Abstract Nonsense

Crushing one theorem at a time

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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[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.


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


  1. […] Surfaces (Pt. III) Point of Post: This is a continuation of this post. […]

    Pingback by Surfaces (Pt. III) « Abstract Nonsense | October 9, 2011 | Reply

  2. […] to prove it’s a surface. Namely, the cone looks like the level surface of a function, and we said that level surfaces were surfaces, right? Well, if you remember correctly the theorem stated that […]

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

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