Different almost-complex structures $Rightarrow$ different complex structures?
Let $(M,J)$ be an almost-complex structure. By definition, if $M$ admits local holomorphic coordinates for $J$ around every point and these patch together to form a holomorphic atlas (complex structure) for $M$, then $J$ is said to be integrable.
I was able to verify that a complex manifold $M$ can admit many almost-complex structures.
But I couldn't verify this: given different integrable almost-complex structures $J_1,J_2$, can we say that they integrate to different complex structures?
differential-geometry complex-geometry connections
add a comment |
Let $(M,J)$ be an almost-complex structure. By definition, if $M$ admits local holomorphic coordinates for $J$ around every point and these patch together to form a holomorphic atlas (complex structure) for $M$, then $J$ is said to be integrable.
I was able to verify that a complex manifold $M$ can admit many almost-complex structures.
But I couldn't verify this: given different integrable almost-complex structures $J_1,J_2$, can we say that they integrate to different complex structures?
differential-geometry complex-geometry connections
1
certainly, right? If the complex structures were identical then the collection of charts would be identical, and the derivative of multiplication-by-$i$ in a chart is how you define the $J_i$.
– Mike Miller
Nov 26 at 20:12
@MikeMiller What I mean by "complex structures ${(U_i,varphi_i)}$ and ${(V_j,psi_j)}$ being the same" is that $varphi_icircpsi_j^{-1}$ is holomorphic for every $i,j$. I don't understand what you said about multiplication by $i$ defining $J_i$, since $J_1,J_2$ are already given.
– rmdmc89
Nov 26 at 20:28
2
You may as well pass to a maximal atlas, which means that both atlases contain exactly the same charts. For your charts to integrate your given almost complex structure means that, on each chart, the map $J$ is the derivative of multiplication-by-$i$.
– Mike Miller
Nov 26 at 20:38
add a comment |
Let $(M,J)$ be an almost-complex structure. By definition, if $M$ admits local holomorphic coordinates for $J$ around every point and these patch together to form a holomorphic atlas (complex structure) for $M$, then $J$ is said to be integrable.
I was able to verify that a complex manifold $M$ can admit many almost-complex structures.
But I couldn't verify this: given different integrable almost-complex structures $J_1,J_2$, can we say that they integrate to different complex structures?
differential-geometry complex-geometry connections
Let $(M,J)$ be an almost-complex structure. By definition, if $M$ admits local holomorphic coordinates for $J$ around every point and these patch together to form a holomorphic atlas (complex structure) for $M$, then $J$ is said to be integrable.
I was able to verify that a complex manifold $M$ can admit many almost-complex structures.
But I couldn't verify this: given different integrable almost-complex structures $J_1,J_2$, can we say that they integrate to different complex structures?
differential-geometry complex-geometry connections
differential-geometry complex-geometry connections
asked Nov 26 at 18:03
rmdmc89
2,0511921
2,0511921
1
certainly, right? If the complex structures were identical then the collection of charts would be identical, and the derivative of multiplication-by-$i$ in a chart is how you define the $J_i$.
– Mike Miller
Nov 26 at 20:12
@MikeMiller What I mean by "complex structures ${(U_i,varphi_i)}$ and ${(V_j,psi_j)}$ being the same" is that $varphi_icircpsi_j^{-1}$ is holomorphic for every $i,j$. I don't understand what you said about multiplication by $i$ defining $J_i$, since $J_1,J_2$ are already given.
– rmdmc89
Nov 26 at 20:28
2
You may as well pass to a maximal atlas, which means that both atlases contain exactly the same charts. For your charts to integrate your given almost complex structure means that, on each chart, the map $J$ is the derivative of multiplication-by-$i$.
– Mike Miller
Nov 26 at 20:38
add a comment |
1
certainly, right? If the complex structures were identical then the collection of charts would be identical, and the derivative of multiplication-by-$i$ in a chart is how you define the $J_i$.
– Mike Miller
Nov 26 at 20:12
@MikeMiller What I mean by "complex structures ${(U_i,varphi_i)}$ and ${(V_j,psi_j)}$ being the same" is that $varphi_icircpsi_j^{-1}$ is holomorphic for every $i,j$. I don't understand what you said about multiplication by $i$ defining $J_i$, since $J_1,J_2$ are already given.
– rmdmc89
Nov 26 at 20:28
2
You may as well pass to a maximal atlas, which means that both atlases contain exactly the same charts. For your charts to integrate your given almost complex structure means that, on each chart, the map $J$ is the derivative of multiplication-by-$i$.
– Mike Miller
Nov 26 at 20:38
1
1
certainly, right? If the complex structures were identical then the collection of charts would be identical, and the derivative of multiplication-by-$i$ in a chart is how you define the $J_i$.
– Mike Miller
Nov 26 at 20:12
certainly, right? If the complex structures were identical then the collection of charts would be identical, and the derivative of multiplication-by-$i$ in a chart is how you define the $J_i$.
– Mike Miller
Nov 26 at 20:12
@MikeMiller What I mean by "complex structures ${(U_i,varphi_i)}$ and ${(V_j,psi_j)}$ being the same" is that $varphi_icircpsi_j^{-1}$ is holomorphic for every $i,j$. I don't understand what you said about multiplication by $i$ defining $J_i$, since $J_1,J_2$ are already given.
– rmdmc89
Nov 26 at 20:28
@MikeMiller What I mean by "complex structures ${(U_i,varphi_i)}$ and ${(V_j,psi_j)}$ being the same" is that $varphi_icircpsi_j^{-1}$ is holomorphic for every $i,j$. I don't understand what you said about multiplication by $i$ defining $J_i$, since $J_1,J_2$ are already given.
– rmdmc89
Nov 26 at 20:28
2
2
You may as well pass to a maximal atlas, which means that both atlases contain exactly the same charts. For your charts to integrate your given almost complex structure means that, on each chart, the map $J$ is the derivative of multiplication-by-$i$.
– Mike Miller
Nov 26 at 20:38
You may as well pass to a maximal atlas, which means that both atlases contain exactly the same charts. For your charts to integrate your given almost complex structure means that, on each chart, the map $J$ is the derivative of multiplication-by-$i$.
– Mike Miller
Nov 26 at 20:38
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1
certainly, right? If the complex structures were identical then the collection of charts would be identical, and the derivative of multiplication-by-$i$ in a chart is how you define the $J_i$.
– Mike Miller
Nov 26 at 20:12
@MikeMiller What I mean by "complex structures ${(U_i,varphi_i)}$ and ${(V_j,psi_j)}$ being the same" is that $varphi_icircpsi_j^{-1}$ is holomorphic for every $i,j$. I don't understand what you said about multiplication by $i$ defining $J_i$, since $J_1,J_2$ are already given.
– rmdmc89
Nov 26 at 20:28
2
You may as well pass to a maximal atlas, which means that both atlases contain exactly the same charts. For your charts to integrate your given almost complex structure means that, on each chart, the map $J$ is the derivative of multiplication-by-$i$.
– Mike Miller
Nov 26 at 20:38