Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This is (a part) of exercise $7c$ page $66$ of DoCarmo's book (Differential geometry of curves and surfaces).

Let $f(x,y,z)=xyz^{2}$. I'm trying to figure out if the preimage of $f$ under $0$ is a regular surface.

Basically the preimage is the union of the three coordinate planes in the $xyz$ plane but how can we prove that there exists a point in where regularity fails?

I don't see this very clear. Can you please help?

share|improve this question
    
+1 for the description of the preimage –  Yuri Vyatkin Oct 22 '11 at 8:47
    
the preimage is a regular surface iff the point is regular (by the implicit function theorem). –  Alexei Averchenko Oct 25 '11 at 5:33
add comment

1 Answer

This exercise asks you for map $$f:\mathbb{R}^3\rightarrow \mathbb{R}:\,f(x,y,z)=xyz^{2}$$ (a) find the critical points and the critical values,

(b) and tell for what values $c$ the set $f(x,y,z)=c$ is a regular surface.

Definition 2 on p. 58 of the book tells you that in order to enumerate all the critical points you need to examine the differential map $df$ which can be represented by a matrix as $$ (f_x,f_y,f_x)=(y z^2,x z^2, 2 x y z)$$ Now you can see that $df:\mathbb{R}^3\rightarrow \mathbb{R}$ is not surjective iff $(y z^2,x z^2, 2 x y z)=(0,0,0)$, so we can write down the set of critical points as $$ \mathcal{K}=\{(x,y,x) \in \mathbb{R}^3|z=0 \text{ OR } (x=o \text{ AND } y=0)\} $$

You can see immediately that the only critical value is $0=f(\mathcal{K})$.

In other words, the regularity fails in the standard way: "the derivative is zero."

This solves part (a).


As for part (b), you now need to argue that the set that you described in your question is not the graph of any function as Proposition 3 on p. 63 of the book suggests. For that it is enough to examine three cases, as shown in this Proposition: $z=f(x,y), y=g(x,z), x=h(y,z)$. It is easy to see that your set cannot be of any of these forms.

A confusion may occur due to an identification of the tangent spaces of $\mathbb{R}^n$ with $\mathbb{R}^n$ (I hope you can see where this happens in the above).

Proposition 2, p. 59, ibid., tells you that all other points will be regular.

Hopefully, this helps.

(Edited to add the correct demonstration of part (b)).

share|improve this answer
    
thanks, sorry but I don't get the line "the regularity fails in the standard way: the derivative is zero" is this a theorem? Can you please explain more? –  user17182 Oct 23 '11 at 22:22
    
By definition, regular points are those that are not critical, whence by critical points we understand points at which the differential is not invertible. The last condition is a generalization of "the derivative is not zero" in one-dimensional case. I meant that. –  Yuri Vyatkin Oct 24 '11 at 19:01
    
to show $f^{-1}(0)$ is not a regular surface we need to show that the definition of regular surface fails, which is this definition: mathworld.wolfram.com/RegularSurface.html , why what you wrote implies that for every open set $W \subset \mathbb{R}^{3}$ containing the preimage of $0$ we cannot find an open set of the plane $U$ and a diffeomorphsim $U \rightarrow W \cap S$? I don't see it. In this case $S=f^{-1}(0)$ –  user17182 Oct 24 '11 at 22:14
    
Ah, thanks, I see your point. What you a referring to is basically the same as Definition 1, p. 52, of the book. I think it is sufficient to use Proposition 3 in this case. Otherwise we would have to cite the entire chapter. I have edited my answer. –  Yuri Vyatkin Oct 25 '11 at 5:17
    
@Yuri, not surjective, it may very well be not invertible (e.g. with Morse functions). –  Alexei Averchenko Oct 25 '11 at 5:42
show 4 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.