Zoran Skoda
affine space

Let VV be a nn-dimensional vector space over a fixed field kk. A set \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 (,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,ba, b \in \mathcal{B}. Then by transitivity of the action, there is an element vVv \in V such that b=a+vb = a + v. By freeness such an element is unique so we denote that unique element by bab-a. Thus a+(ba)=ba + ( b - a ) = b. Other immediate properties are aa=0 a - a = 0, and c+(ba)=b+(ca)c + (b - a) = b + (c - a) what justifies skipping some brackets. A proof of the last equality goes as follows:

c+(ba)=(b+(cb))+(ba)=b+((cb)+(ba))=b+(ca). c + (b - a) = (b + (c - b)) + (b - a) = b + ((c-b) + (b-a)) = b + (c - a).

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

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

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

In other words, (Aμ)(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=pa = p \in \mathcal{B}. On the other hand each element bb \in \mathcal{B} can be represented as p+(bp)p + (b - p) so that if we are given two points p,qp \in \mathcal{B}, q \in \mathcal{B}' and a linear map L:VVL : V \rightarrow V' then !\exists ! affine map A:A : \mathcal{B}\rightarrow \mathcal{B} such that the equation (1) holds and A(p)=qA(p) = q.

See also affine space.

Revised on August 25, 2009 19:55:47 by Zoran Škoda (161.53.130.104)