# Abstract Nonsense

## Instructive Non-Examples

Point of Post: In this post we give some instructive non-examples of surfaces including pointy sets (sets with a singularity), non-two dimensional sets, and ‘thick’ subsets of $\mathbb{R}^3$.

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Motivation

In our last post we defined the notion of surfaces and gave a few examples, coming to the high point when we showed that surfaces are roughly the subsets of $\mathbb{R}^3$ which are locally the graphs of smooth functions. It’s my belief that it’s the non-examples of a subject which offer as much as intuition as these examples. Thus, hopefully these examples, and this ‘characterization’ gives one a good intuition as to precisely what a surface ‘looks like’.

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No Pointy Edges

There is this intuitive notion that a surface shouldn’t have any ‘pointy edges’. While what is meant by this is not very precise hopefully taking a look at example where a topological surface fails to be a surface because of its ‘pointiness’. In particular, let’s look at the single-sheeted cone $z^2=x^2+y^2$ with $z\geqslant 0$.  So, why is this not a surface? Well, before we express why it’s not a surface, let’s discuss what fails with the obvious ways 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 if $f:U\to\mathbb{R}$ is a smooth function, with $U\subseteq\mathbb{R}^3$ open, and $x\in\mathbb{R}$ was a regular value then the level surface $f^{-1}(\{x\})$ was a surface. So, if we hope to prove that the cone is not a surface, this theorem better not apply, and so in particular, the ‘pointy part’ better be a singular value. Indeed, one can quickly check that the vertex of the cone corresponds to $(0,0,0)$ and since $\nabla f(x,y,z)=2(x,y,z)$ we see that, in fact, $\nabla f(0,0,0)=0$ so that $(0,0,0)$ is a singular value, and so the theorem does not apply. This is heartening, because it at least doesn’t immediately disprove our claim that the cone is not a surface. $\text{ }$ So, how do we actually go about ‘proving’ that the cone can’t be a surface? Well, the ideal is simple enough. We shall reduce the problem roughly to the classic fact that $\sqrt{x^2+y^2}$ is not a differentiable function at $(0,0)$. 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 $U$ of $(0,0,0)$, some open subset $V\subseteq\mathbb{R}^2$, and a smooth function $f:V\to\mathbb{R}$ such that $U$ is the ‘graph’ (see remark in the linked post) of $f$, assume without loss of generality that $U=\left\{(x,y,f(x,y)):(x,y)\in V\right\}$. Note that we may assume without loss of generality that $V$ is an open ball around $(0,0)$ since clearly $(0,0)\in V$ and so we may restrict $f$ to an open ball contained in $V$ around $(0,0)$. Note then that for every $(x,y)\in V$ since $(x,y,f(x,y))\in U$ (and $U$ is contained in our cone) we have that $f(x,y)^2=x^2+y^2$. But, since $f(x,y)\geqslant 0$ (since we are dealing with the ‘upper cone’ where $z\geqslant 0$) this says that $f(x,y)=\sqrt{x^2+y^2}$. Uh-oh! We seem to have run into a problem. Indeed, we know that $f$ is smooth on $V$, and so, in particular, differentiable at $(0,0)$. That said, $\sqrt{x^2+y^2}$ is not differentiable at $(0,0)$! Thus, we see that $U$ can’t look like the graph of a function near $(0,0,0)$ and so we may conclude that:

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Theorem: The single-sheeted cone $\left\{(x,y,z)\in\mathbb{R}^3:z^2=x^2+y^2,\; z\geqslant 0\right\}$ is not a surface.

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That said, it’s pretty easy to see that the single-sheeted cone is a topological manifold. Another good example a topological surface which is not a surface is taken by taking the graph of $f(x)=|x^{\frac{1}{3}}|$ and rotating it around the $x$-axis. We shall get a, to abuse the term, ‘vase’ but with a ‘sharp’ edge. This will fail to be a surface since if we try to try to find a chart around a point on the ‘crease’ of this surface (the sharp part) that is the graph of a function, we will run into the same problem that the function will necessarily not be differentiable.

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Too Thick or Too Thin

As the section title should indicate there is a fundamental reason that a subset of $\mathbb{R}^3$ fails to be a surface, not because it’s not differentiable but because it fails to be two dimensional.  There are two obvious ways this can happen, the object can (at some point) be either incontrovertibly $1$ or $3$ dimensional. The first of these is best seen by an example, and the best is qualified by a theorem.

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So, let’s first show what we mean by being ‘too $3$-dimensional’.

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Theorem: Let $\mathcal{S}$ be a surface, then $\mathcal{S}^\circ=\varnothing$ (where $^\circ$ denotes interior in ambient $\mathbb{R}^3$).

