Mixing
Doubt in the definition of principal bundle.
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I am following Kobayashi and Nomizu( Foundations of differential geometry) Volume 1.
In page number 50 while defining principle $G$-bundle $P(M,G)$ they said that the action of the Lie group $G$ is Free and didn't specify that the action is Proper. But they stated $M=P/G$ where $P/G$ is the quotient space of $P$ by the action of $G$. By equality I feel that they have assumed diffeomorphism between $M$ and $P/G$ and hence must have assumed a smooth manifold structure on $P/G$ . The first thing that came to my mind is the Quotient Manifold Theorem(QMT) in order to assume the smooth manifold structure on $P/G$. But in the hypothesis of (QMT) the action of $G$ is assumed to be Proper. So technically we can't apply (QMT) to give a differentiable structure on $P/G$ unless we prove that the the given Free action of $G$ is actually a Proper action.
My questions are following:
1 Do we have to prove that the action of $G$ is Proper or it is assumed to be proper so that we can apply (QMT)?
2 Or we dont have to use (QMT) and somehow in a different way we can give a diffrentiable structure on $P/G$ ?
3 Or they have not assumed any differentiable structure on $P/G$ and by the equality $M= P/G$ they just mean a bijection between two sets and later they gave the smooth manifold structure on $P/G$ using the differentiable structure of $M$?
Thanks in advance if somone can help me out!
differential-geometry lie-groups smooth-manifolds fiber-bundles principal-bundles
edited Feb 4 at 17:00
anomaly
17.8k42666
asked Feb 1 at 2:51
WandereradiWandereradi
838
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add a comment |
$begingroup$
I am following Kobayashi and Nomizu( Foundations of differential geometry) Volume 1.
In page number 50 while defining principle $G$-bundle $P(M,G)$ they said that the action of the Lie group $G$ is Free and didn't specify that the action is Proper. But they stated $M=P/G$ where $P/G$ is the quotient space of $P$ by the action of $G$. By equality I feel that they have assumed diffeomorphism between $M$ and $P/G$ and hence must have assumed a smooth manifold structure on $P/G$ . The first thing that came to my mind is the Quotient Manifold Theorem(QMT) in order to assume the smooth manifold structure on $P/G$. But in the hypothesis of (QMT) the action of $G$ is assumed to be Proper. So technically we can't apply (QMT) to give a differentiable structure on $P/G$ unless we prove that the the given Free action of $G$ is actually a Proper action.
My questions are following:
1 Do we have to prove that the action of $G$ is Proper or it is assumed to be proper so that we can apply (QMT)?
2 Or we dont have to use (QMT) and somehow in a different way we can give a diffrentiable structure on $P/G$ ?
3 Or they have not assumed any differentiable structure on $P/G$ and by the equality $M= P/G$ they just mean a bijection between two sets and later they gave the smooth manifold structure on $P/G$ using the differentiable structure of $M$?
Thanks in advance if somone can help me out!
differential-geometry lie-groups smooth-manifolds fiber-bundles principal-bundles
edited Feb 4 at 17:00
anomaly
17.8k42666
asked Feb 1 at 2:51
WandereradiWandereradi
838
$endgroup$
1
$begingroup$
math.stackexchange.com/questions/560371/…
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– Moishe Kohan
Feb 3 at 4:43
1
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@MoisheCohen Can you please specify what in particular you want to specify in that linked question/answer...
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– Praphulla Koushik
Feb 3 at 16:04
$begingroup$
MoisheCohen@ Thanks... I saw the link but couldn’t understand the relevance to my question..please can you elaborate a little about it?..
$endgroup$
– Wandereradi
Feb 3 at 16:18
add a comment |
$begingroup$
I am following Kobayashi and Nomizu( Foundations of differential geometry) Volume 1.
In page number 50 while defining principle $G$-bundle $P(M,G)$ they said that the action of the Lie group $G$ is Free and didn't specify that the action is Proper. But they stated $M=P/G$ where $P/G$ is the quotient space of $P$ by the action of $G$. By equality I feel that they have assumed diffeomorphism between $M$ and $P/G$ and hence must have assumed a smooth manifold structure on $P/G$ . The first thing that came to my mind is the Quotient Manifold Theorem(QMT) in order to assume the smooth manifold structure on $P/G$. But in the hypothesis of (QMT) the action of $G$ is assumed to be Proper. So technically we can't apply (QMT) to give a differentiable structure on $P/G$ unless we prove that the the given Free action of $G$ is actually a Proper action.
