Mixing
Can't understand the definition of equivalence of topological atlas.
$begingroup$
Wikipedia said
The description of most manifolds requires more than one chart (a single chart is adequate for only the simplest manifolds). A specific collection of charts which covers a manifold is called an atlas. An atlas is not unique as all manifolds can be covered multiple ways using different combinations of charts. Two atlases are said to be equivalent if their union is also an atlas.
I think it means: Let $X$ be a topological space and $Phi_1={phi_alpha:U_{alpha}toBbb R^{n_alpha}midalphain A}$, $Phi_2={phi_beta:U_{beta}toBbb R^{n_beta}midbetain B}$ be two of its(i.e., $X$'s) topological atlas. Then $Phi_1$ and $Phi_2$ are equivalent iff $Phi_1cupPhi_2$ is also a topological atlas of $X$.
My question is: if $Phi_1$ and $Phi_2$ are atlases of $X$, which means they both can "cover" $X$, then it is definitely true that $Phi_1cupPhi_2$ can "cover" $X$, isn't it? I can't see the meaning of such definition of equivalence. Where did I make the mistake?
manifolds differential-topology
asked Jan 30 at 3:36
EricEric
1,803615
$endgroup$
add a comment |
$begingroup$
Wikipedia said
The description of most manifolds requires more than one chart (a single chart is adequate for only the simplest manifolds). A specific collection of charts which covers a manifold is called an atlas. An atlas is not unique as all manifolds can be covered multiple ways using different combinations of charts. Two atlases are said to be equivalent if their union is also an atlas.
I think it means: Let $X$ be a topological space and $Phi_1={phi_alpha:U_{alpha}toBbb R^{n_alpha}midalphain A}$, $Phi_2={phi_beta:U_{beta}toBbb R^{n_beta}midbetain B}$ be two of its(i.e., $X$'s) topological atlas. Then $Phi_1$ and $Phi_2$ are equivalent iff $Phi_1cupPhi_2$ is also a topological atlas of $X$.
My question is: if $Phi_1$ and $Phi_2$ are atlases of $X$, which means they both can "cover" $X$, then it is definitely true that $Phi_1cupPhi_2$ can "cover" $X$, isn't it? I can't see the meaning of such definition of equivalence. Where did I make the mistake?
manifolds differential-topology
asked Jan 30 at 3:36
EricEric
1,803615
$endgroup$
1
$begingroup$
I found two links that may help you: math.stackexchange.com/questions/2930261/… and math.stackexchange.com/questions/1717244/…
$endgroup$
– Sujit Bhattacharyya
Jan 30 at 3:47
1
$begingroup$
They still cover, but do they still have compatible overlaps?
$endgroup$
– Randall
Jan 30 at 3:50
1
$begingroup$
Normally one doesn't speak of a topological atlas for this reason. Instead, the point of talking about atlases is to consider smooth atlases which have an additional requirement which makes things less trivial.
$endgroup$
– Eric Wofsey
Jan 30 at 3:51
add a comment |
$begingroup$
Wikipedia said
The description of most manifolds requires more than one chart (a single chart is adequate for only the simplest manifolds). A specific collection of charts which covers a manifold is called an atlas. An atlas is not unique as all manifolds can be covered multiple ways using different combinations of charts. Two atlases are said to be equivalent if their union is also an atlas.
I think it means: Let $X$ be a topological space and $Phi_1={phi_alpha:U_{alpha}toBbb R^{n_alpha}midalphain A}$, $Phi_2={phi_beta:U_{beta}toBbb R^{n_beta}midbetain B}$ be two of its(i.e., $X$'s) topological atlas. Then $Phi_1$ and $Phi_2$ are equivalent iff $Phi_1cupPhi_2$ is also a topological atlas of $X$.
My question is: if $Phi_1$ and $Phi_2$ are atlases of $X$, which means they both can "cover" $X$, then it is definitely true that $Phi_1cupPhi_2$ can "cover" $X$, isn't it? I can't see the meaning of such definition of equivalence. Where did I make the mistake?
manifolds differential-topology
asked Jan 30 at 3:36
EricEric
1,803615
$endgroup$
Wikipedia said
The description of most manifolds requires more than one chart (a single chart is adequate for only the simplest manifolds). A specific collection of charts which covers a manifold is called an atlas. An atlas is not unique as all manifolds can be covered multiple ways using different combinations of charts. Two atlases are said to be equivalent if their union is also an atlas.
