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Included the title of the linked paper
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Martin Sleziak
Martin Sleziak
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Edit: my previous answer was incorrect

No, this one you cannot do. If one had such an arrangement of circles in the solid torus inside $R^3$, and quotiented by a rotation, then by the equivariant Dehn's lemmaequivariant Dehn's lemma there would be a collection of essential disks in the complement of the split link that are invariant under the symmetry.

Since the link components get permuted, no disk can have a fixed point of the cyclic action. So they are disjoint from the axis of symmetry of the rotation. Hence the components are unknotted in the solid torus, a contradiction

Edit: my previous answer was incorrect

No, this one you cannot do. If one had such an arrangement of circles in the solid torus inside $R^3$, and quotiented by a rotation, then by the equivariant Dehn's lemma there would be a collection of essential disks in the complement of the split link that are invariant under the symmetry.

Since the link components get permuted, no disk can have a fixed point of the cyclic action. So they are disjoint from the axis of symmetry of the rotation. Hence the components are unknotted in the solid torus, a contradiction

Edit: my previous answer was incorrect

No, this one you cannot do. If one had such an arrangement of circles in the solid torus inside $R^3$, and quotiented by a rotation, then by the equivariant Dehn's lemma there would be a collection of essential disks in the complement of the split link that are invariant under the symmetry.

Since the link components get permuted, no disk can have a fixed point of the cyclic action. So they are disjoint from the axis of symmetry of the rotation. Hence the components are unknotted in the solid torus, a contradiction

Fixed the flawed argument.
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Ian Agol
Ian Agol
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Edit: my previous answer was incorrect

No, this one you cannot do. If one had such an arrangement of circles in the solid torus inside $R^3$, and quotiented by the cyclic symmetrya rotation, then by the equivariant Dehn's lemma there would be a collection of essential disks in the complement of the split link that are invariant under the symmetry. All of these disks must meet the symmetry axis since

Since the link is irreducible in the solid toruscomponents get permuted, an theyno disk can only intersect inhave a singlefixed point, since of the cyclic action. So they are invariant under a cyclic groupdisjoint from the axis of symmetry of the rotation. But this implies that Hence the linking number between each component andcomponents are unknotted in the solid torus is non-zero, a contradiction.

No, this one you cannot do. If one had such an arrangement of circles in the solid torus, and quotiented by the cyclic symmetry, then by the equivariant Dehn's lemma there would be a collection of essential disks in the complement of the split link that are invariant under the symmetry. All of these disks must meet the symmetry axis since the link is irreducible in the solid torus, an they can only intersect in a single point, since they are invariant under a cyclic group. But this implies that the linking number between each component and the solid torus is non-zero, a contradiction.

Edit: my previous answer was incorrect

No, this one you cannot do. If one had such an arrangement of circles in the solid torus inside $R^3$, and quotiented by a rotation, then by the equivariant Dehn's lemma there would be a collection of essential disks in the complement of the split link that are invariant under the symmetry.

Since the link components get permuted, no disk can have a fixed point of the cyclic action. So they are disjoint from the axis of symmetry of the rotation. Hence the components are unknotted in the solid torus, a contradiction

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Ian Agol
Ian Agol
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No, this one you cannot do. If one had such an arrangement of circles in the solid torus, and quotiented by the cyclic symmetry, then by the equivariant Dehn's lemma there would be a collection of essential disks in the complement of the split link that are invariant under the symmetry. All of these disks must meet the symmetry axis since the link is irreducible in the solid torus, an they can only intersect in a single point, since they are invariant under a cyclic group. But this implies that the linking number between each component and the solid torus is non-zero, a contradiction.