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:

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


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


Group Action

Let $K$ be a field.

Let $\left({V, +_V, \circ}\right)$ be a vector space over $K$.

Let $\mathcal E$ be a set.

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


Then the ordered pair $\tuple{\mathcal E, \phi}$ 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:

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


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 $p, q \in \mathcal E$ we have:

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

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

\((A4)\)   $:$     \(\displaystyle \forall p \in \mathcal E:\) \(\displaystyle p - p = 0 \)             


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:

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

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

Then:

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

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


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

Then:

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

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:

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

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

Then by $(W1)$:

\(\text {(1)}: \quad\) \(\displaystyle \exists ! q \in \mathcal E\) \(:\) \(\displaystyle q - p = u\)
\(\text {(2)}: \quad\) \(\displaystyle \exists ! r \in \mathcal E\) \(:\) \(\displaystyle r - q = v\)
\(\text {(3)}: \quad\) \(\displaystyle \exists ! s \in \mathcal E\) \(:\) \(\displaystyle s - p = u +_V v\)

Then we have:

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

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

Therefore:

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

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:

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

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 $- : \mathcal E \times \mathcal E \to V$ 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:

\(\displaystyle q - p = v\) \(\iff\) \(\displaystyle p + v = q\) By definition
\(\displaystyle \) \(\iff\) \(\displaystyle p + \paren {q - p} = q\) 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 $p, q \in \mathcal E$ and $v \in V$. Then:

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

$\blacksquare$