Equivalence of Definitions of Affine Space

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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:

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

$\blacksquare$

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