Zoran Skoda affine space

Let VV be a nn-dimensional vector space over a fixed field kk. A set ℬ\mathcal{B} is called an affine space of dimension nn iff it carries a free and transitive action of the additive group of the vector space VV.

Thus an affine space is formally a triple (ℬ,V,μ)(\mathcal{B},V,\mu) where μ\mu is the action. We also write a+v=defμ(v,a)a + v \stackrel{def}{=} \mu(v,a).

Let a,b∈ℬa, b \in \mathcal{B}. Then by transitivity of the action, there is an element v∈Vv \in V such that b=a+vb = a + v. By freeness such an element is unique so we denote that unique element by b−ab-a. Thus a+(b−a)=ba + ( b - a ) = b. Other immediate properties are a−a=0 a - a = 0, and c+(b−a)=b+(c−a)c + (b - a) = b + (c - a) what justifies skipping some brackets. A proof of the last equality goes as follows:

c+(b−a)=(b+(c−b))+(b−a)=b+((c−b)+(b−a))=b+(c−a). c + (b - a) = (b + (c - b)) + (b - a) = b + ((c-b) + (b-a)) = b + (c - a).

For each point a∈ℬa\in \mathcal{B} we define a map ϕ a:ℬ→V\phi_a : \mathcal{B} \rightarrow V by ϕ a(b)=b−a\phi_a(b) = b-a. This map is bijective and therefore there is a unique vector space structure on ℬ\mathcal{B} which makes ϕ a\phi_a an isomorphism of vector spaces. That vector space structure on ℬ\mathcal{B} depends on a∈ℬa \in \mathcal{B}; thus we will denote it by V aV_a. For each pair (a,b)∈ℬ×ℬ(a,b) \in \mathcal{B} \times \mathcal{B} we can therefore define vector space isomorphisms ϕ ab=ϕ a −1∘ϕ b:V b→V a\phi_{ab} = \phi^{-1}_a \circ \phi_b : V_b \rightarrow V_a and ψ ab=ϕ a∘ϕ b −1:V→V\psi_{ab} = \phi_a \circ \phi^{-1}_b : V \rightarrow V.

Let (ℬ,V,μ)(\mathcal{B},V,\mu) and (ℬ′,V′,μ′)(\mathcal{B}',V',\mu') be two affine spaces. A map of sets A:ℬ→ℬ′A : \mathcal{B} \rightarrow \mathcal{B}' is called an affine map if ∃\exists a linear map L:V→V′L : V \rightarrow V' such that

A(a+v)=A(a)+Lv,∀v∈V.(1) A(a + v) = A(a) + L v, \,\,\forall v \in V. \,\,\,\,(1)

In other words, (A∘μ)(v,a)=μ′(Lv,A(a))(A\circ \mu)(v,a) = \mu'(L v,A(a)). That property is satisfied iff it is satisfied for a single a=p∈ℬa = p \in \mathcal{B}. On the other hand each element b∈ℬb \in \mathcal{B} can be represented as p+(b−p)p + (b - p) so that if we are given two points p∈ℬ,q∈ℬ′p \in \mathcal{B}, q \in \mathcal{B}' and a linear map L:V→V′L : V \rightarrow V' then ∃!\exists ! affine map A:ℬ→ℬA : \mathcal{B}\rightarrow \mathcal{B} such that the equation (1) holds and A(p)=qA(p) = q.

See also affine space.

Last revised on August 25, 2009 at 19:55:47. See the history of this page for a list of all contributions to it.