In the real analytic category, are the fibers of a proper submersion isomorphic?On a proof of the existence of tubular neighborhoods.When is a holomorphic submersion with isomorphic fibers locally trivial?Ehresmann's theorem over the $p$-adicsNotions related to De Rham CohomologyEhresmann's fibration theorem in the C1 classComplexifying a real-analytic singularityAre definable sets in an o-minimal expansion of the real field locally analytic?Categorification of covering morphisms
In the real analytic category, are the fibers of a proper submersion isomorphic?
On a proof of the existence of tubular neighborhoods.When is a holomorphic submersion with isomorphic fibers locally trivial?Ehresmann's theorem over the $p$-adicsNotions related to De Rham CohomologyEhresmann's fibration theorem in the C1 classComplexifying a real-analytic singularityAre definable sets in an o-minimal expansion of the real field locally analytic?Categorification of covering morphisms
$begingroup$
Ehresmann's theorem says that a proper smooth submersion is a fiber bundle. The proofs I know rely on the existence of connections locally on the base, and this is furnished by partitions of unity.
This question gives a counterexample in the holomorphic category which is probably classic (elliptic curves), but I don't know any of the story.
I am wondering about the real analytic category. Are the fibers of a proper real analytic submersion isomorphic? If not, will it be locally trivial (in the real analytic category) when they are isomorphic, as in the linked question?
differential-topology fibre-bundles real-analytic-structures
asked Jun 14 at 8:24
ArrowArrow
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add a comment
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$begingroup$
Ehresmann's theorem says that a proper smooth submersion is a fiber bundle. The proofs I know rely on the existence of connections locally on the base, and this is furnished by partitions of unity.
This question gives a counterexample in the holomorphic category which is probably classic (elliptic curves), but I don't know any of the story.
I am wondering about the real analytic category. Are the fibers of a proper real analytic submersion isomorphic? If not, will it be locally trivial (in the real analytic category) when they are isomorphic, as in the linked question?
differential-topology fibre-bundles real-analytic-structures
asked Jun 14 at 8:24
ArrowArrow
4,0881 gold badge15 silver badges52 bronze badges
$endgroup$
$begingroup$
The question in the title is just the opposite than the question in the body of the question. I suggest an edit for the coherence's sake.
$endgroup$
– Francesco Polizzi
Jun 14 at 10:35
$begingroup$
@FrancescoPolizzi done.
$endgroup$
– Arrow
Jun 14 at 10:47
add a comment
|
$begingroup$
Ehresmann's theorem says that a proper smooth submersion is a fiber bundle. The proofs I know rely on the existence of connections locally on the base, and this is furnished by partitions of unity.
This question gives a counterexample in the holomorphic category which is probably classic (elliptic curves), but I don't know any of the story.
I am wondering about the real analytic category. Are the fibers of a proper real analytic submersion isomorphic? If not, will it be locally trivial (in the real analytic category) when they are isomorphic, as in the linked question?
differential-topology fibre-bundles real-analytic-structures
asked Jun 14 at 8:24
ArrowArrow
4,0881 gold badge15 silver badges52 bronze badges
$endgroup$
Ehresmann's theorem says that a proper smooth submersion is a fiber bundle. The proofs I know rely on the existence of connections locally on the base, and this is furnished by partitions of unity.
This question gives a counterexample in the holomorphic category which is probably classic (elliptic curves), but I don't know any of the story.
I am wondering about the real analytic category. Are the fibers of a proper real analytic submersion isomorphic? If not, will it be locally trivial (in the real analytic category) when they are isomorphic, as in the linked question?
differential-topology fibre-bundles real-analytic-structures
differential-topology fibre-bundles real-analytic-structures
asked Jun 14 at 8:24
ArrowArrow
4,0881 gold badge15 silver badges52 bronze badges
asked Jun 14 at 8:24
ArrowArrow
4,0881 gold badge15 silver badges52 bronze badges
edited Jun 14 at 10:47
Arrow
asked Jun 14 at 8:24
ArrowArrow
4,0881 gold badge15 silver badges52 bronze badges
asked Jun 14 at 8:24
ArrowArrow
4,0881 gold badge15 silver badges52 bronze badges
asked Jun 14 at 8:24
ArrowArrow
4,0881 gold badge15 silver badges52 bronze badges
4,0881 gold badge15 silver badges52 bronze badges
$begingroup$
The question in the title is just the opposite than the question in the body of the question. I suggest an edit for the coherence's sake.
$endgroup$
– Francesco Polizzi
Jun 14 at 10:35
$begingroup$
@FrancescoPolizzi done.
