Timeline for (Contradiction) All symplectic manifolds are holomorphic
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 26, 2021 at 22:38 | vote | accept | Anthony D'Arienzo | ||
| Jan 25, 2021 at 14:00 | answer | added | JHM | timeline score: 5 | |
| Jan 24, 2021 at 22:35 | comment | added | Ivan Solonenko | Let me elaborate a bit on @JHM's comment. Suppose you have almost complex manifolds $(M_1, J_1)$ and $(M_2, J_2)$, a diffeomorphism $\varphi \colon M_1 \to M_2$, and a complex isomorphism $F\colon TM_1 \to TM_2$ covering $\varphi$. Although this latter map identifies the a. c. structures, it doesn't identify their Nijenhuis tensors, since the definition of the latter involves the commutator of vector fields. In particular, you may have $(M_1, J_1)$ a complex manifold ($N_{J_1} = 0$) but $(M_2, J_2)$ not. The problem here is that in general $F \ne d\varphi$, so $F$ may not preserve commutators. | |
| Jan 24, 2021 at 22:23 | comment | added | JHM | I was mistaken. It is incorrect to write $\psi_*$, since we are speaking of the restriction of $\psi$ to fibres. What is needed is a self map $f$ of the base $\phi(U)$ for which $f_*\circ j_0 = j_1 \circ f_*$. The bundle morphism from Prop. 2 is identity map $f=id$ on the base. So Prop. 2 does not yield a holomorphic map $f$. | |
| Jan 24, 2021 at 21:59 | comment | added | Anthony D'Arienzo | @JHM I agree that $\psi$ is a linear isomorphism of the fibres. Since the bundle is complex, this isomorphism should be $\mathbb{C}$ linear. Thus, it should intertwine the action of $i$ on the fibres. Isn’t this action the almost complex structure? | |
| Jan 24, 2021 at 20:43 | comment | added | JHM | From Proposition 2, you conclude the tangent bundle of $\phi(U)$ is isomorphic with tangent bundle of $\phi(U)$, i.e. homeomorphism on the base and fibrewise linearly isomorphic. But the isomorphism needs not transport $j_0$ to $j_1$, i.e. inequality generally holds $\psi_* \circ j_0 \neq j_1 \circ \psi_*$ along the fibres. | |
| Jan 24, 2021 at 20:33 | comment | added | dhy | Oops, I misunderstood. I think the actual problem is more fundamental: You indeed get an isomorphism $E_0\cong E_1.$ I don't see how you lift this to an isomorphism $(\phi(U),j_0)\cong(\phi(U),j_1).$ (Indeed, note that the latter cannot exist in general: $j_1$ is integrable and $j_0$ need not be.) | |
| Jan 24, 2021 at 20:16 | comment | added | Anthony D'Arienzo | The Darboux charts are pseudoholomorphic with respect to $j_0$. Shouldn’t this tell us that the transition maps are holomorphic? | |
| Jan 24, 2021 at 20:11 | comment | added | dhy | I don't think this proof guarantees that the transition functions are holomorphic, no? | |
| Jan 24, 2021 at 20:02 | review | First posts | |||
| Jan 24, 2021 at 20:11 | |||||
| Jan 24, 2021 at 20:00 | history | asked | Anthony D'Arienzo | CC BY-SA 4.0 |