# How can I see a visual representation of the difference between Euclidean and Affine geometry?

I have just started studying different types of geometry with the first to I have covered being Euclidean geometry and Affine geometry. I am aware that Euclidean geometry is what we have used the most up until degree-level maths and Affine geometry is to do with the preservation of parallel lines but beyond this I am struggling to visualise the differences between these geometries. I was wondering if anybody knew a good way of explaining this or some resources to aid in my understanding.

## 1 Answer

First of all, it can be very useful to know that there is an even more general geometry including Euclidean and affine geometry: Projective geometry (shortly said: the geometry of perspective).

This said, I think that there at least two ways to consider this issue.

First point of view attached to figures :

• Euclidean geometry: the image of a square can be a/any square.

• Affine geometry: the image of a square is a/any parallelogram.

• Projective geometry: the image of a square is a/any quadrilateral.

Second point of view attached to group of transformations (The "modern" point of view developed by Klein in the so-called Erlangen program):

• Euclidean geometry is characterized by the group of transformations generated by rotations, symmetries, translations, all of them preserving (Eucledean) norm:

$$\begin{cases}x'&= &a x - \varepsilon b y + e\\y'&=& b x + \varepsilon a y +f\ \end{cases} \ \text{with} \ a^2+b^2=1 \ \ \text{or} \ \ \begin{cases}x'&=&x+ e\\y'&=&y +f\ \end{cases}$$

(with $$\varepsilon = +1$$ for a rotation, $$\varepsilon = -1$$ for a symmetry with respect to a straight line ; the second case deals with pure translations).

• Affine geometry is characterized by any transformation of the form :

$$\begin{cases}x'&=&ax+by+e\\y'&=&cx+dy+f \end{cases}$$

• Projective geometry is characterized by any transformation of the form :

$$\begin{cases}x'&=&(ax+by+e)/(gx+hy+i)\\y'&=&(cx+dy+f)/(gx+hy+i) \end{cases}$$

(please note that if, in the common denominator, we take $$f=h=0$$ and $$i=1$$, we are back to an affine transformation).