My questions are following:
1 Do we have to prove that the action of $G$ is Proper or it is assumed to be proper so that we can apply (QMT)?
2 Or we dont have to use (QMT) and somehow in a different way we can give a diffrentiable structure on $P/G$ ?
3 Or they have not assumed any differentiable structure on $P/G$ and by the equality $M= P/G$ they just mean a bijection between two sets and later they gave the smooth manifold structure on $P/G$ using the differentiable structure of $M$?
Thanks in advance if somone can help me out!
differential-geometry lie-groups smooth-manifolds fiber-bundles principal-bundles
edited Feb 4 at 17:00
anomaly
17.8k42666
asked Feb 1 at 2:51
WandereradiWandereradi
838
$endgroup$
I am following Kobayashi and Nomizu( Foundations of differential geometry) Volume 1.
In page number 50 while defining principle $G$-bundle $P(M,G)$ they said that the action of the Lie group $G$ is Free and didn't specify that the action is Proper. But they stated $M=P/G$ where $P/G$ is the quotient space of $P$ by the action of $G$. By equality I feel that they have assumed diffeomorphism between $M$ and $P/G$ and hence must have assumed a smooth manifold structure on $P/G$ . The first thing that came to my mind is the Quotient Manifold Theorem(QMT) in order to assume the smooth manifold structure on $P/G$. But in the hypothesis of (QMT) the action of $G$ is assumed to be Proper. So technically we can't apply (QMT) to give a differentiable structure on $P/G$ unless we prove that the the given Free action of $G$ is actually a Proper action.
My questions are following:
1 Do we have to prove that the action of $G$ is Proper or it is assumed to be proper so that we can apply (QMT)?
2 Or we dont have to use (QMT) and somehow in a different way we can give a diffrentiable structure on $P/G$ ?
3 Or they have not assumed any differentiable structure on $P/G$ and by the equality $M= P/G$ they just mean a bijection between two sets and later they gave the smooth manifold structure on $P/G$ using the differentiable structure of $M$?
Thanks in advance if somone can help me out!
differential-geometry lie-groups smooth-manifolds fiber-bundles principal-bundles
differential-geometry lie-groups smooth-manifolds fiber-bundles principal-bundles
edited Feb 4 at 17:00
anomaly
17.8k42666
asked Feb 1 at 2:51
WandereradiWandereradi
838
edited Feb 4 at 17:00
anomaly
17.8k42666
asked Feb 1 at 2:51
WandereradiWandereradi
838
edited Feb 4 at 17:00
anomaly
17.8k42666
edited Feb 4 at 17:00
anomaly
17.8k42666
edited Feb 4 at 17:00
anomaly
17.8k42666
17.8k42666
asked Feb 1 at 2:51
WandereradiWandereradi
838
asked Feb 1 at 2:51
WandereradiWandereradi
838
asked Feb 1 at 2:51
WandereradiWandereradi
838
838
1
$begingroup$
math.stackexchange.com/questions/560371/…
$endgroup$
– Moishe Kohan
Feb 3 at 4:43
1
$begingroup$
@MoisheCohen Can you please specify what in particular you want to specify in that linked question/answer...
$endgroup$
– Praphulla Koushik
Feb 3 at 16:04
$begingroup$
MoisheCohen@ Thanks... I saw the link but couldn’t understand the relevance to my question..please can you elaborate a little about it?..
$endgroup$
– Wandereradi
Feb 3 at 16:18
add a comment |
1
$begingroup$
math.stackexchange.com/questions/560371/…
$endgroup$
– Moishe Kohan
Feb 3 at 4:43
1
$begingroup$
@MoisheCohen Can you please specify what in particular you want to specify in that linked question/answer...
$endgroup$
– Praphulla Koushik
Feb 3 at 16:04
$begingroup$
MoisheCohen@ Thanks... I saw the link but couldn’t understand the relevance to my question..please can you elaborate a little about it?..
