# Abstract Nonsense

## Surfaces (Pt. III)

Point of Post: This is a continuation of this post.

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Now, one of most fascinating, perhaps non-obvious facts about surfaces is that while every subset of $\mathbb{R}^3$ which looks locally like the graph of a smooth function is is a surface, the converse is, in a sense, also true. In particular, given any surface $\mathcal{S}$ and any point $p$ on the surface, we may find a neighborhood of $p$ which is just the graph of a smooth function $\mathbb{R}^2\to\mathbb{R}$. The first step in this is the following which, after unraveled basically says that every point on a surface has a neighborhood which can be paramaterized by a chart such that the inverse of the chart is nothing more than a projection.

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Theorem: Let $\mathcal{S}$ be a surface, $p\in\mathcal{S}$, and $\varphi:V\to U$ a differentiable chart at $p$. If $\varphi(q)=p$ then there exists a neighborhood $E$ of $q$ in $V$ and an orthogonal projection $\pi:\mathbb{R}^3\to\mathbb{R}^2$ such that $\pi(\varphi(E))=O$ is open in $\mathbb{R}^2$ and $\pi\circ \varphi:E\to O$ is a diffeomorphism.

Proof: Letting $\varphi_1,\varphi_2,\varphi_3$ denote the coordinate functions of $\varphi$ we have that the Jacobian of $\varphi:V\to U$ at $q$ is

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$\text{Jac}_\varphi(q)=\begin{pmatrix}D_1\varphi_1(q) & D_2\varphi_1(q)\\ D_1\varphi_2(q) & D_2\varphi_2(q)\\ D_1\varphi_3(q) & D_2\varphi_3(q)\end{pmatrix}$

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Now, by assumption that $\text{Jac}_\varphi(q)$ is injective we know that it has a minor which is non-zero, we may assume that the minor is created by the first two columns. Obviously, this assumption can be made since otherwise we may just apply a linear isomorphism of $\mathbb{R}^3$ to our map making this conclusion so, but which will not change the conclusion.  Consider then the orthogonal projection $\pi:\mathbb{R}^3\to\mathbb{R}^2$ where we’ve identified $\mathbb{R}^2$ with the $x-y$ plane. We see then that $\pi\circ\varphi:V\to\mathbb{R}^2$ is smooth, and by the chain rule

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$\displaystyle D_{\pi\circ\varphi}(q)=\pi\circ D_\varphi(q)=\begin{pmatrix}D_1\varphi_1(q) & D_2\varphi_1(q)\\ D_1\varphi_2(q) & D_2\varphi_2(q)\end{pmatrix}$

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and so by construction we have that $D_{\pi\circ\varphi}(q)$. The rest is just a literal statement of the inverse function theorem. $\blacksquare$

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So, now we may draw the conclusion we really want.  Namely, think about it. So, since we are able to create a diffeomorphism $\pi\circ\varphi:E\to O$ with $E\subseteq\mathbb{R}^2$ open, and $p\in\varphi(O)$ we obviously have that the map $\psi=\varphi\circ(\pi\circ\varphi)^{-1}:O\to \varphi(E)$ is a differentiable chart at $p$. But! Notice that we then have that $\pi\circ\psi=\text{id}_{O}$. What does this mean? This means that if we write $\psi_1,\psi_2,\psi_3$ as coordinate functions for $\psi$, then

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$(x,y)=\text{id}_O(x,y)=(\pi\circ\psi)(x,y)=\pi(\psi_1(x,y),\psi_2(x,y),\psi_3(x,y))=(\psi_1(x,y),\psi_2(x,y))$

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In other words, $\psi_1,\psi_2$ act trivially on the elements of $O$! Thus,

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$\psi(x,y)=(\psi_1(x,y),\psi_2(x,y),\psi_3(x,y))=(x,y,\psi_3(x,y))$

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Thus, we see that $\psi(O)$ is a neighborhood of $p$ in $\mathcal{S}$ which is just the graph of the smooth function $\psi_3:O\to\mathbb{R}$. Noting that we really can’t conclude, a priori that the ‘function part’ of the graph is in the third coordinate, we just didn’t need it for the problem we may conclude the following:

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Corollary: Let $\mathcal{S}$ be a surface. Then, for each $p\in\mathcal{S}$ there exists an open subset $U\subseteq\mathcal{S}$ containing $p$, an open subset $V\subseteq\mathbb{R}^2$, and a smooth function $f:V\to\mathbb{R}$ such that $U$ is the ‘graph’ of $f$.

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Remark: The scare quotes around graph, are to mean that all the elements of $U$ are of one of the following forms (uniformly, they don’t switch): $(x,y,f(x,y))$, $(x,f(x,y),y)$, $(f(x,y),x,y)$, etc.

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As a last note in this post, we’d like to prove that the transition maps $\tau_{\alpha,\beta}$ between two differentiable charts $(U_\alpha,\varphi_\alpha)$ and $(U_\beta,\varphi_\beta)$ on a surface $\mathcal{S}$ are diffeomorphisms.

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Theorem: Let $\mathcal{S}$ be a surface and let $\left(U_\alpha,\varphi_\alpha\right)$ and $\left(U_\beta,\varphi_\beta\right)$ be two differentiable, intersecting charts on $\mathcal{S}$. Then, the transition map $\tau_{\alpha,\beta}:\varphi_{\alpha}^{-1}(U_\alpha\cap U_\beta)\to U_\alpha\cap U_\beta$ given by $\varphi_\beta^{-1}\circ\varphi_\alpha$ are diffeomorphisms.

Proof: Since $\tau_{\alpha,\beta}$ is obviously a smooth homeomorphism, and the inverse of $\tau_{\alpha,\beta}$ is the transition map $\tau_{\beta,\alpha}$ it suffices to prove that $\tau_{\alpha,\beta}$ is smooth. But this follows immediately since we can for a small neighborhood around $p$ write $\tau_{\alpha,\beta}$ as $\left(\left(\pi\circ\varphi_\beta\right)^{-1}\circ\pi\circ\varphi_\alpha\right)^{-1}$ for some orthogonal projection $\pi$. But, $\pi\circ\varphi_\beta,\pi,$ and $\varphi_\alpha$ are smooth, and so the conclusion follows. $\blacksquare$

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References:

[1] Carmo, Manfredo Perdigão Do. Differential Geometry of Curves and Surfaces. Upper Saddle River, NJ: Prentice-Hall, 1976. Print.

[2] Pressley, Andrew. Elementary Differential Geometry. London: Springer, 2001. Print.

[3]  Montiel, Sebastián, A. Ros, and Donald G. Babbitt. Curves and Surfaces. Providence, RI: American Mathematical Society, 2009. Print.