Title : On knots with trivial Alexander polynomial and the cyclic branched covers
Speaker : Yukihiro Tsutsumi (Sophia University)
Abstract :

Let $K \subset S^3$ be a knot with trivial Alexander polynomial $\Delta_K(t)=1$. Then, the $r$-fold cyclic covering space $M^r_K$ over $S^3$ branched along $K$ is an integral homology sphere. We show that the Casson invariant $\lambda(M^r_K)$ is divisible by $2r$ and there are integers $r_0$ and $k$ such that $\lambda(M^r_K) = 2rk$ for any $r > r_0$. These restrictions are sharp in the following sense: Given mutually distinct $n$ natural numbers $r_1$, $r_2$, \ldots, $r_n$ each greater than one ($r_i > 1$) and $n + 1$ integers $k_1$, $k_2$, \ldots, $k_n$, $k$, there are knots $K$ such that $\Delta_K(t)=1$, $\lambda(M^{r_i}_K) = 2r_i k_i$ and $\lambda(M^{r}_K) = 2 r k$ for any $r (\ge 2) \ne r_i$. The condition that $\Delta_K(t)=1$ is essential. It is well-known that the 5-fold cyclic branched cover $M^5_{3_1}$ for the trefoil knot $3_1$ is the Poincar\'e homology sphere $\Sigma(2,3,5)$. Recall that $\Delta_{3_1}(t) = t - 1 + t^{-1}$ and $\lambda(\Sigma(2,3,5)) = 1$. For the Kinoshita-Terasaka knot $KT$, $\lambda(M^2_{KT}) = 4$, $\lambda(M^{\ge 3}_{KT})=0$. We will also exhibit a sequence of knots $K_m$ such that $K_0 = KT$, $\chi(S^3;(K_m,0)) = \chi(S^3;(KT,0))$ (and thus $\Delta_{K_m}(t) = 1$), $\lambda(M^r_{K_m}) = \lambda(M^r_{KT})$. For the sequence $K_m$, I suspect the following: $J'_{K_m}(-1) = 48$, $J'''_{K_m}(1) = -72$, $J''''_{K_m}(1) = 720$, $J'''''_{K_m}(1) = -14400m - 9120$, where $J(t)$ denotes the Jones polynomial.

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