$endgroup$
– Arrow
Jun 14 at 10:47
add a comment
|
$begingroup$
The question in the title is just the opposite than the question in the body of the question. I suggest an edit for the coherence's sake.
$endgroup$
– Francesco Polizzi
Jun 14 at 10:35
$begingroup$
@FrancescoPolizzi done.
$endgroup$
– Arrow
Jun 14 at 10:47
$begingroup$
The question in the title is just the opposite than the question in the body of the question. I suggest an edit for the coherence's sake.
$endgroup$
– Francesco Polizzi
Jun 14 at 10:35
$begingroup$
The question in the title is just the opposite than the question in the body of the question. I suggest an edit for the coherence's sake.
$endgroup$
– Francesco Polizzi
Jun 14 at 10:35
$begingroup$
@FrancescoPolizzi done.
$endgroup$
– Arrow
Jun 14 at 10:47
$begingroup$
@FrancescoPolizzi done.
$endgroup$
– Arrow
Jun 14 at 10:47
add a comment
|
2 Answers
2
active
oldest
votes
$begingroup$
There exists an analytic Riemann metric on a real-analytic manifold, which follows from embeddability of real analytic manifolds (see The Analytic Embedding of Abstract Real-Analytic Manifolds Charles B. Morrey, Jr.) - maybe can also be proved easier.
So, you can probably just take orthogonal connection.
answered Jun 14 at 8:31
Lev SoukhanovLev Soukhanov
8134 silver badges14 bronze badges
$endgroup$
$begingroup$
So, the answer to the question in the title is yes, right?
$endgroup$
– Francesco Polizzi
Jun 14 at 10:35
2
$begingroup$
Dear @Lev, I just found this comment which remarks the some references assert Ehresmann is false in the real analytic category. Just out of curiosity, have you ever stumbled upon such a claim?
$endgroup$
– Arrow
Jun 14 at 10:50
$begingroup$
Dear @Lev, your answer gives a connection on the tangent bundle of an analytic manifold, but how can it be used to obtain a connection on an analytic submersion?
$endgroup$
– Arrow
Jun 17 at 13:08
$begingroup$
Arrow, considering this comment, I wonder. I hadn't seen such claims, and I have a feeling that Ehresmann should be true in analytic setting from reasons mentioned above. The construction of the connection is similar to the smooth case - we need to chose analytic distribution which is transverse to fibers. Lets just choose it to be orthogonal (so it is even easier than what you've wrote in your answer, I suppose I should've made the proof more clear).
$endgroup$
– Lev Soukhanov
Jun 23 at 21:26
$begingroup$
Dear @LevSoukhanov, I would really appreciate some additional details since I don't see how to construct such an analytic connection on the submersion. (I would like to circumvent the tubular neighborhood proof of Ehresmann sketched in my answer, which feels like transparent to me.)
$endgroup$
– Arrow
Jun 23 at 21:42
|
show 2 more comments
$begingroup$
I think Ehresmann's theorem does hold in the real-analytic category. A proof via tubular neighborhoods can be repeated in the real-analytic context since it makes no use of partitions of unity. Here's a sketch.
By Lev's answer, a real-analytic manifold admits a real-analytic embedding into Euclidean space with its standard real-analytic structure. I think this implies a real-analytic manifold admits a real-analytic Riemannian metric.
The exponential map of this metric is also real-analytic (as a solution to analytic ODE), so an analytically embedded submanifold admits a real-analytic tubular neighborhood. We may then repeat this proof of Ehresmann's theorem using tubular neighborhoods.
Remark. This comment along with several other places in the literature assert Ehresmann's theorem fails in the real-analytic category but provide no examples.
answered Jun 20 at 20:45
ArrowArrow
4,0881 gold badge15 silver badges52 bronze badges
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There exists an analytic Riemann metric on a real-analytic manifold, which follows from embeddability of real analytic manifolds (see The Analytic Embedding of Abstract Real-Analytic Manifolds Charles B. Morrey, Jr.) - maybe can also be proved easier.
So, you can probably just take orthogonal connection.
answered Jun 14 at 8:31
Lev SoukhanovLev Soukhanov
8134 silver badges14 bronze badges
$endgroup$
$begingroup$
So, the answer to the question in the title is yes, right?
$endgroup$
– Francesco Polizzi
Jun 14 at 10:35
2
$begingroup$
Dear @Lev, I just found this comment which remarks the some references assert Ehresmann is false in the real analytic category. Just out of curiosity, have you ever stumbled upon such a claim?
$endgroup$
– Arrow
Jun 14 at 10:50
$begingroup$
Dear @Lev, your answer gives a connection on the tangent bundle of an analytic manifold, but how can it be used to obtain a connection on an analytic submersion?
