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.
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.