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The Question is: Let $W_c = \{ ( x,y,z,w) \in R^4 | xyz = c \}$ and $Y_c = \{ ( x,y,z,w) \in R^4 | xzw = c \}$. For what real numbers $c$ is $Y_c$ a three-manifold? For what pairs $(c1,c2)$ is $W_{c_1} \cap Y_{c_2}$ a two-manifold?

I think that we can say for the first half that $Y_c$ is a three-manifold when $c \neq 0$. This isn't exactly fleshed out, but I can imagine what $xyz = c$ looks like for $c \neq 0$ with none of the variables being $0$, and locally this basically is a piece of $\mathbb{R}^3$. But my intuition is really lacking on the intersection of manifolds, and I want to say something like when both $c_1$ and $c_1$ are non-zero, as their $x$ and $z$ variables are the ones we need to intersect to form a manifold. Is this the right way of thinking about this problem, and what gaps could I fill in to explain this more concisely? Thanks for your help!

Ness
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You need to make sure that $(c_1,c_2)$ is a regular value of $f\colon\mathbb R^4\to\mathbb R^2$ given by $f(x,y,z,w)=(xyz,xzw)$. That is, you need to check that $\text{rank}(Df)=2$ at every point $(x,y,z,w)$ with $f(x,y,z,w)=(c_1,c_2)$. This is equivalent to checking that the gradient vectors of the components are everywhere non-parallel on the level set.

Ted Shifrin
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  • I think I understand the idea for the second part, and now what you have pointed it out, it definitely seems the right way to approach the problem, so thank you! I am curious, is the same idea the right idea for the first half as well? As we're looking for the regular value of $f:\mathbb{R}^4 \rightarrow \mathbb{R}$ with $f(x,y,z,w) = (xzw)$ but checking that the partial derivatives of f evaluated at the point $p$ that allows $f(p) = c$ are zero (and thus a regular value)? Thanks! – Ness Jun 25 '13 at 06:46
  • You need to make sure that the points where all the partials vanish are not on your level set. – Ted Shifrin Jun 25 '13 at 12:20
  • Ah, yes, I misread the definition when I was writing that. So the Jacobian turns out to be $\bigl( \begin{smallmatrix} yz & xz & xy & 0\ zq & 0 & xw & xz \end{smallmatrix} \bigr)$

    So we can see that the 2x2 minor of $\bigl( \begin{smallmatrix} xz & 0\ 0 & xz \end{smallmatrix} \bigr)$

    Gives us us that this is rank 2 when $c_1, c_2$ are not both $0$. Thank you for your help!

     – Ness Jun 25 '13 at 20:48