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Ryan Budney
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I don't know a genuine link to the smooth Poincare conjecture but the link to cobordism for homology spheres is simple. Given a slice disc, construct a branched cover of $D^4$, branched over the slice disc. That gives you a 4-manifold bounding the associated branched cover of the knot in $S^3$. I wouldn't describe it as an approach to determining which homology 3-spheres bound homology 4-balls but it's a natural source of examples, and a linkage. If anything the information seems to flow mostly the other direction. For example, Paolo Lisca's recent paper where he determines precisely which connect-sums of lens spaces bound rational homology balls. As a corollary he deduces the order of 2-bridge knots in the concordance group of knots in $S^3$.

EDIT: Not exactly addressing your question, I think of the slice-ribbon conjecture as a primitive 4-dimensional knotting problem. Given a slice disc you could ask if it's isotopic to a ribbon disc (if the height function on $D^4$ when restricted to the slice disc has only 1-handle and 2-handle attachments, in that order). You can mess up a ribbon disc by taking connect-sums with 2-knots. So modulo connect sums with 2-knots is every slice disc isotopic to a ribbon disc? Perhaps that's too much to ask too, so you can ask the slice-ribbon problem.

2nd edit: As far as I know, the slice-ribbon conjecture has no realmajor consequences. As I describe above, it's more of an ''outer-marker'' type of conjecture. It's a measure of how well we understand knotting of 2-dimensional things in 4-dimensional things.

3rd edit: Here is a type of mild consequence that was pointed out to me recently. In my arXiv preprint on embeddings of 3-manifolds in $S^4$ there's Construction 2.9 which creates embeddings of certain 3-manifolds $M$ in homotopy 4-spheres. The first step is to find a contractible $4$-manifold $W$ that bounds the 3-manifold $M$, then you double $W$ to get a homotopy $S^4$. If the link used in the construction is a ribbon link, the contractible manifold $W$ admits a handle decomposition with one 0-handle, $n$ 1-handles and $n$ 2-handles (for some $n$) and no higher dimensional handles. So the homotopy $S^4$ constructed that contains $M$ is diffeomorphic to $S^4$ provided the corresponding presentation of $\pi_1 W$ is trivializable by Andrews-Curtis moves (handle slides for the handle presentation). This argument will appear in the next draft of the paper, which should appear before January.

I don't know a genuine link to the smooth Poincare conjecture but the link to cobordism for homology spheres is simple. Given a slice disc, construct a branched cover of $D^4$, branched over the slice disc. That gives you a 4-manifold bounding the associated branched cover of the knot in $S^3$. I wouldn't describe it as an approach to determining which homology 3-spheres bound homology 4-balls but it's a natural source of examples, and a linkage. If anything the information seems to flow mostly the other direction. For example, Paolo Lisca's recent paper where he determines precisely which connect-sums of lens spaces bound rational homology balls. As a corollary he deduces the order of 2-bridge knots in the concordance group of knots in $S^3$.

EDIT: Not exactly addressing your question, I think of the slice-ribbon conjecture as a primitive 4-dimensional knotting problem. Given a slice disc you could ask if it's isotopic to a ribbon disc (if the height function on $D^4$ when restricted to the slice disc has only 1-handle and 2-handle attachments, in that order). You can mess up a ribbon disc by taking connect-sums with 2-knots. So modulo connect sums with 2-knots is every slice disc isotopic to a ribbon disc? Perhaps that's too much to ask too, so you can ask the slice-ribbon problem.

2nd edit: As far as I know, the slice-ribbon conjecture has no real consequences. As I describe above, it's more of an ''outer-marker'' type of conjecture. It's a measure of how well we understand knotting of 2-dimensional things in 4-dimensional things.

I don't know a genuine link to the smooth Poincare conjecture but the link to cobordism for homology spheres is simple. Given a slice disc, construct a branched cover of $D^4$, branched over the slice disc. That gives you a 4-manifold bounding the associated branched cover of the knot in $S^3$. I wouldn't describe it as an approach to determining which homology 3-spheres bound homology 4-balls but it's a natural source of examples, and a linkage. If anything the information seems to flow mostly the other direction. For example, Paolo Lisca's recent paper where he determines precisely which connect-sums of lens spaces bound rational homology balls. As a corollary he deduces the order of 2-bridge knots in the concordance group of knots in $S^3$.

