In Section $9$ of Tu's Introduction to Manifolds, we're asked to find all values $c\in\Bbb R$ for which the level set $f^{-1}(c)$ is a regular submanifold when $$f(x,y)=x^3-6xy+y^2.$$By taking each of the partials and setting them to zero, I find that the only critical points are $(0,0)$ and $(6,18)$. But now I'm kind of at a loss. I've been trying to follow an example, and it isn't working out for me... Are there any suggestions for what I should do? My full attempt has been to define $g(x,y)=f(x,y)-c$ (as the example) so that $f^{-1}(c)=g^{-1}(0)$. Now the level set is a regular submanifold if the critical points are not part of the level set, but I'm not really sure how to do this or if this is correct.

I've personally found this to be the least intuitive math course, so as an aside, if anyone wants to suggest a book or notes that might help develop some intuition, I'd appreciate it.

  • $\begingroup$ +1 Nice question. I remember I saw such this problem in this book before "Differentiable Manifolds an Introduction" by Clark. $\endgroup$
    – Mikasa
    Mar 6, 2013 at 5:34
  • $\begingroup$ @JesseMadnick: Are you using the first edition (mine is second)? Otherwise, our page numbers don't match. As for the second comment, I'd have to admit I don't completely understand this material. I've been proving things mostly via direct definition, and the problems have been obvious enough I've not had to worry about much else. $\endgroup$
    – Clayton
    Mar 6, 2013 at 5:46
  • $\begingroup$ Ugh, I must be, yes. I've deleted my comments. $\endgroup$ Mar 6, 2013 at 5:47
  • 1
    $\begingroup$ Regardless of page numbers, what is true is this: you should try to understand the distinctions between "critical point," "critical value," "regular point" and "regular value." The intuition for all these concepts comes from multivariable calculus. And I'm sure there's a theorem in Tu's book (of any edition) that handles most of the problem: namely, if $c$ is a regular value of $f$, then $f^{-1}(c)$ is a regular submanifold. $\endgroup$ Mar 6, 2013 at 5:49

1 Answer 1


Hint: The inverse image of a regular value of $f$ is a regular submanifold. Regular values are just elements of the codomain that aren't critical values, and the set of critical values of $f$ is just the image of the set of critical points of $f$.

For a visual illustration that $0\in\mathbb{R}$, which is the image of the critical point $(0,0)\in\mathbb{R}^2$, is not a regular value, take a look at its inverse image: enter image description here

That self-intersection is the trouble.

For the critical value $-108\in\mathbb{R}$, there is a different "problem", namely that the critical point $(6,18)$ is a local minimum: here is an animation of the preimages of some $-108+\epsilon$:

enter image description here

By the way, I'm a big fan of Lee's Introduction to Smooth Manifolds. There's a second edition that just come out.

Code for first image:

Plot3D[x^3 - 6 x*y + y^2, {x, -40, 40}, {y, -40, 40}, 
 MeshFunctions -> {#3 &}, Mesh -> {{0}}, PlotPoints -> 5, 
 MaxRecursion -> 5]

Code for second image:

 Table[Plot3D[x^3 - 6 x*y + y^2, {x, 5.5, 6.5}, {y, 17.5, 18.5}, 
   MeshFunctions -> {#3 &}, Mesh -> {{z}}, PlotPoints -> 5, 
   MaxRecursion -> 5], {z, -107.95, -108.01, -0.001}], 
 "DisplayDurations" -> {0.075}]
  • $\begingroup$ Zev, I just edited the question, am I thinking at all along the right lines? Either way, I'm not sure what you mean by your last statement about the critical values of $f$ being the image of the set of critical points of $f$. $\endgroup$
    – Clayton
    Mar 6, 2013 at 5:40
  • $\begingroup$ Okay, visually, I see it and I can easily make an argument (thanks for the illustration, by the way). Without the visual aid, I was confident there was a problem at $0$ because it was the image of a critical point. Should I practice trying to visualize things like this, or should I try to stick to theorems (I feel like that latter offers the least amount intuition-devolopment)? $\endgroup$
    – Clayton
    Mar 6, 2013 at 5:58
  • $\begingroup$ Also by a similar reasoning, it seems like $f(6,18)=-108$ implies $-108$ will not be a regular value? $\endgroup$
    – Clayton
    Mar 6, 2013 at 6:00
  • $\begingroup$ The point $(6,18)$ is a singularity and an isolated point of the level set $f(x,y)=108$ . This is a little mysterious since an isolated point is a manifold, but that manifold is of the wrong dimension: zero instead of one. To really dissipate the mystery one should also look at the level set in $\mathbb C^2$ but that requires more sophisticated methods. Beware that using WolframAlpha the isolated point $(6,18)$ is not shown at all on the level set $f(x,y)=108$, which treacherously looks perfectly smooth everywhere. $\endgroup$ Mar 6, 2013 at 8:44

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