# Zoran Skoda affine space

Let $V$ be a $n$-dimensional vector space over a fixed field $k$. A set $\mathcal{B}$ 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 $(\mathcal{B},V,\mu)$ where $\mu$ is the action. We also write $a + v \stackrel{def}{=} \mu(v,a)$.

Let $a, b \in \mathcal{B}$. 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 $a + ( b - a ) = b$. Other immediate properties are $a - a = 0$, and $c + (b - a) = b + (c - a)$ what justifies skipping some brackets. A proof of the last equality goes as follows:

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

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

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

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

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