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In the case of affine varieties we have finitely many (polynomial) equations defining the variety. It is a (smooth) manifold iff it satisfies the Jacobian criterion.

I wonder whether this can be generalized in the following sense:

Is it possible for every smooth submanifold $M\subseteq \mathbb{R}^n$ to find a smooth map $f: \mathbb{R}^k \rightarrow \mathbb{R}^l$ and a regular value $q\in \mathbb{R}^l$ of $f$ such that $M=f^{-1}(q)$?

Severin Schraven
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    I think, but am not sure, that this condition implies that $M$ is stably parallelizable, and most manifolds don't have this property. The simplest example is $\mathbb{RP}^2$, which can be embedded into $\mathbb{R}^4$. – Qiaochu Yuan Jun 02 '16 at 18:58
  • Difficult theorem in real algebraic geometry, we can use polynomials. The hard part is getting your manifold to be the entire variety rather than one connected component. Let me see if I can find a reference, there may be conditions I do not remember, this was a quick conversation with Blaine Lawson some 40 years ago. Oh, regardless, this means diffeomorphic to, not, say, isometric. – Will Jagy Jun 02 '16 at 19:14
  • well, here is a start, https://en.wikipedia.org/wiki/Real_algebraic_geometry .....1973 Tognoli proved that every closed smooth manifold is diffeomorphic to a nonsingular real algebraic set – Will Jagy Jun 02 '16 at 19:17
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    My guess is that a noncompact real algebraic variety has a finite number of topological ends, which rules out some examples, for example the famous Staircase minimal surface of Riemann. https://en.wikipedia.org/wiki/Riemann's_minimal_surface where they use the word "shelf" – Will Jagy Jun 02 '16 at 19:23
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    Compare http://math.stackexchange.com/questions/23764/is-every-embedded-submanifold-globally-a-level-set – Moishe Kohan Jun 03 '16 at 00:26

1 Answers1

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First of all, your condition for smoothness for affine varieties is not quite right: It is sufficient but not necessary. You can say that a real affine variety $X\subset R^n$ is smooth iff for each point $x\in X$ there exists a Zariski open subset $X'\subset X$ containing $x$ such that $X'$ can be defined by a system of polynomial equations satisfying the Jacobian criterion.

Now, suppose that $X\subset R^n$ is a submanifold which is the zero locus of a smooth mapping $f: R^n\to R^k$ such that $0$ is a regular value of $f$. Consider the normal bundle $N(X)$ in $R^n$. You can think of fibers $N_x$ of $N$ as orthogonal complements to $T_xX$. Then for each $x\in X$ the differential $df_x: N_x\to R^k$ is an isomorphism. This defines a trivialization of $N(X)$: $$ (\pi, df): N(X)\to X\times R^k, $$ where $\pi: N(X)\to X$ is the normal bundle. Therefore, the direct sum $T(X)\oplus N(X)$ is the trivial bundle over $X$ (the pull-back of the tangent bundle of $R^n$), which means that $T(X)$ is "stably trivial" and $X$ is "stably parallelizable". However, not every smooth manifold has stably trivial tangent bundle. For instance any nonorientable manifold $X$ gives an example of a manifold which cannot be realized as the zero locus of a smooth map to $R^k$ which has zero as the regular value.

Moishe Kohan
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