# Can any smooth manifold be realized as the zero set of some polynomials?

Is any real smooth manifold diffeomorphic to a real affine algebraic variety? (I.e. is there an "algebraic" Whitney embedding theorem?)

And are all possible ways of realizing a manifold $M$ as an algebraic variety equivalent? I.e. suppose $M$ is diffeomorphic to varieties $V_1$ and $V_2$, are these isomorphic in the algebraic category?

Admittely I'm just asking out of curiosity after reading this question: Can manifolds be uniformly approximated by varieties?

A quick Google search found this paper, where it is stated that the answer to the first question is yes in the compact case (due to Tognoli): this result is called the Nash-Tognoli theorem. In general, the answer is no: a real affine variety has finite-rank homology groups, and it's easy to construct non-compact manifolds for which this is false (e.g. a surface of infinite genus). In fact, apparently there is a bound due to Milnor for the sum of the Betti numbers of a real variety.

The answer to the second question is certainly not: just take two elliptic curves with slightly different $j$-invariants.

• Sorry, what's the paper you link to in the first paragraph? I can't access it... Somehow this made me think of Claire Voisin's recent results: e.g. there are compact Kähler manifolds that don't have the homotopy type of a complex projective manifold (see Q.3.15 and Thm 3.16 on p. 17). This is not strictly related, but still... probably worth mentioning.
– t.b.
May 13, 2011 at 15:48
• Thanks! Your first link leads me to a "file not available", but after typing in my question into google (what I should have done in the first place) I get springerlink.com/content/j53406713j831452 which cites Tognoli for the statement about closed manifolds. May 13, 2011 at 15:48
• The original link wasn't freely available, so I replaced it with a different paper. May 13, 2011 at 15:53

For the sake of completeness, here is the answer covering noncompact manifolds as well:

Definition. 1. A smooth manifold $$M$$ is tame if $$M$$ is diffeomorphic to the interior of a smooth compact manifold $$N$$ with (possibly empty) boundary.

1. A smooth manifold $$M$$ is algebraic if it is diffeomorphic to a nonsingular real-algebraic subset of $${\mathbb R}^n$$ for some $$n$$.

Theorem. A smooth manifold is tame if and only if it is algebraic.

See Corollary 4.3 in

Akbulut, Selman; King, Henry C., The topology of real algebraic sets with isolated singularities, Ann. Math. (2) 113, 425-446 (1981). ZBL0494.57004.