1
$\begingroup$

The only operations defined on points in an affine space are

  • point-vector addition. this yields a new point.
  • point-point subtraction. this yields a vector.

This can be extended to an affine sum, $\sum_i \alpha_i P_i$, where $\sum_i\alpha_i = 1$.

However, it is possible to derive an expression using these operations that should otherwise be undefined.

Imagine points, $P$, $Q$, $R$, and $S$ such that

$ P - Q = R - S $

This leads to

$ P + S = Q + R$

Each side of the equation is now a point-point addition which is not defined. It is also not an affine sum.

How should each side of this equation be interpreted?

$\endgroup$
  • $\begingroup$ On the other hand, $\frac{1}{2} P + \frac{1}{2} S = \frac{1}{2} Q + \frac{1}{2} R$ is a valid equation of affine combinations, and one would hope it could be proved using the axioms of affine space that it's equivalent to $P-Q = R-S$. $\endgroup$ – Daniel Schepler Feb 28 at 1:38
  • $\begingroup$ I also don't have any references, but I think I saw a presentation once that given an affine space $A$ with vector space $V$, you can put a natural vector space structure on $V \sqcup ((\mathbb{R} \setminus \{ 0 \}) \times A)$ where you treat $V$ as being "weight 0" elements and $(\lambda, a)$ where $\lambda \ne 0$ as being "weight $\lambda$" elements. You also embed $A$ into this vector space via $a \mapsto (1, a)$, i.e. as "weight 1" elements. So then, $P+S$ and $Q+R$ become weight 2 elements of that vector space. $\endgroup$ – Daniel Schepler Feb 28 at 1:49
1
$\begingroup$

This leads to weighted points in affine space. The weight of a point must be nonzero and usual affine points have weight one by definition. Given weighted points $\,aP\,$ and $\,b\,Q\,$ their sum is $\,aP+b\,Q\,$ which has weight $\,c:=a+b\,.$ If $\,c\,$ is nonzero then this is the weighted point $\,c\frac{aP+b\,Q}c.\,$ If $\,c=0\,$ then the sum is the vector $\,a(P-Q).\,$ In general a "point" with weight $0$ is just a vector of the underlying vector space. That is, all weighted points have nonzero weights, but if a sum of weighted points has weight zero, then it becomes a vector instead.

I give more details in my answer to MSE question 1059220 "what is the difference between linear transformation and affine transformation?".

$\endgroup$

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.