Affine space

The 2-dimensional plane studied in elementary Euclidean geometry is an example of an affine space.

From elementary geometry we know that any two points P and Q in a plane (a collection of points) can be connected by a line segment. If P and Q are ordered (we say: " P comes before Q "), then the line segment obtains a direction and becomes an arrow pointing from P to Q. An arrow pointing from P to Q can be mapped onto a vector, the difference vector, which for obvious reasons is denoted by ${\overrightarrow {PQ}}$. Sometimes it is stated: "the arrow is a vector", but in the present context it is necessary to carefully distinguish arrows from vectors. If all arrows in a plane can be mapped onto vectors in a 2-dimensional vector space, V₂, the difference space, we may call the plane an affine space of dimension 2, denoted by A₂. Arrows that are mapped onto the same vector in the difference space are said to be parallel, they differ from each other by translation.

In elementary plane geometry, the map of arrows onto vectors is almost always defined by the choice of an origin O, which is a point somewhere in the plane. Clearly, any arbitrary point P is then the head of a unique difference vector ${\overrightarrow {OP}}$. All arrows with tail in O are mapped one-to-one onto a 2-dimensional difference space V₂, with the vector addition in V₂ being in one-to-one correspondence with the parallelogram rule for the addition of arrows in the plane.

Usually one equips the difference space with an inner product, turning the space into an inner product space. Its elements have 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 now be defined as the length of ${\overrightarrow {PQ}}$ in V₂.

Upon formalizing the definition, we replace the dimension 2 by an arbitrary finite dimension n and replace arrows by ordered pairs of points (the "head" and the "tail") in a given point space A. Briefly, A is an affine space of dimension n if there exists a map of the Cartesian product, A × A onto a vector space of dimension n. This map needs to satisfy certain axioms that are treated in the next section.

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′.