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Daniele Zuddas
Daniele Zuddas
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I think that a complete set of moves for cobordisms are the following: Kirby moves (that preserve the 4-manifold), handle trading (dotted 1-handlescircles become 0-framed 2-handles), addition/deletion of pairs of trivial circles with framings $\pm 1$ (corresponding to connected sums with $CP^2 \#\overline{CP}^2$$CP^2 \,\#\, {\overline{CP^2}}$). This could be proved by using Kirby calculus for 3-manifolds (which is exact!) as in Kirby's original paper as follows. Take two cobordant 4-manifolds $M_1$ and $M_2$ with their Kirby diagrams (with one 0-handle and one 4-handle). Up to addition of complementary 2/3-handles, we can assume that they have the same number of 3-handles. Now, removing the 3- and 4-handles, you get 4D 2-handlebodies with the same boundary $\#_k\, (S^1 \times S^2)$ for some $k \geq 0$. Trading 1-handles completely remove them (without changing the boundary and the signature, because it is surgery along embedded circles). By Kirby's theorem, the resulting manifolds can be related by Kirby moves (now only on 2-handles) and addition/deletion of separated $\pm 1$-framed trivial knots. When the latter operation occurs, just compensate by adding another suitable $\mp 1$ trivial circle that will be not involved in subsequent moves. You will end up with the Kirby diagram for $M_2$ (maybe after 1-handle trading) plus some $\pm1$ trivial knots (the ones you added for compensation) and the $\pm 1$'s sum up to 0 since the signature must be the same.

I think that a complete set of moves for cobordisms are the following: Kirby moves (that preserve the 4-manifold), handle trading (dotted 1-handles become 0-framed 2-handles), addition/deletion of pairs of trivial circles with framings $\pm 1$ (corresponding to connected sums with $CP^2 \#\overline{CP}^2$). This could be proved by using Kirby calculus for 3-manifolds (which is exact!) as in Kirby's original paper as follows. Take two cobordant 4-manifolds $M_1$ and $M_2$ with their Kirby diagrams (with one 0-handle and one 4-handle). Up to addition of complementary 2/3-handles, we can assume that they have the same number of 3-handles. Now, removing the 3- and 4-handles, you get 4D 2-handlebodies with the same boundary $\#_k\, (S^1 \times S^2)$ for some $k \geq 0$. Trading 1-handles completely remove them (without changing the boundary and the signature, because it is surgery along embedded circles). By Kirby's theorem, the resulting manifolds can be related by Kirby moves (now only on 2-handles) and addition/deletion of separated $\pm 1$-framed trivial knots. When the latter operation occurs, just compensate by adding another suitable $\mp 1$ trivial circle that will be not involved in subsequent moves. You will end up with the Kirby diagram for $M_2$ (maybe after 1-handle trading) plus some $\pm1$ trivial knots (the ones you added for compensation) and the $\pm 1$'s sum up to 0 since the signature must be the same.

I think that a complete set of moves for cobordisms are the following: Kirby moves (that preserve the 4-manifold), handle trading (dotted circles become 0-framed 2-handles), addition/deletion of pairs of trivial circles with framings $\pm 1$ (corresponding to connected sums with $CP^2 \,\#\, {\overline{CP^2}}$). This could be proved by using Kirby calculus for 3-manifolds (which is exact!) as in Kirby's original paper as follows. Take two cobordant 4-manifolds $M_1$ and $M_2$ with their Kirby diagrams (with one 0-handle and one 4-handle). Up to addition of complementary 2/3-handles, we can assume that they have the same number of 3-handles. Now, removing the 3- and 4-handles, you get 4D 2-handlebodies with the same boundary $\#_k\, (S^1 \times S^2)$ for some $k \geq 0$. Trading 1-handles completely remove them (without changing the boundary and the signature, because it is surgery along embedded circles). By Kirby's theorem, the resulting manifolds can be related by Kirby moves (now only on 2-handles) and addition/deletion of separated $\pm 1$-framed trivial knots. When the latter operation occurs, just compensate by adding another suitable $\mp 1$ trivial circle that will be not involved in subsequent moves. You will end up with the Kirby diagram for $M_2$ (maybe after 1-handle trading) plus some $\pm1$ trivial knots (the ones you added for compensation) and the $\pm 1$'s sum up to 0 since the signature must be the same.

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Daniele Zuddas
Daniele Zuddas
  • 2.3k
  • 13
  • 19

I think that a complete set of moves for cobordisms are the following: Kirby moves (that preserve the 4-manifold), handle trading (dotted 1-handles become 0-framed 2-handles), addition/deletion of pairs of trivial circles with framings $\pm 1$ (corresponding to connected sums with $CP^2 \#\overline{CP}^2$). This could be proved by using Kirby calculus for 3-manifolds (which is exact!) as in Kirby's original paper as follows. Take two cobordant 4-manifolds $M_1$ and $M_2$ with their Kirby diagrams (with one 0-handle and one 4-handle). Up to addition of complementary 2/3-handles, we can assume that they have the same number of 3-handles. Now, removing the 3- and 4-handles, you get 4D 2-handlebodies with the same boundary $\#_k\, (S^1 \times S^2)$ for some $k \geq 0$. Trading 1-handles completely remove them (without changing the boundary and the signature, because it is surgery along embedded circles). By Kirby's theorem, the resulting manifolds can be related by Kirby moves (now only on 2-handles) and addition/deletion of separated $\pm 1$-framed trivial knots. When the latter operation occurs, just compensate by adding another suitable $\mp 1$ trivial circle that will be not involved in subsequent moves. You will end up with the Kirby diagram for $M_2$ (maybe after 1-handle trading) plus some $\pm1$ trivial knots (the ones you added for compensation) and the $\pm 1$'s sum up to 0 since the signature must be the same.