A question concerning the developing map of (G,X) manifolds
$begingroup$
Let $M$ be a $(G,X)$ manifold, that is we have local charts $(U,varphi_U)$ on $M$ with $varphi_U$ a diffeomorphism onto an open subset of $X$ and the transition maps are locally-$G$.
Let $mathfrak{p}:widetilde{M}rightarrow M$ be the universal covering of $M$.
The developing map theorem introduces a local diffeomorphism $dev:widetilde{M}rightarrow X$.
Does the developing map locally commute with the (restricted) covering and local charts, i.e. for $widetilde{U}$ (small enough) $dev|_widetilde{U}=varphi_{U}circmathfrak{p}|_widetilde{U}$ for some $varphi_U$?
dg.differential-geometry
asked 4 hours ago
user135350user135350
111
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$begingroup$
Let $M$ be a $(G,X)$ manifold, that is we have local charts $(U,varphi_U)$ on $M$ with $varphi_U$ a diffeomorphism onto an open subset of $X$ and the transition maps are locally-$G$.
Let $mathfrak{p}:widetilde{M}rightarrow M$ be the universal covering of $M$.
The developing map theorem introduces a local diffeomorphism $dev:widetilde{M}rightarrow X$.
Does the developing map locally commute with the (restricted) covering and local charts, i.e. for $widetilde{U}$ (small enough) $dev|_widetilde{U}=varphi_{U}circmathfrak{p}|_widetilde{U}$ for some $varphi_U$?
dg.differential-geometry
asked 4 hours ago
user135350user135350
111
New contributor
user135350 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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1
$begingroup$
It's hard to make sense of the question (the last sentence). What's "small enough $tilde{U}$"? what's "for some $varphi_U$" when $varphi_U$ is already introduced?
$endgroup$
– YCor
4 hours ago
add a comment |
$begingroup$
Let $M$ be a $(G,X)$ manifold, that is we have local charts $(U,varphi_U)$ on $M$ with $varphi_U$ a diffeomorphism onto an open subset of $X$ and the transition maps are locally-$G$.
Let $mathfrak{p}:widetilde{M}rightarrow M$ be the universal covering of $M$.
The developing map theorem introduces a local diffeomorphism $dev:widetilde{M}rightarrow X$.
Does the developing map locally commute with the (restricted) covering and local charts, i.e. for $widetilde{U}$ (small enough) $dev|_widetilde{U}=varphi_{U}circmathfrak{p}|_widetilde{U}$ for some $varphi_U$?
dg.differential-geometry
asked 4 hours ago
user135350user135350
111
New contributor
user135350 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Let $M$ be a $(G,X)$ manifold, that is we have local charts $(U,varphi_U)$ on $M$ with $varphi_U$ a diffeomorphism onto an open subset of $X$ and the transition maps are locally-$G$.
Let $mathfrak{p}:widetilde{M}rightarrow M$ be the universal covering of $M$.
The developing map theorem introduces a local diffeomorphism $dev:widetilde{M}rightarrow X$.
Does the developing map locally commute with the (restricted) covering and local charts, i.e. for $widetilde{U}$ (small enough) $dev|_widetilde{U}=varphi_{U}circmathfrak{p}|_widetilde{U}$ for some $varphi_U$?
dg.differential-geometry
dg.differential-geometry
asked 4 hours ago
user135350user135350
111
New contributor
user135350 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
user135350user135350
111
New contributor
user135350 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
user135350user135350
111
New contributor
user135350 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
user135350user135350
111
asked 4 hours ago
user135350user135350
111
111
New contributor
user135350 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user135350 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user135350 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
It's hard to make sense of the question (the last sentence). What's "small enough $tilde{U}$"? what's "for some $varphi_U$" when $varphi_U$ is already introduced?
$endgroup$
– YCor
4 hours ago
add a comment |
1
$begingroup$
It's hard to make sense of the question (the last sentence). What's "small enough $tilde{U}$"? what's "for some $varphi_U$" when $varphi_U$ is already introduced?
$endgroup$
– YCor
4 hours ago
1
1
$begingroup$
It's hard to make sense of the question (the last sentence). What's "small enough $tilde{U}$"? what's "for some $varphi_U$" when $varphi_U$ is already introduced?
$endgroup$
– YCor
4 hours ago
$begingroup$
It's hard to make sense of the question (the last sentence). What's "small enough $tilde{U}$"? what's "for some $varphi_U$" when $varphi_U$ is already introduced?