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So, how do we go about proving this? Well, the first step in the proof (which follows a proof in [3] fairly closely) is to prove a certain local representation theorem for surfaces. Namely, we have seen that surfaces are locally the graphs of smooth functions $\mathbb{R}^3\to\mathbb{R}$, which is nice because graphs of functions were one of the two fundamental archetypes we were concerned with. Well, what about the other one, level surfaces of smooth functions (at regular points)? Well, as it turns out, surfaces are also locally the level sets (at regular values) of smooth functions $\mathbb{R}^3\to\mathbb{R}$ ! Indeed:

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Theorem: Let $\mathcal{S}$ be a surface and $p\in\mathcal{S}$. Then, there exists a neighborhood $O$ in $\mathbb{R}^3$ and a smooth function $F:O\to\mathbb{R}$ with $\nabla F(p)\ne (0,0,0)$ such that $F^{-1}(\{0\})=\mathcal{S}\cap O$.

Proof: We know that we can find an open set $V\subseteq\mathbb{R}^2$ and a neighborhood $U$ of $p$ in $\mathcal{S}$ such that, up to a change of coordinates, $U=\left\{(x,y,f(x,y)):(x,y)\in V\right\}$. Let $O=\mathbb{R}\times V$ and let $F:O\to \mathbb{R}$ be given by $(x,y,z)\mapsto z-f(x,y)$. Clearly then $F_z(p)=1$ (so that $\nabla F(p)\ne (0,0,0)$ and evidently $\mathcal{S}\cap O=F^{-1}(\{0\})$. $\blacksquare$

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From this we can prove our previous theorem about surfaces having empty interiors:

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Proof: Suppose that $p\in\mathcal{S}$ was an interior point for $\mathcal{S}$ when thinking of $\mathcal{S}$ as sitting inside the ambient space $\mathbb{R}^3$. We know from the previous theorem that we can find an open set $O\subseteq\mathbb{R}^3$ containing $p$ and a smooth function $F:O\to\mathbb{R}$ having $0$ as a regular value such that $F^{-1}(\{0\})=O\cap\mathcal{S}$, we assume without loss of generality that $F_z(p)\ne 0$. So, define $\widetilde{F}:O\to\mathbb{R}^3$ by $(x,y,z)\mapsto (x,y,F(x,y,z))$. Note then that

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$\displaystyle \text{Jac}_{\widetilde{F}}(p)=\begin{pmatrix}1 & 0 & F_x(p)\\ 0 & 1 & F_y(p)\\ 0 & 0 & F_z(p)\end{pmatrix}$

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thus, since $\det\text{Jac}_{\widetilde{F}}(p)=F_z(p)\ne 0$ we know that $D_{\widetilde{F}}(p)$ is invertible. By the inverse function theorem we know there exists a neighborhood $U$ of $p$ and a neighborhood $V$ of $\widetilde{F}(p)$ such that $\widetilde{F}:U\to V$ is a diffeomorphism. But, this implies that $\widetilde{F}:U\cap\mathcal{S}\to F(U\cap \mathcal{S})=V\cap P$ is a diffeomorphism where $P$ is the $x-y$ plane. Note though, that since $U\cap \mathcal{S}$ is open it contains a neighborhood of $p$ in $\mathbb{R}^3$ and since we know diffeomorphisms are open (since $W$ is open) we can conclude that $\widetilde{F}(p)$ is an interior point in $V\cap P$ when thought of as living in $\mathbb{R}^3$. But, this is impossible since $P$ has no interior points. The conclusion follows. $\blacksquare$

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Thus, things like solids are not surfaces, because they have non-empty interior! This entire theorem makes sense since objects which have interiors are at least intuitively entirely three dimensional (this is actually interesting because it makes the assumption that something can’t be both two and three dimensional, which is the case, but is not eay).

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We would now like to discuss is how something can be too one-dimensional. This is slightly harder to quantify, but what we shall focus on are ‘cut-points’. The fundamental idea is this, suppose that, for example we wanted to prove that a curve is not a surface. Obviously this makes sense since curves are one-dimensional objects, but how do we do this without heavy machinery? Well, the basic idea is that if the curve was a surface, then picking any point we’d see that there is some neighborhood of that point (on the curve) which is homeomorphic to an open ball in $\mathbb{R}^2$. But, the problem is that removing the point in question will disconnect the curve, but the removal of the preimage of the point will not disconnect the disc. This is, in essence, the same elementary argument to prove that $\mathbb{R}\not\approx\mathbb{R}^2$.

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Theorem: Let $S\subseteq\mathbb{R}^3$ have an open subset which contains a cut-point, then $S$ is not a surface.

Proof: Suppose that $S$ has a cut-point $p$. If $S$ was a surface then we could find an open subset of $\mathbb{R}^2$ which is homeomorphic to an open neighborhood of $p$, which we may assume is contained within the neighborhood which has $p$ as the cut point. But, since we can restrict the open set in $\mathbb{R}^2$ to an open ball, we may assume that a neighborhood of $p$ is homeomorphic to an open ball in $\mathbb{R}^2$. But, clearly then this is impossible for if $f:B\to U$ were such a map (where $B$ is an open ball, and $U$ a neighborhood of $p$) then $f:B-\{f^{-1}(p)\}\xrightarrow{\approx}U-\{p\}$, but $B-\{f^{-1}(p)\}$ is connected but $U-\{p\}$ is not. This is a contradiction. $\blacksquare$

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Remark: This actually proves that no topological surface can have a cut point. Note that this does not say that a disconnected subset cannot be a surface!

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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.