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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?

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  • $\begingroup$ 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
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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?".

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