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Given a compact smooth manifold $M \subset R^k$ there is a Polynom $f\in R[x_1,..x_n]$ such that the zero set of $f$ is diffeomorphic to $M$. Can the coefficients of $f$ be pertubated slightly to a Polynomial $g \in Q[x_1,..x_n]$ such that the zero set of $g$ is diffeotopic to $M$? Are their conditions on the homology or homotopy on $M$ such that such a pertubation process is possible / not possible? What happens if Q is replaced by an arbitrary number field K?

edited Jul 27 2011 at 17:33

Ryan Budney
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asked Jul 27 2011 at 15:54

Markus Ulke
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	Why is there a polynomial whose zero set is diffeomorphic to $M?$ there is a real algebraic variety homeomorphic to it, but that's not the same as what you say. The variety might only be one connected component of a zero set of a polynomial... – Igor Rivin Jul 27 2011 at 16:28
	Hi, as far as I know thats the tognoli part of the nash tognoli theorem? Please correct me if i am wrong. A proof can be found in Bochnak Coste Roy (or Akbulut and King) (To get a single polynom from a finite set of polynoms you could sum the squares of the polynoms, (they are real)) – Markus Ulke Jul 27 2011 at 16:34
	I think there is even better Nash theorem that would allow isometric, for a given Embedding, but the distortion is then so weird – Markus Ulke Jul 27 2011 at 16:36

Ben McKay · Accepted Answer · 2011-07-27 17:02:44Z UTC

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Yes: proven in Ballico, E., Tognoli, A., Algebraic models deﬁned over $\mathbb{Q}$ of diﬀerential manifolds. Geom. Dedicata 42 (1992), no. 2, 155–161. In fact, you can get the zero set to be diffeomorphic to $M$, not just diffeotopic.

answered Jul 27 2011 at 17:02

Ben McKay
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Don't you even need orientability? – Mariano Suárez-Alvarez Jul 27 2011 at 17:40

No, you don't, apparently. – Ben McKay Jul 27 2011 at 18:24

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this does imply that there are only contable compact smooth diffeomorphism types. Can this be seen in a more elementary style? – Markus Ulke Jul 27 2011 at 19:02

@Markus: There are various proofs, and I believe there's already an MO thread on that topic but a quick search did not turn it up. One would be to argue that every triangulated manifold has only finitely many smoothings. In dimensions $n \geq 5$ I believe this is a standard theorem in Kirby-Siebenmann smoothing theory. I think you can make a much simpler arguments using the technique Whitehead used to prove all smooth manifolds admit triangulations. – Ryan Budney Jul 27 2011 at 20:29

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@Markus: It follows for example from Cheeger's finiteness theorem (there are only finitely many diff-types of manifolds which admit complete metric with |curvature| $\le 1$ and diameter $\le D$). – Anton Petrunin Jul 28 2011 at 8:40

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Manifolds and Polynomials

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