## 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 .

*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 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’.**

*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 with . 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 is a smooth function, with open, and was a regular value then the level surface 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 and since we see that, in fact, so that 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. 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 is not a differentiable function 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 the ‘graph’ (see remark in the linked post) of , assume without loss of generality that . Note that we may assume without loss of generality that is an open ball around since clearly and so we may restrict to an open ball contained in around . Note then that for every since (and is contained in our cone) we have that . But, since (since we are dealing with the ‘upper cone’ where ) this says that . Uh-oh! We seem to have run into a problem. Indeed, we know that is smooth on , and so, in particular, differentiable at . That said, is not differentiable at ! Thus, we see that can’t look like the graph of a function near and so we may conclude that:**

**Theorem: ***The single-sheeted cone is not a surface.*

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 and rotating it around the -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.

*Too Thick or Too Thin*

As the section title should indicate there is a fundamental reason that a subset of 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 or dimensional. The first of these is best seen by an example, and the best is qualified by a theorem.

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

**Theorem: ***Let be a surface, then (where denotes interior in ambient ).*

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 , 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 ! Indeed:

**Theorem: ***Let be a surface and . Then, there exists a neighborhood in and a smooth function with such that .*

**Proof: **We know that we can find an open set and a neighborhood of in such that, up to a change of coordinates, . Let and let be given by . Clearly then (so that and evidently .

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

**Proof: **Suppose that was an interior point for when thinking of as sitting inside the ambient space . We know from the previous theorem that we can find an open set containing and a smooth function having as a regular value such that , we assume without loss of generality that . So, define by . Note then that

thus, since we know that is invertible. By the inverse function theorem we know there exists a neighborhood of and a neighborhood of such that is a diffeomorphism. But, this implies that is a diffeomorphism where is the plane. Note though, that since is open it contains a neighborhood of in and since we know diffeomorphisms are open (since is open) we can conclude that is an interior point in when thought of as living in . But, this is impossible since has no interior points. The conclusion follows.

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).

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 . 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 .

**Theorem: ***Let have an open subset which contains a cut-point, then is not a surface.*

**Proof: **Suppose that has a cut-point . If was a surface then we could find an open subset of which is homeomorphic to an open neighborhood of , which we may assume is contained within the neighborhood which has as the cut point. But, since we can restrict the open set in to an open ball, we may assume that a neighborhood of is homeomorphic to an open ball in . But, clearly then this is impossible for if were such a map (where is an open ball, and a neighborhood of ) then , but is connected but is not. This is a contradiction.

*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!

**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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