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I know that a given set $M$ is a smooth surface of dimension $k$ in $\mathbb{R}^n$ iff there's a map $r:U\rightarrow\mathbb{R}^n, U\subset \mathbb{R}^k$ is open such that $\forall a\in U, \text{rank}(D_r(a))=k$. How can I determine that a given set is a smooth surface if I am given the set as a level set of a function $\Phi:\mathbb{R}^n\rightarrow\mathbb{R}^{n-k}$? Must I find a parametrization of the set?

For example, how can I tell if $M=\{(x,y,z)\in\mathbb{R}^3|z>y\geq0, x^2-y^2=0\}$ is a smooth surface? I was already given an answer to this one, but it remains unclear how to solve for a more general case. Specifically I would like to know how to tell if a given set is a smooth surface and if it's not a surface how can I prove otherwise, by which tools / theorems?

Edited Progress

I have gathered so far that in order to determine that a given set is a $k$-dimensional surface I must either: (a) find a parametrization of it (b) it's a graph of a function (c) it's a level set of a function. Furthermore, there is no algorithm to prove that a set is not a surface, I'd have to evaluate the set by the definition of a surface. Am I correct in this?

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Your condition on $r$ is not sufficient for $M$ to be smooth: even if it is satisfied $M$ could have the form of a digit $8$ in a plane. You must add the condition that $r$ be a homeomorphism onto its image. For the second question the key word is "submersion". Finally, non-smoothness is usually proved by ad hoc methods, like showing that velocity vectors of differentiable curves through a point vanish: try this on $[0,1]\subset \mathbb R$. –  Georges Elencwajg Oct 8 '11 at 17:40
Thanks! That was very helpful! Though, what do you mean by submersion? –  Donjim Oct 8 '11 at 19:50
Dear Donjim, a differentiable map $\Phi:\mathbb R^n\to \mathbb R^{n-k}$ is a submersion at $x\in \mathbb R^n$ if its differential at $x$ is surjective. If this holds at every $m\in M=\Phi^{-1}(0)\subset \mathbb R^n$, the set $M$ is guaranteed to be a submanifold of dimension $k$ of $\mathbb R^n$.This can be generalized to a $\Phi$ defined only on an open subset of $\mathbb R^n$. –  Georges Elencwajg Oct 8 '11 at 21:51
@GeorgesElencwajg - Thanks again! –  Donjim Oct 9 '11 at 6:49
Let me add the disturbing (but true!) result that every closed set in $\mathbb R^n$ is the zero set of an infinitely differentiable function $f\in \mathcal C^\infty(\mathbb R^n)$. So if you want it to be a submanifold, it is clear that you'll have some requirement on the differential of $f$: that it be a submersion is the pertinent requirement. –  Georges Elencwajg Oct 9 '11 at 8:58

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