$endgroup$
– YCor
4 hours ago
add a comment |
1 Answer
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$begingroup$
The best way to see this is to read the proof of the existence of the developing map.
The $(X,G)$ structure on $M$ induces an $(X,G)$-structure on $tilde M$ its universal cover such that for any $tilde xin tilde M$, write $x=p(tilde x)$ where $p$ is the covering map. Consider a chart $xin U$ such that there exists $tilde U$ which contains $tilde x$ such that the restriction of $p$ to $tilde U$, $p_{tilde U}:tilde Urightarrow U$ is an homeomorphism $phi_{tilde U}=phi_Ucirc p_{tilde U}$.
To construct the developing map, one fixes $tilde x_0in tilde M$ for every $tilde xin hat M$, one considers a path $c:[0,1]rightarrow hat M$ such that $c(0)=tilde x_0$ and $c(1)=tilde x$, then one consider chart $(tilde U_0,tilde phi_0),...(tilde U_n,tildephi_n)$ such that there exists a subdivision $[0=t_0,t_1,...,t_n=1]$ such that $c[t_i,t_{i+1}]subset tilde U_i$ and one sets:
$D(tilde x)=tilde g_0...g_{n-1}tildephi_n(tilde x)$ where $g_iin G$ such that $tilde phi_icirc{tildephi_{i+1}}^{-1}$ is the restriction of $g_i$ to $tildephi_{i+1}(tilde U_{i+1})$.
This shows that if we set $U_n=p(tilde U_n)$, we can define the chart $(U_n,g_0...g_{n-1}phi_n)$ of $x$ and the chart $(tilde U_n,g_0...g_{n-1}tilde phi_n)$ which answer your question.
answered 4 hours ago
Tsemo AristideTsemo Aristide
2,6621616
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$begingroup$
The best way to see this is to read the proof of the existence of the developing map.
The $(X,G)$ structure on $M$ induces an $(X,G)$-structure on $tilde M$ its universal cover such that for any $tilde xin tilde M$, write $x=p(tilde x)$ where $p$ is the covering map. Consider a chart $xin U$ such that there exists $tilde U$ which contains $tilde x$ such that the restriction of $p$ to $tilde U$, $p_{tilde U}:tilde Urightarrow U$ is an homeomorphism $phi_{tilde U}=phi_Ucirc p_{tilde U}$.
To construct the developing map, one fixes $tilde x_0in tilde M$ for every $tilde xin hat M$, one considers a path $c:[0,1]rightarrow hat M$ such that $c(0)=tilde x_0$ and $c(1)=tilde x$, then one consider chart $(tilde U_0,tilde phi_0),...(tilde U_n,tildephi_n)$ such that there exists a subdivision $[0=t_0,t_1,...,t_n=1]$ such that $c[t_i,t_{i+1}]subset tilde U_i$ and one sets:
$D(tilde x)=tilde g_0...g_{n-1}tildephi_n(tilde x)$ where $g_iin G$ such that $tilde phi_icirc{tildephi_{i+1}}^{-1}$ is the restriction of $g_i$ to $tildephi_{i+1}(tilde U_{i+1})$.
This shows that if we set $U_n=p(tilde U_n)$, we can define the chart $(U_n,g_0...g_{n-1}phi_n)$ of $x$ and the chart $(tilde U_n,g_0...g_{n-1}tilde phi_n)$ which answer your question.
answered 4 hours ago
Tsemo AristideTsemo Aristide
2,6621616
$endgroup$
add a comment |
$begingroup$
The best way to see this is to read the proof of the existence of the developing map.
The $(X,G)$ structure on $M$ induces an $(X,G)$-structure on $tilde M$ its universal cover such that for any $tilde xin tilde M$, write $x=p(tilde x)$ where $p$ is the covering map. Consider a chart $xin U$ such that there exists $tilde U$ which contains $tilde x$ such that the restriction of $p$ to $tilde U$, $p_{tilde U}:tilde Urightarrow U$ is an homeomorphism $phi_{tilde U}=phi_Ucirc p_{tilde U}$.
To construct the developing map, one fixes $tilde x_0in tilde M$ for every $tilde xin hat M$, one considers a path $c:[0,1]rightarrow hat M$ such that $c(0)=tilde x_0$ and $c(1)=tilde x$, then one consider chart $(tilde U_0,tilde phi_0),...(tilde U_n,tildephi_n)$ such that there exists a subdivision $[0=t_0,t_1,...,t_n=1]$ such that $c[t_i,t_{i+1}]subset tilde U_i$ and one sets:
$D(tilde x)=tilde g_0...g_{n-1}tildephi_n(tilde x)$ where $g_iin G$ such that $tilde phi_icirc{tildephi_{i+1}}^{-1}$ is the restriction of $g_i$ to $tildephi_{i+1}(tilde U_{i+1})$.
