1. Synthetic and Analytic Geometry

In his Erlangen Program, the great 19th century German Geometer, Felix Klein, proposed that the basis of geometry should be groups of point transformations. Euclid’s geometry was based on axioms that made assertions about primitive geometrical concepts, like point, line, angle etc which were considered to need no definition since "everybody" agreed on what they were anyway. All subsequent geometrical knowledge had to be derived logically from the axioms. We call this an axiomatic system.

Euclid’s program had flaws that were not completely resolved until the end of the 19th century. Applied to geometry, the axiomatic method developed into what is called synthetic geometry, where the "undefined terms" became abstractions, the axioms became sentences relating the undefined terms. As is taught in any course on the subject (such as MA 402 at Illinois), things become interesting only if the undefined terms and their relations are interpreted in a familiar setting, in which the axioms can be checked to be true or not.

In contrast to synthetic geometry, analytic geometry, which derives from the work of Descartes and Fermat, is based on the properties of numbers. You can, if you wish, build up the number system axiomatically (as is done in pute mathematics) or you can just accept numbers as you learned about them in school. Now, geometric objects are sets of numbers, and their relations are expressed in terms of set theory. Of course we still think geometrically about them, but consistency is derived ultimately from the number systems and their algebra. This is the geometry you learned about first in high school, and later added the knowledge of vectors and their properties in calculus.

The dispute as to which approach is "better", the synthetic or the analytic, was resolved by David Hilbert (1900), who proved through logic, that each can be derived from the other. Thus, if the number systems have logical errors, then these will show up in the geometry too, and vice versa. Few mathematicians worry about these issues nowadays, since there are more important things to discover and deeper controversies to resolve.

One of Euclid’s primitives was the idea of congruence. When are two figure the "same" in a geometrical sense. You should recall such criteria as side-angle-side (SAS), which were theorems for Euclid, but his "proof" was not correct. In synthetic geometry, congruence remains undefined, and SAS becomes an axiom.

In analytic geometry, as we shall see below, a congruence is a point transformation (points go to points) which preserves length. More precisely, the distance between any two points remains the same after as it was before the transformation. We define distance in terms of the familiar formula of the Pythagorean theorem, which we express in coordinate and in vector form, where that $ X = (x,y), W =(u,v) $

For $ X = (x,y), W =(u,v) $

$ dist( X, W) = \sqrt{(x-u)^2 + (y-v)^2} $

$ dist( X , W ) = | X - W | $


Question 1.

How are these two definitions related by the dot product?

Answer. [show] ▶

Recall the dot product, $ X \circ W = xu + yv $, permits us to write $ |X| = \sqrt{x^2 + y^2} = \sqrt{ X \circ Y } $. Now expand the RHS of the second line of (1) to obtain the first.

2. Background

Recall that translations and dilatations are point transformations of the Euclidean plane with several important properties. In particular, displacement vectors are preserved (by translations or scaled dilatations. That is, if `tau = tau_A` and `delta=delta_{Q,r}`, and we write the image of a point under a transformation by decorating its name with a superscript, then

`X^tau -W^tau =X-W`

`X^delta - W^delta =r(X-W)`.

Question 2.

Do you recall the formulas for a translation and a dilatation? If you do, or have to look them up, then prove this assertion on your scratch pad by the side of your computer!

Answer. [show] ▶

Do some algebra to see that

$ X^\delta - W^\delta = Q + r(X-Q) - Q - r(W-Q) = r (X-W) $.

The algebra is easier for translations.


From the way these transformations affect displacements we see that translations always preserve distance. So these are definitely isometries. For dilatations $ r = \pm 1 $ will yield isometries. The "plus" case is uninteresting, in the sense that the answer is obvious. Recall that the more interesting "minus" case is for the central reflection, discussed earlier.

2.1. Length, Norm, and Magnitude

Geometrically, these are all the same non-negative real number associated with a vector. We like to use length when we represent the vector as one of a collection of mutually parallel "arrows" in the plane. The norm is a more abstract concept in higher algebra. We tend to use magnitude in the same breath as direction. Any physical object that has magnitude and direction can be represented as a vector. For example, velocity and acceleration have magnitude and direction, and therefore are vector quantities in physics.

The magnitude is intimately connected with the dot product, and in a sense they are equivalent concepts.

Comment on Notation

It is generally inconvenient to use any special symbol for the dot product, unless other products involving vectors are used in the same discussion. It detracts from the ability to scan algebraic manipulations you worked so hard in high school to master.

So we shall, when no ambiguity threatens, just write $ XY $ instead of $ X $ • $ Y $. This allows us to use powers, as in $ X X = X^2 $.

The commutative and distributive properties of numbers carries over to vectors. Just remember that $ XY $ is a number, not a vector. Here is a list of properties you learned in calculus.

