Affine space

The 2- and 3-dimensional point spaces studied in elementary Euclidean geometry are examples of affine spaces, A₂ and A₃, respectively.

Assume that any two points P and Q in a space A can be connected by a line segment in A; this possibility is Axiom 1 of Euclidean geometry. If we order P and Q (we say that P comes before Q), then the line segment obtains a direction (points from P to Q) and has become a vector, the difference vector ${\overrightarrow {PQ}}$. When all difference vectors can be mapped onto vectors of the same n-dimensional vector space V_n, we may call the point space A an affine space of dimension n, written A_n.

Usually one takes as a difference space an inner product space. Its elements have a well-defined length, namely, the square root of the inner product of the vector with itself. The distance between points P and Q in an affine space is then defined as the length of the image of ${\overrightarrow {PQ}}$ in V_n. Difference vectors that are mapped onto the same element of V_n are said to be parallel, they differ from each other by translation.

Formal definition

We will restrict the definition to vector spaces over the field $\mathbb {R}$ of real numbers.

Let V be an n-dimensional vector space and A a set of elements that we will call points. Assume that a relation between points and vectors is defined in the following way:

1. To every ordered pair P, Q of A there is assigned a vector of V, called the difference vector, denoted by ${\overrightarrow {PQ}}\in V$.
2. To every point P of A and every vector ${\vec {v}}$ of V there exists exactly one point Q in A, such that ${\overrightarrow {PQ}}={\vec {v}}$.
3. If P, Q, and R are three arbitrary points in A, then

${\overrightarrow {PQ}}+{\overrightarrow {QR}}={\overrightarrow {PR}}.$

If these three postulates hold, the set A is called an n-dimensional affine space with difference space V.

Immediate consequences:

${\begin{aligned}{\overrightarrow {PP}}+{\overrightarrow {PR}}&={\overrightarrow {PR}}&\Longrightarrow \quad {\overrightarrow {PP}}={\vec {0}}.\qquad \\{\overrightarrow {PQ}}+{\overrightarrow {QP}}&={\vec {0}}\qquad &\Longrightarrow \quad {\overrightarrow {PQ}}=-{\overrightarrow {QP}}.\\\end{aligned}}$

Lemma:

${\overrightarrow {PQ}}={\vec {0}}\quad \Longleftrightarrow \quad P=Q$.

Proof:   If the points coincide, $\,P=Q$, we just saw that the difference vector is the zero vector. Conversely, assume that   $\,P\neq Q$   and   ${\overrightarrow {PQ}}={\vec {0}}$. Then for an arbitrary point $\,R$,

${\vec {0}}={\overrightarrow {PQ}}=-{\overrightarrow {RP}}+{\overrightarrow {RQ}}\quad \Longrightarrow \quad {\overrightarrow {RP}}={\overrightarrow {RQ}},$

which implies that the same vector in V connects $\,R$ in A with two different points, which by postulate 2 is forbidden.

Parallelogram law

Parallelogram law

Consider four points in A:   P₁, P₂, Q₁, and Q₂. Assume that the following difference vectors are equal, i.e., the corresponding line-segments in A are parallel and of equal length:

${\overrightarrow {P_{1}Q_{1}}}={\overrightarrow {P_{2}Q_{2}}}$

then the four points form a parallelogram, that is,

${\overrightarrow {P_{1}P_{2}}}={\overrightarrow {Q_{1}Q_{2}}},$

see the figure.

Proof Subtract the following equations:

${\begin{aligned}{\overrightarrow {P_{1}P_{2}}}&={\overrightarrow {P_{1}Q_{2}}}-{\overrightarrow {P_{2}Q_{2}}}\\{\overrightarrow {Q_{1}Q_{2}}}&={\overrightarrow {P_{1}Q_{2}}}-{\overrightarrow {P_{1}Q_{1}}}.\\\end{aligned}}$

This gives

${\overrightarrow {P_{1}P_{2}}}-{\overrightarrow {Q_{1}Q_{2}}}=-{\overrightarrow {P_{2}Q_{2}}}+{\overrightarrow {P_{1}Q_{1}}}={\vec {0}}.$

Position vector

Choose a fixed point O in the affine space A. Every point P is uniquely determined by the vector ${\vec {v}}\equiv {\overrightarrow {OP}}\in V$ (by the second requirement). The vector ${\vec {v}}$ is the position vector of P with respect to O. After choosing O every point P can be uniquely identified with its corresponding position vector in V. Choice of another point O′ gives a translation of ${\vec {v}}$ by ${\vec {t}}\equiv {\overrightarrow {OO'}}$, for

${\vec {v}}\equiv {\overrightarrow {OP}}={\overrightarrow {OO'}}+{\overrightarrow {O'P}}={\vec {t}}+{\vec {v}}\;'$

with ${\vec {v}}\;'\equiv {\overrightarrow {O'P}}$, the position vector of P with respect to O′.