=== Associativity Axioms implies Weyl's Axioms === Assume the axioms $(A1)$, $(A2)$, $(A3)$. Then for any $p, q \in \mathcal E$ we have: {{begin-eqn}} {{eqn | l = q - p       | r = \paren {p + \paren {q - p} } - p       | c = $(A1)$ }} {{eqn | r = \paren {p - p} +_V \paren {q - p}       | c = $(A3)$ }} {{end-eqn}} Therefore by Identity is Unique applied to the vector space $V$ we have: {{begin-axiom}} {{axiom | n = A4         | q = \forall p \in \mathcal E         | m = p - p = 0 }} {{end-axiom}} 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: {{begin-eqn}} {{eqn | l = q - p       | r = \paren {p + v} - p       | c = Definition of $q$ }} {{eqn | r = \paren {p - p} +_V v       | c = $(A3)$ }} {{eqn | r = v       | c = $(A4)$ }} {{end-eqn}} Now let $r \in \mathcal E$ be any other element such that $v = r - p$. Then: {{begin-eqn}} {{eqn | l = q       | r = p + v       | c = Definition of $q$ }} {{eqn | r = p + \paren {r - p}       | c = Definition of $r$ }} {{eqn | r = r       | c = $(A1)$ }} {{end-eqn}} This shows that $q$ is unique and establishes $(W1)$. Now let $p, q, r \in \mathcal E$ as in $(W2)$. Then: {{begin-eqn}} {{eqn | l = r - p       | r = \paren {q + \paren {r - q} } - p       | c = $(A1)$ }} {{eqn | r = \paren {q - p} +_V \paren {r - q}       | c = $(A3)$ }} {{end-eqn}} 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: {{begin-axiom}} {{axiom | n = RGA1         | q = \forall u, v \in V, p \in \mathcal E         | m = \paren {p + u} + v = p + \paren {u +_V v} }} {{axiom | n = RGA2         | q = \forall p \in \mathcal E         | m =  p + 0_V = p }} {{end-axiom}} To establish $(RGA1)$ let $p \in \mathcal E$ and $u, v \in V$. Then by $(W1)$: {{begin-eqn}} {{eqn | n = 1       | o = :       | l = \exists ! q \in \mathcal E       | r = q - p = u }} {{eqn | n = 2       | o = :       | l = \exists ! r \in \mathcal E       | r = r - q = v }} {{eqn | n = 3       | o = :       | l = \exists ! s \in \mathcal E       | r = s - p = u +_V v }} {{end-eqn}} Then we have: {{begin-eqn}} {{eqn | l = s - q       | r = \paren {s - q} +_V u -_V u       | c = }} {{eqn | r = \paren {s - q} +_V \paren {q - p} -_V u       | c = $(1)$ }} {{eqn | r = \paren {s - p} -_V u       | c = $(W2)$ }} {{eqn | r = u +_V v -_V u       | c = $(3)$ }} {{eqn | r = v       | c = }} {{eqn | r = r - q       | c = $(2)$ }} {{end-eqn}} Therefore by uniqueness in $(W1)$ we must have $r = s$. Therefore: {{begin-eqn}} {{eqn | l = p + \paren {u +_V v}       | r = s       | c = $(3)$ }} {{eqn | r = r       | c = }} {{eqn | r = q + v       | c = $(2)$ }} {{eqn | r = \paren {p + u} + v       | c = $(1)$ }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = p + \paren {v_1 - v_2}       | r = \paren {p + v_1} + \paren {- v_2}       | c = $(RGA1)$ }} {{eqn | r = \paren {p + v_2} + \paren {- v_2}       | c = By hypothesis }} {{eqn | r = p + \paren {v_2 - v_2}       | c = $(RGA1)$ }} {{eqn | r = p + 0_V       | c = $(1)$ }} {{eqn | r = p       | c = $(RGA2)$ }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = q - p = v       | r = p + v = q       | c = By definition       | o = \iff }} {{eqn | r = p + \paren {q - p} = q       | c = By definition       | o = \iff }} {{end-eqn}} 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: {{begin-eqn}} {{eqn | l = \paren {p + u} - q       | r = \paren {\paren {q + \paren {p - q} } + u} - q       | c = $(A1)$ }} {{eqn | r = \paren {q + \paren {\paren {p - q} + u} } - q       | c = $(RGA1)$ }} {{eqn | r = \paren {p - q} + u       | c = $(A1)$ }} {{end-eqn}} {{qed}} Category:Affine Geometry nvf83ruirilhqclx8s5fkm3dqbp6mqk