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I'm studying geometry and am learning about isometries. I'm self-studying with the book Modern Geometry with Applications (Jennings) in case anyone's curious.

The textbook explains that:

A function $f: \Bbb{E}^n \rightarrow \Bbb{E}^n$ is an isometry if for all points $P, Q \in \Bbb{E}^n$,

$$f(P)f(Q) = PQ$$

where "$PQ$" refers to the distance between the two points.

Assume that $f$ is an isometry and that it has an inverse function $f^{-1}$. Show that $f^{-1}$ is also an isometry.

I'm not sure how to even get started. I've been thinking about the definition of what an isometry is, and figured that in order for the inverse of an isometry to also be an isometry then if we were to write $g = f^{-1}$:

$$\forall_{P,\ Q \in \Bbb{E}}\ g(P)g(Q) = PQ$$

but that's where I just started from.

Could anybody provide some tips or hints to move forward? Thanks.

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  • $\begingroup$ Can you please specify what is E^n? $\endgroup$ Apr 9, 2020 at 13:07
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    $\begingroup$ Let $P'=g(P),Q'=g(Q)$. Then $P'Q'=f(P')f(Q')$. But $f(P')=P$, $f(Q')=Q$: so $P'Q'=PQ$, which concludes. $\endgroup$
    – Aphelli
    Apr 9, 2020 at 13:08
  • $\begingroup$ @Akash: $\mathbb E^n$ is a standard alternative notation for $\mathbb R^n$, the Cartesian coordinate model of $\mathbb E$uclidean $n$-space. Of course, ${\mathbb E}$uclid only covered the cases $n=2$ and $3$. $\endgroup$
    – Lee Mosher
    Apr 9, 2020 at 13:42
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    $\begingroup$ @AkashYadav To add to Lee's comment, the difference between $\Bbb{R}^n$ and $\Bbb{E}^n$ is that $\Bbb{R}^n$ comes with a coordinate system and origin point where Euclidean space has no natural coordinates. $\endgroup$
    – Sean
    Apr 9, 2020 at 13:45
  • $\begingroup$ @Mindlack Could I also say that for some metric $d$ in a metric space $(X, d)$, since an isometry $f$ means by definition that $d(P, Q) = d(f(P), f(Q))$ for $P,\ Q \in \Bbb{E}^n$, $d(f^{-1}(P), f^{-1}(Q)) = d(f^{-1}(f(P)), f^{-1}(f(Q))) = d(P, Q)$ and therefore conclude that $f^{-1}$ is an isometry since it preserves distance? I feel this is the same thing as what you wrote just with different notation, but I wanted to be sure. $\endgroup$
    – Sean
    Apr 10, 2020 at 3:35

2 Answers 2

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Let $X, Y$ be arbitrarily chosen from $\mathbb{E}^n$. Because the domain of $g$ is equal to the range of $f$, there exist $P,Q\in\mathbb{E}^n$ such that $X = f(P)$ and $Y = f(Q)$. Likewise $P = g(X)$ and $Q = g(Y)$.

Because $f$ is an isometry, we know that $PQ = f(P)f(Q)$. Substituting in $X$ and $Y$ we have $g(X)g(Y) = XY$. As $X,Y$ are arbitrarily chosen we have shown that $g$ is an isometry.


Not quite yet, we haven't. There's a subtlety: By the book's definition, if it's using standard notation, we can only call a function an "isometry" if its domain is all of $\mathbb{E}^n$. But we don't yet know that the domain of $f^{-1}$ is all of $\mathbb{E}^n$, that is, we don't know that $f$ is onto.

Here is a nice question and answers that hopefully help clear up this last point.

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  • $\begingroup$ This was actually another point of confusion for me, as the proof exercises in the book were to prove that isometries are injective (it's only the first part of chapter 1 so far), but many resources I see on the Internet state that the assumption is that they're bijective. If the standard definition states that the domain of an isometry if all of $\Bbb{E}^n$ and it has an inverse, this leads to the conclusion that isometries are bijective, right? $\endgroup$
    – Sean
    Apr 9, 2020 at 13:48
  • $\begingroup$ Whether "isometry" means bijective depends on context. In this setting I'm not taking that for granted. By your exercise, any isometry always has an inverse on its range. The question is whether the range of $f$ coincides with the codomain $\mathbb{E}^n$. $\endgroup$
    – Neal
    Apr 9, 2020 at 16:54
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At the time of posting there is a preview of the text including the relevant content available on google books: Modern Geometry with Applications, by George A. Jennings

This is what I came up with. The difficulty for me was in determining how much, and what kind of intuition to bring to bear. Since Jennings hasn't discussed inner products nor metrics, they probably shouldn't enter into the solution to the exercise.

Definition 1.2.1 Isometry. A function $f:\mathbf{E}^{n}\to\mathbf{E}^{n}$ is an isometry if, for all points $P$ and $Q$ in $\mathbf{E}^{n},$ $$ f\left(P\right)f\left(Q\right)=PQ. $$

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Exercise 1.2.1 a). Show that every isometry is a one-to-one function. (in other words, show that if $f$ is an isometry and $P$ and $Q$ are points with $P\ne Q$ then $f\left(P\right)\ne f\left(Q\right)$).

b) Assume that $f$ is an isometry and that it has an inverse function $f^{-1}.$ Show that $f^{-1}$also is an isometry.

Based on the intuition of the mythical n-ruler promulgated by Jennings in the second paragraph of section 1.1, we assume that $P=Q\iff PQ=0.$ It then follows immediately from the definition of isometry that $P\ne Q\iff f\left(P\right)\ne f\left(Q\right);$ which is the statement that $f$ is one-to-one. Thus we have addressed part a).

Let $P^{\prime}=f\left(P\right)$ and $Q^{\prime}=f\left(Q\right).$ Clearly if $f\left(P\right)f\left(Q\right)=PQ$ then $f^{-1}\left(P^{\prime}\right)f^{-1}\left(Q^{\prime}\right)=P^{\prime}Q^{\prime},$ as required by the definition of isometry. What remains to be shown is that the domain of $f^{-1}$ is $\mathbb{E}^{n};$ which is equivalent to saying $f\left(\mathbb{E}^{n}\right)=\mathbb{E}^{n}.$ So let's propose that some $P\notin f\left(\mathbb{E}^{n}\right).$ That is, there is no point which $f$ moves to replace $P$ when $P$ is moved to $f\left(P\right).$

Let $P^{\prime}=f\left(P\right)\ne P$ and $PP^{\prime}=r.$ Employing our mythical n-ruler we construct the $n-1$ sphere $\mathcal{S}=\left\{ S\vert SP=r\right\} ,$ centered on $P$ having radius $r.$ The image $\mathcal{S}^{\prime}=f\left(\mathcal{S}\right)$ is the $n-1$ sphere of radius $r$ centered on $P^{\prime}=f\left(P\right).$ ("In short, isometries preserve the size and shape of every geometric figure." Pg. 3) Since $PP^{\prime}=r,$ it follows that $P\in\mathcal{S}^{\prime}\subset f\left(\mathbb{E}^{n}\right),$ contradicting the proposition that $P\notin f\left(\mathbb{E}^{n}\right).$ Thus $f\left(\mathbb{E}^{n}\right)=\mathbb{E}^{n};$ satisfying part b).

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