Affine space

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

We know from high-school geometry that any two points P and Q in a Euclidean plane A₂ can be connected by a line segment. If we order P and Q (we say: " P comes before Q "), then the line segment obtains a direction and becomes an arrow pointing from P to Q. The arrow can be mapped onto a vector, the difference vector ${\overrightarrow {PQ}}$. In the case of a Euclidean plane all arrows can be mapped onto vectors in a 2-dimensional vector space V₂ (the difference space). This implies that we may call the plane A₂ an affine space of dimension 2.

Usually one takes as a difference space V 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 any two points P and Q may then be defined as the length of ${\overrightarrow {PQ}}$ in V₂. Arrows that are mapped onto the same element of V₂ are said to be parallel, they differ from each other by translation.

Upon formalizing the definition in the next section, we replace the dimension 2 by an arbitrary finite dimension n and replace arrows by ordered pairs of points (the head and tail of the arrow).

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

The parallelogram law in the 2-dimensional Euclidean plane.

Consider four points in A:   P₁, P₂, Q₁, and Q₂. Assume that the following difference vectors are equal,

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

then

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

See the figure for a concrete example in which the four points form a parallelogram.

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, an "origin". Every point P is uniquely determined by the vector ${\vec {v}}\equiv {\overrightarrow {OP}}$ (by lemma 1, if simultaneously ${\vec {v}}={\overrightarrow {OQ}}$, it follows that P = Q). 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

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

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