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Motivation

Its a classic set up. Take a metric space $M$, with distance function $d:M\times M\to \mathbb{R}$. The set of isometries of $M$ is the set of functions $f: M \to M$ which preserve distance. This set has much of the structure of a group without additional assumptions; the composition of two isometries gives an isometry, the identity function takes the place of the identity, etc.

Although every isometry must be injective, however, it is not necessarily a bijection, and so might not have an inverse. For example any injective function from a metric space with the discrete metric ($d:M\times M\to \mathbb{R}, d(a,b) = 1$ if $a \neq b$ and $0$ if $a=b$) to itself is an isometry.

As the group of isometries is quite a useful gadget we can get round this. For example in MathWorld an isometry is assumed to be bijective: http://mathworld.wolfram.com/Isometry.html

Yet in Euclidean space, we do not need any additional assumption:

Lemma

Every isometry of $\mathbb{E}^2$, the Euclidean plane, is a surjection.

Proof

On the plane, for example, assume that a point $a$ does not lie in the image of an isometry $T$. Take three distinct points $T(b_1), T(b_2)$ and $T(b_3)$ that do lie in the image (as $T$ is injective they have a unique preimage). Let $d_i = d(b_i,a)$ be the distance between $d$ and $b_i$.

The three circles radius $d_i$ around $T(b_i)$, intersect together only at $a$, as the distances between the $T(b_i)$ and the $B_i$ are the same, the circles radius $d_i$ around $b_i$ will also intersect at a unique point $a'$. The point $T(a')$ must be $a$ as that is the only point that satisfies all the point to point distances, so $a$ does lie in the image of $T$.
$\square$

Question

We can extend this argument to higher dimensional Euclidean spaces, yet it uses non-trivial properties, in particular how circles intersect. Is there a simple propetry of a metric space that ensures that the set of isometries forms a group?

Edit To rather strengthen the question, are there simple properties that are necessary and sufficient.

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It is a fun exercise to prove that every isometry of a compact metric space is surjective. –  Andy Putman Sep 17 '11 at 16:39
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Well... clearly not since a set with the discrete metric is locally compact as a space. Other example: the half line $\mathbb R_{\ge 0}$. –  André Henriques Sep 17 '11 at 18:25
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See Noam Elkies's counterexamples. –  Todd Trimble Sep 17 '11 at 18:28
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Euclidean space $E$ has this other amazing property of isometries: Any isometry defined on a subset of $E$ with values in $E$ has an extension to an isometry of all of $E$ onto all of $E$. –  Gerald Edgar Sep 17 '11 at 19:13
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It appears that there are too many different settings for there to be existing terminology that covers all cases. So, if you want, you are free to define a new property: call a metric space Friendly if every distance-preserving self-map is surjective. I'm not seeing any of the usual topological properties implying this in all cases, and you have the two main relevant cases of examples/counterexamples, real manifolds and various subsets, also Hilbert spaces. So this is new territory and rather cute. –  Will Jagy Sep 17 '11 at 20:27

4 Answers 4

You are asking for conditions on a metric space $X$ for which any distance preserving map $X\to X$ is bijective. (Usually isometry is defined as bijective distance preserving map).

Well, there are arbitrarily bad spaces $X$ such that the only distance preserving map $X\to X$ is identity. In this case distance preserving maps (well, the only one) form the trivial group.

So, I do not see a language which could be used to formulate a necessary and sufficient condition. For sufficient conditions:

  • Compactness;
  • Proper + cocompact isometric group action. (proper=bounded closed sets are compact)
  • Any complete connected space for which the domain invariance theorem holds; in particular complete connected Riemannian manifolds without boundary.
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Every isometry between two complete connected Riemannian manifolds of the same dimension is a bijection.

