Differential Geometry: Isometric Mappings 📂Geometry

Differential Geometry: Isometric Mappings

Definition¹

Suppose a function $f : M \to N$ is given between two surfaces $M, N$. Then $f$ is called an isometry if it satisfies the following conditions:

$f$ is differentiable.
$f$ is bijective.
For every curve $\boldsymbol{\gamma}:[c,d] \to M$, the length of $\boldsymbol{\gamma}$ and the length of $f \circ \boldsymbol{\gamma}$ are the same.

If there exists an isometry $f$ between $M$ and $N$, then $M$ and $N$ are said to be isometric.

Explanation

In simple terms, an isometry is a mapping that preserves the length of a curve on surface $M$ when it is transferred to $N$ or vice versa.

Since we describe geometry through differentiation, being differentiable is a natural condition, and there should be no problem going back and forth between surfaces $M$ and $N$, so it must be bijective. Moreover, since we want to talk about a mapping that preserves distance, the last condition must also be satisfied for it to be called an isometry.

Proposition

Let $f : M \to N$ be an isometry, and $\boldsymbol{\gamma} : [c,d] \to M$ a regular curve. Then, the lengths of $\dfrac{d \boldsymbol{\gamma}}{d t}$ and $\dfrac{d (f \circ \boldsymbol{\gamma})}{d t}$ are the same.

$$ \left| \dfrac{d \boldsymbol{\gamma}}{d t} \right| = \left| \dfrac{d (f \circ \boldsymbol{\gamma})}{d t} \right| $$

Proof

Since $f$ is said to be an isometry, for each $t^{\ast} \in (c,d)$, the lengths of $\boldsymbol{\gamma}$ and $f \circ \boldsymbol{\gamma}$ are the same.

$$ \ell_{[c,t^{\ast}]}(\gamma) = \ell_{[c, t^{\ast}]} (f \circ \gamma), \quad t^{\ast} \in (c,d) $$

$$ \implies \int_{c}^{t^{\ast}} \left| \dfrac{d \boldsymbol{\gamma}}{d t} \right|dt = \int_{c}^{t^{\ast}} \left| \dfrac{d (f \circ \boldsymbol{\gamma})}{d t} \right|dt $$

Differentiating both sides with respect to $t^{\ast}$ gives:

$$ \left| \dfrac{d \boldsymbol{\gamma}}{d t} \right| = \left| \dfrac{d (f \circ \boldsymbol{\gamma})}{d t} \right| $$

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