Is it always true that a tubular neighborhood of a knot $K \subset M$, where $M$ is a generic smooth 3-manifold, is diffeomorphic to $S^1 \times B^2$ (if you prefer diffeomorphic to $S^1 \times \mathbb{R}^2$, I just find $B^2$ more intuitive), where $B^2$ denotes the interior of $D^2$? If yes, why? I know that all tubular neighborhoods are unique up to isotopy, but I cannot figure out if this helps me.
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Yes, if orientable. https://en.wikipedia.org/wiki/Tubular_neighborhood – Moishe Kohan May 19 '19 at 18:23
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Thanks for your comment! In the while I came up with a sketch of proof: since the existence of the tubular neighborhood is proved with the normal bundle and all tubular neighborhoods are diffeomorphic (even though I am not sure whether this is true) the problem can be rephrased as "proving that the normal bundle of $S^1$ embedded into an orientable manifold is trivial", which is true because every oriented bundle over $S^1$ is trivial (see this question). Is this correct? – Nicolò Cavalleri May 19 '19 at 19:25
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Yes, this is correct. – Moishe Kohan May 19 '19 at 19:26