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Hallo,

Let $(M,g)$ be a Riemannian $k$-dim real analytic submanifold of $\mathbb{R}^{n}$. Is it true that $M$ in $\mathbb{R}^{n}$ looks locally (in a small neigbourhood around some point in $M$) as the zero set of some polynomials? If yes, why ? Are there any references?

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In the sense that analytic functions look locally like polynomials, it's true, in the sense that analytic functions have Taylor approximations. Is that what you're interested in? Your question seems rather vague. –  Ryan Budney Dec 6 '12 at 8:13
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You have to specify exactly what do you mean by "looks locally". –  Alexandre Eremenko Dec 6 '12 at 13:05
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2 Answers

up vote 5 down vote accepted

I agree with the comments it would be very nice if you make your question more precise, what do you mean by "looks locally as"?

By the way, this question reminds me two beautiful results of S. Akbulut and H. King which they proved in "On approximating submanifolds by algebraic sets and a solution to the Nash conjecture." Invent. Math. 107 (1992), no. 1, 87–98.

I cite the MathSciNet review:

"Theorem I: Let $M\subset{\bf R}^n$ be a compact smooth submanifold; then $M$ is $\epsilon$-isotopic, for any $\epsilon>0$, to a nonsingular real algebraic set of ${\bf R}^{n+1}$.

Theorem II: If $M\subset{\bf R}^n$ is a compact smooth submanifold, then $M$ is $\epsilon$-isotopic, for any $\epsilon>0$, to the set of nonsingular points of a real algebraic subset of ${\bf R}^n$."

These are global results, they tell you that compact smooth submanifolds are very close (in the isotopic sense as close as you want) to real algebraic subsets.

Edit: I am not an expert in real algebraic geometry and I don't know if $M$ is $\epsilon$-isotopic, for any $\epsilon>0$, to a nonsingular real algebraic subset of ${\bf R}^n$ (this is a difficult problem).
What I know is another result of S. Akbulut and H. King in "Transcendental submanifolds of ${\bf R}^n$." Comment. Math. Helv. 68 (1993), no. 2, 308–318. In this paper they prove that there exist manifolds $M\subset\mathbb{R}^n$ which can not be isotoped to the real parts of nonsingular complex algebraic subvarieties of $\mathbb{C}P^n$.

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It would be interesting to know an example where (in Theorem 1) $n$ is not enough and one actually needs the nonsingular algebraic subset to wind around $M$ in the $(n+1)$-th dimension. –  Qfwfq Dec 6 '12 at 22:02
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No, this is not true. A counterexample is the graph of the exponential function: $M=\{(x,e^x)/X\in\mathbb R\}.$ If there would be a not identically zero polynomial $P=P(x,y)=\sum_{k,l}a_{k,l}x^k y^l$ such that $M$ is locally the set of zeros of $P$, this would even be true globally,as we are dealing with analytic functions. This cannot be true as not all coefficients of the (converging) series (in $x$) of $P(x,e^x)$ vanish.

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I guess that “looks localy as...” must be understood as “locally isomorphic as Riemannian manifold to...” –  Lierre Dec 6 '12 at 7:48
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Lierre: but then it's trivially true as manifolds (with any restriction on them) look locally like the zero set of a linear function. –  Ryan Budney Dec 6 '12 at 8:11
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@Ryan: Not isometrically, though. But anyway, the author should make the question more precise. –  Alexander Shamov Dec 6 '12 at 8:34
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