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3
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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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5
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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). |
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4
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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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