the notion of complete atlas
The following picture comes from foundations of differential geometry volume I.
According to the definition of complete atlas, would it be more precise to call it maximal atlas?
Secondly, the statement
Every atlas of $M$ compatible with $Gamma$ is contained in a unique atlas of $M$ compatible with $Gamma$.
can be proved routinely:
Let $mathscr A={mbox{all the atlases of }M mbox{ compatible with }Gammambox{ that contains }A}$, I guess it is a set.
For each chain $mathscr A_m$ in $mathscr A$, let $A_m=cup mathscr A_m$, then it is in $mathscr A$: Suppose $(U_1,phi_1)in A_1in mathscr A_m, (U_2,phi_2)in A_2in mathscr A_m$ and $A_1subset A_2$, then $phi_2circphi_1^{-1}:phi_1(U_1cap U_2)tophi_2(U_1cap U_2) $ is an element of $Gamma$ whenever $U_1cap U_2$ is nonempty, since $(U_1,phi_1)$ and $(U_2,phi_2)$ are both in $A_2$. Hence $A_m$ is an bound of the chain $mathscr A_m$. By Zorn's lemma, there is a maximal element in $mathscr A$.
Can someone verify for me if the proof is correct?
differential-geometry notation manifolds
edited Nov 19 '18 at 5:08
Brahadeesh
6,11742360
asked Nov 19 '18 at 2:41
C.Ding
1,3931321
|
show 1 more comment
The following picture comes from foundations of differential geometry volume I.
According to the definition of complete atlas, would it be more precise to call it maximal atlas?
Secondly, the statement
Every atlas of $M$ compatible with $Gamma$ is contained in a unique atlas of $M$ compatible with $Gamma$.
can be proved routinely:
Let $mathscr A={mbox{all the atlases of }M mbox{ compatible with }Gammambox{ that contains }A}$, I guess it is a set.
For each chain $mathscr A_m$ in $mathscr A$, let $A_m=cup mathscr A_m$, then it is in $mathscr A$: Suppose $(U_1,phi_1)in A_1in mathscr A_m, (U_2,phi_2)in A_2in mathscr A_m$ and $A_1subset A_2$, then $phi_2circphi_1^{-1}:phi_1(U_1cap U_2)tophi_2(U_1cap U_2) $ is an element of $Gamma$ whenever $U_1cap U_2$ is nonempty, since $(U_1,phi_1)$ and $(U_2,phi_2)$ are both in $A_2$. Hence $A_m$ is an bound of the chain $mathscr A_m$. By Zorn's lemma, there is a maximal element in $mathscr A$.
Can someone verify for me if the proof is correct?
differential-geometry notation manifolds
edited Nov 19 '18 at 5:08
Brahadeesh
6,11742360
asked Nov 19 '18 at 2:41
C.Ding
1,3931321
4
Are you asking something?
– Will M.
Nov 19 '18 at 2:45
Oh, I'm asking if it is more precise to call the complete atlas maximal atlas, and if my proof is right.
– C.Ding
Nov 19 '18 at 2:49
2
I knew it as "Saturated Atlas" but sure, maximal atlas is better. Your proof is correct but it is unnecessary to appeal to Zorn's lemma. Just take the union of all atlases compatible with $A$ to get the maximal atlas.
– Will M.
Nov 19 '18 at 2:51
@WillM. But I used the condition of the chain in my proof, how to prove without that?
– C.Ding
Nov 19 '18 at 2:58
@WillM How to prove $phi_2circphi_1^{-1}in Gamma$ without using $A_1subset A_2$?
– C.Ding
Nov 19 '18 at 3:40
|
show 1 more comment
The following picture comes from foundations of differential geometry volume I.
According to the definition of complete atlas, would it be more precise to call it maximal atlas?
Secondly, the statement
Every atlas of $M$ compatible with $Gamma$ is contained in a unique atlas of $M$ compatible with $Gamma$.
can be proved routinely:
Let $mathscr A={mbox{all the atlases of }M mbox{ compatible with }Gammambox{ that contains }A}$, I guess it is a set.
For each chain $mathscr A_m$ in $mathscr A$, let $A_m=cup mathscr A_m$, then it is in $mathscr A$: Suppose $(U_1,phi_1)in A_1in mathscr A_m, (U_2,phi_2)in A_2in mathscr A_m$ and $A_1subset A_2$, then $phi_2circphi_1^{-1}:phi_1(U_1cap U_2)tophi_2(U_1cap U_2) $ is an element of $Gamma$ whenever $U_1cap U_2$ is nonempty, since $(U_1,phi_1)$ and $(U_2,phi_2)$ are both in $A_2$. Hence $A_m$ is an bound of the chain $mathscr A_m$. By Zorn's lemma, there is a maximal element in $mathscr A$.
Can someone verify for me if the proof is correct?
differential-geometry notation manifolds
edited Nov 19 '18 at 5:08
Brahadeesh
6,11742360
asked Nov 19 '18 at 2:41
C.Ding
1,3931321
The following picture comes from foundations of differential geometry volume I.
According to the definition of complete atlas, would it be more precise to call it maximal atlas?
Secondly, the statement
Every atlas of $M$ compatible with $Gamma$ is contained in a unique atlas of $M$ compatible with $Gamma$.
can be proved routinely:
Let $mathscr A={mbox{all the atlases of }M mbox{ compatible with }Gammambox{ that contains }A}$, I guess it is a set.
