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Let $X$ be a closed smooth oriented 4-manifold with $H_1(X;\Bbb Z)=0$. Then for a smoothly embedded orientable surface $F\subset X$ which is characteristic, there is a well-defined invariant $\text{Arf}(F)\in \Bbb Z/2$ which only depends on the homology class of $F$. (Its definition is quite long to write it here, so I'll leave a reference: https://www.maths.ed.ac.uk/~v1ranick/papers/matumoto5.pdf, pp.120-121.)

Suppose $X'$ is a manifold homeomorphic to $X$, and the homology class $[F]$ corresponds to $[F']$. Then is it true that $\text{Arf}_X(F)=\text{Arf}_{X'}(F')$?

The definition of $\text{Arf}(F)$ uses the smooth structure of $X$ (mildly), so this does not seem to be obvious if $X'$ is not diffeomorphic to $X$

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  • $\begingroup$ Can the definition not be made in the topological category? How is the smooth structure used? $\endgroup$
    – Anthony Conway
    Commented Feb 19, 2023 at 23:00
  • $\begingroup$ @AnthonyConway $F$ should be smoothly embedded in $X$. If $X\to X'$ is merely a homeomorphism, then the corresponding embedding $F\to X'$ need not be smooth $\endgroup$
    – blancket
    Commented Feb 20, 2023 at 0:07
  • $\begingroup$ I meant: can you not define this Arf invariant for a locally flatly embedded surface? $\endgroup$
    – Anthony Conway
    Commented Feb 20, 2023 at 2:51
  • $\begingroup$ @AnthonyConway I think I can. Thanks, I'll try to figure $\endgroup$
    – blancket
    Commented Feb 20, 2023 at 4:03

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