For vectors $ X,Y,Z, O $ and scalars $ r, s, 1, 0 $ we have the

2.2. Bilinearity Properties of the Dot Product

Formula Property name

$ X Y = Y X $

commutative for dot product

$ X(Y+Z)= XY + XZ $

distributive for dot product

$ (rX)Y = r(XY) $

associative for scalar product

$ (r+s)X = rX + sX $

distributive for scalar product

$ 1X = X $

scalar product with 1

$ 0X = O $

scalar product with 0

Note the associative property does not hold: $ (XY)Z \ne X(YZ) $. The LHS is a vector parallel to $ Z $, and the RHS has the same direction as $ X $. Also note that when it is important to distinguish between the scalar zero $ 0 $ and the zero vector, $ O $, we follow the Tondeur’s convention of using the same letter for the point int he plane, and the vector defined by the displacement from the origin to that point. When these distinctions are not obvious from the context, and it is important to emphasize them, you can underline vectors, or stick a little arrow on top of them. The less notational fuss the better, especially in emails.

For a displacement vector `B-A`, we have `|B-A|^2=` `(B-A)(B-A)`. By the bilinearity of the dot product, we can write this as `(B-A)^2=B^2-2AB+A^2`. From this we see we could "define" the dot product itself in terms of norms.

The length of a displacement vector between two points has the following properties:

Note that these are precisely the properties of the distance between two points. Property L3 is called the triangle inequality because any side of a triangle is no longer than the sum of the other two sides.

Question 3.

When is the triangle inequality an equality?

Answer. [show] ▶

When the three points are collinear, and $ B $ lies between $ A $ and $ B $.

3. Definition of an Isometry

So, to preserve distances we want to preserve the length of displacement vectors. Any point transformation that preserves it will be called an isometry. More precisely, we have the

Question 4.

Why is this not the same as saying that $ (X^\alpha - Y^\alpha) = (X - Y)$ or that `|(X-Y)^alpha|=|X-Y|`?

Answer. [show] ▶

The first would say that $ \alpha $ preserves the entire displacement vector, body and soul, namely the direction as well as the magnitude. The second confuses the displacement of the images of two points, with the image of the displacement of two points. The displacement $ X-Y $ is a vector. A transformation is defined on points. So $ (X-Y)^\alpha $ can only mean the image of the point $ (X-Y) $.

A counterexample for the second goes as follows. Let $ X^\alpha = X +D $ be a translation which is not the identity. Then $ \| (X-Y)^\alpha\| = \| X-Y+D\| $. Now take the special case that $ X = Y $ and conclude that $\| D\| = 0 $, contrary to our assumption.

Translations preserve displacement vectors, and central reflections reverse displacement vectors. Thus both of these point transformations are isometries.

`X^sigma - Y^sigma = -(X-Y)`

`|X^sigma - Y^sigma| = |X-Y|`.

Question 5.

Let `\delta = delta_{Q,r}` be a dilatation. When is `delta` an isometry?

Answer. [show] ▶

Since `B^delta - A^delta =r(B-A) rArr |B^delta - A^delta|=|r(B-A)|=|r|\ |B-A|`, `delta` is an isometry if and only if `r=+-1`.

Another example of an isometry is a rotation `rho_{Q,theta}` about the point `Q` by `theta` degrees. This is intuitively obvious, since rotations are rigid motions, like translations. We will study rotations more thoroughly in a subsequent lesson.

We end this section with a proposition that has an easy proof.

Proof. Suppose `alpha` and `beta` are two isometries and let `gamma=beta alpha` be their composition. Then for any pair of points,

`|X^gamma - Y^gamma|\ ` =

`|beta alpha(X) - beta alpha(Y)|`


`|beta(X^alpha) - beta(Y^alpha)|`


`|X^alpha - Y^alpha|`

(`beta` is an isometry )



(`alpha` is an isometry).


Note that we pay a price for using the convenient superscript notation for point transformations. Compositions in functional notations have to be read from right to left, while in the exponent they read from left to right: $ \alpha \beta (X) = (X^\beta)^\alpha $.

Moreover, above we defined isometries to be point transformations, and hence we assume they are 1:1 and onto although we will prove that this is a consequence of the distance preserving property. We could, with Tondeur, just assume that isometries are transformations of the plane that takes points to points, and derive the bijectivity.

Recall that for subsets of the point transformations to be a groups, in addition to closure which we have just proved, we only need to check that inverses preserve distance.

Question 6.

Let $ \omega = \alpha^{-1} $ be the inverse of an isometry. Why is $ \|X^\omega - Y^\omega \| = \| X - Y \| $ ?

Answer. [show] ▶

$ \| X - Y \| =_{bb{1}} \| \alpha (X^\omega) - \alpha (Y^\omega) \| =_{bb{2}} \|X^\omega - Y^\omega \| $

where $ =_{bb{1}} $ is true because $ \alpha \omega = \iota $, and $ =_{bb{2}} $ follows because $ \alpha $ is given to be an isometry.

We conclude this lesson by observing that we have proved this theorem.


In Tondeur’s text the proof of this theorem is considerably more complicated than we have given here. This is true only because we make the initial stipulation that an isometry is already a point-transformation, and that such transformations are 1:1 and onto. See Tondeur’s argument that it is, in fact, not necessary to make this assumption. But it shortens the exposition.