For each chain $mathscr A_m$ in $mathscr A$, let $A_m=cup mathscr A_m$, then it is in $mathscr A$: Suppose $(U_1,phi_1)in A_1in mathscr A_m, (U_2,phi_2)in A_2in mathscr A_m$ and $A_1subset A_2$, then $phi_2circphi_1^{-1}:phi_1(U_1cap U_2)tophi_2(U_1cap U_2) $ is an element of $Gamma$ whenever $U_1cap U_2$ is nonempty, since $(U_1,phi_1)$ and $(U_2,phi_2)$ are both in $A_2$. Hence $A_m$ is an bound of the chain $mathscr A_m$. By Zorn's lemma, there is a maximal element in $mathscr A$.
Can someone verify for me if the proof is correct?
differential-geometry notation manifolds
differential-geometry notation manifolds
edited Nov 19 '18 at 5:08
Brahadeesh
6,11742360
asked Nov 19 '18 at 2:41
C.Ding
1,3931321
edited Nov 19 '18 at 5:08
Brahadeesh
6,11742360
asked Nov 19 '18 at 2:41
C.Ding
1,3931321
edited Nov 19 '18 at 5:08
Brahadeesh
6,11742360
edited Nov 19 '18 at 5:08
Brahadeesh
6,11742360
edited Nov 19 '18 at 5:08
Brahadeesh
6,11742360
6,11742360
asked Nov 19 '18 at 2:41
C.Ding
1,3931321
asked Nov 19 '18 at 2:41
C.Ding
1,3931321
asked Nov 19 '18 at 2:41
C.Ding
1,3931321
1,3931321
4
Are you asking something?
– Will M.
Nov 19 '18 at 2:45
Oh, I'm asking if it is more precise to call the complete atlas maximal atlas, and if my proof is right.
– C.Ding
Nov 19 '18 at 2:49
2
I knew it as "Saturated Atlas" but sure, maximal atlas is better. Your proof is correct but it is unnecessary to appeal to Zorn's lemma. Just take the union of all atlases compatible with $A$ to get the maximal atlas.
– Will M.
Nov 19 '18 at 2:51
@WillM. But I used the condition of the chain in my proof, how to prove without that?
– C.Ding
Nov 19 '18 at 2:58
@WillM How to prove $phi_2circphi_1^{-1}in Gamma$ without using $A_1subset A_2$?
– C.Ding
Nov 19 '18 at 3:40
|
show 1 more comment
4
Are you asking something?
– Will M.
Nov 19 '18 at 2:45
Oh, I'm asking if it is more precise to call the complete atlas maximal atlas, and if my proof is right.
– C.Ding
Nov 19 '18 at 2:49
2
I knew it as "Saturated Atlas" but sure, maximal atlas is better. Your proof is correct but it is unnecessary to appeal to Zorn's lemma. Just take the union of all atlases compatible with $A$ to get the maximal atlas.
– Will M.
Nov 19 '18 at 2:51
@WillM. But I used the condition of the chain in my proof, how to prove without that?
– C.Ding
Nov 19 '18 at 2:58
@WillM How to prove $phi_2circphi_1^{-1}in Gamma$ without using $A_1subset A_2$?
– C.Ding
Nov 19 '18 at 3:40
4
4
Are you asking something?
– Will M.
Nov 19 '18 at 2:45
Are you asking something?
– Will M.
Nov 19 '18 at 2:45
Oh, I'm asking if it is more precise to call the complete atlas maximal atlas, and if my proof is right.
– C.Ding
Nov 19 '18 at 2:49
Oh, I'm asking if it is more precise to call the complete atlas maximal atlas, and if my proof is right.
– C.Ding
Nov 19 '18 at 2:49
2
2
I knew it as "Saturated Atlas" but sure, maximal atlas is better. Your proof is correct but it is unnecessary to appeal to Zorn's lemma. Just take the union of all atlases compatible with $A$ to get the maximal atlas.
– Will M.
Nov 19 '18 at 2:51
I knew it as "Saturated Atlas" but sure, maximal atlas is better. Your proof is correct but it is unnecessary to appeal to Zorn's lemma. Just take the union of all atlases compatible with $A$ to get the maximal atlas.
– Will M.
Nov 19 '18 at 2:51
@WillM. But I used the condition of the chain in my proof, how to prove without that?
– C.Ding
Nov 19 '18 at 2:58
@WillM. But I used the condition of the chain in my proof, how to prove without that?
– C.Ding
Nov 19 '18 at 2:58
@WillM How to prove $phi_2circphi_1^{-1}in Gamma$ without using $A_1subset A_2$?
– C.Ding
Nov 19 '18 at 3:40
@WillM How to prove $phi_2circphi_1^{-1}in Gamma$ without using $A_1subset A_2$?
– C.Ding
Nov 19 '18 at 3:40
|
show 1 more comment
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4
Are you asking something?
– Will M.
Nov 19 '18 at 2:45
Oh, I'm asking if it is more precise to call the complete atlas maximal atlas, and if my proof is right.
– C.Ding
Nov 19 '18 at 2:49
2
I knew it as "Saturated Atlas" but sure, maximal atlas is better. Your proof is correct but it is unnecessary to appeal to Zorn's lemma. Just take the union of all atlases compatible with $A$ to get the maximal atlas.
– Will M.
Nov 19 '18 at 2:51
@WillM. But I used the condition of the chain in my proof, how to prove without that?
– C.Ding
Nov 19 '18 at 2:58
@WillM How to prove $phi_2circphi_1^{-1}in Gamma$ without using $A_1subset A_2$?
– C.Ding
Nov 19 '18 at 3:40