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Chris Gerig
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The 4-sphere $S^4$ cannot be a symplectic manifold. In particular, it does not admit an atlas whose transition maps are symplectomorphisms $(\mathbb{R}^4,\omega_0)\to(\mathbb{R}^4,\omega_0)$$(\mathbb{R}^4,\omega_\text{std})\to(\mathbb{R}^4,\omega_\text{std})$. A theorem of Eliashberg-Gromov statesproved that if the $C^0$-limit of symplectomorphisms is a diffeomorphism, then it is a symplectomorphism. We should thencan now ask about the next best thing:

Does $S^4$ have a symplectohomeomorphic structure, i.e. are there charts (equipped with the standard symplectic form on $\mathbb{R}^4$) whose transition maps are homeomorphisms that are $C^0$-limits of symplectomorphisms?

(This question should be credited tomay have originated from Alan Weinstein, who I heardtold me it from after conversations aboutconcerning symplectic rigidity.)

The 4-sphere $S^4$ cannot be a symplectic manifold. In particular, it does not admit an atlas whose transition maps are symplectomorphisms $(\mathbb{R}^4,\omega_0)\to(\mathbb{R}^4,\omega_0)$. A theorem of Eliashberg-Gromov states that if the $C^0$-limit of symplectomorphisms is a diffeomorphism, then it is a symplectomorphism. We should then ask about the next best thing:

Does $S^4$ have a symplectohomeomorphic structure, i.e. are there charts (equipped with the standard symplectic form on $\mathbb{R}^4$) whose transition maps are homeomorphisms that are $C^0$-limits of symplectomorphisms?

(This question should be credited to Alan Weinstein, who I heard it from after conversations about symplectic rigidity.)

The 4-sphere cannot be a symplectic manifold. In particular, it does not admit an atlas whose transition maps are symplectomorphisms $(\mathbb{R}^4,\omega_\text{std})\to(\mathbb{R}^4,\omega_\text{std})$. Eliashberg-Gromov proved that if the $C^0$-limit of symplectomorphisms is a diffeomorphism, then it is a symplectomorphism. We can now ask about the next best thing:

Does $S^4$ have a symplectohomeomorphic structure, i.e. are there charts (equipped with the standard symplectic form on $\mathbb{R}^4$) whose transition maps are homeomorphisms that are $C^0$-limits of symplectomorphisms?

(This question may have originated from Alan Weinstein, who told me it after conversations concerning symplectic rigidity.)

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Chris Gerig
  • 17.5k
  • 2
  • 71
  • 116

Does $S^4$ have a "symplecto-homeomorphic" structure?

The 4-sphere $S^4$ cannot be a symplectic manifold. In particular, it does not admit an atlas whose transition maps are symplectomorphisms $(\mathbb{R}^4,\omega_0)\to(\mathbb{R}^4,\omega_0)$. A theorem of Eliashberg-Gromov states that if the $C^0$-limit of symplectomorphisms is a diffeomorphism, then it is a symplectomorphism. We should then ask about the next best thing:

Does $S^4$ have a symplectohomeomorphic structure, i.e. are there charts (equipped with the standard symplectic form on $\mathbb{R}^4$) whose transition maps are homeomorphisms that are $C^0$-limits of symplectomorphisms?

(This question should be credited to Alan Weinstein, who I heard it from after conversations about symplectic rigidity.)