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Suppose I have a differentiable function $f: {\rm I\!R}^m \rightarrow {\rm I\!R}^n$. I obtain a level surface of $f$ by taking the set of points where $f$ is some constant vector: $S = \{\mathbf{x} | f(\mathbf{x}) = \mathbf{c}\}$.

Since $f$ is differentiable, is it the case that $S$ is itself "smooth" in a sense? Specifically, does there exist a differentiable $g: {\rm I\!R}^k \rightarrow {\rm I\!R}^m$ so that the image of $g$ is $S$? If so, is there a standard reference for this factoid? If not, what is a counterexample? Also if not, is there a condition on $f$ that will produce a "smooth" $S$ (like being infinitely differentiable, or some such)?

Him
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This is not true in general. Take $f(x,y,z)=x^2+y^2-z^2$. It’s zeroth level is a cone which is not smooth at its vertex. You have to assume that at the given point the gradient does not vanish to guarantee that the level surface is smooth (use the implicit function theorem)

GReyes
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  • "You have to assume that at the given point the gradient does not vanish" can you elaborate on this point a bit? – Him Jan 02 '20 at 17:12
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    Yes. If you know that $\nabla f(P)$ is not zero at some point $P$ on your surface level then by the implicit function theorem you can express one of the variables as a function of the others having the same smoothness as $f$. In the example you see that $\nabla f(O)=0$ where O is the origin but it works for any other point of the surface $f=0$ – GReyes Jan 02 '20 at 17:30
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    This is for a function taking real values $n=1$. If $n$ Is larger than 1 and less than $m$ you need some $n\times n$ minor of the Jacobian matrix different from zero and then you get an $m-n$ dimensional smooth manifold. – GReyes Jan 02 '20 at 17:49
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This is not true. Consider the case where $m=n=1$, and choose $f(x)=\sin(x)$ and $c=0$. Then $S$ is disconnected, and the image of a continuous function with domain $\mathbb{R}^k$ is connected.

user722227
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