# Abstract Nonsense

## Surfaces (Pt. II)

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

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So, we have proven that the unit sphere $\mathbb{S}^2$ is a surface. That said, it should be intuitively obvious that any sphere of any radius and any center should be a surface. In fact, it should be the case that any ‘smooth’ deformation of $\mathbb{S}^2$ into an ellipsoid, or something of the sort, should also be a surface. The operative term in this last sentence was ‘smooth’.  Indeed, it’s evident that if one has a surface $\mathcal{S}$ which can be thought of, at least locally, as the smooth deformation of $\mathbb{R}^2$, then evidently applying a smooth deformation $f$ to $\mathcal{S}$ will result in something that locally is just a smooth deformation of $\mathbb{R}^2$–namely each point locally looks like you took $\mathbb{R}^2$ smoothly deformed it into a piece of $\mathcal{S}$, and then smoothly deformed that piece of $\mathcal{S}$ into the piece of $f(\mathcal{S})$. In other words, it seems intuitive that if we compose chart maps for $\mathcal{S}$ with $f$ we should get maps that deform $\mathbb{R}^2$ smoothly into pieces of $f(\mathcal{S})$. This intuition is formalized by the following theorem:

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Theorem: Let $V_1,V_2\subseteq\mathbb{R}^3$ be open. Suppose that $f:V_1\to V_2$ is a diffeomorphism (i.e. a bijective $C^\infty$ function with $C^{\infty}$ inverse). If $\mathcal{S}_1\subseteq V_1$ is a surface then $f(\mathcal{S}_1)=\mathcal{S}_2$ is a surface.

Proof: Let $f(p)\in \mathcal{S}_2$ be arbitrary. Since $\mathcal{S}_1$ is a surface we may choose a smooth chart $(U,\varphi)$ with $p\in U$. Consider then the map $f\circ\varphi:\varphi^{-1}(U)\to f(U)$. Obviously $f(U)$ is open in $\mathcal{S}_2$ (since $f$ is a homeomorphism), $f\circ\varphi$ is a homeomorphism (since they are both homeomorphisms), $f\circ \varphi$ is smooth (since they’re both smooth) and $f(p)\in f(U)$. Thus, if we can verify that $D_{f\circ\varphi}(x)$ is injective for all $x\in \varphi^{-1}(U)$ we’ll be done. That said, from the chain rule we know that $D_{f\circ\varphi}(x)=D_f(\varphi(x))\circ D_\varphi(x)$ and since $D_f(\varphi(x))$ and $D_\varphi(x)$ are both injective ($D_\varphi(x)$ by assumption, and $D_f(\varphi(x))$ (actually is invertible) by assumption that $f$ is a diffeomorphism)  we may conclude that $D_{f\circ\varphi}(x)$ is injective. The conclusion follows. $\blacksquare$

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This tells us that, in fact we were correct, that things like ellipsoids and spheres of arbitrary radius and center are surfaces, being just images of $\mathbb{S}^2$ under affine transformations.

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So, we have shown our one claim in the motivation, namely that the graph of a smooth function is a surface. We now show the other claim about level sets of curves. For the setup of this theorem we recall that we call, for a function $f:U\to\mathbb{R}$ where $U\subseteq\mathbb{R}^3$ is open, the point $x\in\mathbb{R}$ a regular value if for every $p\in U$ with $f(p)=x$ one has that $D_f(p)$ is non-zero, which, recalling that $D_f(p)(q)=\nabla f(p)\cdot q$, is equivalent to $\nabla f(p)\ne 0$. With this in mind we can state the following:

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Theorem: Let $U\subseteq\mathbb{R}^3$ be open and $f:U\to\mathbb{R}$ a smooth function. Then, if $c\in\mathbb{R}$ is a regular value for $f$ one has that $\mathcal{S}=f^{-1}(\{c\})$ is a surface.

Proof: Let $p\in\mathcal{S}$ be arbitrary. Since $p$ is a regular point for $f$ we know from the implicit function theorem that since $\nabla f(p)\ne 0$ we may find  neighborhood $X$ of $p$ and open sets $O\subseteq\mathbb{R}^2$, $V\subseteq\mathbb{R}$ with a smooth function $g:O\to V$ with $X=\Gamma_g$ (the graph) from where we may easily conclude by our previous comments about graphs that $\mathcal{S}$ has a differentiable chart at $p$. Since $p$ was arbitrary the conclusion follows. $\blacksquare$

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Said roughly, we used the implicit function theorem to say that every surface which is the level set of a smooth function at a regular point, locally looks like the graph of a function and thus is a surface.

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This gives us convenient ways of proving a lot of objects are surfaces. For example:

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Example: Let’s now show that the torus with tube radius $r$, and distance from center of tube to center of hole equal to $R$ with $R\ne r$ is a surface (centered at the origin). Indeed, one can quickly check that this torus is given implicitly by the level set $\left\{(x,y,z)\in\mathbb{R}^3:\left(R-\sqrt{x^2+y^2}\right)^2+z^2=r^2\right\}$. Thus, if we can show that $\nabla f(x,y,z)\ne 0$ (where $f(x,y,z)=\left(R-\sqrt{x^2+y^2}\right)^2+z^2$) for any $(x,y,z)\in\mathbb{R}^3$ with $f(x,y,z)=r^2$ then we may conclude from the previous theorem. To do this we calculate that

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$\displaystyle \nabla f(x,y,z)=\left(2x\left(1-\frac{R}{\sqrt{x^2-y^2}}\right),2y\left(\frac{R}{\sqrt{x^2-y^2}}-1\right),2z\right)$

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Clearly then we see that $\nabla f(x,y,z)=0$ if and only if $(x,y,z)=(0,0,0)$ but $f(0,0,0)=R^2$ and since we assumed that $r\ne R$ we see that $f(0,0,0)\ne r^2$. Thus, we see that for every point $(x,y,z)\in\mathbb{R}^3$ such that $f(x,y,z)=r^2$ one has that $\nabla f(x,y,z)\ne 0$ and so the torus is a surface by the previous theorem.

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Clearly we know then that an analogous torus centered at any point is a surface, being the image of an example of the tori discussed above, by an affine transformation.

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Remark: It’s a curiosity that when $R=r$ we don’t get a surface. In particular, we see that it doesn’t satisfy the ‘smooth’ condition (it’s clearly still a topological surface). There is a very visually obvious reason for this. If one graphs tori with $r=R$ then one gets what is called a ‘horned torus’. This is a torus which doesn’t have a hole, and which curves up ‘inside itself’, ending in a sharp corner. A good picture of this can be found here.

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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.