$endgroup$
– Arrow
Jun 17 at 13:08
$begingroup$
Arrow, considering this comment, I wonder. I hadn't seen such claims, and I have a feeling that Ehresmann should be true in analytic setting from reasons mentioned above. The construction of the connection is similar to the smooth case - we need to chose analytic distribution which is transverse to fibers. Lets just choose it to be orthogonal (so it is even easier than what you've wrote in your answer, I suppose I should've made the proof more clear).
$endgroup$
– Lev Soukhanov
Jun 23 at 21:26
$begingroup$
Dear @LevSoukhanov, I would really appreciate some additional details since I don't see how to construct such an analytic connection on the submersion. (I would like to circumvent the tubular neighborhood proof of Ehresmann sketched in my answer, which feels like transparent to me.)
$endgroup$
– Arrow
Jun 23 at 21:42
|
show 2 more comments
$begingroup$
There exists an analytic Riemann metric on a real-analytic manifold, which follows from embeddability of real analytic manifolds (see The Analytic Embedding of Abstract Real-Analytic Manifolds Charles B. Morrey, Jr.) - maybe can also be proved easier.
So, you can probably just take orthogonal connection.
answered Jun 14 at 8:31
Lev SoukhanovLev Soukhanov
8134 silver badges14 bronze badges
$endgroup$
$begingroup$
So, the answer to the question in the title is yes, right?
$endgroup$
– Francesco Polizzi
Jun 14 at 10:35
2
$begingroup$
Dear @Lev, I just found this comment which remarks the some references assert Ehresmann is false in the real analytic category. Just out of curiosity, have you ever stumbled upon such a claim?
$endgroup$
– Arrow
Jun 14 at 10:50
$begingroup$
Dear @Lev, your answer gives a connection on the tangent bundle of an analytic manifold, but how can it be used to obtain a connection on an analytic submersion?
$endgroup$
– Arrow
Jun 17 at 13:08
$begingroup$
Arrow, considering this comment, I wonder. I hadn't seen such claims, and I have a feeling that Ehresmann should be true in analytic setting from reasons mentioned above. The construction of the connection is similar to the smooth case - we need to chose analytic distribution which is transverse to fibers. Lets just choose it to be orthogonal (so it is even easier than what you've wrote in your answer, I suppose I should've made the proof more clear).
$endgroup$
– Lev Soukhanov
Jun 23 at 21:26
$begingroup$
Dear @LevSoukhanov, I would really appreciate some additional details since I don't see how to construct such an analytic connection on the submersion. (I would like to circumvent the tubular neighborhood proof of Ehresmann sketched in my answer, which feels like transparent to me.)
$endgroup$
– Arrow
Jun 23 at 21:42
|
show 2 more comments
$begingroup$
There exists an analytic Riemann metric on a real-analytic manifold, which follows from embeddability of real analytic manifolds (see The Analytic Embedding of Abstract Real-Analytic Manifolds Charles B. Morrey, Jr.) - maybe can also be proved easier.
So, you can probably just take orthogonal connection.
answered Jun 14 at 8:31
Lev SoukhanovLev Soukhanov
8134 silver badges14 bronze badges
$endgroup$
There exists an analytic Riemann metric on a real-analytic manifold, which follows from embeddability of real analytic manifolds (see The Analytic Embedding of Abstract Real-Analytic Manifolds Charles B. Morrey, Jr.) - maybe can also be proved easier.
So, you can probably just take orthogonal connection.
answered Jun 14 at 8:31
Lev SoukhanovLev Soukhanov
8134 silver badges14 bronze badges
answered Jun 14 at 8:31
Lev SoukhanovLev Soukhanov
8134 silver badges14 bronze badges
answered Jun 14 at 8:31
Lev SoukhanovLev Soukhanov
8134 silver badges14 bronze badges
answered Jun 14 at 8:31
Lev SoukhanovLev Soukhanov
8134 silver badges14 bronze badges
8134 silver badges14 bronze badges
$begingroup$
So, the answer to the question in the title is yes, right?
$endgroup$
– Francesco Polizzi
Jun 14 at 10:35
2
$begingroup$
Dear @Lev, I just found this comment which remarks the some references assert Ehresmann is false in the real analytic category. Just out of curiosity, have you ever stumbled upon such a claim?
$endgroup$
– Arrow
Jun 14 at 10:50
$begingroup$
Dear @Lev, your answer gives a connection on the tangent bundle of an analytic manifold, but how can it be used to obtain a connection on an analytic submersion?
