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Are conformal maps between Riemannian manifolds real-analytic?
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This is a cross-post.
Let $M,N$ be oriented smooth ($C^infty$) $n$-dimensional Riemannian manifolds, and let $f:M to N$ be a smooth orientation-preserving weakly* conformal map.
Do there exist real-analytic structures on $M,N$ that make $f$ real-analytic?
I only assume the metrics are $C^infty$.
Every smooth manifold has a unique real-analytic structure (up to diffeomorphism) compatible with its smooth structure.
A reasonable starting point would be to know whether every $C^infty$ conformal map between real-analytic manifolds with real-analytic metrics is real-analytic. (I don't know a reference for that; anyway, what I am asking seems harder).
*A weakly conformal map is a map whose differential at every point is either conformal or zero. (This is equivalent to $df^Tdf =(det df)^frac2n , textId_TM$).
Motivation:
I am trying to understand if smooth weakly conformal maps whose differential vanishes at a point are constant (for dimensions $n ge 3$). This seems to be the case for analytic maps, hence my interest in the possible analyticity of such maps.
For the Euclidean case, this follows directly by Liouville's theorem:
For $n=2$, every such map is complex-analytic. Let $Omega subseteq mathbbR^n$ be an open subset, $n ge 3$, and let $f:Omega to mathbbR^n$ be a smooth conformal map. By Liouville's theorem, $f$ is of the form
$$ f(x)=b+alphafrac1A(x-a),$$
where $A$ is an orthogonal matrix, and $epsilon in 0,2, b in mathbbR^n,alpha in mathbbR,a in mathbbR^n setminus Omega$.
So, up to translations and dilations,
$ f(x)=fracA x$ (where $ 0 notin Omega$) which is real-analytic as a multiplication of two analytic maps. ($1/x^2$ is analytic on $mathbbR setminus 0$).
ap.analysis-of-pdes riemannian-geometry conformal-geometry analytic-functions real-analytic-structures
asked 4 hours ago
Asaf Shachar
2,1561744
add a comment |Â
up vote
6
down vote
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This is a cross-post.
Let $M,N$ be oriented smooth ($C^infty$) $n$-dimensional Riemannian manifolds, and let $f:M to N$ be a smooth orientation-preserving weakly* conformal map.
Do there exist real-analytic structures on $M,N$ that make $f$ real-analytic?
I only assume the metrics are $C^infty$.
Every smooth manifold has a unique real-analytic structure (up to diffeomorphism) compatible with its smooth structure.
A reasonable starting point would be to know whether every $C^infty$ conformal map between real-analytic manifolds with real-analytic metrics is real-analytic. (I don't know a reference for that; anyway, what I am asking seems harder).
*A weakly conformal map is a map whose differential at every point is either conformal or zero. (This is equivalent to $df^Tdf =(det df)^frac2n , textId_TM$).
Motivation:
I am trying to understand if smooth weakly conformal maps whose differential vanishes at a point are constant (for dimensions $n ge 3$). This seems to be the case for analytic maps, hence my interest in the possible analyticity of such maps.
For the Euclidean case, this follows directly by Liouville's theorem:
For $n=2$, every such map is complex-analytic. Let $Omega subseteq mathbbR^n$ be an open subset, $n ge 3$, and let $f:Omega to mathbbR^n$ be a smooth conformal map. By Liouville's theorem, $f$ is of the form
$$ f(x)=b+alphafrac1A(x-a),$$
where $A$ is an orthogonal matrix, and $epsilon in 0,2, b in mathbbR^n,alpha in mathbbR,a in mathbbR^n setminus Omega$.
So, up to translations and dilations,
$ f(x)=fracA x$ (where $ 0 notin Omega$) which is real-analytic as a multiplication of two analytic maps. ($1/x^2$ is analytic on $mathbbR setminus 0$).
ap.analysis-of-pdes riemannian-geometry conformal-geometry analytic-functions real-analytic-structures
asked 4 hours ago
Asaf Shachar
2,1561744
1
While there is an analytic structure on the manifold, the Riemannian metric will not necessarily be real analytic. Does it bother you?
– Piotr Hajlasz
2 hours ago
add a comment |Â
up vote
6
down vote
favorite
up vote
6
down vote
favorite
This is a cross-post.
Let $M,N$ be oriented smooth ($C^infty$) $n$-dimensional Riemannian manifolds, and let $f:M to N$ be a smooth orientation-preserving weakly* conformal map.
Do there exist real-analytic structures on $M,N$ that make $f$ real-analytic?
I only assume the metrics are $C^infty$.
Every smooth manifold has a unique real-analytic structure (up to diffeomorphism) compatible with its smooth structure.
