0
$\begingroup$

An isometry between two metric spaces $(X_1, d_1)$ and $(X_2,d_2)$ is a map $f: X_1 \to X_2$ such that $d_1( y , z) = d_2(f(y),f(z))$ for all $y,z \in X_1$.

Yet on the other hand, two metric spaces are isometric if and only if there exists a bijective isometry between them. And the isometry group consists of all bijective isometries of the space.

(An isometry, as defined above, only has to be injective, not necessarily surjective.)

Question: Would it make more sense to call bijective isometries (as defined above) "isometries" and to call (possibly non-surjective) isometries something else?

Or would it make more sense to invent a new term for bijective isometries, the same way there is a special term for continuous maps with continuous inverse (homeomorphisms)?

Note: To prevent this question from being (too) opinion-based, answers which give pointers to references discussing these definitions would be preferred. I am not necessarily opposed to answers which give the answerer's personal opinion, but such answers might cause the question to be closed as being "primarily opinion-based".

The lack of separate terms for bijective and non-bijective isometries seems to lead to confusion when studying metric spaces (see e.g. here) which is probably unnecessary. It would be like using the same term for continuous maps and homeomorphisms -- possible to distinguish between the two uses depending on the context, but still confusing.

$\endgroup$
5
  • 3
    $\begingroup$ My personal preference is to use "isometric embedding" (or a "distance-preserving map") when surjectivity is not assumed and "isometry" in the surjective case. $\endgroup$ Apr 30 '17 at 12:53
  • 2
    $\begingroup$ I use exactly the same conventions as @MoisheCohen. In fact, I am surprised at the implication that there might be textbooks which do not require an isometry to be bijective. Can you give an example of such a textbook? $\endgroup$
    – Lee Mosher
    Apr 30 '17 at 13:05
  • 2
    $\begingroup$ @LeeMosher: Sadly, wikipedia is to blame. $\endgroup$ Apr 30 '17 at 13:13
  • $\begingroup$ @LeeMosher Yes, Moishe Cohen is right -- the first paragraph uses the convention you both mention, but then does not assume surjectivity here: en.wikipedia.org/wiki/Isometry#Formal_definitions That being said, I was not familiar with the term "isometric embedding" before, so this information is still quite helpful to me. $\endgroup$ Apr 30 '17 at 13:14
  • $\begingroup$ If I may ask, how could an isometry fail being surjective in the first place? $\endgroup$ Aug 3 '19 at 23:32
3
$\begingroup$

My personal preference is to use "isometric embedding" (as used, say, in this article), or a "distance-preserving map" (as in "Metric Spaces, Convexity and Nonpositive Curvature" by A. Papadopoulos), when surjectivity is not assumed and "isometry" in the surjective case.

See also page 75 of Burago, Burago, Ivanov, "A course in metric geometry". They use the terminology "isometry" in the bijective case and "isometry onto its image" without the surjectivity assumption. I find "isometric embedding" more concise. It is not optimal since the terminology an "isometric embedding" is used in Riemannian geometry for maps $f: (M_1,g_1)\to (M_2,g_2)$ such that $f^*g_2=g_1$. My take on the latter is that one should treat metric geometry and Riemannian geometry as separate (but related) areas of mathematics, which are then entitled to use the same name for different notions (akin to the fact that the name "normal" has different meaning in different areas of mathematics). When dealing with Riemannian manifolds which, at the same time are regarded as metric spaces, I would suggest using "distance preserving map" to avoid ambiguity.

$\endgroup$
1
  • $\begingroup$ This makes a lot of sense to me. In particular, "isometric embedding" seems to give exactly the right intuition -- such a map is always injective, and any injective map is bijective onto its image, i.e. an "isometry" on its image. Also the comments about what to do in the case of Riemannian-manifolds is also very helpful -- this is another question I have had, but thankfully now I do not need to ask it. Thank you for your help! $\endgroup$ Apr 30 '17 at 13:17

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.