This shows that if we set $U_n=p(tilde U_n)$, we can define the chart $(U_n,g_0...g_{n-1}phi_n)$ of $x$ and the chart $(tilde U_n,g_0...g_{n-1}tilde phi_n)$ which answer your question.
answered 4 hours ago
Tsemo AristideTsemo Aristide
2,6621616
$endgroup$
add a comment |
$begingroup$
The best way to see this is to read the proof of the existence of the developing map.
The $(X,G)$ structure on $M$ induces an $(X,G)$-structure on $tilde M$ its universal cover such that for any $tilde xin tilde M$, write $x=p(tilde x)$ where $p$ is the covering map. Consider a chart $xin U$ such that there exists $tilde U$ which contains $tilde x$ such that the restriction of $p$ to $tilde U$, $p_{tilde U}:tilde Urightarrow U$ is an homeomorphism $phi_{tilde U}=phi_Ucirc p_{tilde U}$.
To construct the developing map, one fixes $tilde x_0in tilde M$ for every $tilde xin hat M$, one considers a path $c:[0,1]rightarrow hat M$ such that $c(0)=tilde x_0$ and $c(1)=tilde x$, then one consider chart $(tilde U_0,tilde phi_0),...(tilde U_n,tildephi_n)$ such that there exists a subdivision $[0=t_0,t_1,...,t_n=1]$ such that $c[t_i,t_{i+1}]subset tilde U_i$ and one sets:
$D(tilde x)=tilde g_0...g_{n-1}tildephi_n(tilde x)$ where $g_iin G$ such that $tilde phi_icirc{tildephi_{i+1}}^{-1}$ is the restriction of $g_i$ to $tildephi_{i+1}(tilde U_{i+1})$.
This shows that if we set $U_n=p(tilde U_n)$, we can define the chart $(U_n,g_0...g_{n-1}phi_n)$ of $x$ and the chart $(tilde U_n,g_0...g_{n-1}tilde phi_n)$ which answer your question.
answered 4 hours ago
Tsemo AristideTsemo Aristide
2,6621616
$endgroup$
The best way to see this is to read the proof of the existence of the developing map.
The $(X,G)$ structure on $M$ induces an $(X,G)$-structure on $tilde M$ its universal cover such that for any $tilde xin tilde M$, write $x=p(tilde x)$ where $p$ is the covering map. Consider a chart $xin U$ such that there exists $tilde U$ which contains $tilde x$ such that the restriction of $p$ to $tilde U$, $p_{tilde U}:tilde Urightarrow U$ is an homeomorphism $phi_{tilde U}=phi_Ucirc p_{tilde U}$.
To construct the developing map, one fixes $tilde x_0in tilde M$ for every $tilde xin hat M$, one considers a path $c:[0,1]rightarrow hat M$ such that $c(0)=tilde x_0$ and $c(1)=tilde x$, then one consider chart $(tilde U_0,tilde phi_0),...(tilde U_n,tildephi_n)$ such that there exists a subdivision $[0=t_0,t_1,...,t_n=1]$ such that $c[t_i,t_{i+1}]subset tilde U_i$ and one sets:
$D(tilde x)=tilde g_0...g_{n-1}tildephi_n(tilde x)$ where $g_iin G$ such that $tilde phi_icirc{tildephi_{i+1}}^{-1}$ is the restriction of $g_i$ to $tildephi_{i+1}(tilde U_{i+1})$.
This shows that if we set $U_n=p(tilde U_n)$, we can define the chart $(U_n,g_0...g_{n-1}phi_n)$ of $x$ and the chart $(tilde U_n,g_0...g_{n-1}tilde phi_n)$ which answer your question.
answered 4 hours ago
Tsemo AristideTsemo Aristide
2,6621616
edited 4 hours ago
answered 4 hours ago
Tsemo AristideTsemo Aristide
2,6621616
answered 4 hours ago
Tsemo AristideTsemo Aristide
2,6621616
answered 4 hours ago
Tsemo AristideTsemo Aristide
2,6621616
2,6621616
add a comment |
add a comment |
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$begingroup$
It's hard to make sense of the question (the last sentence). What's "small enough $tilde{U}$"? what's "for some $varphi_U$" when $varphi_U$ is already introduced?
$endgroup$
– YCor
4 hours ago