I know that there is a theorem that a non-empty level set of a regular value of a smooth function $f:M\rightarrow\mathbb{R}$ on a smooth manifold is a regular submanifold (or embedded submanifold) of codimension 1.

Now I wonder if there is also a condition in which the converse holds true. I mean suppose $c\in\mathbb{R}$ is a critical value of $f$ is there a condition under which you know the level set of $c$ is not a regular submanifold.

If $g:M\rightarrow\mathbb{R}$ is defined by mapping all of $M$ to $0$ then it seems to me that $0$ is a critical value of $g$. Now $g^{-1}(0)=M$ so definitely a regular submanifold of $M$. So I know that it is at least not true without an extra condition.

Also if the function is second degree polynomial from $\mathbb{R}$ to $\mathbb{R}$ then the level set of a critical value is just a point which would again be a regular submanifold.

If there is no such general condition on the function $f$ or the critical value then how in general does one go about showing that a critical level set is or is not a regular submanifold?

Thanks in advance

  • $\begingroup$ By the way, I would love a reference for the statement in the first theorem, or even better, a textbook which I can pass to a student of mine who needs to know these things. $\endgroup$ Oct 13, 2012 at 23:59
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    $\begingroup$ You probably should say amend the statement of the theorem to say that a non-empty level set of a regular value of a smooth function f:M→ℝ on a smooth manifold is a regular submanifold of codimension one. (That takes care of the problem with f identically zero.) $\endgroup$ Oct 14, 2012 at 0:16
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    $\begingroup$ The most immediate converse is that a manifold with a trivial normal bundle is the level-set corresponding to a regular value. If you have a non-trivial normal bundle, you will need to allow for more degenerate functions. Bott-type Morse functions are a standard generalization. $\endgroup$ Oct 14, 2012 at 1:28
  • $\begingroup$ @Andrej If you or your student Google submersion theorem or constant rank theorem you or they should find it. Or look in any basic differential geometry text under those names. $\endgroup$ Oct 14, 2012 at 12:39
  • $\begingroup$ @Andrej I found this theorem in "An Introduction to Manifolds" by Loring W. Tu. In the capter on submanifolds. The constant rank is indeed a generalization of this theorem. $\endgroup$ Oct 14, 2012 at 14:23

1 Answer 1


Suppose that $0$ is a regular value of $f$. Then $0$ is a critical value of $g=f^2$, yet the level set $g^{-1}(0)$ is a regular submanifold. In this case all the points on $g^{-1}(0)$ are critical points of $g$.

The general answer is difficult. You need to assume something about $f$. A natural assumption would be that the critical points of $f$ are isolated and of finite type. Near such points one can find local coordinates so that in these coordinate $f$ looks like a polynomial. (This is a generalization of the classical Morse lemma due to Tougeron.) In such cases you need to understand the zero sets of real polynomials which can be challenging. A nice place to consult for such issues is the book by Arnold, Gussein-Zade and Varchenko on singularities of differentiable mappings.

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    $\begingroup$ A short version of this answer could be: if at a critical point the second derivative is non-degenerate but non-definite, then you know by Morse lemma that the level set is not a regular submanifold. This gives you a simple condition (that captures only a few cases, but is generic). $\endgroup$ Oct 14, 2012 at 7:52
  • $\begingroup$ Thank you, I didn't think to check the book by Arnold, Gussein-Zade and Varchenko even though I had it lying around the house! $\endgroup$ Oct 14, 2012 at 14:22

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