# Proving that a construct is a geometry

So here's a question about geometry that I came across recently and have been scratching my head over it.

Let S be the set of Euclidean transformations that are compositions of two reflections, i.e, S = {r1 of r2: r1 and r2 are reflections in the Euclidean Plane}.

Prove that (Complex, S) is a geometry and determine which Euclidean transformations are in S. Also give an example of a pair of shapes that are congruent in standard Euclidean Geometry but are not congruent in the new geometry, or prove that no such pair exists.

I've tried many things such as formal proofs, can I have some help/a solution? Thanks in advance!

• What is your definition of "a geometry"? – Noah Schweber Jul 4 '18 at 2:28
• According to Klein, geometry can be viewed as the action of a group on a space, be it smooth or finite. That is, a geometry on a set X is a triple (X,G,A), where G is a group with action A on X. – Garage Sensor Jul 4 '18 at 2:45

Some background things

By "Complex" I assume you mean the space of the proposed geometry is $\mathbb C$. As the symmetries in question are Euclidean, I will drop the extra structure and use $\mathbb R^2$ instead of $\mathbb C$ right up to the end.

By Klein's definition, we want a Lie group $G$ which acts on $\mathbb R^2$, and we also want $\mathbb R^2 = G/H$ for some closed Lie subgroup $H$ of $G$. Intuitively, the group $G$ comprises the symmetries of your space, and $H$ comprises everything that should be invariant under the symmetries.

For example, I'll start by building the well known Klein Geometry for $\mathbb R^2$. $E(2)$ is the group of all isometries of the plane: (reflections, translations, rotations, and glide reflections). When $E(2)$ acts on $\mathbb R^2$, it keeps distances and angles invariant. Thus we "quotient out by everything in $E(2)$ that does nothing", which is the group $O(2)$ of orthogonal transformations (rotations, rotations composed with reflections). In other words, the space of the geometry is the group $E(2)/O(2) \cong \mathbb R^2$.

Now let's look at your case. $S$ is comprised of translations (when $r_1$ and $r_2$ are parallel) and rotations (when $r_1$ and $r_2$ are not parallel). This is a subgroup of $E(2)$ from above. (Exercise.) It is also closed (in the topological sense): Let $M \in E(2)\setminus S$. Let $I \in S$ be the identity, and consider a neighbourhood $\mathcal N \subset S$ of $I$. (We can do this because the determinant of $I$ is $1$.) But now the set $\{P\circ M : P \in \mathcal N\}$ is open and belongs to $E(2)\setminus S$, so $S$ is closed. Finally, every closed subgroup of $GL(2)$ is a Lie subgroup (by a useful theorem).
$S$ is a Lie group (that acts on $\mathbb R^2$) as required.
What are the invariants of this group action? Like the case where $E(2)$ is the group action, the invariants are distances and angles. The subgroup of $S$ that captures these invariants is the collection of all rotations, which I denote $O^+$. Similar reasoning to above shows that $O^+$ is a closed Lie subgroup of $S$. Thus I claim that $S/O^+$ is isomorphic to $\mathbb R^2$, so $S$ acting on $\mathbb R^2$ determines a geometry as required.
To see the isomorphism, consider the map $\varphi: S/O^+ \to \mathbb R^2$ defined by $$\varphi (M + O^+) = M(0)$$ for all $M \in S$. (Exercise - show that it is truly an isomorphism. The key is that the stabilizer of $0$ is precisely $O^+$.) And we're done!
The complex plane $\mathbb C$ equipped with $S$ is a geometry, because $S$ is a lie group, and $O^+$ is a closed lie subgroup of $S$ such that $S/O^+ \cong \mathbb R^2 \cong \mathbb C$