Regular neighbourhoods of non-orintable surfaces in $S^4$

Question

Suppose that $F \subset S^4$ is a non-orientable surface. Let $N \subset S^4$ be a regular neighbourhood of $F$. Clearly, the boundary of $N$ is a circle bundle over $F$. Which is its Euler number? My guess is that it is equal to zero, I wish to know a proof...

Answer 1

This is an answer which I am not comfortable assessing the correctness of as I cannot find all the relevant sections of Whitney's paper and I don't understand Massey's.

This apparently a somewhat well-known question of Hassler Whitney. It seems the Euler number can be nonzero. I think the Euler number (with twisted coefficients) here https://projecteuclid.org/download/pdf_1/euclid.pjm/1102978058 agrees with the Seifert invariant definition. The proof critically uses the famous Atiyah-Singer index theorem.

This apparently unfinished preprint claims to prove that the Euler numbers from Massey's paper are precisely the ones realized by any embeddings of such Seifert fibred spaces into $S^4$ (without the condition that the embedding comes from the regular neighborhood of a surface). Both sources claim that the Euler numbers realized are $$\{2\chi-4,2\chi,2\chi+4,\dots,4-2\chi\}$$ where $\chi$ is the Euler characteristic of the surface.