For a given circle we have a corresponding equation to generate it. For a given ellipsoid we also can write a corresponding equation for it. In general, can we write for any given manifold an equation to characterize it? I googled this but I did not get useful information.
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Are you referring to a metric? – Doryan Miller Aug 25 '14 at 02:44
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Sorry? What does it mean by "referring to a metric"? @Doryan Miller – Yes Aug 25 '14 at 02:47
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To be precise, are you asking whether any submanifold of $\mathbb R^n$ is the zero set of some polynomial? Elementary function? Smooth function? – Anthony Carapetis Aug 25 '14 at 02:47
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Ah yes. Is any submanifold of $\mathbb{R}^{n}$ the zero set of some polynomial? – Yes Aug 25 '14 at 02:49
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4@Brian Polynomials are not enough. For example, the sine curve on a plane is not the zero set of any polynomial (reason: it intersects real line infinitely often). Zero set of a smooth function: yes, but this means less than you might imagine. It's a theorem of Whitney that every closed set is the zero set of some smooth function. – Aug 25 '14 at 02:56
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Disregard my comment. The edits clarified what you're asking. – Doryan Miller Aug 25 '14 at 02:59
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@Thursday: Thank you. I am not sure, do you mean that any Euclidean submanifold is the zero set of a smooth function? Is there a theorem for this? – Yes Aug 25 '14 at 02:59
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1You might be interested in reading about algebraic manifolds. – Anthony Carapetis Aug 25 '14 at 03:04
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1@Thursday: Is that really the reason? Isn't the entire real line in the zero set of the polynomial $f(x,y)=xy$? – MPW Aug 25 '14 at 03:04
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2@MPW You can have the entire line, but you can't have an infinite proper subset thereof. The restriction of polynomial to a line is either identically zero, or has finitely many zeros. – Aug 25 '14 at 03:08
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@Thursday: Interesting, I did not know that. +1, then. – MPW Aug 25 '14 at 03:18
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Is the open unit disk a submanifold of $\Bbb R^2$? – Quang Hoang Aug 25 '14 at 04:41
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No, certainly not. Consider the curve $\{ (x,y): y = e^x \}$. This is a submanifold of $\mathbb{R}^2$, but it is not the zero set of a polynomial equation.
mweiss
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Because there is no polynomial function that produces the same graph as an exponential function. – mweiss Aug 25 '14 at 03:24
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4Unconvinced. The curve $y=1/(1+x^2)$ is not the graph of any polynomial, but it's the zero set of $p(x,y)=x^2y+y-1$. – Aug 25 '14 at 03:27
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