Can we define a metric on affine space?

Question

I tried to find a nontrivial metric defined constructively on the affine space, but I could not get any result (except physics articles). I believe there is function analogous to metric that can be defined on $A\times A$ to $\mathbb{R}$ or some field $\mathbb{F}$ where $A$ is affine space. Tell me if you have any ideas.

Answer 1

An affine space is a set $A$ equipped with a map $$+: A \times V \to A,$$ where $V$ is a vector space (thought of as adding a vector to a point in affine space), satisfying 1) $a + 0 = a$

2) $(a + v) + w = a + (v+w)$

3) For every $a,b \in A$, there is a unique $v$ with $b + v = a$.

From this we can cook up a map $$-: A \times A \to V;$$ by definition, $a-b$ is the unique vector $v$ from (3) above; that is, $a-b$ is the unique vector which satisfies $$b + (a-b) = a.$$

This satisfies $(a-b) + (b-c) = a-c$ and $a-a = 0$.

Thus if $V$ is a normed vector space, then this naturally gives rise to a metric on $A$, as described in the comments: $$d(a,b) = \|a-b\|_V.$$ This is the most natural way to induce a metric on affine space: from a norm on a vector space. That this is a metric follow from the properties of the previous line, and the fact that $\|\cdot\|_V$ is a norm on $V$.

Answer 2

An affine space $A$ is equivalently described as a ternary algebra with an operation: $$(a,b,c) ∈ A×A×A ↦ a - b + c ∈ A,$$ satisfying the identities $$ a - a + c = c, \\ a - c + c = a, \\ a - b + (c - d + e) = (a - b + c) - d + e, \\ a - b + c = c - b + a, $$ and an operation: $$(a,r,b) ∈ A×ℝ×A ↦ [a,r,b] ∈ A,$$ satisfying the identities $$ [a,0,b] = a, \\ [a,1,b] = b, \\ [a,rt(1-t),[b,s,c]] = [[a,rt(1-s),b],t,[a,rs(1-t),c]]. $$ As a vector operation, the latter would correspond to $[a,r,b] = (1 - r)a + r b$.

As a footnote: this also works for affine spaces over any field other than $ℝ$ except the two and three element field. For the three element field, the above axioms define a commutative quandle.

The first operation characterizes Abelian torsors (or torsors, if excluding the identity $a - b + c = c - b + a$). It is actually redundant, since one could define it in terms of the second as: $$a - b + c = \left[\left[b,\frac{1}{1 - s},a\right],s,\left[b,\frac{1}{s},c\right]\right] \hspace 1em (s ≠ 0, 1).$$

The vector space $dA$ associated with the affine space $A$ may be characterized as the set of pairs $a - b$, for $(a,b) ∈ A×A$, subject to the equivalence relation $(a - b + c) - d = a - (d - c + b)$. All of the pairs $a - a$ are provably equivalent to one another: $$a - a = (b - b + a) - a = b - (a - a + b) = b - b,$$ and the sum operation is read directly off from the equivalence relation, itself: $$(a - b) + (c - d) = (a - b + c) - d = a - (d - c + b).$$ The addition operation is commutative, associative and has the zero and inverse property, with $-(a - b) = (b - a)$. Products with scalars are defined by $r(a - b) = [c,r,a] - [c,r,b]$ and is provably independent of $c$.

In addition (pun) there is also an add-on operation $$(a - b, c) ∈ dA×A ↦ a - b + c ∈ A,$$ which characterizes the action of the vector space $dA$ on the affine space $A$.

A norm over $dA$ can be characterized as a distance function over $A$ given by: $$(a,b) ∈ A×A ↦ ab ∈ ℝ,$$ satisfying the properties: $$aa = 0, \hspace 1em a ≠ b → ab > 0, \hspace 1em ab = ba, \hspace 1em ac ≤ ab + bc,$$ satisfying the additional property: $$[a,r,b][a,r,c] = |r| bc,$$ independently of $a$. The corresponding norm over $dA$ is then given by: $$a - b ∈ dA ↦ |a - b| = ab ∈ ℝ,$$ and is invariant under the equivalence; i.e. $(a - b + c)d = a(d - c + b)$ is provable.