Let $\Sigma \subset \mathbb R^3$ be a regular surface and $\gamma: (- \varepsilon, \varepsilon) \to \mathbb R^3$ be a smooth curve with $\mathrm{im}( \gamma) \subset \Sigma$, and $\gamma (0) = p$. Let $\sigma : U \subset \mathbb R^2 \to V \subset \Sigma$ be a parameterization of $\Sigma$ near $p$ (so it's smooth as a map to $\mathbb R^3$). Define the map $\Gamma: (- \varepsilon, \varepsilon) \to \mathbb R^2$ via $\Gamma(t) = \sigma^{-1}(\gamma(t))$, i.e. $\Gamma$ is the curve in the parameterization plane corresponding to $\gamma$ on the surface.
Is $\Gamma$ necessarily a smooth map to $\mathbb R^2$? Can I write, say, $D_0 \gamma = D_{\sigma^{-1} (p)} \sigma \circ D_0 \Gamma$ by the chain rule?
Best Answer
An equivalent definition for a regular surface is as such: $\Sigma$ is a regular surface iff for all $p \in \Sigma$ there exists an open neighborhood $T$ of $p$ in $\mathbb R^3$, an open neighborhood $W$ of the origin in $\mathbb R^3$, and a diffeomorphism $g: T \to W$ such that $g(T \cap \Sigma) = W \cap (\mathbb R^2 \times \{ 0 \})$.
Let's WLOG show that $\Gamma$ is smooth at $0 \in (-\varepsilon, \varepsilon)$. For $t$ near $0$ (i.e. $\gamma(t)$ near $p$), we have $$ \Gamma = \sigma^{-1} \circ \gamma = (g \circ \sigma)^{-1} \circ (g \circ \gamma). $$ $g\circ\sigma$ is a smooth map in $\mathbb R^3$ landing in $\mathbb R^2$, so we can treat it as a smooth map to $\mathbb R^2$. As a map to $\mathbb R^2$, it's a local diffeomorphism at $\sigma^{-1}(p)$ since $$ D_{\sigma^{-1}(p)}(g \circ \sigma) = \underbrace{D_p g}_{\text{isometry $\mathbb R^3 \to \mathbb R^3$}} \circ \underbrace{D_{\sigma^{-1}(p)} \sigma}_{\text{rank 2}} $$ has rank $2$, so is invertible. Hence there exists a neighborhood about $\sigma^{-1} (p)$ on which $g \circ \sigma$ is a diffeomorphism onto its image, i.e. $(g \circ \sigma)^{-1}$ is smooth in a neighborhood about $g(p)$.
Moreover, $g\circ \gamma$ is a composition of smooth maps, so is smooth. Hence, there is a neighborhood about $0$ in which $\Gamma$ is a composition of smooth maps, so is smooth.
Please let me know if you spot a mistake.