Affine Space clarification

Question

An affine space is a set $A$ together with a vector space $V$ with a regular action of $V$ on $A$.

Can someone please explain to me why the plane $P_{2}$ in this answer is indeed an affine space?

Since we are in $\mathbb{R}^{3}$, the group action is translations. A regular action is an action which is both transitive and free. That means there is exactly one group element $g \in G$ s.t $g * x = y$ for $x,y \in P_2$.

So to check if that plane is Affine, we have to check if this is true: There is exactly one $g \in G$ s.t. $g * x = y$ for $x,y \in A = \{(x,y,1) \mid x,y \in \mathbb{R}\}$. Is that correct? I know that plane is $\mathbb{R} \times \mathbb{R} \times \{0\}$ with a translation $(0,0,1)$. BUT, I don't really know how to verify the regular property. Can someone show me how to do that?

Answer 1

Let $V=\mathbb{R}^2$, and the action of $V$ on $A$ is given by $$(a,b)\cdot (x,y,1) := (a+x, b+y, 1).$$ Given $(x,y,1), (x',y',1) \in A$, take $a=x-x', b=y-y'$. Then $(a,b) \in V$ and $$(a,b)\cdot (x',y',1)=(x,y,1).$$ This shows that the action of $V$ on $A$ is regular.

Note: This answer was implicit in the comment of rschwieb in the link given in the question.