Equivalence of Definitions of Affine Space

Associativity Axioms implies Weyl's Axioms
Assume the axioms $(A1)$, $(A2)$, $(A3)$.

Then for any $p, q \in \EE$ we have:

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

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

We must show there exists a unique $q \in \EE$ such that:
 * $v = q - p$

Let:
 * $q = p + v$

Then:

Now let $r \in \EE$ be arbitrary such that:
 * $v = r - p$

Then:

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

Now let:
 * $p, q, r \in \EE$

as in $(W2)$.

Then:

This establishes $(W2)$.

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

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


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

where $q \in \EE$ is the unique point such that:
 * $v = q - p$

given by $(W1)$.

We must verify:

To establish $(RGA1)$ let $p \in \EE$ 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 $p \in \EE$ and choose any other point $q \in \EE$.

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:


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

Let $v \in V$ be any vector such that:
 * $p + v = p$

that is:
 * $p - p = v$

We have shown for $(RGA2)$ that:
 * $p - p = 0_V$

and $-$ is a mapping which associates to any $p, q \in \EE$ 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:


 * $\forall p, q \in \EE: \exists v \in V: p + v = q$

For any $p, q \in \EE$ 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 $\phi: \EE \times V \to \EE$ be a free and transitive group action of $\struct {V, +_V}$ on $\EE$.

For $\tuple {p, v} \in \EE \times V$ write $p + v = \map \phi {p, v}$.

For any points $p, q \in \EE$, 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:
 * $- : \EE \times \EE \to V$

that associates to $\tuple {p, q} \in \EE \times \EE$ 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 $p, q \in \EE$ and $v \in V$.

Then: