It seems that isometry does not have to be bijective (surjective) as I learnt from my lecture. But why on https://en.wikipedia.org/wiki/Isometry_(Riemannian_geometry) defined isometry as diffeomorphism? So it is bijective? A bit confusing.

Which is right?

  • $\begingroup$ Wikipedia is right. Do you have a precise statement of what your lecturer said? $\endgroup$
    – gj255
    Jan 8, 2017 at 19:22
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    $\begingroup$ @gj255 The embedding $P:\mathbb{R}^2\rightarrow\mathbb{R}^3$ with $P(x) = (x_1,x_2,0)$ is an isometric embedding. It is not surjective. Maybe, one strictly distinguishes between an isometry and an isometric embedding. This could be the source of confusion. $\endgroup$
    – Tobias
    Jan 8, 2017 at 19:46
  • $\begingroup$ Or even here: en.wikipedia.org/wiki/Isometry section: Formal definitions. it seems that they attribute "bijective" to isometry, so there can be a non-bijective? $\endgroup$
    – Upc
    Jan 8, 2017 at 19:58
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    $\begingroup$ Your link en.wikipedia.org/wiki/Isometry is informative with respect to your question. There they say "usually assumed to be bijective". This already implies that there are different versions in use. I recommend sticking to the version of your teacher and keeping in mind that there might be a different version in literature. $\endgroup$
    – Tobias
    Jan 8, 2017 at 20:50
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    $\begingroup$ If $M$ is a complete, connected Riemannian manifold and $N$ is connected, then a local isometry $f:M \to N$ is bijective: How to go from local to global isometry. Without completeness of $M$, you have examples such as $x \mapsto x + 1$ on the set of positive reals. $\endgroup$ Jan 8, 2017 at 22:56

1 Answer 1


No, it is not accurate. Typically, a general manifold may be not equipped with an metric structure. If two manifolds are diffemorphic and isometric, we can say the local diffemorphic group is unitary.


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