Proof Since $S^3 \times I$ is simply connected, $\pi _1(\overline {W}) = \mathbb {Z}/p\mathbb {Z} = H_1(\overline {W})$ , and the maps

$$\begin{align*}H_1(S^3/\rho) \to H_1(\overline{W}) \mbox{ and } H_1(S^3/\rho') \to H_1(\overline{W}) \end{align*}$$

induced by inclusion are isomorphisms. Since W is connected, the same is true for $H_0$ so that $\overline {W}$ is a homology cobordism.

Proof Apply Theorem 2.5 to the quotient of the equivariant cobordism, and use the homeomorphism classification of lens spaces [Reference BrodyBro60].

Proof Let S be an equivariant surface for K with respect to an order p diffeomorphism $\rho :B^4 \to B^4$ . By classical Smith theory [Reference SmithSmi41], $\rho $ has a contractible fixed-point set F. Furthermore, since the fixed set of a self-diffeomorphism is a submanifold, F is a single point. Similarly, the fixed-point set of $\rho |_S$ is also the single point F. Removing an equivariant neighborhood $N(F)$ from $B^4$ leaves an $S^3$ boundary component containing an unknot U. Hence, we have $(S - N(F)) \subset S^3 \times I$ , which is preserved by the free $\mathbb {Z}/p\mathbb {Z}$ action and $\partial (S - N(F)) = K \cup U$ . That is an equivariant cobordism between K and U. By Corollary 2.6, U comes with a freely $(p,q)$ -periodic symmetry.

On the other hand, suppose that we have an equivariant cobordism between K and U. By Proposition 2.2, U can be taken to be the lift of a core of a handlebody in a genus 1 Heegaard decomposition of $L(p,q)$ , and in particular, the cone of $(S^3,U)$ is a smooth equivariant disk in $B^4$ . Gluing this to the cobordism gives an equivariant surface for K.

Proof Suppose $K \subset S^3$ is a freely $(p,q)$ -periodic knot which bounds a surface $S \subset B^4$ which is invariant under an order p diffeomorphism $\rho :B^4 \to B^4$ with $\rho |_{S^3}$ the free $\mathbb {Z}/p\mathbb {Z}$ action preserving K. Then, by Lemma 2.7, there is an equivariant cobordism from K to a freely $(p,q)$ -periodic unknot $U_{p,q}$ . By Lemma 2.4, the quotient of this cobordism is a cobordism of the quotient knots $\overline {K}$ and $\overline {U}_{p,q}$ in a homology cobordism of lens spaces. In particular, there is a map $f:H_1(L(p,q)) \to H_1(L(p,q))$ with $[f(\overline {K})] = [\overline {U}_{p,q}]$ which is induced by a homology cobordism of lens spaces. By Theorem 2.5, $f = \pm $ Id. Hence, $\overline {K}$ represents the same class as an unknot in $H_1(L(p,q))$ . Then, by Proposition 2.2, we have that $[\overline {K}]$ is simple.

On the other hand, suppose that $[\overline {K}]$ is simple. Then, by Proposition 2.2, $\overline {K}$ is in the same homology class as the core $\overline {U}$ of a handlebody in the genus 1 Heegaard decomposition of $L(p,q)$ , which lifts to a freely periodic unknot U. Hence, there is a surface $\overline {S}$ in $L(p,q)$ with boundary $\overline {K} \cup \overline {U}$ , and lifting $\overline {S}$ to $S^3$ gives us an equivariant surface S with boundary $K \cup U$ . We can then take the cone of $(S^3,U)$ with the free symmetry to obtain a smooth equivariant disk in $B^4$ . Gluing this to S gives us an equivariant surface for K.

Proof It is clear that a tangle with one strand is homologous to the core c of a genus 1 Heegaard splitting for $L(p,q)$ . It follows immediately from the definition of a lens space that the other core is $q^{-1} \cdot c \in H_1(L(p,q))$ so that a diagram with $q^{-1}$ strands is homologous to the other core.

Proof By [Reference MilnorMS86, Theorem 2.2] and [Reference SchreierSch24] (see also [Reference KawauchiKaw96, Exercise 10.6.4]), the freely periodic symmetry preserves a Seifert fibered structure on the exterior of $T(r,s)$ , which is a circle bundle over $D(r,s)$ , the disk with cone points of orders r and s. For $p>2$ , this orbifold has no order p symmetries. For the involution reflecting across an arc containing the two cone points, we must also reflect the fibers to get an orientation-preserving symmetry of the exterior of $T(r,s)$ , but this has global fixed points. Thus, our free symmetry acts as identity on the base orbifold, and therefore as a rotation on the fibers. In particular, there is a unique symmetry for each p.

When p has a common factor with r or s, then the order gcd $(p,rs)$ subgroup has a fixed circle in $S^3$ , and so there is no free symmetry of order p. On the other hand, the symmetry is free when gcd $(p,rs) = 1$ . We will describe these free symmetries explicitly. Consider the standard tangle diagram for $T(r,s)$ with r strands. Then there are s total twists, which can be grouped as $\pm q$ full twists and p concatenated tangles of n twists each (see Figure 2 for an example). Hence, $s = \pm q \cdot r + n \cdot p$ from which we deduce that $\pm q \equiv sr^{-1}$ (mod p). However, a free $(p,q)$ -symmetry can also be regarded as a free $(p,q^{-1})$ -symmetry (cf. the homeomorphism classification of lens spaces). In particular, the diagram with s strands has $\pm q \equiv rs^{-1}$ (mod p).

Proof Consider the freely $(p,q)$ -periodic diagram for $T(r,s)$ with r strands, which consists of s total twists: $\pm q$ full twists and p tangles of n twists each. By Theorems 1.2 and 1.3, $T(r,s)$ is an equivariant boundary if and only if $r \equiv \pm 1$ or $r \equiv \pm q^{-1}$ (mod p). By Theorem 2.12, $\pm q \equiv sr^{-1}$ (mod p) so that p is a divisor of $r-1,r+1,s-1,$ or $s+1$ .

Proof Let n be the genus of S so that $\chi (S) = 1-2n$ . Since S has exactly one fixed point F (see the proof of Lemma 2.7), $\chi (S - F) = -2n$ . Quotienting $S-F$ by the free $\mathbb {Z}/p\mathbb {Z}$ action leaves a surface $\overline {S}$ with $\chi (\overline {S}) = -2n/p$ . However, since $\overline {S}$ has exactly two boundary components, it must have an even Euler characteristic and so $p|n$ .

Question 3.4 What tools can give a lower bound on the genus of an equivariant surface for a freely periodic knot?

Question 3.5 Does there exist a freely periodic slice knot which does not bound any (smooth) equivariant disk?