Compatibility of coordinate charts
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I know that this is probably not the conventional way to define things, but my lecturer defined coordinate charts on a topological manifolds as maps mapping from a Cartesian space into the manifold.
Now he defined compatibility by saying:
Two charts $(U, phi), (V, psi)$ on X (the manifold) are compatible if
$$phi^{-1} circ psi: phi^{-1}(psi(V)) rightarrow psi^{-1}(phi(U))$$ is a smooth map of open subsets of $R^n$.
I see that this is a sensible definition, although I'm a bit unsure if this definition implies that the the other coordinate change map is also smooth in the same way. Is this true? How does this follow?
geometry manifolds
asked Dec 15 '18 at 13:45
gengen
4572521
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add a comment |
$begingroup$
I know that this is probably not the conventional way to define things, but my lecturer defined coordinate charts on a topological manifolds as maps mapping from a Cartesian space into the manifold.
Now he defined compatibility by saying:
Two charts $(U, phi), (V, psi)$ on X (the manifold) are compatible if
$$phi^{-1} circ psi: phi^{-1}(psi(V)) rightarrow psi^{-1}(phi(U))$$ is a smooth map of open subsets of $R^n$.
I see that this is a sensible definition, although I'm a bit unsure if this definition implies that the the other coordinate change map is also smooth in the same way. Is this true? How does this follow?
geometry manifolds
asked Dec 15 '18 at 13:45
gengen
4572521
$endgroup$
add a comment |
$begingroup$
I know that this is probably not the conventional way to define things, but my lecturer defined coordinate charts on a topological manifolds as maps mapping from a Cartesian space into the manifold.
Now he defined compatibility by saying:
Two charts $(U, phi), (V, psi)$ on X (the manifold) are compatible if
$$phi^{-1} circ psi: phi^{-1}(psi(V)) rightarrow psi^{-1}(phi(U))$$ is a smooth map of open subsets of $R^n$.
I see that this is a sensible definition, although I'm a bit unsure if this definition implies that the the other coordinate change map is also smooth in the same way. Is this true? How does this follow?
geometry manifolds
asked Dec 15 '18 at 13:45
gengen
4572521
$endgroup$
I know that this is probably not the conventional way to define things, but my lecturer defined coordinate charts on a topological manifolds as maps mapping from a Cartesian space into the manifold.
Now he defined compatibility by saying:
Two charts $(U, phi), (V, psi)$ on X (the manifold) are compatible if
$$phi^{-1} circ psi: phi^{-1}(psi(V)) rightarrow psi^{-1}(phi(U))$$ is a smooth map of open subsets of $R^n$.
I see that this is a sensible definition, although I'm a bit unsure if this definition implies that the the other coordinate change map is also smooth in the same way. Is this true? How does this follow?
geometry manifolds
geometry manifolds
asked Dec 15 '18 at 13:45
gengen
4572521
asked Dec 15 '18 at 13:45
gengen
4572521
asked Dec 15 '18 at 13:45
gengen
4572521
asked Dec 15 '18 at 13:45
gengen
4572521
asked Dec 15 '18 at 13:45
gengen
4572521
4572521
add a comment |
add a comment |
1 Answer
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$begingroup$
I know that this is probably not the conventional way to define things
It is, except your recounting of it here has left off a couple details:
my lecturer defined coordinate charts on a topological manifolds as maps mapping from a Cartesian space into the manifold.
More accurately, your lecturer defined coordinate charts as homeomorphisms between open sets of $Bbb R^n$ and of the manifold. (It is possible that your lecturer defines the topology on the manifold by requiring it makes the selected charts to be homeomorphisms, but that approach requires more axioms, so it is far more common just to assume that the manifold has a topology that the charts must be compatible with.)
Therefore, if $phi, psi$ are charts, $phi^{-1}$ and $psi^{-1}$ are both homeomorphisms as well. If the images of $phi$ and $psi$ overlap, then $f = phi^{-1} circ psi$ is a homeomorphism between two open sets in $Bbb R^n$. In particular, $f$ is injective. If $f$ is smooth, then by the Inverse Function Theorem, $f^{-1}$ will be as well, because "smoothness" not only requires the derivatives to exist (to whatever level of smoothness you desire), but also to be non-singular. Thus $phi$ is compatible with $psi$ if and only if $psi$ is compatible with $phi$.
answered Dec 15 '18 at 21:47
Paul SinclairPaul Sinclair
19.7k21442
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$begingroup$
I know that this is probably not the conventional way to define things
It is, except your recounting of it here has left off a couple details:
my lecturer defined coordinate charts on a topological manifolds as maps mapping from a Cartesian space into the manifold.
More accurately, your lecturer defined coordinate charts as homeomorphisms between open sets of $Bbb R^n$ and of the manifold. (It is possible that your lecturer defines the topology on the manifold by requiring it makes the selected charts to be homeomorphisms, but that approach requires more axioms, so it is far more common just to assume that the manifold has a topology that the charts must be compatible with.)
