Isometry of Riemannian manifolds

Remarks

Example: let $A \in O(n + 1)$ be an orthogonal matrix. The map $x \mapsto Ax$ restricts to a diffeomorphism of the unit sphere $S^n$ with differential $v \mapsto Av$. Since $\ip{Av}{Aw} = \ip{v}{w}$, it is an isometry of the round sphere.

Non-example: the map $x \mapsto 2x$ of $\R^n$ is a diffeomorphism, but $\ip{2v}{2w} = 4 \ip{v}{w}$, so it does not preserve the metric.

Let $M$ and $N$ have the same dimension. A smooth map $\varphi$ with $\varphi^* h = g$, not required to be a diffeomorphism, is a local isometry. The covering map $\R \to S^1$, $t \mapsto (\cos t, \sin t)$, is an example. An isometry preserves the Riemannian distance; conversely, by the Myers–Steenrod theorem, a distance-preserving bijection between connected Riemannian manifolds is an isometry in this sense.