# Affine space $-$ Understanding basic example

I'm learning affine geometry and I'm having a hard time understanding a basic example related to the definition of an affine space and the notion of an action.

Before giving the example which causes me problems I'm just going to restate the definition of an affine space just so we can refer to it later:

Definition. An affine space is a triple $(E, \vec E, +)$ consisting of a nonempty set $E$ (of points), a vector space $\vec E$ (of translations, or free vectors), and an action $+ : E \times \vec E \rightarrow E$, satisfying the following conditions.

$(A1)$ $a + 0 = a$, for every $a \in E$.

$(A2)$ $(a + u) + \vec v = a + (u + v)$, for every $a \in E$, and every $u, v \in \vec E$.

$(A3)$ For any two points $a, b \in E$, there is a unique $u \in \vec E$ such that $a + u = b$.

I like the following description which says that, intuitively, we can think of the elements of $\vec E$ as forces moving the points in $E$, considered as physical particles. The effect of applying a force (free vector) $u \in \vec E$ to a point $a \in E$ is a translation. By this, we mean that for every force $u \in \vec E$ , the action of the force $u$ is to “move” every point $a \in E$ to the point $a + u \in E$ obtained by the translation corresponding to $u$ viewed as a vector.

Given the above description I think a understand the definition of an affine space reasonably well. However my comprehension of the definition is not as good when I look at the following basic example:

Consider the subset $\mathscr l$ of $\mathbb A^2$ consisting of all points $(x, y)$ satisfying the equation $x + y −1 = 0$. The set $\mathscr l$ is the line of slope $−1$ passing through the points $(1,0)$ and $(0,1)$. No problem until here. The line $\mathscr l$ can be made into an official affine space by defining the action $+ : \mathscr l \times \mathbb R \to \mathscr l$ of $\mathbb R$ on $\mathscr l$ defined such that for every point $(x, 1 − x)$ on $\mathscr l$ and any $u ∈ \mathbb R$, $$(x,1−x)+u = (x+u,1−x−u).$$ It is easily verified that this action makes $\mathscr l$ into an affine space.

The last part is not at all obvious to me. Here are a couple of questions to begin with:

• $(Q1)$ Why do they choose $\mathbb R$ as the vector space acting on $\mathscr l$ and not $\mathbb R^2$? Is it possible to define an action with $\mathbb R^2$ here? Vectors in $\mathbb R^2$ will act as forces (translations) on the points of $\mathscr l$. Since $\mathbb R$ consists of points on the real line I don't see how they are actually acting on $\mathscr l$.
• $(Q2)$ How do they actually construct the action? Is there a general method to define an action?
• $(Q3)$ Is there a geometric interpretation as to what is actually going on here? I don't see how the real line $\mathbb R$ is related to $\mathscr l$.
• $(Q4)$ I've read that an affine space is a way of defining a vector space structure on a set of points, without making a commitment to a fixed origin. I don't really understand this and how I can relate this statement to this problem.

(Q1) ... Since $\mathbb R$ consists of points on the real line I don't see how they are actually acting on $\mathscr l$.
The quoted $+ : \mathscr l \times \mathbb R \to \mathscr l$, defined by $(x,1−x)+u = (x+u,1−x−u)$, completes the definition of the affine space $(\mathscr l, \mathbb R, +)$. You should verify that $(\mathscr l, \mathbb R, +)$ indeed satisfies the definition of an affine space, by verifying axioms A1, A2, and A3.
For example, to verify A1, show that for every $(x,1−x)\in\mathscr l$, we have $(x,1−x)+0=(x,1−x)$.
(Q1) Why do they choose $\mathbb R$ as the vector space acting on $\mathscr l$ and not $\mathbb R^2$? Is it possible to define an action with $\mathbb R^2$ here? Vectors in $\mathbb R^2$ will act as forces (translations) on the points of $\mathscr l$.
Since $\mathscr l$ has the same cardinality as $\mathbb R^n$ for any $n\geq1$, technically, yes, we can define an action with $\mathbb R^n$. But it won't look the way you expect. If you try the naive action with $\mathbb R^2$, it will either fail to be closed, or it will violate A3. Try it!