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 $\left({V, +_V, \circ}\right)$ 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:
| \((\text W 1)\) | $:$ | \(\ds \forall p \in \EE: \forall v \in V: \exists ! q \in \EE:\) | \(\ds v = q - p \) | |||||
| \((\text W 2)\) | $:$ | \(\ds \forall p, q, r \in \EE:\) | \(\ds \paren{r - q} +_V \paren{q - p} = r - p \) |
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 $p, q \in \mathcal E$ 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:
| \((A4)\) | $:$ | \(\ds \forall p \in \mathcal E:\) | \(\ds p - p = 0 \) |
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:
| \(\ds q - p\) | \(=\) | \(\ds \paren {p + v} - p\) | Definition of $q$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {p - p} +_V v\) | $(A3)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds v\) | $(A4)$ |
Now let $r \in \mathcal E$ be any other element 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 $p, q, r \in \mathcal E$ 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 $\phi: \mathcal E \times V \to \mathcal E$ 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:
| \((RGA1)\) | $:$ | \(\ds \forall u, v \in V, p \in \mathcal E:\) | \(\ds \paren {p + u} + v = p + \paren {u +_V v} \) | |||||
| \((RGA2)\) | $:$ | \(\ds \forall p \in \mathcal E:\) | \(\ds p + 0_V = p \) |
To establish $(RGA1)$ let $p \in \mathcal E$ and $u, v \in V$.
Then by $(W1)$:
| \(\text {(1)}: \quad\) | \(\ds \exists ! q \in \mathcal E\) | \(:\) | \(\ds q - p = u\) | |||||||||||
| \(\text {(2)}: \quad\) | \(\ds \exists ! r \in \mathcal E\) | \(:\) | \(\ds r - q = v\) | |||||||||||
| \(\text {(3)}: \quad\) | \(\ds \exists ! s \in \mathcal E\) | \(:\) | \(\ds s - p = u +_V v\) |
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 $p \in \mathcal E$ 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:
| \(\ds p + \paren {v_1 - v_2}\) | \(=\) | \(\ds \paren {p + v_1} + \paren {- v_2}\) | $(RGA1)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {p + v_2} + \paren {- v_2}\) | 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 $- : \mathcal E \times \mathcal E \to V$ 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:
| \(\ds q - p = v\) | \(\iff\) | \(\ds p + v = q\) | By definition | |||||||||||
| \(\ds \) | \(\iff\) | \(\ds p + \paren {q - p} = q\) | 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:
| \(\ds \paren {p + u} - q\) | \(=\) | \(\ds \paren {\paren {q + \paren {p - q} } + u} - q\) | $(A1)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {q + \paren {\paren {p - q} + u} } - q\) | $(RGA1)$ | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {p - q} + u\) | $(A1)$ |
$\blacksquare$