A reasonable starting point would be to know whether every $C^infty$ conformal map between real-analytic manifolds with real-analytic metrics is real-analytic. (I don't know a reference for that; anyway, what I am asking seems harder).
*A weakly conformal map is a map whose differential at every point is either conformal or zero. (This is equivalent to $df^Tdf =(det df)^frac2n , textId_TM$).
Motivation:
I am trying to understand if smooth weakly conformal maps whose differential vanishes at a point are constant (for dimensions $n ge 3$). This seems to be the case for analytic maps, hence my interest in the possible analyticity of such maps.
For the Euclidean case, this follows directly by Liouville's theorem:
For $n=2$, every such map is complex-analytic. Let $Omega subseteq mathbbR^n$ be an open subset, $n ge 3$, and let $f:Omega to mathbbR^n$ be a smooth conformal map. By Liouville's theorem, $f$ is of the form
$$ f(x)=b+alphafrac1A(x-a),$$
where $A$ is an orthogonal matrix, and $epsilon in 0,2, b in mathbbR^n,alpha in mathbbR,a in mathbbR^n setminus Omega$.
So, up to translations and dilations,
$ f(x)=fracA x$ (where $ 0 notin Omega$) which is real-analytic as a multiplication of two analytic maps. ($1/x^2$ is analytic on $mathbbR setminus 0$).
ap.analysis-of-pdes riemannian-geometry conformal-geometry analytic-functions real-analytic-structures
asked 4 hours ago
Asaf Shachar
2,1561744
This is a cross-post.
Let $M,N$ be oriented smooth ($C^infty$) $n$-dimensional Riemannian manifolds, and let $f:M to N$ be a smooth orientation-preserving weakly* conformal map.
Do there exist real-analytic structures on $M,N$ that make $f$ real-analytic?
I only assume the metrics are $C^infty$.
Every smooth manifold has a unique real-analytic structure (up to diffeomorphism) compatible with its smooth structure.
A reasonable starting point would be to know whether every $C^infty$ conformal map between real-analytic manifolds with real-analytic metrics is real-analytic. (I don't know a reference for that; anyway, what I am asking seems harder).
*A weakly conformal map is a map whose differential at every point is either conformal or zero. (This is equivalent to $df^Tdf =(det df)^frac2n , textId_TM$).
Motivation:
I am trying to understand if smooth weakly conformal maps whose differential vanishes at a point are constant (for dimensions $n ge 3$). This seems to be the case for analytic maps, hence my interest in the possible analyticity of such maps.
For the Euclidean case, this follows directly by Liouville's theorem:
For $n=2$, every such map is complex-analytic. Let $Omega subseteq mathbbR^n$ be an open subset, $n ge 3$, and let $f:Omega to mathbbR^n$ be a smooth conformal map. By Liouville's theorem, $f$ is of the form
$$ f(x)=b+alphafrac1A(x-a),$$
where $A$ is an orthogonal matrix, and $epsilon in 0,2, b in mathbbR^n,alpha in mathbbR,a in mathbbR^n setminus Omega$.
So, up to translations and dilations,
$ f(x)=fracA x$ (where $ 0 notin Omega$) which is real-analytic as a multiplication of two analytic maps. ($1/x^2$ is analytic on $mathbbR setminus 0$).
ap.analysis-of-pdes riemannian-geometry conformal-geometry analytic-functions real-analytic-structures
ap.analysis-of-pdes riemannian-geometry conformal-geometry analytic-functions real-analytic-structures
asked 4 hours ago
Asaf Shachar
2,1561744
asked 4 hours ago
Asaf Shachar
2,1561744
asked 4 hours ago
Asaf Shachar
2,1561744
asked 4 hours ago
Asaf Shachar
2,1561744
asked 4 hours ago
Asaf Shachar
2,1561744
2,1561744
1
While there is an analytic structure on the manifold, the Riemannian metric will not necessarily be real analytic. Does it bother you?
– Piotr Hajlasz
2 hours ago
add a comment |Â
1
While there is an analytic structure on the manifold, the Riemannian metric will not necessarily be real analytic. Does it bother you?
– Piotr Hajlasz
2 hours ago
1
1
While there is an analytic structure on the manifold, the Riemannian metric will not necessarily be real analytic. Does it bother you?
– Piotr Hajlasz
2 hours ago
While there is an analytic structure on the manifold, the Riemannian metric will not necessarily be real analytic. Does it bother you?
– Piotr Hajlasz
2 hours ago
add a comment |Â
1 Answer
1
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up vote
6
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As essentially follows from your argument in the comment to Piotr Hajlasz's answer (now deleted), for $fcolon Mto N$ to be a counterexample, it cannot be a diffeomorphism. Because then you could just pick any analytic structure on $N$, pull it back to $M$ by $f$ and make $f$ trivially analytic.
