2
$\begingroup$

Sorry if this is a beginner question but I have been trying to find a good definition of an affine space and can't seem to find one that makes intuitive sense. Hoping that one could help explain what an affine space is after defining it mathematically. I understand its relation to a Euclidean space at a high level at least.

Some resources I checked out include:

The last one is the closest to one that makes sense, but still not quite getting it:

A real affine space is a triple $(A, V, φ)$ where $A$ is a set of points, $V$ is a real vector space and $φ: A × A \rightarrow V$ is a map verifying:

  1. $∀P ∈ A$ and $∀u ∈ V$ there exists a unique $Q ∈ A$ such that $φ(P, Q) = u$.
  2. $φ(P, Q) + φ(Q, R) = φ(P, R)$ for every $P, Q, R ∈ A$.

My translation of that would be:

A real affine space is a set of points over a real vector space and a function mapping combinations of the points to the vector space, where the function satisfies:

  • For any point in the vector space and point in the set of points there is another point in the set of points such that the function maps the vector space and the other point to the first point.
  • The function is transitive.

That doesn't seem quite right, and still don't understand the meaning of that middle part.

$\endgroup$
  • $\begingroup$ So, what is your question ? Are thee axioms (1-2) complete ? Are they indendant ? Do they define an affine space ? what is your question ? $\endgroup$ – Duchamp Gérard H. E. Mar 7 '18 at 4:37
  • $\begingroup$ My question is what is a better definition of an affine space that makes sense. $\endgroup$ – Lance Pollard Mar 7 '18 at 4:38
  • $\begingroup$ I would think of an affine space as a 'vector space without a distinguished origin'. $\endgroup$ – copper.hat Mar 7 '18 at 4:52
  • $\begingroup$ A "better" definition how? What makes one definition better than another? $\endgroup$ – Morgan Rodgers Mar 7 '18 at 5:50
1
$\begingroup$

Given your function $\phi$ and a vector $v \in V$, we get a function "move in the $v$ direction"

$$ m_v(a): A \to A $$

by the rule $m_v(a)$ is the unique $b$ such that $\phi(b, a) = v$.

This rule is additive in the sense that $m_v + m_{v'} = m_{v+v'}$, and $m_{0}$ is the identity map (don't move). Conversely, given a collection $m_v$ of bijections indexed by $V$ satisfying this axiom, you get a subtraction map $\phi$ (as in your definition) by the rule $\phi(b, a)$ is the unique $v$ such that $m_v(a) = b$.

So your definition is completely equivalent to having "movement operators" on $A$ for every $v \in V$.

It may help to think of the one-dimensional case. It makes sense to move ten meters north. It does not make sense to "be ten meters north."

$\endgroup$
1
$\begingroup$

All depends the point of view you like (at that is a matter of taste), IMHO, the "most connected" (with the rest of mathematics) is that of an orbit under translations (see hunter's answer). Then you have (a) a group of translations $(T_v)_{v\in V}$ (the additive group $(V,+)$) on a set $A$ (see group actions here), but (b) there must be only one orbit (i.e. the action must be transitive) and (c) faithful. These structures, very important in mathematics, are called principal homogeneous space or torsors and can be thought as "the group (here $V,+$) without priviledged origin or unit" (see Eric Towers's answer).

$\endgroup$
0
$\begingroup$

It might help to read $\varphi$ as "the vector to get from here to there". For instance, $\varphi(P,Q)$ is the vector to get from $P$ to $Q$. In a linear space, every point is also a displacement from the origin (a vector). If we want a homogeneous space, we must throw out the origin, as it is "special". Then we don't have a linear space any more and we have to talk about what displacements do to each point in the space. Condition (1) says from each point, every vector ("displacement") takes you to a unique point. Condition (2) says the vector to get from $P$ to $Q$ added to the vector to get from $Q$ to $R$ is the (unique) vector to get from $P$ to $R$. (That is, "displacement sums" behave just like vector sums in $V$.)

(One can also think of elements of $V$ as translations of the entire space $A$. The correct first model of this is translations of Euclidean space, with which you have considerable direct experience. Notice how reality doesn't come with a special point labelled "the origin"?)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.