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Let $M$ be the set of points $(x,y,z)$ in $\mathbb{R}^3$ such that $x^2+y^2+z^2=1$ and $x^2=yz^2$.

The point $(0,-1,0)$ is removed.

The question is: after removing a second point (to determine), why is this a manifold?

I can argue that each of the above, the sphere and the $x^2=yz^2$, are manifolds since they are level sets of smooth functions. But for the intersection I'm not sure what to do.

Any suggestion is welcome!

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By the way: the solution set of $x^2 = yz^2$ is not a manifold. Note that to conclude that a level set is a manifold you need that the value of the function on that level set to be a regular value. But $(0,-1,0)$ is a critical point. –  Willie Wong Jan 3 '13 at 17:21
    
Yes but that point is removed by assumption, as stated. –  Isa Jan 3 '13 at 17:39
    
The point being that $\{x² = yz²\}$, even restricted to $\mathbb{R}^3 \setminus \{(0,-1,0)\}$ is not a manifold for the reason you stated. In particular, all the singular points $(0,y,0)$ with $y \neq -1$ are in this set. (Look at what you wrote in the second to last paragraph.) –  Willie Wong Jan 4 '13 at 8:18
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1 Answer 1

Hint: the intersection of two transverse manifolds is a manifold.

So where are the two surfaces $\{x^2 + y^2 + z^2 = 1\}$ and $\{x^2 = yz^2\}$ tangent to each other?

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Thanks! Unfortunately I'm not familiar with the concept of transversality, I'm looking for an alternative proof. However, to answer your question, I'd say the surfaces are tangent to each other when the tangent planes are the same, i.e. when the gradient of both level sets are equal? –  Isa Jan 3 '13 at 16:38
    
@Isa: not equal; parallel. Now the gradients are $(2x, 2y, 2z)$ and $(2x, -z^2, -2yz)$. So you can solve for where the two are parallel using that. But if you don't know about transversality, what do you know? (This is not meant to be snarky, but it may be easier for us to find an answer using techniques you are familiar with if you tell us which techniques you know.) –  Willie Wong Jan 3 '13 at 17:09
    
Maybe instead you want to use some version of the regular value theorem? For that you still need to consider the differential maps. –  Willie Wong Jan 3 '13 at 17:22
    
Well I have a theorem saying that a level set defined by a smooth function f: $${(x,y,z)|f(x,y,z)=constant- and- f- regular- at- (x,y,z)}$$ is a manifold. But here I have 2 level sets to combine (constant= respectively 1 and 0) so I'm a bit unsure of what to do, because applying the theorem to each of them gives me 2 manifolds, not 1. –  Isa Jan 3 '13 at 17:35
    
In your theorem the smooth function $f$ can be manifold valued. In particular, it can take values in $\mathbb{R}^2$. Just take $$ f(x,y,z) = \begin{pmatrix} x^2 + y^2 + z^2 \\ x^2 - yz^2 \end{pmatrix} $$ and consider the "level set" $f^{-1}((1,0))$. Now, $(1,0)$ is not a regular value, so you need to remove from its preimage all points that are critical. –  Willie Wong Jan 4 '13 at 8:21
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