$endgroup$
– Wandereradi
Feb 3 at 16:18
1
1
$begingroup$
math.stackexchange.com/questions/560371/…
$endgroup$
– Moishe Kohan
Feb 3 at 4:43
$begingroup$
math.stackexchange.com/questions/560371/…
$endgroup$
– Moishe Kohan
Feb 3 at 4:43
1
1
$begingroup$
@MoisheCohen Can you please specify what in particular you want to specify in that linked question/answer...
$endgroup$
– Praphulla Koushik
Feb 3 at 16:04
$begingroup$
@MoisheCohen Can you please specify what in particular you want to specify in that linked question/answer...
$endgroup$
– Praphulla Koushik
Feb 3 at 16:04
$begingroup$
MoisheCohen@ Thanks... I saw the link but couldn’t understand the relevance to my question..please can you elaborate a little about it?..
$endgroup$
– Wandereradi
Feb 3 at 16:18
$begingroup$
MoisheCohen@ Thanks... I saw the link but couldn’t understand the relevance to my question..please can you elaborate a little about it?..
$endgroup$
– Wandereradi
Feb 3 at 16:18
add a comment |
2 Answers
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If you read closely what they write on page 50, you will notice that they make four requirements on the action of $G$ on $P$:
(0) You are given [ahead of time, as a part of a principal fiber bundle data] a smooth manifold $M$. [This part precedes their definition.]
(1) You are given a smooth free action $Gtimes Pto P$ of a Lie group $G$ on a smooth manifold $P$.
(2) The quotient space $P/G$, equipped with quotient topology [they did not specify this but this is the default assumption when dealing with quotient spaces of topological spaces] is homeomorphic to $M$. [Note that at this point it is meaningless to say "diffeomorphic" since $P/G$ is just a topological space.]
(3) The projection map $f: Pto P/Gto M$ [where the latter map is the homeomorphism in (2)] is locally a product in the smooth category, i.e. for every $xin M$ there is a neighborhood $U$ of $x$ in $M$ such that $f^{-1}(U)$ is $G$-equivariantly diffeomorphic to $Gtimes U$ where $G$ acts on itself via left multiplication and acts trivially on $U$.
I will refer to this as the KN-definition (of a principal fiber bundle).
This is a definition and you do not need to prove anything about it, so the answer to your question 1 is negative; the answer to your question 2 is that a smooth structure on $P/G$ is given by the requirements 0 and 2; the answer to your question 3 is also negative. In particular, the QMT (or the "slice theorem") is never relevant.
What I will prove now is that one can modify their definition so that the connection to the QMT/slice theorem is apparent.
Lemma A. KN definition implies that the action of $G$ on $P$ is proper.
This lemma will be a corollary of a stronger statement where we weaken the KN definition and still get properness of the action:
Lemma B. Suppose that $P$ is a Hausdorff locally compact topological space, $Gtimes Pto P$ is a continuous action of a topological group such that the quotient space $B=P/G$ is Hausdorff and the quotient map $q: Pto B$ is locally trivial in the topological sense: For every $xin B$ there is a neighborhood $U$ of $x$ in $B$ such that $q^{-1}(U)$ is $G$-equivariantly diffeomorphic to $Gtimes U$ where $G$ acts on itself via left multiplication and acts trivially on $U$. Then the action of $G$ on $P$ is proper. [Note that we have no manifold assumptions anywhere in this theorem, in particular, $G$ is not assumed to be a Lie group.]
Proof. Since $P$ is locally compact, it suffices to check that any two distinct points $x, yin P$ have neighborhoods $U_x, U_y$ in $P$ such that
$$
G_{U_x,U_y}:= {gin G: gU_xcap U_yne emptyset}
$$
is relatively compact in $G$.