$endgroup$
– Arrow
Jun 17 at 13:08
$begingroup$
Arrow, considering this comment, I wonder. I hadn't seen such claims, and I have a feeling that Ehresmann should be true in analytic setting from reasons mentioned above. The construction of the connection is similar to the smooth case - we need to chose analytic distribution which is transverse to fibers. Lets just choose it to be orthogonal (so it is even easier than what you've wrote in your answer, I suppose I should've made the proof more clear).
$endgroup$
– Lev Soukhanov
Jun 23 at 21:26
$begingroup$
Dear @LevSoukhanov, I would really appreciate some additional details since I don't see how to construct such an analytic connection on the submersion. (I would like to circumvent the tubular neighborhood proof of Ehresmann sketched in my answer, which feels like transparent to me.)
$endgroup$
– Arrow
Jun 23 at 21:42
|
show 2 more comments
$begingroup$
So, the answer to the question in the title is yes, right?
$endgroup$
– Francesco Polizzi
Jun 14 at 10:35
2
$begingroup$
Dear @Lev, I just found this comment which remarks the some references assert Ehresmann is false in the real analytic category. Just out of curiosity, have you ever stumbled upon such a claim?
$endgroup$
– Arrow
Jun 14 at 10:50
$begingroup$
Dear @Lev, your answer gives a connection on the tangent bundle of an analytic manifold, but how can it be used to obtain a connection on an analytic submersion?
$endgroup$
– Arrow
Jun 17 at 13:08
$begingroup$
Arrow, considering this comment, I wonder. I hadn't seen such claims, and I have a feeling that Ehresmann should be true in analytic setting from reasons mentioned above. The construction of the connection is similar to the smooth case - we need to chose analytic distribution which is transverse to fibers. Lets just choose it to be orthogonal (so it is even easier than what you've wrote in your answer, I suppose I should've made the proof more clear).
$endgroup$
– Lev Soukhanov
Jun 23 at 21:26
$begingroup$
Dear @LevSoukhanov, I would really appreciate some additional details since I don't see how to construct such an analytic connection on the submersion. (I would like to circumvent the tubular neighborhood proof of Ehresmann sketched in my answer, which feels like transparent to me.)
$endgroup$
– Arrow
Jun 23 at 21:42
$begingroup$
So, the answer to the question in the title is yes, right?
$endgroup$
– Francesco Polizzi
Jun 14 at 10:35
$begingroup$
So, the answer to the question in the title is yes, right?
$endgroup$
– Francesco Polizzi
Jun 14 at 10:35
2
2
$begingroup$
Dear @Lev, I just found this comment which remarks the some references assert Ehresmann is false in the real analytic category. Just out of curiosity, have you ever stumbled upon such a claim?
$endgroup$
– Arrow
Jun 14 at 10:50
$begingroup$
Dear @Lev, I just found this comment which remarks the some references assert Ehresmann is false in the real analytic category. Just out of curiosity, have you ever stumbled upon such a claim?
$endgroup$
– Arrow
Jun 14 at 10:50
$begingroup$
Dear @Lev, your answer gives a connection on the tangent bundle of an analytic manifold, but how can it be used to obtain a connection on an analytic submersion?
$endgroup$
– Arrow
Jun 17 at 13:08
$begingroup$
Dear @Lev, your answer gives a connection on the tangent bundle of an analytic manifold, but how can it be used to obtain a connection on an analytic submersion?
$endgroup$
– Arrow
Jun 17 at 13:08
$begingroup$
Arrow, considering this comment, I wonder. I hadn't seen such claims, and I have a feeling that Ehresmann should be true in analytic setting from reasons mentioned above. The construction of the connection is similar to the smooth case - we need to chose analytic distribution which is transverse to fibers. Lets just choose it to be orthogonal (so it is even easier than what you've wrote in your answer, I suppose I should've made the proof more clear).
$endgroup$
– Lev Soukhanov
Jun 23 at 21:26
$begingroup$
Arrow, considering this comment, I wonder. I hadn't seen such claims, and I have a feeling that Ehresmann should be true in analytic setting from reasons mentioned above. The construction of the connection is similar to the smooth case - we need to chose analytic distribution which is transverse to fibers. Lets just choose it to be orthogonal (so it is even easier than what you've wrote in your answer, I suppose I should've made the proof more clear).
$endgroup$
– Lev Soukhanov
Jun 23 at 21:26
$begingroup$
Dear @LevSoukhanov, I would really appreciate some additional details since I don't see how to construct such an analytic connection on the submersion. (I would like to circumvent the tubular neighborhood proof of Ehresmann sketched in my answer, which feels like transparent to me.)
