# Equivalence of Definitions of Affine Space

## Theorem

The following definitions of the concept of Affine Space are equivalent:

### Associativity Axioms

Let $K$ be a field.

Let $\struct {V, +_V, \circ}$ be a vector space over $K$.

Let $\EE$ be a set on which two mappings are defined:

$+ : \EE \times V \to \EE$
$- : \EE \times \EE \to V$

satisfying the following associativity conditions:

 $(\text A 1)$ $:$ $\ds \forall p, q \in \EE:$ $\ds p + \paren {q - p} = q$ $(\text A 2)$ $:$ $\ds \forall p \in \EE: \forall u, v \in V:$ $\ds \paren {p + u} + v = p + \paren {u +_V v}$ $(\text A 3)$ $:$ $\ds \forall p, q \in \EE: \forall u \in V:$ $\ds \paren {p - q} +_V u = \paren {p + u} - q$

Then the ordered triple $\struct {\EE, +, -}$ is an affine space.

### Group Action

Let $K$ be a field.

Let $\struct {V, +_V, \circ}$ be a vector space over $K$.

Let $\EE$ be a set.

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

Then the ordered pair $\struct {\EE, \phi}$ is an affine space.

### Weyl's Axioms

Let $K$ be a field.

Let $\struct{V, +_V, \circ}$ be a vector space over $K$.

Let $\EE$ be a set on which a mapping is defined:

$- : \EE \times \EE \to V$

satisfying the following associativity conditions:

 $(\text W 1)$ $:$ $\ds \forall p \in \EE: \forall v \in V: \exists ! q \in \EE:$ $\ds v = q - p$ $(\text W 2)$ $:$ $\ds \forall p, q, r \in \EE:$ $\ds \paren{r - q} +_V \paren{q - p} = r - p$

Then the ordered pair $\tuple {\EE, -}$ is an affine space.

## Proof

### Associativity Axioms implies Weyl's Axioms

Assume the axioms $(A1)$, $(A2)$, $(A3)$.

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

 $\ds q - p$ $=$ $\ds \paren {p + \paren {q - p} } - p$ $(A1)$ $\ds$ $=$ $\ds \paren {p - p} +_V \paren {q - p}$ $(A3)$

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

 $(A4)$ $:$ $\ds \forall p \in \EE:$ $\ds p - p = 0$

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:

 $\ds q - p$ $=$ $\ds \paren {p + v} - p$ Definition of $q$ $\ds$ $=$ $\ds \paren {p - p} +_V v$ $(A3)$ $\ds$ $=$ $\ds v$ $(A4)$

Now let $r \in \EE$ be arbitrary such that:

$v = r - p$

Then:

 $\ds q$ $=$ $\ds p + v$ Definition of $q$ $\ds$ $=$ $\ds p + \paren {r - p}$ Definition of $r$ $\ds$ $=$ $\ds r$ $(A1)$

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

Now let:

$p, q, r \in \EE$

as in $(W2)$.

Then:

 $\ds r - p$ $=$ $\ds \paren {q + \paren {r - q} } - p$ $(A1)$ $\ds$ $=$ $\ds \paren {q - p} +_V \paren {r - q}$ $(A3)$

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:

 $(RGA1)$ $:$ $\ds \forall u, v \in V, p \in \EE:$ $\ds \paren {p + u} + v = p + \paren {u +_V v}$ $(RGA2)$ $:$ $\ds \forall p \in \EE:$ $\ds p + 0_V = p$

To establish $(RGA1)$ let $p \in \EE$ and $u, v \in V$.

Then by $(W1)$:

 $\text {(1)}: \quad$ $\ds \exists ! q \in \EE: \,$ $\ds q - p$ $=$ $\ds u$ $\text {(2)}: \quad$ $\ds \exists ! r \in \EE: \,$ $\ds r - q$ $=$ $\ds v$ $\text {(3)}: \quad$ $\ds \exists ! s \in \EE: \,$ $\ds s - p$ $=$ $\ds u +_V v$

Then we have:

 $\ds s - q$ $=$ $\ds \paren {s - q} +_V u -_V u$ $\ds$ $=$ $\ds \paren {s - q} +_V \paren {q - p} -_V u$ $(1)$ $\ds$ $=$ $\ds \paren {s - p} -_V u$ $(W2)$ $\ds$ $=$ $\ds u +_V v -_V u$ $(3)$ $\ds$ $=$ $\ds v$ $\ds$ $=$ $\ds r - q$ $(2)$

Therefore by uniqueness in $(W1)$ we must have:

$r = s$

Therefore:

 $\ds p + \paren {u +_V v}$ $=$ $\ds s$ $(3)$ $\ds$ $=$ $\ds r$ $\ds$ $=$ $\ds q + v$ $(2)$ $\ds$ $=$ $\ds \paren {p + u} + v$ $(1)$

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:

 $\ds p + \paren {v_1 - v_2}$ $=$ $\ds \paren {p + v_1} + \paren {- v_2}$ $(RGA1)$ $\ds$ $=$ $\ds \paren {p + v_2} + \paren {- v_2}$ by hypothesis $\ds$ $=$ $\ds p + \paren {v_2 - v_2}$ $(RGA1)$ $\ds$ $=$ $\ds p + 0_V$ $(1)$ $\ds$ $=$ $\ds p$ $(RGA2)$

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:

 $\ds q - p$ $=$ $\ds v$ $\ds \leadstoandfrom \ \$ $\ds p + v$ $=$ $\ds q$ By definition $\ds \leadstoandfrom \ \$ $\ds p + \paren {q - p}$ $=$ $\ds q$ By definition

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:

 $\ds \paren {p + u} - q$ $=$ $\ds \paren {\paren {q + \paren {p - q} } + u} - q$ $(A1)$ $\ds$ $=$ $\ds \paren {q + \paren {\paren {p - q} + u} } - q$ $(RGA1)$ $\ds$ $=$ $\ds \paren {p - q} + u$ $(A1)$

$\blacksquare$