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Let $f\colon\mathbb{R}^n\to\mathbb{R}$ be a function with $n\geq 2$, and define $S_{f}=\{(x_1,x_2, \ldots x_n) \in\mathbb{R}^n \mid f(x_1,x_2, \ldots x_n)=1\} = f^{-1}(1)$.

Is $S_f$ path-connected when $f$ is $C^{\infty}$? If not, what about if $f \in \mathbb{Q}[x_1,\ldots,x_n]$?

Context: I had to establish that the set of diagonal square matrices with $\det=1$ was path-connected, which I did, but it got me thinking on more general lines.

By the way, please answer (if possible) such that one needs as little prerequisites as possible to understand.

Added: Though Eric has already answered the question. There's a natural next question to ask: For what kinds of functions is this set path-connected?

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I think the level sets of holomorphic functions are connected (Maybe it's plurisubharmonic functions? I can't remember a reference for this off the top of my head), but then you have to replace $\mathbb R$ by $\mathbb C$ (and this is for functions on $\mathbb C^n$ with $n \geq 2$, obviously). – Gunnar Þór Magnússon Dec 27 '10 at 20:52
up vote 4 down vote accepted

Not in general. Let $f:R \to R, f(x) = x^2 - 1$. Then $S = \{\pm \sqrt{2}\}$ which is not path-connected.

EDIT: for the $n=2$ case you can take an equation defining a hyperbola, e.g. $x^2 - y^2$. The set where this function is 1 has two connected components.

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So sorry, should have added $n \geq 2$ – Dactyl Dec 27 '10 at 19:18
Ah ok. I just edited to give a counterexample for $n=2$. – Eric O. Korman Dec 27 '10 at 19:22
Thanks so much. So, I was going about this the wrong way. – Dactyl Dec 27 '10 at 19:30

This properly belongs as a comment to Eric's answer, but I don't have enough reputation to post one. More generally, defining $f:\mathbb{R}^n\to\mathbb{R}$ by $f(x_1,\ldots,x_n) = x_1^2-1$ gives $S$ with two connected components, the affine hyperplanes $\{\pm \sqrt 2\}\times\mathbb{R}^{n-1}$.

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I suppose you mean $\{ \pm \sqrt{2} \} \times \mathbb{R}^{n-1}$ =) – Adrián Barquero Dec 27 '10 at 19:29
Yes, thank you. Of course my gut instinct is to think of zero loci of polynomials rather than... $1$ loci :) – Brad Dec 27 '10 at 19:35

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