Therefore, if $phi, psi$ are charts, $phi^{-1}$ and $psi^{-1}$ are both homeomorphisms as well. If the images of $phi$ and $psi$ overlap, then $f = phi^{-1} circ psi$ is a homeomorphism between two open sets in $Bbb R^n$. In particular, $f$ is injective. If $f$ is smooth, then by the Inverse Function Theorem, $f^{-1}$ will be as well, because "smoothness" not only requires the derivatives to exist (to whatever level of smoothness you desire), but also to be non-singular. Thus $phi$ is compatible with $psi$ if and only if $psi$ is compatible with $phi$.
answered Dec 15 '18 at 21:47
Paul SinclairPaul Sinclair
19.7k21442
$endgroup$
add a comment |
$begingroup$
I know that this is probably not the conventional way to define things
It is, except your recounting of it here has left off a couple details:
my lecturer defined coordinate charts on a topological manifolds as maps mapping from a Cartesian space into the manifold.
More accurately, your lecturer defined coordinate charts as homeomorphisms between open sets of $Bbb R^n$ and of the manifold. (It is possible that your lecturer defines the topology on the manifold by requiring it makes the selected charts to be homeomorphisms, but that approach requires more axioms, so it is far more common just to assume that the manifold has a topology that the charts must be compatible with.)
Therefore, if $phi, psi$ are charts, $phi^{-1}$ and $psi^{-1}$ are both homeomorphisms as well. If the images of $phi$ and $psi$ overlap, then $f = phi^{-1} circ psi$ is a homeomorphism between two open sets in $Bbb R^n$. In particular, $f$ is injective. If $f$ is smooth, then by the Inverse Function Theorem, $f^{-1}$ will be as well, because "smoothness" not only requires the derivatives to exist (to whatever level of smoothness you desire), but also to be non-singular. Thus $phi$ is compatible with $psi$ if and only if $psi$ is compatible with $phi$.
answered Dec 15 '18 at 21:47
Paul SinclairPaul Sinclair
19.7k21442
$endgroup$
add a comment |
$begingroup$
I know that this is probably not the conventional way to define things
It is, except your recounting of it here has left off a couple details:
my lecturer defined coordinate charts on a topological manifolds as maps mapping from a Cartesian space into the manifold.
More accurately, your lecturer defined coordinate charts as homeomorphisms between open sets of $Bbb R^n$ and of the manifold. (It is possible that your lecturer defines the topology on the manifold by requiring it makes the selected charts to be homeomorphisms, but that approach requires more axioms, so it is far more common just to assume that the manifold has a topology that the charts must be compatible with.)
Therefore, if $phi, psi$ are charts, $phi^{-1}$ and $psi^{-1}$ are both homeomorphisms as well. If the images of $phi$ and $psi$ overlap, then $f = phi^{-1} circ psi$ is a homeomorphism between two open sets in $Bbb R^n$. In particular, $f$ is injective. If $f$ is smooth, then by the Inverse Function Theorem, $f^{-1}$ will be as well, because "smoothness" not only requires the derivatives to exist (to whatever level of smoothness you desire), but also to be non-singular. Thus $phi$ is compatible with $psi$ if and only if $psi$ is compatible with $phi$.
answered Dec 15 '18 at 21:47
Paul SinclairPaul Sinclair
19.7k21442
$endgroup$
I know that this is probably not the conventional way to define things
It is, except your recounting of it here has left off a couple details:
my lecturer defined coordinate charts on a topological manifolds as maps mapping from a Cartesian space into the manifold.
More accurately, your lecturer defined coordinate charts as homeomorphisms between open sets of $Bbb R^n$ and of the manifold. (It is possible that your lecturer defines the topology on the manifold by requiring it makes the selected charts to be homeomorphisms, but that approach requires more axioms, so it is far more common just to assume that the manifold has a topology that the charts must be compatible with.)
Therefore, if $phi, psi$ are charts, $phi^{-1}$ and $psi^{-1}$ are both homeomorphisms as well. If the images of $phi$ and $psi$ overlap, then $f = phi^{-1} circ psi$ is a homeomorphism between two open sets in $Bbb R^n$. In particular, $f$ is injective. If $f$ is smooth, then by the Inverse Function Theorem, $f^{-1}$ will be as well, because "smoothness" not only requires the derivatives to exist (to whatever level of smoothness you desire), but also to be non-singular. Thus $phi$ is compatible with $psi$ if and only if $psi$ is compatible with $phi$.
answered Dec 15 '18 at 21:47
Paul SinclairPaul Sinclair
19.7k21442
answered Dec 15 '18 at 21:47
Paul SinclairPaul Sinclair
19.7k21442
answered Dec 15 '18 at 21:47
Paul SinclairPaul Sinclair
19.7k21442
answered Dec 15 '18 at 21:47
Paul SinclairPaul Sinclair
19.7k21442
19.7k21442
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add a comment |
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