There exist many constructions of infinite families of knots with the same Alexander polynomial. However, alternating knots seem very special. While there are also many result on restricting the form the Alexander polynomial of an alternating knot can have, is it known whether infinitely many alternating knots can have the same Alexander polynomial?



Can infinitely many alternating knots have the same Alexander polynomial?
No, there cannot exist infinitely many alternating knots with the same Alexander polynomial.


To see why, suppose for the contrary that $K$ belongs to an infinite family $\{K\_n\}\_{n\in\mathbb{Z}}$ of alternating knots with $\Delta\_{K\_n}(t)=\Delta\_K(t)$. Immediately we have 
$$ \det(K\_n ) = |\Delta\_{K\_n} (−1)| = |\Delta\_K (−1)| = \det(K ). $$
for all $n$. Because these knots are alternating, each one admits a reduced alternating diagram, say $D\_n$ with crossing number $c(D\_n)$. The Bankwitz Theorem implies that $c(K\_n) \leq \det(K\_n) = \det(K)$. However, there are only finitely many knots of a given crossing number, and in particular $c(K\_n)$ grows arbitrarily large with $n$, which contradicts that $c(K\_n) \leq \det(K)$.


For this reason, there also cannot exist infinitely many alternating knots with the isomorphic (bigraded) $\widehat{HFK}(S^3, K)$, either.


In general though, I think these types of botany questions for the Alexander polynomial are interesting both on their own, and in the context of knot Floer homology. Some nice open botany and geography questions were recently mentioned in a paper of Hedden and Watson (see <http://arxiv.org/abs/1404.6913>).