Consider first the case when $q(x)ne q(y)$. Since $B$ is Hausdorff and $q(x)ne q(y)$, these points of $B$ have disjoint neighborhoods $V_x, V_y$. Therefore, $U_x:= q^{-1}(V_x)$, $U_y:= q^{-1}(V_y)$ are also disjoint. Since $G U_x=U_x$, $G U_y=U_y$,
$$
G_{U_x,U_y}= emptyset.
$$Suppose now that $q(x)=q(y)$, i.e. $Gx=Gy$. Thus, without loss of generality, we may assume that $x=y$. Local triviality of the projection
$$
q: Pto B
$$
means that there exists a neighborhood $V$ of $q(x)$ in $B$ such that $q^{-1}(V)$ is $G$-equivariantly homeomorphic to the product $Gtimes V$ (via a homeomorphism $f$). Then $f(x)= (g,q(x))$ for some $gin G$. Since $G$ acts on itself properly by left multiplication (I will leave this as an exercise), we can take as the required neighborhood of $x$ the subset $f^{-1}(Ktimes V)$, where $K$ is a relatively compact neighborhood of $gin G$. qed
It is clear that Lemma B implies Lemma A since manifolds are Hausdorff (at least, according to the standard definition).
Now, let us connect this discussion to the QMT and slice theorem proven in
R. Palais, On the existence of slices for actions of non-compact Lie groups. Ann. of Math. (2) 73 (1961) 295-323
Proposition C. Suppose that $P$ is a smooth manifold, $Gtimes Pto P$ is a smooth free action of a Lie group which satisfies the assumptions of Lemma B (i.e. the quotient $B=P/G$ is Hausdorff and the projection $q: Pto B$ is locally trivial). Then $B$ has structure of a smooth manifold $M$ such that the quotient map $Pto B=M$ is a principal fiber bundle in the KN sense. Moreover such a smooth structure on $M$ is unique. [Answering the question that you asked in a comment.]
Proof. By Lemma B, the action of $G$ on $P$ is proper. Therefore, the QMT (or slice theorem) applies and, therefore, every $G$-orbit $Gysubset P$ has a $G$-invariant neighborhood $Vsubset P$ which is $G$-equivariantly diffeomorphic to the product $Gtimes U$, where $U$ is an open subset in $R^n$ and $n=dim(P)-dim(G)$. Equivariance of this diffeomorphism yields a homeomorphism $h: V/Gto U$. Thus, we obtain an atlas on $B$ given by the maps $h: V/Gto Usubset R^n$. The fact that transition maps are smooth follows comes from the fact that the equivariant maps $tilde{h}: Vto Gtimes U$ were diffeomorphisms. Thus, $B$ is now equipped with the structure of a smooth manifold $M$. (Hausdorfness of $B$ was an assumption and 2nd countability of $B$ follows from 2nd countability of $P$.)
The projection map $q: Pto B=M$ is smooth by the construction. Since the maps $tilde{h}$ are diffeomorphisms, we also obtain that the local trivializations appearing in Part 3 of KN-definition are diffeomorphisms. This proves the existence part of the proposition.
Let us prove the uniqueness part. Suppose that $Wsubset B$ is an open subset, $h: Wto Usubset R^n, h': Wto U'subset R^n$ are charts of two smooth atlases on $B$ satisfying the KN-requirements, i.e. there exist $G$-equivariant diffeomorphisms
$$
tilde{h}: q^{-1}(W)to Gtimes U, tilde{h}': q^{-1}(W)to Gtimes U'
$$
which project to the maps $h, h'$. I claim that the transition maps $h'circ h^{-1}$ is smooth, hence, atlases are equivalent. Indeed, the composition
$$
tilde{h}'circ tilde{h}^{-1}: Gtimes Uto Gtimes U'
$$
is smooth (since $tilde{h}, tilde{h}'$ are diffeomorphisms) and, due to its $G$-equivariance, has the form
$$
(g,u)mapsto (phi(g), h'circ h^{-1}(u)).
$$
This implies smoothness of $h'circ h^{-1}$. qed
One last thing, I initially thought that in Lemma B it suffices to assume that the quotient space $B$ is Hausdorff to conclude properness of the action (i.e. that the local triviality part is redundant). I now realized that I was wrong:
Example. There is an example (derived from the Reeb foliation) of a smooth free action of $G={mathbb R}$ on $P={mathbb R}^2- {(0,0)}$ such that the quotient space $P/G$ is homeomorphic to $[0,infty)$ (and, hence, is Hausdorff) but the action of $G$ on $P$ is not proper.