I think it means: Let $X$ be a topological space and $Phi_1={phi_alpha:U_{alpha}toBbb R^{n_alpha}midalphain A}$, $Phi_2={phi_beta:U_{beta}toBbb R^{n_beta}midbetain B}$ be two of its(i.e., $X$'s) topological atlas. Then $Phi_1$ and $Phi_2$ are equivalent iff $Phi_1cupPhi_2$ is also a topological atlas of $X$.
My question is: if $Phi_1$ and $Phi_2$ are atlases of $X$, which means they both can "cover" $X$, then it is definitely true that $Phi_1cupPhi_2$ can "cover" $X$, isn't it? I can't see the meaning of such definition of equivalence. Where did I make the mistake?
manifolds differential-topology
manifolds differential-topology
asked Jan 30 at 3:36
EricEric
1,803615
asked Jan 30 at 3:36
EricEric
1,803615
asked Jan 30 at 3:36
EricEric
1,803615
asked Jan 30 at 3:36
EricEric
1,803615
asked Jan 30 at 3:36
EricEric
1,803615
1,803615
1
$begingroup$
I found two links that may help you: math.stackexchange.com/questions/2930261/… and math.stackexchange.com/questions/1717244/…
$endgroup$
– Sujit Bhattacharyya
Jan 30 at 3:47
1
$begingroup$
They still cover, but do they still have compatible overlaps?
$endgroup$
– Randall
Jan 30 at 3:50
1
$begingroup$
Normally one doesn't speak of a topological atlas for this reason. Instead, the point of talking about atlases is to consider smooth atlases which have an additional requirement which makes things less trivial.
$endgroup$
– Eric Wofsey
Jan 30 at 3:51
add a comment |
1
$begingroup$
I found two links that may help you: math.stackexchange.com/questions/2930261/… and math.stackexchange.com/questions/1717244/…
$endgroup$
– Sujit Bhattacharyya
Jan 30 at 3:47
1
$begingroup$
They still cover, but do they still have compatible overlaps?
$endgroup$
– Randall
Jan 30 at 3:50
1
$begingroup$
Normally one doesn't speak of a topological atlas for this reason. Instead, the point of talking about atlases is to consider smooth atlases which have an additional requirement which makes things less trivial.
$endgroup$
– Eric Wofsey
Jan 30 at 3:51
1
1
$begingroup$
I found two links that may help you: math.stackexchange.com/questions/2930261/… and math.stackexchange.com/questions/1717244/…
$endgroup$
– Sujit Bhattacharyya
Jan 30 at 3:47
$begingroup$
I found two links that may help you: math.stackexchange.com/questions/2930261/… and math.stackexchange.com/questions/1717244/…
$endgroup$
– Sujit Bhattacharyya
Jan 30 at 3:47
1
1
$begingroup$
They still cover, but do they still have compatible overlaps?
$endgroup$
– Randall
Jan 30 at 3:50
$begingroup$
They still cover, but do they still have compatible overlaps?
$endgroup$
– Randall
Jan 30 at 3:50
1
1
$begingroup$
Normally one doesn't speak of a topological atlas for this reason. Instead, the point of talking about atlases is to consider smooth atlases which have an additional requirement which makes things less trivial.
$endgroup$
– Eric Wofsey
Jan 30 at 3:51
$begingroup$
Normally one doesn't speak of a topological atlas for this reason. Instead, the point of talking about atlases is to consider smooth atlases which have an additional requirement which makes things less trivial.
$endgroup$
– Eric Wofsey
Jan 30 at 3:51
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
It's not the covering that's the (potential) problem, it's the compatibility, but actually this won't be an issue for topological atlases.
Let $U_{alpha_1alpha_2} = U_{alpha_1} cap U_{alpha_2}$. For the two charts $phi_{alpha_1}, phi_{alpha_2}$ to be compatible, you need $$tau_{1,2}=phi_{alpha_2} circ phi_{alpha_1}^{-1}: phi_{alpha_1}(U_{alpha_1alpha_2}) to phi_{alpha_2}(U_{alpha_1alpha_2})$$ to be continuous (in fact a homeomorphism, but swapping the indices gives the continuous inverse). But this is automatic, since both $phi_{alpha_1}$ and $phi_{alpha_2}$ are generally required to be homeomorphisms.
