Abstract

For knots in $S^3$, it is well-known that the Alexander polynomial of a ribbon knot factorizes as $f(t)f(t^{-1})$ for some polynomial $f(t)$. By contrast, the Alexander polynomial of a ribbon 2-knot in $S^4$ is not even symmetric in general. Via an alternative notion of ribbon 2-knots, we give a topological condition on a 2-knot that implies the factorization of the Alexander polynomial.

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