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Let $H$ be a shape, i.e., a non-empty compact set in $\mathbb{R}^2$. Denote by $\text{Sym}(H)$ the set of isometries preserving $H$ and by $\text{Sym}^+(H)$ the set of positive orientation-preserving isometries preserving $H$. Prove that $\text{Sym}^+(H)$ is an abelian group.

If $f\in\text{Sym}(H)$ then $f(H)=H, A_f$ is an orthogonal matrix, i.e $A\in O(2)$.
If $f\in\text{Sym}^+(H)$ then $f(H)=H, A_f=\begin{pmatrix} \cos x & -\sin x\\ \sin x & \cos x\end{pmatrix}$.
To prove $\text{Sym}^+(H)$ is an abelian group, we need to prove it's indeed a group and $fg=gf\quad\forall f,g\in\text{Sym}^+(H)$. But how can I prove this? Could someone show me?

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  • $\begingroup$ Essentially, such an isometry has to fix the center of mass of $H,$ but that is messy to define for sets with no interior points. $\endgroup$
    – Thomas Andrews
    Commented Jul 3 at 22:37
  • $\begingroup$ You can restrict tot the case when $H$ is convex, I think. I might be wrong. $\endgroup$
    – Thomas Andrews
    Commented Jul 3 at 22:39
  • $\begingroup$ Can you not compute $f \circ g = g \circ f $ ? $\endgroup$
    – Noctis
    Commented Jul 3 at 23:19
  • $\begingroup$ First prove that $\text{Sym}^{+}$ is a group. Next, it's a subgroup of $SO(2)$, which is the group of rotations. Do you see why this group (and so its subgroups) are abelian? $\endgroup$
    – Qiaochu Yuan
    Commented Jul 3 at 23:19
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    $\begingroup$ Yeah, once you have a fixed point $O$, each symmetry $f$ gives you the linear transformation $x \mapsto f(x + O) - O$ - and this makes an injective homomorphism $\operatorname{Sym}(H) \to O(2)$. $\endgroup$
    – arkeet
    Commented Jul 3 at 23:56

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