With that in mind, one can think of counterexamples already in 1-dimension. For instance let $(N,g_N) = (M,g_M) = (mathbbR,dx^2)$, where $dx^2$ is the metric that appears flat with respect to some global coordinate $x$. Let
$$
f(x) = begincases
(x+epsilon) e^-1/(x+epsilon)^2 & x<-epsilon \
0 & -epsilonle x le epsilon \
(x-epsilon) e^-1/(x-epsilon)^2 & epsilon < x
endcases .
$$
It is locally a diffeomorphism if you consider it as a map from $mathbbRsetminus [-epsilon,epsilon]$ to $mathbbRsetminus 0$. But there is no pair of analytic structures on $mathbbR$ that would make $fcolon mathbbR to mathbbR$ analytic. If $f$ were analytic (with respect to any analytic atlas), then it could not vanish on any open set, while not being globally zero. But since we are in 1-dimension, any map is weakly conformal. So then $fcolon M to N$ is weakly conformal, but can never be made analytic.
Higher dimensional examples could be generated from this one by using rotational symmetry and basically reducing to the 1-dimensional case.
answered 2 hours ago
Igor Khavkine
11.4k23470
2
Could you spell out the higher dimensional argument in your last paragraph? It can't work in dimension $2$ because of Louisville's theorem and I spent some time trying to make a construction like this work in dimension 3 and failing when I thought about the OP's earlier question.
– David E Speyer
2 hours ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
As essentially follows from your argument in the comment to Piotr Hajlasz's answer (now deleted), for $fcolon Mto N$ to be a counterexample, it cannot be a diffeomorphism. Because then you could just pick any analytic structure on $N$, pull it back to $M$ by $f$ and make $f$ trivially analytic.
With that in mind, one can think of counterexamples already in 1-dimension. For instance let $(N,g_N) = (M,g_M) = (mathbbR,dx^2)$, where $dx^2$ is the metric that appears flat with respect to some global coordinate $x$. Let
$$
f(x) = begincases
(x+epsilon) e^-1/(x+epsilon)^2 & x<-epsilon \
0 & -epsilonle x le epsilon \
(x-epsilon) e^-1/(x-epsilon)^2 & epsilon < x
endcases .
$$
It is locally a diffeomorphism if you consider it as a map from $mathbbRsetminus [-epsilon,epsilon]$ to $mathbbRsetminus 0$. But there is no pair of analytic structures on $mathbbR$ that would make $fcolon mathbbR to mathbbR$ analytic. If $f$ were analytic (with respect to any analytic atlas), then it could not vanish on any open set, while not being globally zero. But since we are in 1-dimension, any map is weakly conformal. So then $fcolon M to N$ is weakly conformal, but can never be made analytic.
Higher dimensional examples could be generated from this one by using rotational symmetry and basically reducing to the 1-dimensional case.
answered 2 hours ago
Igor Khavkine
11.4k23470
2
Could you spell out the higher dimensional argument in your last paragraph? It can't work in dimension $2$ because of Louisville's theorem and I spent some time trying to make a construction like this work in dimension 3 and failing when I thought about the OP's earlier question.
– David E Speyer
2 hours ago
add a comment |Â
up vote
6
down vote
As essentially follows from your argument in the comment to Piotr Hajlasz's answer (now deleted), for $fcolon Mto N$ to be a counterexample, it cannot be a diffeomorphism. Because then you could just pick any analytic structure on $N$, pull it back to $M$ by $f$ and make $f$ trivially analytic.
With that in mind, one can think of counterexamples already in 1-dimension. For instance let $(N,g_N) = (M,g_M) = (mathbbR,dx^2)$, where $dx^2$ is the metric that appears flat with respect to some global coordinate $x$. Let
$$
f(x) = begincases
(x+epsilon) e^-1/(x+epsilon)^2 & x<-epsilon \
0 & -epsilonle x le epsilon \
(x-epsilon) e^-1/(x-epsilon)^2 & epsilon < x
endcases .
$$
It is locally a diffeomorphism if you consider it as a map from $mathbbRsetminus [-epsilon,epsilon]$ to $mathbbRsetminus 0$. But there is no pair of analytic structures on $mathbbR$ that would make $fcolon mathbbR to mathbbR$ analytic. If $f$ were analytic (with respect to any analytic atlas), then it could not vanish on any open set, while not being globally zero. But since we are in 1-dimension, any map is weakly conformal. So then $fcolon M to N$ is weakly conformal, but can never be made analytic.
