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Let $M$ be a smooth manifold (over $\mathbb R$) and $N \hookrightarrow M$ a closed embedding. Locally near any point in $N$, I can find coordinates $x^1,\dots,x^{\dim M}$ on $M$ so that $N$ is the vanishing locus of $x^{1+\dim N},\dots,x^{\dim M}$. My question is essentially whether I can do this globally. More precisely, I would like:

Does there exist a smooth vector bundle $V \to M$ with $\operatorname{rank} V = \dim M - \dim N$ and a global section $s\in \Gamma(V)$ so that $N = \lbrace s = 0 \rbrace$ and this critical locus is nondegenerate in the following sense: for $n\in N\hookrightarrow M$, the matrix $\mathrm d s: \mathrm T_n M \to \mathrm T_{(n,0)}V = V_n$ is full rank (i.e. it has kernel only $\mathrm T_nN \hookrightarrow \mathrm T_nM$)?

My intuition is strongly "yes", but I have been unable to come up with a construction of $V,s$. Note that, in $C^\infty$-land, every closed set is the vanishing locus of a single smooth function, but said function tends to vanish to very high (often infinite) order on the closed set. On the other hand, for my question, one cannot hope to win with only trivializable bundles. For example, when $M = S^1$ is a circle and $N$ is an odd number of points, the vector bundle $V$ is necessarily the "other" rank-$1$ vector bundle, i.e. the non-orientable one.

The reason I would like this is because the data $(V,s)$ is essentially a "projective resolution" of the embedding $N \hookrightarrow M$. From this point of view, I can see larger more general constructions that will achieve what I need for my application, but (being larger and more general) those constructions are harder to compute with.

Provided that $(V,s)$ does exist, my follow-up question is how unique it is (at best, unique up to unique isomorphism, but this cannot be right if $M$ has interesting topology far away from the image of $N$, so instead you expect that the choices of $(V,s)$ should be parametrized by some measure of the relative topology)?

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up vote 12 down vote accepted

The answer in general is no. Take e.g. $M=\mathbb{R}^n$. There are no non-trivial vector bundles on it. So, if $N$ is the zero locus of a transversal section of a bundle $E$ (which is necessarily trivial), then using the exact sequence $$0\to T N\to T M|_N\to \nu_M N\to 0,$$ where $\nu_M N$ is the normal bundle of $N$ in $M$, and the fact that $\nu_M N\cong E|_N$ we conclude that the tangent Stiefel-Whitney classes of $N$ are all zero.

[upd: the same strategy works for $M=S^n$; the tangent Stiefel-Whitney class of $S^n$ is zero; the top Stiefel-Whitney class of $E$ need not be zero, but it gives zero when restricted to $N$.]

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Very nice. Why the asterisks on the $T$'s denoting tangent bundles? –  Georges Elencwajg Jun 2 '11 at 16:54
    
George -- you are right, thanks. Will correct this. –  algori Jun 2 '11 at 16:58
4  
I guess I just like asterisks. –  algori Jun 2 '11 at 17:01
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