Abstract Nonsense

Crushing one theorem at a time

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.

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October 9, 2011 - Posted by | Differential Geometry | , , , , , , , , ,

1 Comment »

  1. […] at . Indeed, we assume for a second that the single-sheeted cone is a surface. We know then from previous theorems that there exists some neighborhood of , some open subset , and a smooth function such that is […]

    Pingback by Instructive Non-Examples « Abstract Nonsense | October 14, 2011 | Reply


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