5.6. Barycenters

Conceiving affine and vector spaces as ℝ-duality systems, reduces the issue of formalizing affine and linear geometries, to the following questions:
1. Describing ℝ with its algebraic structures, and its axioms (including second-order ones);
2. We shall accept all duality morphisms. Works of functional analysis involving more primitive structures to restrict the class of morphisms, are beyond the scope of this exposition.
3. Specify the axioms (criteria to accept duality systems as spaces): E and E' will be made algebras by stability axioms E ∈ Sub ℝE' and E' ∈ Sub ℝE, from different algebraic structures on ℝ which we have to describe. That will be all for us.

Properties of trajectories have generalizations to other arities, as follows. Consider a concrete category having for every n∈ℕ a space with a basis of n elements. They form together a language L that is a clone of operations which means, generalizing the concept of monoid, that all operations defined by L-terms also belong to L....

Vector spaces

Linear geometry can be conceived as equivalent to affine geometry with an additional constant symbol 0 called origin or zero (no additional axiom could be relevant since affine groups have only one orbit):
• A vector space is an affine space E with a chosen point 0E called "origin"
• A linear map between vector spaces E and F is an affine map f such that f(0E) = 0F.
• A subspace of a vector space E is an affine subspace of E containing 0E (so by definition, each subspace is an affine subspace but not conversely)
• Subspaces can be used in the role of directions, as any direction contains a unique subspace.
This does not formalize linear geometry, as we did not formalize affine geometry here assumed. Optimal formalizations of linear geometry do not even go this way, but logically come first, as affine geometry can be more usefully formalized using it.
Indeed, affine subspaces F of vector spaces E are still affine spaces: affine structures of F are definable from linear structures of E with the choice of F, in the sense that the group GF⊂GLE of automorphisms of E which preserve F, is morphed by restriction into Aff F.
Now for this to form an equivalent presentation of the affine geometry of F, this morphism from GF to Aff F must be bijective:
• 0EF (gives surjectivity : the choice of 0E is not structuring for F).
• Then F must be an hyperplane (to give injectivity).

Basis of affine spaces

Now let us start a proper formalization of affine and linear geometries in terms of basis.
A fundamental feature of affine geometry is that for any n∈ℕ, n-dimensional spaces have basis of n+1 elements, and any point belongs to some basis. Then a possible presentation of linear geometry (but not the most formally convenient),

The barycenter of points x and y with coefficients (1-a) and a is the image of a by the unique affine map from ℝ to M which sends 0 to x and 1 to y.

Middles of segments are defined from barycenters.

k-dimensional subspaces are generated by a set of k+1 points but not generated by any set of k points or less.

A subset F of an affine space E is a subspace of E, if and only if ∀a,bF, ababF.
An equivalent condition for FE to be an hyperplane is ∃aE\F, E = Fa, which implies ∀aE\F, E = Fa.
Generally ∀FS(E), ∀aE\F, Dim(Fa) = Dim(F)+1.

Vector spaces in duality

A pair of dual vector spaces, is a particular case of duality system.
It is a pair of sets (E,E'), where the elements of E are called vectors, those of E' are called covectors (or linear forms) together with an operation of scalar product from E×E' to ℝ : for each u∈E and x∈E', we have u⋅x∈ℝ, and that is subject to the following axioms:

We have a zero element in E, whose scalar product with any element of E' is zero; and a zero element of E', whose scalar product by any element of E is zero. We shall abusively denote the zeros of all sets by the same symbol 0, letting their precise identity be given by the operations where they are used, as this won't cause any ambiguity:

• ∀x∈E', 0⋅x = 0
• ∀u∈E', u⋅0 = 0

We define an operation of addition in E  as the addition of the scalar products with any element of E' ; and the same in E'. Formally:

• ∀u,v∈E, ∀x∈E', (u+v)⋅x = u⋅x + v⋅x
• ∀x,y∈E', ∀u∈E, u⋅(x+y) = u⋅x + u⋅y

In other words : for all u,v∈E, the function from E' to R defined by (x ↦ u⋅x + v⋅x) "belongs to" (or : is represented through the scalar product by) an element of E that is denoted u+v.

We define an operation of multiplication of elements of E by any real number, as multiplying by the same number, the scalar product with any element of E'. And the same in E'. Formally :

• ∀a∈ℝ, ∀u∈E, ∀x∈E', (au)⋅x = a(u⋅x) = u⋅(ax)

Remark : the zero element of E can as well be obtained from any element u of E by the multiplication by 0∈ℝ: 0 = 0u. But declaring the constant 0∈E further says that E is nonempty.
The above definitions of the operations in E can also be equivalently expressed in the following two ways

• E represents (though the scalar product) a vector subspace of the space of functions from E' to ℝ. (stable by addition and multiplication by a scalar).
• E is a vector space, and the elements of E' represent linear functions from E to ℝ.

(the below will be reworked later)

A subspace is a subalgebra (i.e. stable) for addition and multiplication by a number.
Note that (as for any duality system) from any pair (E,E') with an operation from E×E' to ℝ we can produce a pair of spaces in duality by replacing E by the vector space generated by its image in ℝE', and the same for E'. This can be done in any order without affecting the result and indeed gives a pair of dual vector spaces by doing this replacement just once on each side, because this replacement for E does not affect the linear relations in E' (relations in E' that are true in ℝE remain true when replacing E by the vector space generated by its image in ℝE'), thanks to the algebraic relations between addition and multiplication (commutativity, associativity, distributivity).

If (E,E') is a pair of dual spaces and E is finite-dimensional then E' is the space of all linear forms on E and has the same dimension.
There are counter-examples in the infinite-dimensional case. For example take E = E' = the set of continuous maps from [-1,1] to ℝ, and the operation of integral of the product of these functions.
Then the Dirac mass in 0 (or in any other point of [0,1], which maps any f in E to f(0), is outside E'. It may still be understood as a limit of a sequence of elements of E'.