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 verify:

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.

Group Action implies Associativity Axioms
Let $\phi: \mathcal E \times V \to \mathcal E$ be a free and transitive group action of $\struct{V, +_V}$ on $\mathcal E$.

For $\paren {p, v} \in \mathcal E \times V$ write $p + v = \phi \paren {p, v}$.

For any points $p, q \in \mathcal E$, by the definition of a transitive group action there exists $v \in V$ such that $p + v = q$.

Now let us show that the vector $v$ with this property is unique.

If $p + v_1 = p + v_2$ then:

Now by the definition of a free group action $p + \paren {v_1 - v_2} = 0$ implies that $v_1 - v_2 = 0$.

That is $v_1 = v_2$, which shows that there is a unique vector $v$ such that $p + v = q$.

Therefore we can define a mapping $- : \mathcal E \times \mathcal E \to V$ that associates to $\paren {p, q} \in \mathcal E \times \mathcal E$ the unique vector $v = q - p \in V$ such that $p + v = q$.

Now that the mappings $+$ and $-$ are defined we verify $(A1)$, $(A2)$ and $(A3)$

First:

This establishes $(A1)$.

Now $(A2)$ is:


 * $p + \paren {u + v} = \paren {p + u} + v$

But this is simply the statement $(RGA1)$ of a group action.

Finally for $(A3)$, let $p, q \in \mathcal E$ and $v \in V$. Then: