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 $a+v\stackrel{\mathrm{def}}{=}\mu (v,a)$.

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 ${\varphi}_{a}:\mathcal{B}\to V$ by ${\varphi}_{a}(b)=b-a$. This map is bijective and therefore there is a unique vector space structure on $\mathcal{B}$ which makes ${\varphi}_{a}$ 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 ${\varphi}_{\mathrm{ab}}={\varphi}_{a}^{-1}\circ {\varphi}_{b}:{V}_{b}\to {V}_{a}$ and ${\psi}_{\mathrm{ab}}={\varphi}_{a}\circ {\varphi}_{b}^{-1}:V\to V$.

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

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

In other words, $(A\circ \mu )(v,a)=\mu \prime (Lv,A(a))$. 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}\prime $ and a linear map $L:V\to V\prime $ then $\exists !$ affine map $A:\mathcal{B}\to \mathcal{B}$ such that the equation (1) holds and $A(p)=q$.

See also affine space.

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