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I am confuse with these concepts: Isometry and Immersion. Let $M$ and $N$ be riemannian manifolds.

If $f:M\to N$ is a smooth isometry and will it be a immersion...

If $g$ is a immersion then i know that it need not be isometry... but whether converse is true?

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An immersion is a map such that its differential at every point is injective. An isometry is a map such that its differential at every point is a linear isometry between the corresponding tangent spaces. Since every linear isometry is injective, the result follows.

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An isometry requires a bit more than that (you've described an isometric immersion). – yasmar Mar 17 '12 at 22:02

A smooth isometry will always be an immersion. To see this, let $p\in M$. Let $v\in T_p M$ be a nonzero vector. The goal is to show $d_p f \ v\neq 0$. But what do you know about $\|v\|$? What do you know about $\langle d_p f \ v, d_p f \ v\rangle = \|d_p f \ v\|^2$?

An immersion need not be an isometry: For example, putting any metric on $S^1$ at all, any distance preserving immersion $S^1\rightarrow \mathbb{R}^2$ will be a Riemannian immersion, but cannot be an isometry for dimension reasons.

If you want an example where the 2 spaces have the same dimension, given any 2 different manifolds $M$ and $N$ with $\pi:M\rightarrow N$ a covering map, any (complete) metric on $N$ pulls back to a (complete) metric on $M$. With these 2 metrics, the map $\pi$ is a Riemannian immersion, but not an isometry.

In fact, at least when $M$ and $N$ are of the same dimension and given complete metrics, this is the only way to have Riemannian immersions which are not isometries.

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