I do not know if there are similar examples (a nonproper smooth free action of a Lie group on a smooth manifold, $Gtimes Xto X$) where $X/G$ is a topological manifold (without boundary). Here is what I do know:
Theorem. Suppose that $G$ is a Lie group with finitely many connected components acting smoothly and freely on a smooth manifold $X$ such that $B=X/G$ is a topological manifold $M$ which admits a smooth structure such that the quotient map $Xto M$ is a submersion. Then the action of $G$ on $X$ is proper and, hence, $Gtimes Xto B$ is a principal fiber bundle.
answered Feb 3 at 23:34
Moishe KohanMoishe Kohan
48.5k344110
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1
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Moishe Cohen@ Thanks a lot! ...I did not think before that local triviality will imply properness of the action. I will try to prove this implication !
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– Wandereradi
Feb 4 at 0:39
1
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@ Moishe Cohen Sir, one thing is confusing me. Local triviality (3rd requirement) assumes the existence of $pi$ ,a differentiable map from $P$ to $M$ which is taken to be the canonical projection when we identify $M$ with $P/G$(which is the 2nd requirement). So the paradox is if I assume local triviality imply properness and hence $P/G$ has the differentiable structure by QMT then while identifying $M$ and $P/G$ we just used set bijection (not diffeomorphism) as while identifying we dont have the privilege of local triviality.. So I couldn't understand how local trivial will imply proper?
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– Wandereradi
Feb 4 at 15:01
1
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"There is some smooth structure on $P/G$ and a diffeomorphism $Mrightarrow P/G$.. Then, local triviality condition implies that the action $G$ on $P$ is proper... Because the action of $G$ on $P$ is proper (it was free to start with), the quotient space $P/G$ has a unique smooth structure and all that... By uniqueness (unique smooth structure on $P/G$ such that $Prightarrow P/G$ is a submerison) in Quotient manifold theorem, it happens that, the some structure that we have started with is same to the structre coming from quotient manifold theorem".. Is this what you are saying Sir?
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– Praphulla Koushik
Feb 4 at 16:42
1
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"One does not need to know that $B=P/G$ is a manifold, only that it is Hausdorff" How do we even know that it Hausdorff... Is it that we are assuming? You are also saying $B$ is homeomorphic to a smooth manifold... This is something I did not expect :) Not diffeomorphic but homeomorphic... Can you please give a reference where one can see this explanation...
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– Praphulla Koushik
Feb 4 at 18:41
1
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@PraphullaKoushik: Manifolds are normally assumed to be Hausdorff (and 2nd countable). In fact, I just realized that one does not need local triviality to conclude properness of the action, all you need to assume that $P/G$ is Hausdorff.
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– Moishe Kohan
Feb 5 at 16:16
|
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Prof. Jack Lee proved here that action of $G$ on $P$ is proper in a principal $G$ bunlde. The problem there is, the user who asked the question does not mention what definition of principal bundle he is using. I only see that he is using local trivialization. He is also assuming that the quotient has a smooth structure.
I am only giving link to the answer. I will make this answer self explanatory when I get time.
answered Feb 5 at 5:26
Praphulla KoushikPraphulla Koushik
203119
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$begingroup$
If you read closely what they write on page 50, you will notice that they make four requirements on the action of $G$ on $P$:
(0) You are given [ahead of time, as a part of a principal fiber bundle data] a smooth manifold $M$. [This part precedes their definition.]
(1) You are given a smooth free action $Gtimes Pto P$ of a Lie group $G$ on a smooth manifold $P$.
(2) The quotient space $P/G$, equipped with quotient topology [they did not specify this but this is the default assumption when dealing with quotient spaces of topological spaces] is homeomorphic to $M$. [Note that at this point it is meaningless to say "diffeomorphic" since $P/G$ is just a topological space.]
(3) The projection map $f: Pto P/Gto M$ [where the latter map is the homeomorphism in (2)] is locally a product in the smooth category, i.e. for every $xin M$ there is a neighborhood $U$ of $x$ in $M$ such that $f^{-1}(U)$ is $G$-equivariantly diffeomorphic to $Gtimes U$ where $G$ acts on itself via left multiplication and acts trivially on $U$.