The trouble comes in when you want to look at atlases with more structure. For example, a "smooth atlas" is one where you require the transition maps $tau_{i,j}$ to be smooth, not just continuous. Then there is the possiblity that two charts will be incompatible.
answered Jan 30 at 3:50
Nathaniel MayerNathaniel Mayer
1,863516
$endgroup$
$begingroup$
Thanks! Very clear. Instantly clarify my confusions. Good teacher!
$endgroup$
– Eric
Jan 30 at 13:43
add a comment |
$begingroup$
Wikipedia's brief overview is not actually the definition of an atlas (or, well, I suppose it might be the definition of a "topological atlas", but no one ever uses those). The full definition of a (smooth) atlas has a crucial extra requirement: for each $phi:Utomathbb{R}^n$ and $psi:Vtomathbb{R}^n$ in the atlas, the "transition map" $psicircphi^{-1}:phi(Ucap V)topsi(Ucap V)$ is required to be smooth. Here $phi(Ucap V)$ and $psi(Ucap V)$ are open subsets of $mathbb{R}^n$, so it makes sense to talk about a smooth (i.e., $C^infty$) function between them.
So, if $Phi_1$ and $Phi_2$ are atlases, the union $Phi_1cupPhi_2$ may not be an atlas, since the transition map between a chart in $Phi_1$ and a chart in $Phi_2$ may not be smooth.
answered Jan 30 at 3:59
Eric WofseyEric Wofsey
192k14217350
$endgroup$
$begingroup$
Thanks a lot! Btw did you mean that the word "topological atlas" was not common in literature?
$endgroup$
– Eric
Jan 30 at 13:46
add a comment |
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2 Answers
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2 Answers
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$begingroup$
It's not the covering that's the (potential) problem, it's the compatibility, but actually this won't be an issue for topological atlases.
Let $U_{alpha_1alpha_2} = U_{alpha_1} cap U_{alpha_2}$. For the two charts $phi_{alpha_1}, phi_{alpha_2}$ to be compatible, you need $$tau_{1,2}=phi_{alpha_2} circ phi_{alpha_1}^{-1}: phi_{alpha_1}(U_{alpha_1alpha_2}) to phi_{alpha_2}(U_{alpha_1alpha_2})$$ to be continuous (in fact a homeomorphism, but swapping the indices gives the continuous inverse). But this is automatic, since both $phi_{alpha_1}$ and $phi_{alpha_2}$ are generally required to be homeomorphisms.
The trouble comes in when you want to look at atlases with more structure. For example, a "smooth atlas" is one where you require the transition maps $tau_{i,j}$ to be smooth, not just continuous. Then there is the possiblity that two charts will be incompatible.
answered Jan 30 at 3:50
Nathaniel MayerNathaniel Mayer
1,863516
$endgroup$
$begingroup$
Thanks! Very clear. Instantly clarify my confusions. Good teacher!
$endgroup$
– Eric
Jan 30 at 13:43
add a comment |
$begingroup$
It's not the covering that's the (potential) problem, it's the compatibility, but actually this won't be an issue for topological atlases.
Let $U_{alpha_1alpha_2} = U_{alpha_1} cap U_{alpha_2}$. For the two charts $phi_{alpha_1}, phi_{alpha_2}$ to be compatible, you need $$tau_{1,2}=phi_{alpha_2} circ phi_{alpha_1}^{-1}: phi_{alpha_1}(U_{alpha_1alpha_2}) to phi_{alpha_2}(U_{alpha_1alpha_2})$$ to be continuous (in fact a homeomorphism, but swapping the indices gives the continuous inverse). But this is automatic, since both $phi_{alpha_1}$ and $phi_{alpha_2}$ are generally required to be homeomorphisms.
The trouble comes in when you want to look at atlases with more structure. For example, a "smooth atlas" is one where you require the transition maps $tau_{i,j}$ to be smooth, not just continuous. Then there is the possiblity that two charts will be incompatible.
answered Jan 30 at 3:50
Nathaniel MayerNathaniel Mayer
1,863516
$endgroup$
$begingroup$
Thanks! Very clear. Instantly clarify my confusions. Good teacher!
$endgroup$
– Eric
Jan 30 at 13:43
add a comment |
$begingroup$
It's not the covering that's the (potential) problem, it's the compatibility, but actually this won't be an issue for topological atlases.
