[Math] Isometry of $\mathbb R^2$ that preserves orientation is either a rotation or a translation

geometric transformationgeometry

I want to show that if $\gamma$ is an isometry of $\mathbb{R}^2$ that preserves orientation, then it is either a translation or a rotation about a point.

If $\gamma$ is an isometry of $\mathbb{R}^2$ then it is of the form $\vec{\gamma}(\vec{x})=A\vec{x}+\vec{b}$ for some orthogonal matrix $A$ and $\vec{b} \in \mathbb{R}^2$. Since $\gamma$ preserves orientation we have that $\det A=1$.

If $A=I$, then clearly $\gamma$ is a translation $\vec{x} \mapsto \vec{x} + \vec{b}$.

I'm stuck with the case $A \neq I$, though. How do I proceed?

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Orthogonal matrices with determinant 1 are exactly rotation matrices (about the origin). Hence $\vec{\gamma}(\vec{x})=A\vec{x}+\vec{b}$ is a rotation plus a translation, i.e. a rotation about the point which has the coordinates of the vector $\vec{b}$.

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[Math] Showing that a rigid motion preserves distances

Let $x,y \in \mathbb{R}^n$ and let your affine transformation be $f : v \mapsto Av + b$ for some $b \in \mathbb{R}^m$. Then $$\lVert f(x) - f(y) \rVert = \lVert (Ax+b)-(Ay+b) \rVert = \lVert A(x-y) \rVert$$ Now use a familiar property of orthogonal matrices in a vector space to show that this is equal to $\lVert x-y \rVert$.

Hint: If $v$ and $w$ are vectors then $v \cdot w = v^Tw$, and $\lVert v \rVert = \sqrt{v \cdot v}$.

[Math] Isometries of $\mathbb C^2$

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Cpx}{\mathbf{C}}$When you speak without qualification of an isometry with respect to the standard Hermitian norm, you're speaking of an isometry of Euclidean $4$-space. The story of Euclidean isometries is more-or-less uniform independently of dimension, so let's speak of Euclidean $n$-space with $n$ an arbitrary positive integer. Here are some useful factlets:

If an isometry $F$ fixes the origin and the standard basis vectors, then $F$ is the identity.

Sketch of proof: This is clear if $n = 1$. To argue inductively on the dimension, decompose a point of $\Reals^{m+1}$ as $(x, x_{m}) = ((x_{1}, \dots, x_{m}), x_{m+1})$, and show $F$ preserves both $\|x\|$ (so $F(x) = x$ by the inductive hypothesis) and the $(m + 1)$th coordinate $x_{m+1}$.
If an isometry $F$ fixes the origin, there exists a unique $n \times n$ (real) orthogonal matrix $A$ such that $F(x) = Ax$ for all $x$ in $\Reals^{n}$.

Sketch: By the isometry condition and the polarization identity $u \cdot v = \frac{1}{4}(\|u + v\|^{2} - \|u - v\|^{2})$, the points $F(e_{i})$ constitute an orthonormal set in $\Reals^{n}$. Let $A$ be the unique orthogonal matrix satisfying $F(e_{i}) = Ae_{i}$ for each $i$, and apply the first item to $A^{-1} \circ F$. (This answers your first question affirmatively.)
If $F$ is an isometry, there exists a unique $n \times n$ orthogonal matrix and a unique $b$ in $\Reals^{n}$ such that $$ F(x) = Ax + b\quad\text{for all $x$ in $\Reals^{n}$.} $$ (You already have this proof. :)
If $p$ is a point of $\Reals^{n}$ and $u$ is a unit vector, then reflection in the hyperplane through $p$ orthogonal to $u$ is the mapping $$ R(x) = x - 2[(x - p) \cdot u] u, $$ namely, $x$ minus "twice the $u$-component of $x - p$". For example, if $u = e_{n}$ and $p$ is the origin, then $$ R(x_{1}, \dots, x_{n-1}, x_{n}) = (x_{1}, \dots, x_{n-1}, x_{n}) - 2x_{n} e_{n} = (x_{1}, \dots, x_{n-1}, -x_{n}). $$ The Euclidean group is generated by the set of reflections. In fact, if $F$ is a Euclidean isometry, there exist at most $(n + 1)$ reflections (not uniquely-defined!) whose composition is $F$.

A non-trivial translation is a composition of two reflections. (Why?)

A rotation is the composition of at most $n$ reflections, and is always the composition of an even number of reflections (because a reflection reverses orientation while a rotation preserves orientation). Thus, for example, every rotation of Euclidean $3$-space is a composition of at most two reflections, while a rotation of Euclidean $4$- or $5$-space is a composition of at most four reflections. (To prove this inductively, note that if $u$ and $v$ are arbitrary unit vectors, there exists a reflection exchanging $u$ and $v$.)

If, on the other hand, you want to speak only of Euclidean isometries of $\Cpx^{m} \simeq \Reals^{2m}$ for which the "rotation part" is complex-linear, you have the much smaller "holomorphic isometry group" corresponding to the unitary group $U(m) \subset SO(2m)$.

Thinking of $\Cpx^{2}$, the set of complex lines (a special class of real $2$-plane) is the well-known complex projective line, topologically a $2$-sphere, while the set of (oriented) real $2$-planes is topologically a product $S^{2} \times S^{2}$. This shows geometrically how small $U(2)$ is compared to $SO(4)$,

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[Math] Showing that a rigid motion preserves distances

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