EDIT: Not exactly addressing your question, I think of the slice-ribbon conjecture as a primitive 4-dimensional knotting problem. Given a slice disc you could ask if it's isotopic to a ribbon disc (if the height function on $D^4$ when restricted to the slice disc has only 1-handle and 2-handle attachments, in that order). You can mess up a ribbon disc by taking connect-sums with 2-knots. So modulo connect sums with 2-knots is every slice disc isotopic to a ribbon disc? Perhaps that's too much to ask too, so you can ask the slice-ribbon problem.

2nd edit: As far as I know, the slice-ribbon conjecture has no major consequences. As I describe above, it's more of an ''outer-marker'' type of conjecture. It's a measure of how well we understand knotting of 2-dimensional things in 4-dimensional things.

3rd edit: Here is a type of mild consequence that was pointed out to me recently. In my arXiv preprint on embeddings of 3-manifolds in $S^4$ there's Construction 2.9 which creates embeddings of certain 3-manifolds $M$ in homotopy 4-spheres. The first step is to find a contractible $4$-manifold $W$ that bounds the 3-manifold $M$, then you double $W$ to get a homotopy $S^4$. If the link used in the construction is a ribbon link, the contractible manifold $W$ admits a handle decomposition with one 0-handle, $n$ 1-handles and $n$ 2-handles (for some $n$) and no higher dimensional handles. So the homotopy $S^4$ constructed that contains $M$ is diffeomorphic to $S^4$ provided the corresponding presentation of $\pi_1 W$ is trivializable by Andrews-Curtis moves (handle slides for the handle presentation). This argument will appear in the next draft of the paper, which should appear before January.

2nd edit
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Ryan Budney
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I don't know a genuine link to the smooth Poincare conjecture but the link to cobordism for homology spheres is simple. Given a slice disc, construct a branched cover of $D^4$, branched over the slice disc. That gives you a 4-manifold bounding the associated branched cover of the knot in $S^3$. I wouldn't describe it as an approach to determining which homology 3-spheres bound homology 4-balls but it's a natural source of examples, and a linkage. If anything the information seems to flow mostly the other direction. For example, Paolo Lisca's recent paper where he determines precisely which connect-sums of lens spaces bound rational homology balls. As a corollary he deduces the order of 2-bridge knots in the concordance group of knots in $S^3$.

EDIT: Not exactly addressing your question, I think of the slice-ribbon conjecture as a primitive 4-dimensional knotting problem. Given a slice disc you could ask if it's isotopic to a ribbon disc (if the height function on $D^4$ when restricted to the slice disc has only 1-handle and 2-handle attachments, in that order). You can mess up a ribbon disc by taking connect-sums with 2-knots. So modulo connect sums with 2-knots is every slice disc isotopic to a ribbon disc? Perhaps that's too much to ask too, so you can ask the slice-ribbon problem.

2nd edit: As far as I know, the slice-ribbon conjecture has no real consequences. As I describe above, it's more of an ''outer-marker'' type of conjecture. It's a measure of how well we understand knotting of 2-dimensional things in 4-dimensional things.

I don't know a genuine link to the smooth Poincare conjecture but the link to cobordism for homology spheres is simple. Given a slice disc, construct a branched cover of $D^4$, branched over the slice disc. That gives you a 4-manifold bounding the associated branched cover of the knot in $S^3$. I wouldn't describe it as an approach to determining which homology 3-spheres bound homology 4-balls but it's a natural source of examples, and a linkage. If anything the information seems to flow mostly the other direction. For example, Paolo Lisca's recent paper where he determines precisely which connect-sums of lens spaces bound rational homology balls. As a corollary he deduces the order of 2-bridge knots in the concordance group of knots in $S^3$.

EDIT: Not exactly addressing your question, I think of the slice-ribbon conjecture as a primitive 4-dimensional knotting problem. Given a slice disc you could ask if it's isotopic to a ribbon disc (if the height function on $D^4$ when restricted to the slice disc has only 1-handle and 2-handle attachments, in that order). You can mess up a ribbon disc by taking connect-sums with 2-knots. So modulo connect sums with 2-knots is every slice disc isotopic to a ribbon disc? Perhaps that's too much to ask too, so you can ask the slice-ribbon problem.

