Let $S$ be a submanifold of a real smooth manifold $M$. By a splitting of the normal bundle of $S$ I mean a sub-bundle $V$ of $TM|\_S$ such that $V\oplus TS = TM|\_S$.


*Question:* Given such a splitting, can I always find local coordinates $x\_1,\dots,x\_s,y\_1,\ldots, y\_r$ on $M$ around each point $p\in S$, such that the submanifold $S$ is given by $y\_1=0,\ldots, y\_r=0$ and $V$ is given by the kernels of $dx\_1,\ldots,dx\_s$?



Are all splittings of the normal bundle to a submanifold locally isomorphic?
Yes, one can always do this. Start with a $p$-centered local coordinate system $(x^\sigma,y^\rho)$ on an open $p$-neighborhood $U\subset M$ such that $S\cap U$ is given by $y^\rho=0$ $(1\le\rho\le r)$. Then there will exist functions $F^\sigma\_\rho(x)$ such that $V$ along $S\cap U$ is given by the equations
$$
\mathrm{d}x^\sigma + F^\sigma\_\rho(x)\,\mathrm{d}y^\rho = 0
$$
(summation convention assumed). This is the same as
$$
\mathrm{d}(x^\sigma+F^\sigma\_\rho(x)y^\rho) - y^\rho\,\mathrm{d}(F^\sigma\_\rho(x)) = 0.
$$
Since $y^\rho=0$ along $S$, this says that $V$ along $S$ is defined by
$$
\mathrm{d}(x^\sigma+F^\sigma\_\rho(x)y^\rho) = 0,
$$
so take $\bar x^\sigma = x^\sigma+F^\sigma\_\rho(x)y^\rho$, and let $(\bar x^\sigma,y^\rho)$ be your new coordinate system. (One can easily check that these *are* local coordinates in an open neighborhood of $S\cap U\subset U$.)


**Added Remark:** Just a comment on your terminology: Literally, what you are defining is not a *splitting of the normal bundle of $S$*, but rather a *choice of a normal bundle along $S$*, and you are asking whether there are any local invariants that distinguish such choices. As the above shows, the answer is 'no'. In fact, there is even a global version: For two normal bundles $V\_1$ and $V\_2$ along a closed submanifold $S$, there is a diffeomorphism of $M$ that fixes $S$ and carries $V\_1$ to $V\_2$. 


Note also that, in the holomorphic category (in contrast to the smooth case), normal bundles need not be unique in this sense, and may not even exist.


