by infinite sums that converge to a
limit, that could be used to generalize the concept).
- 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
- 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
But let us show that it makes sense in all other cases, that is
For this, choose a way to "cut the graph" at least once at some
finite-dimensional part of every loop : either
- Whatever tensor expression only using finite-dimensional
- 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).
As each of these vertices is replaced by a linear by a linear
combination of disconnected graphs, this produces a big linear
combination of graphs where all loops are cut (this combination does
not depend on the order between vertices to which the decomposition
is applied, thanks to the commutativity and associativity of
addition). Thus a well-defined result according to the above
- 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)
How transformation formulas are obtained
The expression of how components of tensors are transformed during
changes of coordinates (or transformations of the composing spaces),
can be naturally obtained by the following tools:
Any family of n vectors in a space E can be formalized as an element
of ℝn⊗E. It is a basis when it is inversible. In this
case, its inverse, that belongs to ℝn⊗E', defines the
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 "ℝn as a representation of E
in this basis", thus distinct from its dual "ℝn as a
representation of E' in the dual basis".
(to be continued)
Quantum Mechanics is an application of the Penrose diagrams
notation for tensors, to the case of quantum mechanics.