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I would appreciate it if someone could explain the difference(s) between a $C^\infty$ and a $C^\omega$ surface embedded in $\mathbb{R}^3$. I ran across these terms in M. Berger's Geometry Revealed book (p.387). The context is: There are examples of two different $C^\infty$ compact surfaces that are isometric, but no known examples for "two real analytic (class $C^\omega$)" surfaces which are isometric. Thanks!

Clarification. Thanks to Mariano and Willie for trying to help---I appreciate that! It is difficult to be clear when you are confused :-). Let me try two more specific questions: (1) Where does the $\omega$ enter into the definition of $C^\omega$? Presumably $\omega$ is the first infinite ordinal. (2) What I'm really after is the geometric "shape differences" between $C^\infty$ and $C^\omega$. The non-analytic but smooth functions I know smoothly join, say, an exponential to a straight line, but geometrically they look just like smooth functions. I guess I don't understand what the constraints of real-analyticity imply geometrically. Maybe that's why this isometric question Berger mentioned is unsolved?!

Addendum. Here are Ryan's two functions:
Two functions
Left: $C^\infty$ but not $C^\omega$. Right: $C^\omega$.

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What kind of difference you have in mind? As you know, "smooth" is much,. much more flexible than "analytic"... – Mariano Suárez-Alvarez Jan 24 '11 at 21:49
I am pretty sure you know the difference between smooth and analytic functions, so do you mind clarifying the question a bit on what you are looking for exactly? Intrinsically a manifold in class $C^k$ is defined by an atlas whose transition functions between charts are $C^k$. Extrinsically, a class $C^k$ surface in $\mathbb{R}^3$ is locally the level-set of a $C^k$ function. See (Somehow I don't think this completely answers your question, but I am not really sure what you are asking for.) – Willie Wong Jan 24 '11 at 21:52
+1 for the images. You can really see the 'rigidity' that timur talks about in the analytic surface! Almost as if it's shrink-wrapped plastic being pushed out by a finger, and it's straining under the high tension. – AndrewG Jan 5 '13 at 20:27
up vote 14 down vote accepted

To Clarification (1): I think $\omega$ is there just to indicate that the smoothness is "more than $C^\infty$", and I don't think it has any substantial connection to the first infinite ordinal.

(2): Analytic surfaces are much more rigid than smooth surfaces. For example, you can choose any open set on a smooth surface and do surgery so that only the chosen set is affected. For analytic surfaces you knock on a point, you disturb the entire surface. Smooth surfaces are made of sheet metal, while analytic surfaces are ceramic. I reserve rubber for topological surfaces.

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Thanks, timur! Your "knocking on a point" intuition is helpful! – Joseph O'Rourke Jan 25 '11 at 12:59
+1 for those last two sentences – Jesse Madnick Sep 13 '12 at 5:07

$C^\omega$ means analytic and $C^\infty$ means infinitely-differentiable. So for example, the union of the two graphs:

$$ z = e^{\frac{1}{x^2+y^2-1}}, \text{ for } x^2+y^2<1$$


$$ z=0, \text{ for } x^2+y^2 \geq 1 $$

is a $C^\infty$-surface in $\mathbb R^3$, but it's not a $C^\omega$-surface.

On the other hand, $z = \frac{1}{1+x^2+y^2}$ is an $C^\omega$-surface in $\mathbb R^3$ and the graph "looks" similar, in that they're both isolated hill-tops.

Were you looking for something more specific than that?

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Thanks, Ryan, that makes it more 'real' (if I may use that word in this context!). – Joseph O'Rourke Jan 25 '11 at 0:23
@Joseph: Perhaps 'tangible' might be better =) – barf Jun 19 '11 at 1:39

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