# Zoran Skoda affine space

Let $V$ be a $n$-dimensional vector space over a fixed field $k$. A set $ℬ$ is called an affine space of dimension $n$ iff it carries a free and transitive action of the additive group of the vector space $V$.

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

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

$c+\left(b-a\right)=\left(b+\left(c-b\right)\right)+\left(b-a\right)=b+\left(\left(c-b\right)+\left(b-a\right)\right)=b+\left(c-a\right).$c + (b - a) = (b + (c - b)) + (b - a) = b + ((c-b) + (b-a)) = b + (c - a).

For each point $a\in ℬ$ we define a map ${\varphi }_{a}:ℬ\to V$ by ${\varphi }_{a}\left(b\right)=b-a$. This map is bijective and therefore there is a unique vector space structure on $ℬ$ which makes ${\varphi }_{a}$ an isomorphism of vector spaces. That vector space structure on $ℬ$ depends on $a\in ℬ$; thus we will denote it by ${V}_{a}$. For each pair $\left(a,b\right)\in ℬ×ℬ$ we can therefore define vector space isomorphisms ${\varphi }_{\mathrm{ab}}={\varphi }_{a}^{-1}\circ {\varphi }_{b}:{V}_{b}\to {V}_{a}$ and ${\psi }_{\mathrm{ab}}={\varphi }_{a}\circ {\varphi }_{b}^{-1}:V\to V$.

Let $\left(ℬ,V,\mu \right)$ and $\left(ℬ\prime ,V\prime ,\mu \prime \right)$ be two affine spaces. A map of sets $A:ℬ\to ℬ\prime$ is called an affine map if $\exists$ a linear map $L:V\to V\prime$ such that

$A\left(a+v\right)=A\left(a\right)+Lv,\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\forall v\in V.\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left(1\right)$A(a + v) = A(a) + L v, \,\,\forall v \in V. \,\,\,\,(1)

In other words, $\left(A\circ \mu \right)\left(v,a\right)=\mu \prime \left(Lv,A\left(a\right)\right)$. That property is satisfied iff it is satisfied for a single $a=p\in ℬ$. On the other hand each element $b\in ℬ$ can be represented as $p+\left(b-p\right)$ so that if we are given two points $p\in ℬ,q\in ℬ\prime$ and a linear map $L:V\to V\prime$ then $\exists !$ affine map $A:ℬ\to ℬ$ such that the equation (1) holds and $A\left(p\right)=q$.