Imagine a vector space where two points $P$ and $P^{\prime }$ exist. Then, there’s a unique translation of the plane that maps $P$ to $P^{\prime }$. This means that the space of translations in the plane can be identified with a set of vectors that exist in the plane. The composition of translations corresponds to the addition of vectors, e.g., $\overrightarrow{P{P}^{″}}=\overrightarrow{P{P}^{\prime }}+\overrightarrow{P^{\prime }{P}^{″}}$.

An affine space is a space where translation is defined. Formally, an affine space is a set $E$ (of points) that admits a free transitive action of a vector space $\overrightarrow{E}$ (of translations) whose action results in an element of the set $E$. That is, there’s a map $E\times \overrightarrow{E}\to E:(a,\mathbf{v})\mapsto a+\mathbf{v}$ such that

1. The zero vector acts as an identity, i.e., for all $a\in E$, $a+\mathbf{0}=a$.
2. Addition of vectors corresponds to translations, i.e., for all $a\in E$ and $\mathbf{u}\mathbf{,}\mathbf{v}\in \overrightarrow{E}$, $x+(\mathbf{u}+\mathbf{v})=(x+\mathbf{u})+\mathbf{v}$.
3. For any $a,b\in E$, there’s a unique free vector $\mathbf{u}\in \overrightarrow{E}$ such that $a + \mathbf{u} = b.

The affine space is commonly represented by the triple $⟨E,\overrightarrow{E},+⟩$, where $E$ is a set of points, $\overrightarrow{E}$ is a vector space acting on $E$, and $+$ is an action $E\times \overrightarrow{E}\to E$.

Consider a subset $L$ of ${\mathbb{A}}^2$ consisting of points satisfying

Where any point has the form $(x,f(x))=(x,2+x)$, the line can be made into an affine space by defining $+:L\times V\to L$ (note that $V$ is a vector space) such that for any $u\in V$

For example, the point $(-2,0)$ added to the vector $u=[1,1]$ results in the point $(-1,1)$, which belongs to the set $L$. Note that for the example above, the vector space $V$ has only vectors parallel to $u=[1,1]$.

Chasles’s Identity

Given any three points $a,b,c\in E$, we know that $c=a+\mathbf{ac}$, $b=a+\mathbf{ab}$, and $c=b+\mathbf{bc}$ by Axiom 3. Therefore,

And thus

Which is known as Chasles’s identity

Affine combinations

Consider ${\mathbb{R}}^2$ as an affine space with its origin at $(0,0)$ and basis vectors ${\mathbf{b}}_{\mathbf{1}}=[1,0]$ and ${\mathbf{b}}_{\mathbf{2}}=[0,1]$. Given any two points $a,b\in {\mathbb{R}}^2$ with coordinates $a=(a_1,a_2)$ and $b=(b_1,b_2)$, we can define the affine combination $\lambda a+\mu b$ as the point with coordinates

Let $\lambda =1,\mu =1$, $a=(-1,1)$, and $b=(2,2)$. Then $a+b=(1,1)$.

If we change the coordinate system to have an origin at $(1,1)$ with the same basis vectors, then the coordinates of the given points are $a=(-2,-2)$ and $b=(1,1)$. The linear combination is then $a+b=(-1,-1)$, which is the same as the point $(0,0)$ of the first coordinate system. Therefore, $a+b$ corresponds to two different points depending on the coordinate system used.

A restriction is needed for affine combinations to make sense: the scalars must add up to 1.

Lemma: Given an affine space $E,\overrightarrow{E},+$, let $a_i,i\in I$ be a family of points in $E$ and let ${\lambda }_i,i\in I$ be a family of scalars. Then, for any two points $a,b\in E$, the following properties hold:

To prove $\text{(1)}$, we apply Chasles’s identity:

For $\text{(2)}$, we also have:

Formally, for any family of points $a_i,i\in I$ in $E$ and for any family ${\lambda }_i,i\in I$ of scalars such that $\sum _{i\in I}{\lambda }_i=1$, the point

is independent of $a\in E$ and is called the barycenter or affine combination of the points $a_i$ with weights ${\lambda }_i$. It is denoted as

Affine maps

An affine map between two affine spaces $X$ and $Y$ is a map $f:X\to Y$ that preserves affine combinations, i.e.,