4.2.5 Angle-Preserving Mappings

The euclidean motions we have just described as $v\mapsto R\cdot v + w$, where $R$ is a rotation or reflection and $w$ is a translation vector, are length and angle preserving. Now we try to identify those mappings which are only angle preserving.

Lemma (Linear conformal mappings).
Let $f:\protect\mathbb{R}^n \to \protect\mathbb{R}^m$ be linear. Then the following statements are equivalent:

1. $f$ respects angles, i.e. is CONFORMAL;
2. $\exists \lambda >0: \langle f(x),f(y) \rangle=\lambda \langle x,y \rangle$ for all $x,y\in \protect\mathbb{R}^n$;
3. $\exists \mu >0$: $\mu \, f$ is an isometry.

Proof. $(2\Leftrightarrow 3)$ is obvious with $\lambda \,\mu ^2=1$.

$(1\Leftarrow 2)$ let $\alpha$ be the angle between $x$ and $y$ and $\alpha '$ the one between $f(x)$ und $f(y)$. Then

$$\cos\alpha '=\frac{\langle f(x)\vert f(y) \rangle}{\Vert f(x)\Ver... ...y \rangle}{\sqrt \lambda \Vert x\Vert\sqrt \lambda \Vert y\Vert}= \cos\alpha .$$

Thus $\alpha =\alpha '$, and $f$ respects angles.

$(1\Rightarrow 2)$ We define $\lambda (v)\geq 0$ implicitly by $\langle f(v)\vert f(v) \rangle=: \lambda (v)\langle v\vert v\rangle$.

Let $v,w$ be orthonormal vectors, then $(v+w)\perp(v-w)$. Since $f$ is conform, we have:

$$0=\langle f(v+w)\vert f(v-w) \rangle = \langle f(v)\vert f(v) \rangle-\langle f(w)\vert f(w) \rangle =\lambda (v)-\lambda (w).$$

Thus $\lambda$ constant on the orthonormal basis $(e_1,\ldots,e_n)$ von $\protect\mathbb{R}^n$. We put $\lambda :=\lambda (e_1)=\ldots=\lambda (e_n)$. For an arbitrary vector $v\in \protect\mathbb{R}^n$ we have:

$$\phantom{\Rightarrow \quad} v=\sum _{i=1}^n v^i e_i$$

$$\Rightarrow \quad \lambda (v)\sum (v^i)^2=\lambda (v)\left \langle \sum v^i e_i\vert\sum v^i e_i \right \rangle=$$

$$\qquad=\left\langle f\left({\sum _i v^i e_i}\right), f\left({\sum _j v^j e_j}\right)\right\rangle$$

$$\qquad=\sum _{i,j} v^iv^j\underbrace{\langle f(e_i)\vert f(e_j) \rangle} _{\lambda \delta _{i,j}} = \lambda \sum _i (v^i)^2$$

$$\Rightarrow \quad \lambda (v)=\lambda \quad\forall v\in \protect\mathbb{R}^n.$$

The lemma now follows from:

$$2\langle v\vert w \rangle =\langle v+w\vert v+w \rangle-\langle v\vert v \rangle- \langle w\vert w \rangle$$

$$2\langle f(v)\vert f(w)\rangle =\langle f(v)+f(w)\vert f(v)+f(w) \rangle- \langle f(v)\vert f(v) \rangle - \langle f(w)\vert f(w) \rangle$$

$$=\lambda \langle v+w\vert v+w \rangle-\lambda \langle v\vert v \rangle- \lambda \langle w\vert w \rangle$$

$$=2\lambda \langle v\vert w \rangle.\quad\qed$$