$endgroup$
– Arrow
Jun 23 at 21:42
$begingroup$
Dear @LevSoukhanov, I would really appreciate some additional details since I don't see how to construct such an analytic connection on the submersion. (I would like to circumvent the tubular neighborhood proof of Ehresmann sketched in my answer, which feels like transparent to me.)
$endgroup$
– Arrow
Jun 23 at 21:42
|
show 2 more comments
$begingroup$
I think Ehresmann's theorem does hold in the real-analytic category. A proof via tubular neighborhoods can be repeated in the real-analytic context since it makes no use of partitions of unity. Here's a sketch.
By Lev's answer, a real-analytic manifold admits a real-analytic embedding into Euclidean space with its standard real-analytic structure. I think this implies a real-analytic manifold admits a real-analytic Riemannian metric.
The exponential map of this metric is also real-analytic (as a solution to analytic ODE), so an analytically embedded submanifold admits a real-analytic tubular neighborhood. We may then repeat this proof of Ehresmann's theorem using tubular neighborhoods.
Remark. This comment along with several other places in the literature assert Ehresmann's theorem fails in the real-analytic category but provide no examples.
answered Jun 20 at 20:45
ArrowArrow
4,0881 gold badge15 silver badges52 bronze badges
$endgroup$
add a comment
|
$begingroup$
I think Ehresmann's theorem does hold in the real-analytic category. A proof via tubular neighborhoods can be repeated in the real-analytic context since it makes no use of partitions of unity. Here's a sketch.
By Lev's answer, a real-analytic manifold admits a real-analytic embedding into Euclidean space with its standard real-analytic structure. I think this implies a real-analytic manifold admits a real-analytic Riemannian metric.
The exponential map of this metric is also real-analytic (as a solution to analytic ODE), so an analytically embedded submanifold admits a real-analytic tubular neighborhood. We may then repeat this proof of Ehresmann's theorem using tubular neighborhoods.
Remark. This comment along with several other places in the literature assert Ehresmann's theorem fails in the real-analytic category but provide no examples.
answered Jun 20 at 20:45
ArrowArrow
4,0881 gold badge15 silver badges52 bronze badges
$endgroup$
add a comment
|
$begingroup$
I think Ehresmann's theorem does hold in the real-analytic category. A proof via tubular neighborhoods can be repeated in the real-analytic context since it makes no use of partitions of unity. Here's a sketch.
By Lev's answer, a real-analytic manifold admits a real-analytic embedding into Euclidean space with its standard real-analytic structure. I think this implies a real-analytic manifold admits a real-analytic Riemannian metric.
The exponential map of this metric is also real-analytic (as a solution to analytic ODE), so an analytically embedded submanifold admits a real-analytic tubular neighborhood. We may then repeat this proof of Ehresmann's theorem using tubular neighborhoods.
Remark. This comment along with several other places in the literature assert Ehresmann's theorem fails in the real-analytic category but provide no examples.
answered Jun 20 at 20:45
ArrowArrow
4,0881 gold badge15 silver badges52 bronze badges
$endgroup$
I think Ehresmann's theorem does hold in the real-analytic category. A proof via tubular neighborhoods can be repeated in the real-analytic context since it makes no use of partitions of unity. Here's a sketch.
By Lev's answer, a real-analytic manifold admits a real-analytic embedding into Euclidean space with its standard real-analytic structure. I think this implies a real-analytic manifold admits a real-analytic Riemannian metric.
The exponential map of this metric is also real-analytic (as a solution to analytic ODE), so an analytically embedded submanifold admits a real-analytic tubular neighborhood. We may then repeat this proof of Ehresmann's theorem using tubular neighborhoods.
Remark. This comment along with several other places in the literature assert Ehresmann's theorem fails in the real-analytic category but provide no examples.
answered Jun 20 at 20:45
ArrowArrow
4,0881 gold badge15 silver badges52 bronze badges
answered Jun 20 at 20:45
ArrowArrow
4,0881 gold badge15 silver badges52 bronze badges
answered Jun 20 at 20:45
ArrowArrow
4,0881 gold badge15 silver badges52 bronze badges
answered Jun 20 at 20:45
ArrowArrow
4,0881 gold badge15 silver badges52 bronze badges
4,0881 gold badge15 silver badges52 bronze badges
add a comment
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$begingroup$
The question in the title is just the opposite than the question in the body of the question. I suggest an edit for the coherence's sake.
$endgroup$
– Francesco Polizzi
Jun 14 at 10:35
$begingroup$
@FrancescoPolizzi done.
$endgroup$
– Arrow
Jun 14 at 10:47