Let $P_n$ be a fixed regular convex $n$-gon in the plane. For a metric space $M$ we denote by $\text{Isom}(M)$ the set of distance-preserving maps $M \to M$. How can I show that $$ D_n := \left\{\, f \in \text{Isom}(\mathbb{R}^2) : f(P_n) = P_n \right\} \cong \text{Isom}(P_n) $$ as groups where $\mathbb{R}^2$ is given the euclidean and $P_n$ the induced metric? Also is there a general criteria or classification for subsets like $P_n \subseteq \mathbb{R}^2$ for which these two groups are isomorphic?


I assume that you aren't talking about abstract isomorphism, but rather that each isometry of $\mathbf{P}_n$ extends uniquely to an isometry of $\mathbf{R}^2$. The answer is contained in the following proposition.

Let $S \subseteq T \subseteq \mathbf{R}^2$, and let $f \colon S \to \mathbf{R}^2$ be a distance preserving map. Assume that $S$ is not contained in a line. Then $f$ extends uniquely to a distance preserving map from $T$ into the plane.

Here is an outline of the proof.

For uniqueness, start by taking $A, B, C \in S$ that are not aligned. Show that their images $A', B', C',$ are not aligned. Now if you take any point $M$ in $T$, and call its image $M'$, the distances $A'M'$, $B'M'$ and $C'M'$ are uniquely determined. Show that there is at most one point $M'$ in the plane for which $A'M'$, $B'M'$ and $C'M'$ take the prescribed values.

For existence, first assume that $S = \{A, B, C\}$, $T = \mathbf{R}^2$, let $A'B'C'$ be any triangle with the same side lengths as $ABC$, and let $f$ be the mapping taking $A$, $B$ and $C$ to $A'$, $B'$ and $C'$, respectively. Then first let $\phi_1$ be the reflection taking $A$ to $A'$. Write $B_1 = \phi_1(B)$. Let $\phi_2$ be the reflection with respect to the bisector of the angle $B'A'B_1$. Then since $\phi_2$ takes ray $A'B_1$ to ray $A'B'$, and the distances $A'B_1$ and $A'B'$ are equal, we have $\phi_2(B_1) = B'$. Thus $\phi_2 \circ \phi_1$ takes $A$ to $A'$ and $B$ to $B'$. Finally, let $C_2 = \phi_2 \circ \phi_1 (C)$. We have $A'C_2 = A'C'$, and $B'C_2 = B'C'$. This shows that the points $C_2$ and $C'$ are either identical or symmetric about the line $A'B'$. In the former case, $\phi_2 \circ \phi_1$ is the desired extension. In the latter, use $\phi_3 \circ \phi_2 \circ \phi_1$, where $\phi_3$ is reflection with respect to $A'B'$.

For the general case, let $f \colon S \to \mathbf{R}^2$ be a distance-preserving map. Let $A, B, C$ be three non-aligned points in $S$, and let $f_0$ denote the restriction of $f$ to $S_0 = \{A, B, C\}$. Then by the existence part above, $f_0$ has an extension $f_1$ to all of $\mathbf{R}^2$. Since $f_1|_S$ and $f$ both extend $f_0$ from $S_0$ to $S$, we must have $f_1|_S = f$ by the uniqueness part. Therefore $f_1$ extends $f$ to the entire plane.

  • $\begingroup$ This looks very close to what i want but I'm a bit confused. So the groups $D_n$ and $\text{Isom}(P_n)$ from my initial question are not isomorphic? Maybe only bijective? $\endgroup$ – legacytron Jul 6 '15 at 6:52
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    $\begingroup$ Every element of the second group extends uniquely to an element of the first, and this correspondence is compatible with the operation of composition, so yes, the groups are isomorphic. What I meant by my first sentence is that for some subsets of the plane, it is at least conceivable that some isomorphism might exist other than the natural one by restriction, but you would probably not be interested in those kinds of "accidental" isomorphisms. I assumed you were only interested in isomorphisms via restriction of the isometries to the smaller set. $\endgroup$ – Keith Jul 6 '15 at 7:19
  • $\begingroup$ Thanks for explaining, now everything makes sense. You are right in that I was actually interested in isomorphisms via restrictions. $\endgroup$ – legacytron Jul 6 '15 at 21:12

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