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I was studying differentiable manifolds (an introduction) and found the following example, but I am confused.

Example The function \begin{align} f: &\mathbb{R}^{3} \to \mathbb{R}, \\ f: &(x,y,z) \mapsto x^{3} + 2 y^{3} + z^{3} + 6 x^{2} y - 1 \end{align} defines the structure of a $ C^{\infty} $-manifold on $ {f^{-1}}(0) $.

Thank you.

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This may help you: math.stackexchange.com/questions/262211/… –  Manoel Feb 7 '13 at 5:09
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Welcome to math.SE: since you are new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. If this is homework, please add the homework tag; people will still help, so don't worry. Also, many find the use of imperative ("Prove", "Solve", etc.) to be rude when asking for help; please consider rewriting your post. –  Michael Albanese Feb 7 '13 at 5:09
    
Thanks for the guidance @MichaelAlbanese. –  Nathan Marke Feb 7 '13 at 5:19
    
Thanks for the hint @Manoel. –  Nathan Marke Feb 7 '13 at 5:20
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Have you taken a look at the Implicit Function Theorem yet? –  Haskell Curry Feb 7 '13 at 5:27

2 Answers 2

up vote 4 down vote accepted

Firstly, compute the Jacobian matrix of $ f $: \begin{align} \forall (x,y,z) \in \mathbb{R}^{3}: \quad [\mathbf{D}(f)](x,y,z) &= \left[ \matrix{({\partial_{x}} f)(x,y,z) & ({\partial_{y}} f)(x,y,z) & ({\partial_{z}} f)(x,y,z)} \right] \\ &= \left[ \matrix{3 x^{2} + 12xy & 6 x^{2} + 6 y^{2} & 3 z^{2}} \right]. \end{align} Observe that for $ (a,b,c) \in {f^{\leftarrow}}[\{ 0 \}] $, we cannot have $ b = c = 0 $, so either

  1. $ \left[ 6 a^{2} + 6 b^{2} \right] $ is a non-singular $ (1 \times 1) $-submatrix of $ [\mathbf{D}(f)](a,b,c) $, or

  2. $ \left[ 3 c^{2} \right] $ is a non-singular $ (1 \times 1) $-submatrix of $ [\mathbf{D}(f)](a,b,c) $.

In Case 1, the Implicit Function Theorem says that there exist an open neighborhood $ U $ of $ (a,c) $, an open neighborhood $ V $ of $ b $, and a unique continuously differentiable function $ g: U \to V $ such that \begin{align} &\{ (x,y,z) \in \mathbb{R}^{3} ~|~ (x,z) \in U ~ \land ~ y = g(x,z) \} \\ = &\{ (x,y,z) \in \mathbb{R}^{3} ~|~ (x,z) \in U ~ \land ~ y \in V ~ \land ~ f(x,y,z) = 0 \}. \end{align} We thus have a surface-parametrization of an open neighborhood of $ (a,b,c) $ in $ {f^{\leftarrow}}[\{ 0 \}] $, which is of the form $$ \forall (x,z) \in U: \quad (x,z) \longmapsto (x,g(x,z),z) \subseteq {f^{\leftarrow}}[\{ 0 \}]. $$

By a similar argument, we can reach the same conclusion for Case 2.

To establish the smoothness of the function $ g $, we can apply the Analytic Implicit Function Theorem, whose proof may be found in this set of notes.


Conclusion: $ {f^{\leftarrow}}[\{ 0 \}] $ is a $ C^{\infty} $-manifold of dimension $ 2 $.

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Guillemin and Pollack would go about this by defining a regular value: A regular value of a map $f:X\to Y$ is any point $y\in Y$ such that the derivative of $f$ is nonsingular at points mapping to $y$. Then they show that the preimage $f^{-1}(y)$ is a submanifold of $X$. (Differential Topology, p. 21)

So to solve your question this way, you just need to compute the Jacobian of $f$ and find when it is singular, then you show that none of these points are in $f^{-1}(0)$. Then you know $0$ is a regular value and $f^{-1}(0)$ is a manifold.

Note that the regular value method subsumes the machinery used in the other answer (i.e. arguments like that are used to prove the things leading up to the regular value theorem.)

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