# Representation of an oriented manifold as a set of common zeroes of smooth functions

Let $M$ be an arbitrary oriented smooth manifold of dimension $m$. Is it always diffeomorphic to a sumbanifold in ${\mathbb R}^n$ (with some $n$) defined as a set $X$ of common zeroes of $n-m$ smooth functions $f_1,...,f_{n-m}$ (defined on an open set $U\subseteq {\mathbb R}^n$ and having linearly independent differentials in each point $x\in X$: $df_1(x)\wedge...\wedge df_{n-m}(x ) \ne 0$)?

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If you had such functions then the tangent bundle of the manifold would be stably trivial, which is not always the case. The lowest dimension where there is a countexample is 4. Thus, $\mathbb{CP}^2$ can indeed be embedded in Euclidean space, as can any compact manifold, but its tangent bundle is not stably trivial, in the sense that one cannot add a trivial bundle to the tangent bundle to get another trivial bundle. This follows from characteristic class theory, see e.g. the classic text Milnor, Stasheff, Characteristic classes.
I am not sure what level text you are studying now, so I would add that the gradient vector fields of your functions $f_i$ would give a trivialisation of the normal bundle of the submanifold, and the ambient space $\mathbb{R}^n$ gives a trivialisation of the sum of the tangent bundle and the normal bundle.
Is ${\mathbb C}{\mathbb P}^2$ orientable? – Sergei Akbarov Jun 21 '13 at 8:43
Yes. Moreover it is simply connected ($\pi_1(\mathbb{CP}^2)=0$). – Mikhail Katz Jun 21 '13 at 8:44
Dear user72694, I think the lowest dimension for a counterexample is $2$, not $4$. Indeed the Stiefel-Whitney class ot the tangent bundle to $\mathbb R \mathbb P^2$ is $1+x+x^2$, so that this tangent bundle cannot be stably trivial. – Georges Elencwajg Jun 21 '13 at 9:12
Those manifolds, which can be represented as zeroes of $f_i$ like I described, do they have special name? – Sergei Akbarov Jun 21 '13 at 9:16