If I have a 4-manifold M and a smooth closed curve C $\subset$ M, it appears that any sufficciently small, closed tubular neighbourhood of C in M is homeomorphic to $S^1 \times E^3$ if M is orientable.
How would i go about proving this, and in what way does it matter, that M has to be orientable?
The tubular neighborhood will be an orientable vector bundle of rank $3$ over $S^1$. Vector bundles of rank $3$ over $S^1$ are classified by elements of $\pi_0 O(3)$, which has two elements, an orientable one (the trivial bundle) and a non-orientable one. Since your bundle is orientable, it has to be the trivial bundle.