Let $U_{alpha_1alpha_2} = U_{alpha_1} cap U_{alpha_2}$. For the two charts $phi_{alpha_1}, phi_{alpha_2}$ to be compatible, you need $$tau_{1,2}=phi_{alpha_2} circ phi_{alpha_1}^{-1}: phi_{alpha_1}(U_{alpha_1alpha_2}) to phi_{alpha_2}(U_{alpha_1alpha_2})$$ to be continuous (in fact a homeomorphism, but swapping the indices gives the continuous inverse). But this is automatic, since both $phi_{alpha_1}$ and $phi_{alpha_2}$ are generally required to be homeomorphisms.
The trouble comes in when you want to look at atlases with more structure. For example, a "smooth atlas" is one where you require the transition maps $tau_{i,j}$ to be smooth, not just continuous. Then there is the possiblity that two charts will be incompatible.
answered Jan 30 at 3:50
Nathaniel MayerNathaniel Mayer
1,863516
$endgroup$
It's not the covering that's the (potential) problem, it's the compatibility, but actually this won't be an issue for topological atlases.
Let $U_{alpha_1alpha_2} = U_{alpha_1} cap U_{alpha_2}$. For the two charts $phi_{alpha_1}, phi_{alpha_2}$ to be compatible, you need $$tau_{1,2}=phi_{alpha_2} circ phi_{alpha_1}^{-1}: phi_{alpha_1}(U_{alpha_1alpha_2}) to phi_{alpha_2}(U_{alpha_1alpha_2})$$ to be continuous (in fact a homeomorphism, but swapping the indices gives the continuous inverse). But this is automatic, since both $phi_{alpha_1}$ and $phi_{alpha_2}$ are generally required to be homeomorphisms.
The trouble comes in when you want to look at atlases with more structure. For example, a "smooth atlas" is one where you require the transition maps $tau_{i,j}$ to be smooth, not just continuous. Then there is the possiblity that two charts will be incompatible.
answered Jan 30 at 3:50
Nathaniel MayerNathaniel Mayer
1,863516
answered Jan 30 at 3:50
Nathaniel MayerNathaniel Mayer
1,863516
answered Jan 30 at 3:50
Nathaniel MayerNathaniel Mayer
1,863516
answered Jan 30 at 3:50
Nathaniel MayerNathaniel Mayer
1,863516
1,863516
$begingroup$
Thanks! Very clear. Instantly clarify my confusions. Good teacher!
$endgroup$
– Eric
Jan 30 at 13:43
add a comment |
$begingroup$
Thanks! Very clear. Instantly clarify my confusions. Good teacher!
$endgroup$
– Eric
Jan 30 at 13:43
$begingroup$
Thanks! Very clear. Instantly clarify my confusions. Good teacher!
$endgroup$
– Eric
Jan 30 at 13:43
$begingroup$
Thanks! Very clear. Instantly clarify my confusions. Good teacher!
$endgroup$
– Eric
Jan 30 at 13:43
add a comment |
$begingroup$
Wikipedia's brief overview is not actually the definition of an atlas (or, well, I suppose it might be the definition of a "topological atlas", but no one ever uses those). The full definition of a (smooth) atlas has a crucial extra requirement: for each $phi:Utomathbb{R}^n$ and $psi:Vtomathbb{R}^n$ in the atlas, the "transition map" $psicircphi^{-1}:phi(Ucap V)topsi(Ucap V)$ is required to be smooth. Here $phi(Ucap V)$ and $psi(Ucap V)$ are open subsets of $mathbb{R}^n$, so it makes sense to talk about a smooth (i.e., $C^infty$) function between them.
So, if $Phi_1$ and $Phi_2$ are atlases, the union $Phi_1cupPhi_2$ may not be an atlas, since the transition map between a chart in $Phi_1$ and a chart in $Phi_2$ may not be smooth.
answered Jan 30 at 3:59
Eric WofseyEric Wofsey
192k14217350
$endgroup$
$begingroup$
Thanks a lot! Btw did you mean that the word "topological atlas" was not common in literature?
