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:

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

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:

\(\displaystyle \paren {p + u} - q\)

\(=\)

\(\displaystyle \paren {\paren {q + \paren {p - q} } + u} - q\)

$(A1)$

\(\displaystyle \)

\(=\)

\(\displaystyle \paren {q + \paren {\paren {p - q} + u} } - q\)

$(RGA1)$

\(\displaystyle \)

\(=\)

\(\displaystyle \paren {p - q} + u\)

$(A1)$

$\blacksquare$