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Let $f:\mathbb{R}^3\to \mathbb{R}$ be defined by $f(x,y,z)=x^4+y^6+z^8$. Let $M=f^{−1}(1)$.

Is $M$ is diffeomorphic to a sphere $S^2$?

I tried to solve this problem, but I realized that I have no tools to solve it.

The constant rank theorem tells me $M$ is a smooth 2-dimensional manifold, but does not tell me how it looks like.

And more generally, when is $N=\{x,y,z\in\mathbb{R}^3\mid ax^n+by^m+cz^l=1\}$ diffeomorphic to a sphere? What tools can I use to solve this problem?

Thank you for reading. Hoping get some shedding light in your reply.

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You can find some good starting points on how to format mathematics on the site here. This AMS reference is very useful. If you need to format more advanced things, there are many excellent references on LaTeX on the internet, including StackExchange's own TeX.SE site. –  Zev Chonoles Feb 21 '13 at 12:46
+1. How eager I am to solve this problem, but cant' manage the way. ;-) –  B. S. Feb 21 '13 at 12:47
I have not given enough thought to the following to include it as an answer, but it seems like something along the following lines could work for the first question (although I am sure that there are more elegant approaches). The gradient $\nabla f = \langle 4x^3, 6y^5, 8z^7\rangle$ is orthogonal to the level set $N$. Now consider the vector field $\mathbf{v} = \langle -6y^5, 4x^3, 0\rangle$ defined on $\mathbb{R}^{3}$. Regarding $\mathbf{v}$ as a vector field on $N$, then $\mathbf{v}$ is a tangent vector field and the zeros of $\mathbf{v}$ are isolated . . . –  THW Feb 21 '13 at 19:27
. . . at the points $P(0,0,1)$ and $Q(0,0,-1)$. Couldn't one then compute the Euler characteristic of $N$ via the indices of the zeros? I might be overlooking something that prevents this (or maybe it is a miserable exercise), but it at least seems viable. –  THW Feb 21 '13 at 19:31

2 Answers 2

up vote 11 down vote accepted

I'll recast the problem in more general terms. We are given a compact submanifold $M\subset \mathbb R^n\setminus \{0\}$ which intersects every ray $\{t x:t>0\}$, $x\in\mathbb R^n\setminus\{0\}$, exactly once, and transversely. The claim is that $M$ is diffeomorphic to $\mathbb S^{n-1}$.

Transverse intersection of submanifolds $M_1,M_2$ means that at every point $p\in M_1\cap M_2$ the union of tangent spaces $T_pM_1$ and $T_pM_2$ spans the tangent space of the ambient manifold ($\mathbb R^n$ for us). In our situation this requirement amounts to $T_pM$ not containing the vector pointing from $p$ to the origin. And if $M$ is defined by equation $f=c$, this can be rephrased again by saying that $\nabla f(x)$ is never orthogonal to $x$; the latter is easy to check in your example.

Consider the radial map $G( x)=\dfrac{ x}{| x|}$ which radially projects $\mathbb R^n\setminus\{0\}$ onto the sphere $\mathbb S^{n-1}$. This is a submersion: a smooth surjective map such that the rank of differential is equal to dimension of the target space. Indeed, the derivative matrix of $G$ is $\dfrac{|x|^2I-x\otimes x}{|x|^3}$, which has one-dimensional kernel: namely, the vectors collinear to $x$.

Let $g=G_{|M}$, the restriction of $G$ to $M$. The differential also restricts, and by the transversality condition the differential of $g$ has trivial kernel. Since $g:M\to\mathbb S^{n-1}$ is a bijection by assumption, and both spaces are compact, we conclude that $g$ is a homeomorphism. Having invertible derivative at every point, it is also a diffeomorphism.

The above applies to $ax^n+by^m+cz^l=1$ whenever this surface is compact. When it is not compact, it can't be homeomorphic to the sphere, let alone diffeomorphic to it.

Added remark: I think it suffices to assume that $M$ is a compact submanifold whose intersection with every ray $\{t x:t>0\}$, $x\in\mathbb R^n\setminus\{0\}$ is transverse when it is nonempty. Indeed, this assumption implies, via the above argument, that $g:M\to\mathbb S^{n-1}$ is an open map. Hence $g(M)$ is open in $\mathbb S^{n-1}$, but being also compact, it must coincide with $\mathbb S^{n-1}$. It remains to show that $g$ is injective, but I'm drawing a blank here.

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Very nice! (and some more characters) –  Jason DeVito Feb 21 '13 at 19:32
Yes, this is quite nice. Thanks for putting this in a more general setting. –  THW Feb 21 '13 at 19:42
With regards to your added remark: First, one must assume $M$ is connected, for other wise a dsjoint union of cocentric spheres gives a counterexample. Adding this hypothesis (and that $n>2$), $g$ must be $1-1$: Since $g$ is open, it's a submersion. By Ehresmann's theorem, $g$ gives $M$ the structure of a fiber bundle over $S^{n-1}$. By dimension count, the fiber must be discrete, so $M$ is a covering of $S^{n-1}$. At least when $n > 2$, this is enough to guarantee that $M$ is diffeomorphic to $S^{n-1}$. If $n = 2$, this breaks down, but probably one can do it with barehands. –  Jason DeVito Feb 21 '13 at 20:11
(Actually, I think there should be a much easier argument than invoking Ehresmann's theorem, but at least you know your intuition is good!) –  Jason DeVito Feb 21 '13 at 20:13

Morse Theory is a very suitable tool to solve this kind of problems. We shall find a Morse function on $M$ and we can compute the indices of its critical points. Take $$ \begin{array}{rccc} h:&M&\longrightarrow&\mathbb{R}\\ &(x,y,z)&\longmapsto & x \end{array}. $$

Note that $T_{(x,y,z)}M=\ker d_{(x,y,z)}f$. Since $$ df=4x^3dx+6y^5dy+8z^7dz $$ its kernel is generated by $$ 3y^5\frac{\partial}{\partial x}-2x^3\frac{\partial}{\partial y}\ \mbox{and}\ 2z^7\frac{\partial}{\partial x}-x^3\frac{\partial}{\partial z},\ \mbox{at points with}\ x\neq 0 $$ $$ \frac{\partial}{\partial x}\ \mbox{and}\ 4z^7\frac{\partial}{\partial y}-3y^5\frac{\partial}{\partial z},\ \mbox{at points with}\ x=0 $$ Now, we have $dh=dx$ which applied to the vector fields above gives the functions $3y^5,2z^7$ at points with $x\neq 0$ and the constants $1,0$ at points with $x=0$. From this it follows that $(x,y,z)\in M$ is a critical point of $h$ if and only if $y=z=0$. So $h$ has exactly two critical points, which are $(1,0,0)$ and $(-1,0,0)$.

Finally, since $M$ is compact --I assume you know how to prove this-- and admits a Morse function with only two critical points, it must be a sphere (in this very simple case there's no need to compute the indices of the critical points, although they are 0 and 2, of course).

As for the general case $ax^n+by^m+cz^l=1$ you can carry out a similar computation and decide what surface you get in terms of critical points and its indices. However you'll need to be careful with compactness (what happens if one of the exponents is odd?) and probably you'd like to assume $abc\neq 0$.

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