Equivalence of Definitions of Affine Space

Associativity Axioms implies Weyl's Axioms
Assume the axioms $(A1)$, $(A2)$, $(A3)$. Then for any $p, q \in \mathcal E$ we have:

Therefore by Identity is Unique applied to the vector space $V$ we have:

Now let $p \in \mathcal E$, $v \in V$ as in $(W1)$.

We must show there exists a unique $q \in \mathcal E$ such that $v = q - p$.

Let $q = p + v$. Then:

Now let $r \in \mathcal E$ be any other element such that $v = r - p$. Then:

This shows that $q$ is unique and establishes $(W1)$.

Now let $p, q, r \in \mathcal E$ as in $(W2)$. Then:

This establishes $(W2)$.

Weyl's Axioms implies Group Action
Assume the axioms $(W1)$, $(W2)$.

Let $\phi: \mathcal E \times V \to \mathcal E$ be the group action defined by:


 * $\forall \tuple {p, v} \in \mathcal E \times V: p + v := \map \phi {p, v} = q$

where $q \in \mathcal E$ is the unique point such that $v = q - p$ given by $(W1)$.

We must verfiy:

To establish $(RGA1)$ let $p \in \mathcal E$ and $u, v \in V$. Then by $(W1)$

Then we have:

Therefore by uniqueness in $(W1)$ we must have $r = s$. Therefore:

Now to establish $(RGA2)$ let $p \in \mathcal E$ and choose any other point $q \in \mathcal E$.

Then by $(W2)$:


 * $q - p = \paren {q - p} +_V \paren {p - p}$

So $\paren {p - p} = 0_V$, or $p + 0_V = p$, which establishes $(RGA2)$.

Next we must show that the action is free, that is:


 * $\forall v \in V: \forall p \in \mathcal E : p + v = p \implies v = 0_V$

Let $v \in V$ be any vector such that $p + v = p$, i.e. $p - p = v$.

We have shown for $(RGA2)$ that $p - p = 0_V$, and $-$ is a mapping which associates to any $p, q \in \mathcal E$ a unique point in $q - p \in V$.

It follows that $v = 0_V$, i.e. the action $+$ is free.

Finally we show that the action is transitive, that is:


 * $\forall p, q \in \mathcal E \ \exists v \in V : p + v = q$.

For any $p, q \in \mathcal E$ we let $v = q - p$.

By the definition of the action $+$ this means that $p + v = q$, which shows that the action is transitive.