Higher dimensional examples could be generated from this one by using rotational symmetry and basically reducing to the 1-dimensional case.
answered 2 hours ago
Igor Khavkine
11.4k23470
2
Could you spell out the higher dimensional argument in your last paragraph? It can't work in dimension $2$ because of Louisville's theorem and I spent some time trying to make a construction like this work in dimension 3 and failing when I thought about the OP's earlier question.
– David E Speyer
2 hours ago
add a comment |Â
up vote
6
down vote
up vote
6
down vote
As essentially follows from your argument in the comment to Piotr Hajlasz's answer (now deleted), for $fcolon Mto N$ to be a counterexample, it cannot be a diffeomorphism. Because then you could just pick any analytic structure on $N$, pull it back to $M$ by $f$ and make $f$ trivially analytic.
With that in mind, one can think of counterexamples already in 1-dimension. For instance let $(N,g_N) = (M,g_M) = (mathbbR,dx^2)$, where $dx^2$ is the metric that appears flat with respect to some global coordinate $x$. Let
$$
f(x) = begincases
(x+epsilon) e^-1/(x+epsilon)^2 & x<-epsilon \
0 & -epsilonle x le epsilon \
(x-epsilon) e^-1/(x-epsilon)^2 & epsilon < x
endcases .
$$
It is locally a diffeomorphism if you consider it as a map from $mathbbRsetminus [-epsilon,epsilon]$ to $mathbbRsetminus 0$. But there is no pair of analytic structures on $mathbbR$ that would make $fcolon mathbbR to mathbbR$ analytic. If $f$ were analytic (with respect to any analytic atlas), then it could not vanish on any open set, while not being globally zero. But since we are in 1-dimension, any map is weakly conformal. So then $fcolon M to N$ is weakly conformal, but can never be made analytic.
Higher dimensional examples could be generated from this one by using rotational symmetry and basically reducing to the 1-dimensional case.
answered 2 hours ago
Igor Khavkine
11.4k23470
As essentially follows from your argument in the comment to Piotr Hajlasz's answer (now deleted), for $fcolon Mto N$ to be a counterexample, it cannot be a diffeomorphism. Because then you could just pick any analytic structure on $N$, pull it back to $M$ by $f$ and make $f$ trivially analytic.
With that in mind, one can think of counterexamples already in 1-dimension. For instance let $(N,g_N) = (M,g_M) = (mathbbR,dx^2)$, where $dx^2$ is the metric that appears flat with respect to some global coordinate $x$. Let
$$
f(x) = begincases
(x+epsilon) e^-1/(x+epsilon)^2 & x<-epsilon \
0 & -epsilonle x le epsilon \
(x-epsilon) e^-1/(x-epsilon)^2 & epsilon < x
endcases .
$$
It is locally a diffeomorphism if you consider it as a map from $mathbbRsetminus [-epsilon,epsilon]$ to $mathbbRsetminus 0$. But there is no pair of analytic structures on $mathbbR$ that would make $fcolon mathbbR to mathbbR$ analytic. If $f$ were analytic (with respect to any analytic atlas), then it could not vanish on any open set, while not being globally zero. But since we are in 1-dimension, any map is weakly conformal. So then $fcolon M to N$ is weakly conformal, but can never be made analytic.
Higher dimensional examples could be generated from this one by using rotational symmetry and basically reducing to the 1-dimensional case.
answered 2 hours ago
Igor Khavkine
11.4k23470
answered 2 hours ago
Igor Khavkine
11.4k23470
answered 2 hours ago
Igor Khavkine
11.4k23470
answered 2 hours ago
Igor Khavkine
11.4k23470
11.4k23470
2
Could you spell out the higher dimensional argument in your last paragraph? It can't work in dimension $2$ because of Louisville's theorem and I spent some time trying to make a construction like this work in dimension 3 and failing when I thought about the OP's earlier question.
– David E Speyer
2 hours ago
add a comment |Â
2
Could you spell out the higher dimensional argument in your last paragraph? It can't work in dimension $2$ because of Louisville's theorem and I spent some time trying to make a construction like this work in dimension 3 and failing when I thought about the OP's earlier question.
– David E Speyer
2 hours ago
2
2
Could you spell out the higher dimensional argument in your last paragraph? It can't work in dimension $2$ because of Louisville's theorem and I spent some time trying to make a construction like this work in dimension 3 and failing when I thought about the OP's earlier question.
– David E Speyer
2 hours ago
Could you spell out the higher dimensional argument in your last paragraph? It can't work in dimension $2$ because of Louisville's theorem and I spent some time trying to make a construction like this work in dimension 3 and failing when I thought about the OP's earlier question.
– David E Speyer
2 hours ago
add a comment |Â
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1
While there is an analytic structure on the manifold, the Riemannian metric will not necessarily be real analytic. Does it bother you?
– Piotr Hajlasz
2 hours ago