$k$-cobordism for links in $S\sp 3$

We give an explicit finite set of (based) links which generates, under connected sum, the $k$-cobordism classes of links. We show that the union of these generating sets, $2 \leq k < \infty$, is not a generating set for $\omega$-cobordism classes or even $\infty$-cobordism classes.

For $2$-component links in ${S^3}$ we define $(2,k)$-corbordism and show that the concordance invariants ${\beta ^i},i \in {\mathbb{Z}^+}$, previously defined by the author, are invariants under $(2,i + 1)$-cobordism. Moreover we show that the $(2,k)$-cobordism classes of links (with linking number 0) is a free abelian group of rank $k - 1$, detected precisely by ${\beta ^1} \times \cdots \times {\beta^{k - 1}}$. We write down a basis. The union of these bases $(2 \leq k < \infty)$ is not a generating set for $(2,\infty)$ or $(2,\omega)$-cobordism classes. However, we can show that $\prod _{i = 1}^\infty {\beta ^i}(\;)$ is an isomorphism from the group of $(2,\infty)$-cobordism classes to the subgroup $\mathcal{R} \subset \prod _{i = 1}^\infty \mathbb{Z}$ of linearly recurrent sequences, so a basis exists by work of T. Jin.