I will refer to this as the KN-definition (of a principal fiber bundle).
This is a definition and you do not need to prove anything about it, so the answer to your question 1 is negative; the answer to your question 2 is that a smooth structure on $P/G$ is given by the requirements 0 and 2; the answer to your question 3 is also negative. In particular, the QMT (or the "slice theorem") is never relevant.
What I will prove now is that one can modify their definition so that the connection to the QMT/slice theorem is apparent.
Lemma A. KN definition implies that the action of $G$ on $P$ is proper.
This lemma will be a corollary of a stronger statement where we weaken the KN definition and still get properness of the action:
Lemma B. Suppose that $P$ is a Hausdorff locally compact topological space, $Gtimes Pto P$ is a continuous action of a topological group such that the quotient space $B=P/G$ is Hausdorff and the quotient map $q: Pto B$ is locally trivial in the topological sense: For every $xin B$ there is a neighborhood $U$ of $x$ in $B$ such that $q^{-1}(U)$ is $G$-equivariantly diffeomorphic to $Gtimes U$ where $G$ acts on itself via left multiplication and acts trivially on $U$. Then the action of $G$ on $P$ is proper. [Note that we have no manifold assumptions anywhere in this theorem, in particular, $G$ is not assumed to be a Lie group.]
Proof. Since $P$ is locally compact, it suffices to check that any two distinct points $x, yin P$ have neighborhoods $U_x, U_y$ in $P$ such that
$$
G_{U_x,U_y}:= {gin G: gU_xcap U_yne emptyset}
$$
is relatively compact in $G$.
Consider first the case when $q(x)ne q(y)$. Since $B$ is Hausdorff and $q(x)ne q(y)$, these points of $B$ have disjoint neighborhoods $V_x, V_y$. Therefore, $U_x:= q^{-1}(V_x)$, $U_y:= q^{-1}(V_y)$ are also disjoint. Since $G U_x=U_x$, $G U_y=U_y$,
$$
G_{U_x,U_y}= emptyset.
$$Suppose now that $q(x)=q(y)$, i.e. $Gx=Gy$. Thus, without loss of generality, we may assume that $x=y$. Local triviality of the projection
$$
q: Pto B
$$
means that there exists a neighborhood $V$ of $q(x)$ in $B$ such that $q^{-1}(V)$ is $G$-equivariantly homeomorphic to the product $Gtimes V$ (via a homeomorphism $f$). Then $f(x)= (g,q(x))$ for some $gin G$. Since $G$ acts on itself properly by left multiplication (I will leave this as an exercise), we can take as the required neighborhood of $x$ the subset $f^{-1}(Ktimes V)$, where $K$ is a relatively compact neighborhood of $gin G$. qed
It is clear that Lemma B implies Lemma A since manifolds are Hausdorff (at least, according to the standard definition).
Now, let us connect this discussion to the QMT and slice theorem proven in
R. Palais, On the existence of slices for actions of non-compact Lie groups. Ann. of Math. (2) 73 (1961) 295-323
Proposition C. Suppose that $P$ is a smooth manifold, $Gtimes Pto P$ is a smooth free action of a Lie group which satisfies the assumptions of Lemma B (i.e. the quotient $B=P/G$ is Hausdorff and the projection $q: Pto B$ is locally trivial). Then $B$ has structure of a smooth manifold $M$ such that the quotient map $Pto B=M$ is a principal fiber bundle in the KN sense. Moreover such a smooth structure on $M$ is unique. [Answering the question that you asked in a comment.]
Proof. By Lemma B, the action of $G$ on $P$ is proper. Therefore, the QMT (or slice theorem) applies and, therefore, every $G$-orbit $Gysubset P$ has a $G$-invariant neighborhood $Vsubset P$ which is $G$-equivariantly diffeomorphic to the product $Gtimes U$, where $U$ is an open subset in $R^n$ and $n=dim(P)-dim(G)$. Equivariance of this diffeomorphism yields a homeomorphism $h: V/Gto U$. Thus, we obtain an atlas on $B$ given by the maps $h: V/Gto Usubset R^n$. The fact that transition maps are smooth follows comes from the fact that the equivariant maps $tilde{h}: Vto Gtimes U$ were diffeomorphisms. Thus, $B$ is now equipped with the structure of a smooth manifold $M$. (Hausdorfness of $B$ was an assumption and 2nd countability of $B$ follows from 2nd countability of $P$.)
