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fixed statement of symplectic BMY
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Tim Perutz
Tim Perutz
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Symplectic geography in 4 dimensions can be mapped using Chern number coordinates $(c_1^2,c_2)$. The part of the plane where $c_1^2 > 4c_2$ is uncharted. It's unknown whether there are any symplectic 4-manifolds in this region, besides blow-ups of ruled surfaces, though by the Bogomolov-Miyaoka-Yau inequality and the Kodaira-Enriques classification, there are no complex surfaces.

I can't answer the question but I'll point out that a symplectic 4-manifold with $b_-=0$ and $b_+>1$ necessarily lies in this unknown region - in particular, it's not Kaehler.

To see this, rewrite $c_1^2-4c_2$ in terms of Euler characteristic $\chi$ and signature $\sigma$ as $(2\chi+3\sigma)-4\chi$. For a symplectic manifold with $b_-=0$, this quantity equals $4b_1+b_2-4$ and is positive unless $b_1=0$ and $b_2= 1$ or $3$; I use the parity argument mentioned in Paul's comment. If $b_1=0$ and $b_2=3$ then the intersection form on $H_2/tors.$ is $\mathbb{Z}^3$, the unique rank 3 positive-definite unimodular lattice. So $c_1^2=15$ is the sum of three squares; but it's not.

Symplectic geography in 4 dimensions can be mapped using Chern number coordinates $(c_1^2,c_2)$. The part of the plane where $c_1^2 > 4c_2$ is uncharted. It's unknown whether there are any symplectic 4-manifolds in this region, though by the Bogomolov-Miyaoka-Yau inequality and the Kodaira-Enriques classification, there are no complex surfaces.

I can't answer the question but I'll point out that a symplectic 4-manifold with $b_-=0$ and $b_+>1$ necessarily lies in this unknown region - in particular, it's not Kaehler.

To see this, rewrite $c_1^2-4c_2$ in terms of Euler characteristic $\chi$ and signature $\sigma$ as $(2\chi+3\sigma)-4\chi$. For a symplectic manifold with $b_-=0$, this quantity equals $4b_1+b_2-4$ and is positive unless $b_1=0$ and $b_2= 1$ or $3$; I use the parity argument mentioned in Paul's comment. If $b_1=0$ and $b_2=3$ then the intersection form on $H_2/tors.$ is $\mathbb{Z}^3$, the unique rank 3 positive-definite unimodular lattice. So $c_1^2=15$ is the sum of three squares; but it's not.

Symplectic geography in 4 dimensions can be mapped using Chern number coordinates $(c_1^2,c_2)$. The part of the plane where $c_1^2 > 4c_2$ is uncharted. It's unknown whether there are any symplectic 4-manifolds in this region, besides blow-ups of ruled surfaces, though by the Bogomolov-Miyaoka-Yau inequality and the Kodaira-Enriques classification, there are no complex surfaces.

I can't answer the question but I'll point out that a symplectic 4-manifold with $b_-=0$ and $b_+>1$ necessarily lies in this unknown region - in particular, it's not Kaehler.

To see this, rewrite $c_1^2-4c_2$ in terms of Euler characteristic $\chi$ and signature $\sigma$ as $(2\chi+3\sigma)-4\chi$. For a symplectic manifold with $b_-=0$, this quantity equals $4b_1+b_2-4$ and is positive unless $b_1=0$ and $b_2= 1$ or $3$; I use the parity argument mentioned in Paul's comment. If $b_1=0$ and $b_2=3$ then the intersection form on $H_2/tors.$ is $\mathbb{Z}^3$, the unique rank 3 positive-definite unimodular lattice. So $c_1^2=15$ is the sum of three squares; but it's not.

Source Link
Tim Perutz
Tim Perutz
  • 13.2k
  • 1
  • 53
  • 79

Symplectic geography in 4 dimensions can be mapped using Chern number coordinates $(c_1^2,c_2)$. The part of the plane where $c_1^2 > 4c_2$ is uncharted. It's unknown whether there are any symplectic 4-manifolds in this region, though by the Bogomolov-Miyaoka-Yau inequality and the Kodaira-Enriques classification, there are no complex surfaces.

I can't answer the question but I'll point out that a symplectic 4-manifold with $b_-=0$ and $b_+>1$ necessarily lies in this unknown region - in particular, it's not Kaehler.

To see this, rewrite $c_1^2-4c_2$ in terms of Euler characteristic $\chi$ and signature $\sigma$ as $(2\chi+3\sigma)-4\chi$. For a symplectic manifold with $b_-=0$, this quantity equals $4b_1+b_2-4$ and is positive unless $b_1=0$ and $b_2= 1$ or $3$; I use the parity argument mentioned in Paul's comment. If $b_1=0$ and $b_2=3$ then the intersection form on $H_2/tors.$ is $\mathbb{Z}^3$, the unique rank 3 positive-definite unimodular lattice. So $c_1^2=15$ is the sum of three squares; but it's not.