Affine spaces are useful to describe certain geometric structures. Basically, the main operation is that given to affine vectors $\mathbf{a}$ and $\mathbf{b}$ and a scalar $\lambda$ (please correct me if I am not using the right terminology).

One can generate a third vector by:

$\mathbf{c} = \lambda \mathbf{a} + (1 - \lambda) \mathbf{b}$ or $\mathbf{c} = d_1 \mathbf{a} + d_2 \mathbf{b}$ where $d_1 + d_2 = 1$.

I find this notation confusing because in some sense it is redundant and gives the impression of a linear space operation. Is there a better notation in usage for this operation?

For example something like:

$\mathbf{c} = \mathbf{a} | \lambda | \mathbf{b}$

$\mathbf{c} = (\mathbf{a} \rightarrow \lambda \rightarrow \mathbf{b}$)

$\mathbf{c} = \mathbf{a} \stackrel{ \lambda }{\longrightarrow} \mathbf{b}$

As @Hurkyl pointed out, there are more general affine functions involving more than one point, and the notation doesn't help too much here since for example.

$ ((\mathbf{a}|\lambda_1|\mathbf{b})|\lambda_2|\mathbf{c}) = (\mathbf{a} | \lambda_2 \lambda_1 | (\mathbf{b} | \frac{\lambda_2 - \lambda_2\lambda_1}{1-\lambda_2\lambda_1} | \mathbf{c}))$


2 Answers 2


I'm not understanding how this gives anything redundant. What is it repeating?

"It gives the impression of a linear space operation": First of all, an affine space is a linear space. Furthermore, I would guess that you are talking about linear spaces in the sense of a vector space over a (possibly skew) field $\mathbb{F}$, which are exactly the examples of affine spaces. The translates of subspaces in the vector space are the subspaces of the affine space (so the vectors, translates of the trivial vector subspace $\{0\}$, are the points of the affine space). In particular, the points on the line through points $\mathbf{a}$ and $\mathbf{b}$ are given by $\{ \lambda \mathbf{a} + (1-\lambda) \mathbf{b} \ : \ \lambda \in \mathbb{F}\}$.

Your notation at the bottom is.... very confusing. And it actually is redundant, since we already have a way of writing this that is not significantly longer than what you suggest.

  • $\begingroup$ $\lambda$ repeats twice. Points well taken. $\endgroup$
    – alfC
    Aug 30, 2018 at 21:35

Let $\mathbb{A}$ be an affine space, and let $\mathbb{V}$ be the associated vector space.

There are lots more operations on affine spaces than just that one:

  • If $\sum_i d_i = 1$ and $P_i \in \mathbb{A}$, then you have a well-defined affine combination $\sum_i d_i P_i \in \mathbb{A}$
  • If $\sum_i d_i = 0$ and $P_i \in \mathbb{A}$, then you have a well-defined linear combination $\sum_i d_i P_i \in \mathbb{V}$

and all of these operations satisfy all of the identities implied by the choice of notation, such as

$$ 3(2P - Q) - 2R = 6P - 3Q - 2R = 6P - 5\left(\frac{3}{5} Q + \frac{2}{5} R \right)$$

picking a different notation would greatly obscure all of these identities and make calculation difficult. For example, writing $f_\lambda(P,Q)$ for $\lambda P +(1-\lambda) Q$, you'd have

$$ f_3(f_2(P, Q), R) = f_6(P, f_{3/5}(Q, R))$$

Furthermore, given any choice of origin $O \in \mathbb{A}$, you can identify the point $P \in \mathbb{A}$ with the vector $P-O \in \mathbb{V}$.

Doing so, affine combinations are linear space operations: for example,

$$\lambda P + (1-\lambda) Q = R \qquad \Longleftrightarrow \qquad \lambda (P-O) + (1-\lambda) (Q-O) = R-O $$

where the left side is the affine space operation and the right side is all calculation in the associated vector space.

The affine combinations are distinguished in that they don't depend on the choice of $O$.


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