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Sam Nead
Sam Nead
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Every (closed, connected, oriented) three-manifold contains a fibred knot. That is, every three-manifold has an open book decomposition with non-empty, connected binding and connected, oriented, compact page. This is due to Winkelnkemper or Gonzalez-Acuna - see page 625 of this book.

Edited

So, $M = \Sigma \times S^1$ has the "obvious" open book decomposition (if we allow empty bindings). By the above $M$ also has (infinitely many) open book decompositions with non-empty bindings. I very much doubt that there is a preferred one among the latter.

Edited

Every (closed, connected, oriented) three-manifold contains a fibred knot. That is, every three-manifold has an open book decomposition with non-empty, connected binding and connected, oriented, compact page. This is due to Winkelnkemper or Gonzalez-Acuna - see page 625 of this book.

So, $M = \Sigma \times S^1$ has the "obvious" open book decomposition (if we allow empty bindings). By the above $M$ also has (infinitely many) open book decompositions with non-empty bindings. I very much doubt that there is a preferred one among the latter.

Every (closed, connected, oriented) three-manifold contains a fibred knot. That is, every three-manifold has an open book decomposition with non-empty, connected binding and connected, oriented, compact page. This is due to Winkelnkemper or Gonzalez-Acuna - see page 625 of this book.

Edited

So, $M = \Sigma \times S^1$ has the "obvious" open book decomposition (if we allow empty bindings). By the above $M$ also has (infinitely many) open book decompositions with non-empty bindings. I very much doubt that there is a preferred one among the latter.

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Source Link
Sam Nead
Sam Nead
  • 28.2k
  • 5
  • 72
  • 131

Edited

Every (closed, connected, oriented) three-manifold contains a fibred knot. That is, every three-manifold has an open book decomposition with non-empty, connected binding and connected, oriented, compact page. This is due to Winkelnkemper or Gonzalez-Acuna - see page 625 of this book.

So, even your example of a product$M = \Sigma \times S^1$ has the "obvious" open book decomposition (of a closed surface with a circleif we allow empty bindings). By the above $M$ also has some non-trivial(infinitely many) open book decompositiondecompositions with non-empty bindings. It just I very much doubt that there is not the "obvious"a preferred one among the latter.

Every (closed, connected, oriented) three-manifold contains a fibred knot. That is, every three-manifold has an open book decomposition with non-empty, connected binding and connected, oriented, compact page. This is due to Winkelnkemper or Gonzalez-Acuna - see page 625 of this book.

So, even your example of a product (of a closed surface with a circle) has some non-trivial open book decomposition. It just is not the "obvious" one.

Edited

Every (closed, connected, oriented) three-manifold contains a fibred knot. That is, every three-manifold has an open book decomposition with non-empty, connected binding and connected, oriented, compact page. This is due to Winkelnkemper or Gonzalez-Acuna - see page 625 of this book.

So, $M = \Sigma \times S^1$ has the "obvious" open book decomposition (if we allow empty bindings). By the above $M$ also has (infinitely many) open book decompositions with non-empty bindings. I very much doubt that there is a preferred one among the latter.

Source Link
Sam Nead
Sam Nead
  • 28.2k
  • 5
  • 72
  • 131

Every (closed, connected, oriented) three-manifold contains a fibred knot. That is, every three-manifold has an open book decomposition with non-empty, connected binding and connected, oriented, compact page. This is due to Winkelnkemper or Gonzalez-Acuna - see page 625 of this book.

So, even your example of a product (of a closed surface with a circle) has some non-trivial open book decomposition. It just is not the "obvious" one.