Knot surgery $4$-manifolds $E(n)_K$ without $1$- and $3$-handles

Abstract

In this article, we demonstrate that for any positive integer $n$, the knot surgery $4$-manifold $E(n)_K$ has a handle decomposition without $1$- and $3$-handles. Here, $K$ represents either a fibered two-bridge knot $C(2ε_1, 2ε_2,\cdots, 2ε_{2g})$ ($ε_i \in \{ 1, -1\}$) in Conway's notation or a Stallings knot $K_m$ ($m \in \mathbb{Z}$).

Figures (7)

• Figure 1: Vanishing cycles on $E(n)_K$
• Figure 2: A $2$-bridge knot $C(n_1, n_2, \cdots, n_k)$ (for $k$ odd and $k$ even, respectively)
• Figure 3: Curves for $\phi_{D(\epsilon_1, \epsilon_2,\cdots, \epsilon_{2k})}$
• Figure 4: Handle decomposition of $\Sigma_{2g(K) + n-1}^1$: core $\alpha_i$ and co-core $\alpha_i^*$ of each $2$-dimensional $1$-handle (ordered set of oriented arcs $\{ \alpha_i, \alpha_i^*\}$ give the positive orientation of the fiber surface.)
• Figure 5: Decomposition of $B_i$ on $\Sigma_{2g(K) + n-1}$ (simple closed curve in red color is a decomposition of $B_2 \simeq \beta_2*\gamma*\alpha_{4g(K)-1}$)

Theorems & Definitions (20)

• Definition 2.1: BZ:03
• Lemma 2.2: BZ:03Kawauchi:1996
• proof
• Definition 3.1
• Definition 3.2
• Remark 3.3
• Remark 3.4
• Lemma 3.5
• proof
• Lemma 3.6