$endgroup$
– Eric
Jan 30 at 13:46
add a comment |
$begingroup$
Wikipedia's brief overview is not actually the definition of an atlas (or, well, I suppose it might be the definition of a "topological atlas", but no one ever uses those). The full definition of a (smooth) atlas has a crucial extra requirement: for each $phi:Utomathbb{R}^n$ and $psi:Vtomathbb{R}^n$ in the atlas, the "transition map" $psicircphi^{-1}:phi(Ucap V)topsi(Ucap V)$ is required to be smooth. Here $phi(Ucap V)$ and $psi(Ucap V)$ are open subsets of $mathbb{R}^n$, so it makes sense to talk about a smooth (i.e., $C^infty$) function between them.
So, if $Phi_1$ and $Phi_2$ are atlases, the union $Phi_1cupPhi_2$ may not be an atlas, since the transition map between a chart in $Phi_1$ and a chart in $Phi_2$ may not be smooth.
answered Jan 30 at 3:59
Eric WofseyEric Wofsey
192k14217350
$endgroup$
$begingroup$
Thanks a lot! Btw did you mean that the word "topological atlas" was not common in literature?
$endgroup$
– Eric
Jan 30 at 13:46
add a comment |
$begingroup$
Wikipedia's brief overview is not actually the definition of an atlas (or, well, I suppose it might be the definition of a "topological atlas", but no one ever uses those). The full definition of a (smooth) atlas has a crucial extra requirement: for each $phi:Utomathbb{R}^n$ and $psi:Vtomathbb{R}^n$ in the atlas, the "transition map" $psicircphi^{-1}:phi(Ucap V)topsi(Ucap V)$ is required to be smooth. Here $phi(Ucap V)$ and $psi(Ucap V)$ are open subsets of $mathbb{R}^n$, so it makes sense to talk about a smooth (i.e., $C^infty$) function between them.
So, if $Phi_1$ and $Phi_2$ are atlases, the union $Phi_1cupPhi_2$ may not be an atlas, since the transition map between a chart in $Phi_1$ and a chart in $Phi_2$ may not be smooth.
answered Jan 30 at 3:59
Eric WofseyEric Wofsey
192k14217350
$endgroup$
Wikipedia's brief overview is not actually the definition of an atlas (or, well, I suppose it might be the definition of a "topological atlas", but no one ever uses those). The full definition of a (smooth) atlas has a crucial extra requirement: for each $phi:Utomathbb{R}^n$ and $psi:Vtomathbb{R}^n$ in the atlas, the "transition map" $psicircphi^{-1}:phi(Ucap V)topsi(Ucap V)$ is required to be smooth. Here $phi(Ucap V)$ and $psi(Ucap V)$ are open subsets of $mathbb{R}^n$, so it makes sense to talk about a smooth (i.e., $C^infty$) function between them.
So, if $Phi_1$ and $Phi_2$ are atlases, the union $Phi_1cupPhi_2$ may not be an atlas, since the transition map between a chart in $Phi_1$ and a chart in $Phi_2$ may not be smooth.
answered Jan 30 at 3:59
Eric WofseyEric Wofsey
192k14217350
answered Jan 30 at 3:59
Eric WofseyEric Wofsey
192k14217350
answered Jan 30 at 3:59
Eric WofseyEric Wofsey
192k14217350
answered Jan 30 at 3:59
Eric WofseyEric Wofsey
192k14217350
192k14217350
$begingroup$
Thanks a lot! Btw did you mean that the word "topological atlas" was not common in literature?
$endgroup$
– Eric
Jan 30 at 13:46
add a comment |
$begingroup$
Thanks a lot! Btw did you mean that the word "topological atlas" was not common in literature?
$endgroup$
– Eric
Jan 30 at 13:46
$begingroup$
Thanks a lot! Btw did you mean that the word "topological atlas" was not common in literature?
$endgroup$
– Eric
Jan 30 at 13:46
$begingroup$
Thanks a lot! Btw did you mean that the word "topological atlas" was not common in literature?
$endgroup$
– Eric
Jan 30 at 13:46
add a comment |
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1
$begingroup$
I found two links that may help you: math.stackexchange.com/questions/2930261/… and math.stackexchange.com/questions/1717244/…
$endgroup$
– Sujit Bhattacharyya
Jan 30 at 3:47
1
$begingroup$
They still cover, but do they still have compatible overlaps?
$endgroup$
– Randall
Jan 30 at 3:50
1
$begingroup$
Normally one doesn't speak of a topological atlas for this reason. Instead, the point of talking about atlases is to consider smooth atlases which have an additional requirement which makes things less trivial.
$endgroup$
– Eric Wofsey
Jan 30 at 3:51