I don't know a genuine link to the smooth Poincare conjecture but the link to cobordism for homology spheres is simple. Given a slice disc, construct a branched cover of $D^4$, branched over the slice disc. That gives you a 4-manifold bounding the associated branched cover of the knot in $S^3$. I wouldn't describe it as an approach to determining which homology 3-spheres bound homology 4-balls but it's a natural source of examples, and a linkage. If anything the information seems to flow mostly the other direction. For example, Paolo Lisca's recent paper where he determines precisely which connect-sums of lens spaces bound rational homology balls. As a corollary he deduces the order of 2-bridge knots in the concordance group of knots in $S^3$.

EDIT: Not exactly addressing your question, I think of the slice-ribbon conjecture as a primitive 4-dimensional knotting problem. Given a slice disc you could ask if it's isotopic to a ribbon disc (if the height function on $D^4$ when restricted to the slice disc has only 1-handle and 2-handle attachments, in that order). You can mess up a ribbon disc by taking connect-sums with 2-knots. So modulo connect sums with 2-knots is every slice disc isotopic to a ribbon disc? Perhaps that's too much to ask too, so you can ask the slice-ribbon problem.

2nd edit: As far as I know, the slice-ribbon conjecture has no real consequences. As I describe above, it's more of an ''outer-marker'' type of conjecture. It's a measure of how well we understand knotting of 2-dimensional things in 4-dimensional things.

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Ryan Budney
  • 44.4k
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  • 245

I don't know a genuine link to the smooth Poincare conjecture but the link to cobordism for homology spheres is simple. Given a slice disc, construct a branched cover of $D^4$, branched over the slice disc. That gives you a 4-manifold bounding the associated branched cover of the knot in $S^3$. I wouldn't describe it as an approach to determining which homology 3-spheres bound homology 4-balls but it's a natural source of examples, and a linkage. If anything the information seems to flow mostly the other direction. For example, Paolo Lisca's recent paper where he determines precisely which connect-sums of lens spaces bound rational homology balls. As a corollary he deduces the order of 2-bridge knots in the concordance group of knots in $S^3$.

EDIT: Not exactly addressing your question, I think of the slice-ribbon conjecture as a primitive 4-dimensional knotting problem. Given a slice disc you could ask if it's isotopic to a ribbon disc (if the height function on $D^4$ when restricted to the slice disc has only 1-handle and 2-handle attachments, in that order). You can mess up a ribbon disc by taking connect-sums with 2-knots. So modulo connect sums with 2-knots is every slice disc isotopic to a ribbon disc? Perhaps that's too much to ask too, so you can ask the slice-ribbon problem.

I don't know a genuine link to the smooth Poincare conjecture but the link to cobordism for homology spheres is simple. Given a slice disc, construct a branched cover of $D^4$, branched over the slice disc. That gives you a 4-manifold bounding the associated branched cover of the knot in $S^3$. I wouldn't describe it as an approach to determining which homology 3-spheres bound homology 4-balls but it's a natural source of examples, and a linkage. If anything the information seems to flow mostly the other direction. For example, Paolo Lisca's recent paper where he determines precisely which connect-sums of lens spaces bound rational homology balls. As a corollary he deduces the order of 2-bridge knots in the concordance group of knots in $S^3$.

I don't know a genuine link to the smooth Poincare conjecture but the link to cobordism for homology spheres is simple. Given a slice disc, construct a branched cover of $D^4$, branched over the slice disc. That gives you a 4-manifold bounding the associated branched cover of the knot in $S^3$. I wouldn't describe it as an approach to determining which homology 3-spheres bound homology 4-balls but it's a natural source of examples, and a linkage. If anything the information seems to flow mostly the other direction. For example, Paolo Lisca's recent paper where he determines precisely which connect-sums of lens spaces bound rational homology balls. As a corollary he deduces the order of 2-bridge knots in the concordance group of knots in $S^3$.

EDIT: Not exactly addressing your question, I think of the slice-ribbon conjecture as a primitive 4-dimensional knotting problem. Given a slice disc you could ask if it's isotopic to a ribbon disc (if the height function on $D^4$ when restricted to the slice disc has only 1-handle and 2-handle attachments, in that order). You can mess up a ribbon disc by taking connect-sums with 2-knots. So modulo connect sums with 2-knots is every slice disc isotopic to a ribbon disc? Perhaps that's too much to ask too, so you can ask the slice-ribbon problem.

Source Link
Ryan Budney
  • 44.4k
  • 2
  • 139
  • 245
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