0
$\begingroup$

I have a question regarding isometric spaces and isometries on these spaces in general.

Say I have two metric spaces, two sets X and Y each with their own intrinsic metric, which are isometric (1), ie. there is a distance preserving, bijective function $\phi: X\rightarrow Y$ (and vice versa in fact, in my case).

Let $I(X)$ and $I(Y)$ be the two sets of isometries $\phi_X :X \rightarrow X$ and $\phi_Y: Y\rightarrow Y$ respectively. With composition, these are obviously groups.

Now, in my book they state that, because of (1), the two groups $I(X)$ and $I(Y)$ are isomomorphic, but I have trouble understanding why this is. How would I even come up with a group homeomorphism between the two groups to begin with? And what are the consequences of such a fact? Does it mean that the isometries behave in the same way in either of the two spaces?

$\endgroup$
1
$\begingroup$

Take the map$$\begin{array}{rccc}\Psi\colon&I(X)&\longrightarrow&I(Y)\\&f&\mapsto&\phi\circ f\circ\phi^{-1},\end{array}$$whose inverse is$$\begin{array}{rccc}\Psi^{-1}\colon&I(Y)&\longrightarrow&I(X)\\&f&\mapsto&\phi^{-1}\circ f\circ\phi.\end{array}$$

$\endgroup$
  • $\begingroup$ Ah, I see, thank you so much! $\endgroup$ – Mursten May 29 at 7:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.