Image a vector space where two points $$P$$ and $$P'$$ exist, then there's a unique translation of the plane that maps $$P$$ to $$P'$$ which means that the space of translations in the plane can be identified with a set of vectors that exist in the plane, composition of translation correspond to addition of vectors e.g. $$\v{PP''} = \v{PP'} + \v{P'P''}$$

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 $$\v{E}$$ (of translations) whose action results in an element of the set $$E$$, that is there's a map $$E \times \v{E} \rightarrow E: (a,\mathbf{v}) \mapsto a + \mathbf{v}$$ 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 correspond to translations i.e. for all $$a \in E$$ and $$\mathbf{u,v} \in \v{E}$$, $$x + (\mathbf{u} + \mathbf{v}) = (x + \mathbf{u}) + \mathbf{v}$$
3. For any $$a,b \in E$$ there's a unique free vector $$\mathbf{u} \in \v{E}$$ such that $$a + \mathbf{u} = b$$

The affine space is commonly represented with the triple $$\left \langle E, \v{E}, + \right \rangle$$ where $$E$$ is a set of points, $$\v{E}$$ a vector space acting on $$E$$ and an action $$+: E \times \v{E} \rightarrow E$$

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

$-x + y - 2 = 0$

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

$(x, 2 + x) + u = (x + u, 2 + x + u)$

For example the point $$(-2,0)$$ added with 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 parallels 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 the axiom 3, therefore

$c = b + \mathbf{bc} = (a + \mathbf{ab}) + \mathbf{bc} = a + (\mathbf{ab} + \mathbf{bc})$

And thus

$\mathbf{ab} + \mathbf{bc} = \mathbf{ac}$

Which is known as Chasles's identity

## Affine combinations

Consider $$\mathbb{R}^2$$ an affine space with its origin at $$(0,0)$$ and basis vectors $$\mathbf{b_1} = [1, 0]$$ and $$\mathbf{b_2} = [0,1]$$, 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 of coordinates

$(\lambda a_1 + \mu b_1, \lambda a_2 + \mu b_2)$

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 and the restriction is that the scalar add up to 1

Lemma: Given an affine space $$E,v{E},+$$, let $$a_i, i \in I$$ be a family of points in $$E$$ and let $$\lambda_i, i \in I$$ a family of scalars then any two points $$a,b \in E$$ the following properties hold

$\begin{equation} \label{lemma-1} a + \sum_{i \in I} \lambda_i \mathbf{aa_i} = b + \sum_{i \in I} \lambda_i \mathbf{ba_i} \quad \text{if \sum_{i \in I} \lambda_i = 1} \end{equation}$

and

$\begin{equation} \label{lemma-2} \sum_{i \in I} \lambda_i \mathbf{aa_i} = \sum_{i \in I} \lambda_i \mathbf{ba_i} \quad \text{if \sum_{i \in I} \lambda_i = 0} \end{equation}$

To prove \eqref{lemma-1} we apply Chasles's identity

\begin{align*} a + \sum_{i \in I} \lambda_i \mathbf{aa_i} &= a + \sum_{i \in I} \lambda_i (\mathbf{ab} + \mathbf{ba_i}) \\ &= a + (\sum_{i \in I} \lambda_i) \mathbf{ab} + \sum_{i \in I} \lambda_i \mathbf{ba_i} \\ &= a + \mathbf{ab} + \sum_{i \in I} \lambda_i \mathbf{ba_i} \quad \text{since \sum_{i \in I} \lambda_i = 1} \\ &= b + \sum_{i \in I} \lambda_i \mathbf{ba_i} \quad \text{since b = a + \mathbf{ab}} \\ \end{align*}

For \eqref{lemma-2} we also have

\begin{align*} \sum_{i \in I} \lambda_i \mathbf{aa_i} &= \sum_{i \in I} \lambda_i (\mathbf{ab} + \mathbf{ba_i}) \\ &= (\sum_{i \in I} \lambda_i) \mathbf{ab} + \sum_{i \in I} \lambda_i \mathbf{ba_i} \\ &= \sum_{i \in I} \lambda_i \mathbf{ba_i} \quad \text{since \sum_{i \in I} \lambda_i = 0} \\ \end{align*}

Formally for any family of points $$a_i, i \in I$$ in $$E$$, for any family $$\lambda_i, i \in I$$ of scalars such that $$\sum_{i \in I} \lambda_i = 1$$ the point

$\begin{equation} \label{affine-combination} x = a + \sum_{i \in I} \lambda_i \mathbf{aa_i} \end{equation}$

Is independent of $$a \in E$$ and is called the barycenter or affine combination of the points $$a_i$$ with weights $$\lambda_i$$, and is denoted as

$\sum_{i \in I} \lambda_i a_i$

## Affine maps

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

$f \left (\sum_{i \in I} \lambda_i a_i \right ) = \sum_{i \in I} \lambda_i f(a_i)$

References
• Bærentzen, J. A., Gravesen, J., Anton François, & Aanæs, H. (2012). Guide to computational geometry processing: foundations, algorithms, and methods. London: Springer.