Sketchy proof: Let $f:M\to N$ be an isometry between manifolds as above. I want to show that $f$ is surjective. Given $z\in N$, I want to show that $z\in im(f)$. Pick a point $x\in M$, and let $y:=f(x)\in N$. Pick a curve $\gamma$ connecting $y$ to $z$. Since $df$ is everywhere invertible, the map $f$ is a local homeomorphism, and so there is at most one lift of the curve $\gamma\subset N$ to a curve $\tilde \gamma\subset M$. Since the metric in $M$ is complete, the curve $\tilde \gamma$ does exist. Its end point is the desired preimage of $z$.  $\square$

Analysis of what's used in the above proof:
    • Every isometry $M\to N$ is a local homeomorphism.
    • $N$ is path connected.
    • The metric on $M$ is complete.
...and let's not forget:
    • $M$ is non-empty.

Another example where you get your desired result: regular trees.

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There is a school of thought that says that connected spaces are nonempty by definition. (I couldn't tell from what you wrote whether you are of that school!) This is like declaring that a prime must be a non-unit; for more on this, see ncatlab.org/nlab/show/connected+space#definitions_9 –  Todd Trimble Sep 17 '11 at 19:54
    
Yes. I belong to that school of thought. For me, a space $X$ is connected iff it satisfies the following two conditions: any map from $\{x\in\mathbb R^1:\|x\|=1\}$ to $X$ extends to $\{x\in\mathbb R^1:\|x\|\le1\}$ and any map from $\{x\in\mathbb R^0:\|x\|=1\}$ to $X$ extends to $\{x\in\mathbb R^0:\|x\|\le1\}$. However, I also think that these confusing conventions should always be made explicit. –  André Henriques Sep 17 '11 at 20:08

Essentially all you need is surjectivity, which is usually part of the definition of being an isometry. However, if you prefer to keep the definition as you stated above, there are classes of metric spaces for which every isometry is surjective:

Your example above extends to all Euclidean spaces: in this case, you can classify the group of isometries; every isometry is the composition of an orthogonal transformation and a translation.

If $M$ is a Riemannian manifold, then the Riemannian metric induces a metric $d$ on $M$, by defining $d(p,q)$ as the infimum of the length of all piecewise smooth curves joining $p$ and $q$. In this case, the group of isometries of $(M,d)$ is the same as the group of isometries of $M$ as a Riemannian manifold, and is a Lie group by a theorem of Myers and Steenrod.

Counterexamples to the general case abound: take $M$ to be a Hilbert space, say $l^2$ and take the unilateral shift: $$ \phi(x_1,x_2, \dots)= (0,x_1,x_2, \dots ) $$ which is an isometry, but not surjective.

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Much simpler counterexamples: the positive or nonnegative reals, or the natural numbers, with the shift $x \mapsto x+1$; or any infinite space with the discrete metric and any map at all that's injective but not surjective. –  Noam D. Elkies Sep 17 '11 at 16:37
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I agree that surjectivity is all that is required for the isometry; however the question asks what properties of the metric space make it unecessary to assume that isometries are surjective. –  Edmund Harriss Sep 17 '11 at 16:46
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The Riemannian manifolds in the answer are tacitly assumed to be connected. –  Andreas Blass Sep 17 '11 at 18:10

Not sure how to comment, but it seems that some concept of dimension is important here. Things clearly break down for infinite dimensional spaces, but a finite dimensional space cannot have a lower dimensional subspace that is isomorphic.

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this is quite false. as was remarked in one of the earlier comments a half line (which is one-dimensional by any measure) admits a non-surjective self isometry. –  Vitali Kapovitch Dec 27 '11 at 15:01
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That is unduly harsh. I wanted to make this a comment, as it is clearly not a full answer, but cannot as I am new to the site. In the example you give the non-surjective isometry is not to a lower dimensional subspace, such examples need to be considered, but it does not make anything I said false. –  Davey Dec 27 '11 at 19:46
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sorry, I misread your statement and missed the part about the lower dimension. Then what you say is correct. –  Vitali Kapovitch Dec 28 '11 at 3:35

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