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I was trying to prove that the horn torus and the spindle torus are not manifolds by definition(locally diffeomorphic to some Euclidean space.). I have no idea how to do this, but I attempted it in the following way:

I failed to show that a "slice" of manifold is a manifold itself. By a slice I mean, if the manifold $X\in \mathbf R^n$, then we set the coordinates $x_i=0$ for some $i$ where $0\leq i\leq n$. I feel this would work but I have no idea how to prove it.

Then I look at the cross section for the tori I mentioned(this is equivalent to taking a slice.). For the horn torus, you have two circles touching each other. And the fore spindle torus, you have two circles intersecting each other. Since a circle is locally diffeomorphic to $\mathbf R^1$, the two circles better have to be diffeomorphic to $R^1$. Otherwise we get the result we want.

I tried to proof that a neighbourhood around the point they touch(or intersect) cannot be locally diffeomorphic to $\mathbf R^1$. I tried to do it by contradiction. However I'm stuck on finding a contradiction...

Any thoughts?

P.S. Under the request of Sam Lisi, here are the definitions or horn and spindle tori:

A torus is the set of points in $\mathbf R^3$ at a distance $b$ from the circle of radius a in the $xy$ plane. This is like you put a circle with radius $b$ in the $yz$ plane centred at $(a,0,0)$. Then you make it orbit around the origin and you get a torus.

A horn torus is when $a=b$. If you take a cross-section, you'll find that it's two circles touching each other at one point. When $a<b$, it's called a spindle torus. The cross-section would look like two circles intersecting with each other at two pionts.

There are pictures in this wikipedia article that might help you visualise the horn and spindle tori: http://en.wikipedia.org/wiki/Torus

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Can you define what the spindle torus and horn torus are? –  Sam Lisi May 31 '13 at 11:54
    
Hi Sam, I have added the definitions to the question. Sorry to confuse you. –  Evariste May 31 '13 at 12:05
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This "slice" idea is not the right approach. You can easily construct a submanifold of $\mathbb{R}^3$ so that a coordinate slice (e.g. $z = const$) is not a submanifold for some suitable choice of constant. (Simple example: take a standard torus and choose a slice that gives you a figure 8). In your horn and spindle tori, you have points of self-intersection. Look at the neighbourhood of one of them. Does it look like a neighbourhood in a Euclidean space? –  Sam Lisi May 31 '13 at 12:31
    
Thanks Sam. I didn't quite get your argument for taking a standard torus and choose a slice that gives a figure 8. I thought you would never get a figure 8 when $a>b$. Also by looking at the neighbourhood, it looks like two planes intersecting with each other, which is definitely not a Euclidean space. But I have no idea how to put it into mathematical language. –  Evariste May 31 '13 at 12:45
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Imagine standing the standard torus on its end, and consider slices with horizontal planes. At the top you have just a point, and as you move down you get ellipse-like curves until you get to the top of the "hole" of the torus (the next critical point of height). At that height, the slice is a figure-8. –  Ted Shifrin May 31 '13 at 12:55

1 Answer 1

up vote 2 down vote accepted

Often the best way to show a $k$-dimensional subset $M$ of $\mathbb R^n$ is not a submanifold is this: Show that there's some point $p\in M$ so that no neighborhood is a graph of a smooth function over any of the standard coordinate $k$-planes.

In the case of your horn torus, at that central pinch point, the vertical line test will fail for any choice of the 3 standard coordinate ($2$-)planes ($xy$, $yz$, $xz$). Indeed, by symmetry, it fails for every plane, but you don't need to show this. (Intuitively, if the surface were to have a tangent plane at this point, by rotational symmetry, it would have infinitely many.)

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Hi Prof. Shifrin, thanks a lot for the comments! The first paragraph of your answer makes perfect sense. However I don't quite get what you mean by a "vertical line test". The one I know is "If a curve is a graph of a function, then every vertical line will intersect the curve at at most one point.". I haven't found any source that explains how this transforms to manifolds. –  Evariste May 31 '13 at 13:21
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I mean exactly the same thing. Without even worrying about smoothness, it's not locally the graph of a function because for every $(x,y)$, there's two $z$ values. ... By the way, for fun, apply this to the cuspidal cubic $y^2=x^3$ at the origin :) –  Ted Shifrin May 31 '13 at 13:25
    
Cool! Just one more question: why would vertical line test for the 3 standard coordinate be enough to rule out the possibility of locally being a function? I was thinking of this example: If we take a parabola, rotate it by some angle. Is it possible to have it intersecting with both $x$ and $y$ axes for two times(therefore checking the standard coordinates is not enough?)? Sorry I can't prove or disprove this(my bad)... –  Evariste May 31 '13 at 13:45
    
Well... I think it's a bad counter-example. A parabola can never be like that... But I can't really justify why having $n$ vertical line tests wrt $n$ standard coordinates would always suffice. –  Evariste May 31 '13 at 13:47
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It's a good exercise using the Inverse Function Theorem. Basically, the proof is this: The tangent plane $T_pM$, being a $k$-dimensional subspace, projects isomorphically onto some coordinate $k$-plane [if you put basis vectors for $T_pM$ into a matrix as columns, some $k$ rows must be linearly independent], and therefore locally the same is true for the manifold. Prove it :P –  Ted Shifrin May 31 '13 at 14:03

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