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Let $f: M \rightarrow N$ be a smooth map between two submanifolds of $\mathbb{R}^{m}$, $\mathbb{R}^{n}$ respectively. Sard's famous theorem asserts that the set of critical values $C$ of $f$ has measure zero.

My question is: Does every null set in $\mathbb{R}^n$ arise as the set of critical values of some smooth map $f$ as above?

As a start: For $n=1$ and $C \subset \mathbb{R}$ countable, I think one can construct such a map. Namely, let $M=\coprod_{c \in C} \mathbb{R}$ be the disjoint union of $|C|$ copies of $\mathbb{R}$ and $f: M \rightarrow \mathbb{R}$ be defined by $(x,c) \mapsto x^2+c$. $M$ is a one-dimensional real manifold (it is $2$nd countable, Hausdorff and carries a natural smooth structure coming from the one on $\mathbb{R}$). Hence, by Whitney, it can be embedded as a submanifold of some $\mathbb{R}^m$. Moreover, the set of critical values of $f$ is exactly $C$.

But this approach does not seem to work in general. If $C \subset \mathbb{R}$ is an uncountable null set, for example, then $M=\coprod_{c \in C} \mathbb{R}$ is not $2$nd countable, hence is not a smooth manifold in the usual sense and Whitney's theorem does not apply.

Any help towards an answer to my question is much appreciated! In particular, references are also welcome.

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    $\begingroup$ If the (uncountable) Cantor set $C$ were the critical locus of a smooth function $f:\mathbb R\to \mathbb R$, then we could construct another smooth function $\tilde f:\mathbb R^2\to\mathbb R$ defined by $\tilde f(x,y)=f(x)+f(y)$, and its critical locus, a null set by Sard's Theorem, would necessarily contain as a subset the sum $C+C=[0,2]$, contradiction. $\endgroup$ – Brenin Oct 9 '12 at 22:45
  • $\begingroup$ @atricolf: nice! $\endgroup$ – Nils Matthes Oct 10 '12 at 17:54
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The answer is no.

Sard gave a refined version of his theorem in 1965. It states if

$$f : U \to \mathbb R^m$$

with $U \subset \mathbb R^n$ and $f$ is $C^k$ for $k \geq \max(n-m+1, 1)$, and if we let

$$ A_r = \{ p \in U : Df_p \text{ has rank } \leq r \}$$

Then $f(A_r)$ has Hausdorff dimension $\leq r$.

So the the case you're interested in, the critical values not only has measure zero, but the Hausdorff dimension is at most $\min(m-1,n-1)$.

You can of course realize this number fairly easily.

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    $\begingroup$ Thank you! So for example the "standard" (ternary) Cantor set cannot be the set of critical values of a smooth function $f: U \rightarrow \mathbb{R}$ (as its Hausdorff dimension is bigger than zero), right? $\endgroup$ – Nils Matthes Oct 10 '12 at 17:54
  • $\begingroup$ Right you are. .. $\endgroup$ – Ryan Budney Oct 10 '12 at 20:38

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