The projection map $q: Pto B=M$ is smooth by the construction. Since the maps $tilde{h}$ are diffeomorphisms, we also obtain that the local trivializations appearing in Part 3 of KN-definition are diffeomorphisms. This proves the existence part of the proposition.
Let us prove the uniqueness part. Suppose that $Wsubset B$ is an open subset, $h: Wto Usubset R^n, h': Wto U'subset R^n$ are charts of two smooth atlases on $B$ satisfying the KN-requirements, i.e. there exist $G$-equivariant diffeomorphisms
$$
tilde{h}: q^{-1}(W)to Gtimes U, tilde{h}': q^{-1}(W)to Gtimes U'
$$
which project to the maps $h, h'$. I claim that the transition maps $h'circ h^{-1}$ is smooth, hence, atlases are equivalent. Indeed, the composition
$$
tilde{h}'circ tilde{h}^{-1}: Gtimes Uto Gtimes U'
$$
is smooth (since $tilde{h}, tilde{h}'$ are diffeomorphisms) and, due to its $G$-equivariance, has the form
$$
(g,u)mapsto (phi(g), h'circ h^{-1}(u)).
$$
This implies smoothness of $h'circ h^{-1}$. qed
One last thing, I initially thought that in Lemma B it suffices to assume that the quotient space $B$ is Hausdorff to conclude properness of the action (i.e. that the local triviality part is redundant). I now realized that I was wrong:
Example. There is an example (derived from the Reeb foliation) of a smooth free action of $G={mathbb R}$ on $P={mathbb R}^2- {(0,0)}$ such that the quotient space $P/G$ is homeomorphic to $[0,infty)$ (and, hence, is Hausdorff) but the action of $G$ on $P$ is not proper.
I do not know if there are similar examples (a nonproper smooth free action of a Lie group on a smooth manifold, $Gtimes Xto X$) where $X/G$ is a topological manifold (without boundary). Here is what I do know:
Theorem. Suppose that $G$ is a Lie group with finitely many connected components acting smoothly and freely on a smooth manifold $X$ such that $B=X/G$ is a topological manifold $M$ which admits a smooth structure such that the quotient map $Xto M$ is a submersion. Then the action of $G$ on $X$ is proper and, hence, $Gtimes Xto B$ is a principal fiber bundle.
answered Feb 3 at 23:34
Moishe KohanMoishe Kohan
48.5k344110
$endgroup$
1
$begingroup$
Moishe Cohen@ Thanks a lot! ...I did not think before that local triviality will imply properness of the action. I will try to prove this implication !
$endgroup$
– Wandereradi
Feb 4 at 0:39
1
$begingroup$
@ Moishe Cohen Sir, one thing is confusing me. Local triviality (3rd requirement) assumes the existence of $pi$ ,a differentiable map from $P$ to $M$ which is taken to be the canonical projection when we identify $M$ with $P/G$(which is the 2nd requirement). So the paradox is if I assume local triviality imply properness and hence $P/G$ has the differentiable structure by QMT then while identifying $M$ and $P/G$ we just used set bijection (not diffeomorphism) as while identifying we dont have the privilege of local triviality.. So I couldn't understand how local trivial will imply proper?
$endgroup$
– Wandereradi
Feb 4 at 15:01
1
$begingroup$
"There is some smooth structure on $P/G$ and a diffeomorphism $Mrightarrow P/G$.. Then, local triviality condition implies that the action $G$ on $P$ is proper... Because the action of $G$ on $P$ is proper (it was free to start with), the quotient space $P/G$ has a unique smooth structure and all that... By uniqueness (unique smooth structure on $P/G$ such that $Prightarrow P/G$ is a submerison) in Quotient manifold theorem, it happens that, the some structure that we have started with is same to the structre coming from quotient manifold theorem".. Is this what you are saying Sir?
$endgroup$
– Praphulla Koushik
Feb 4 at 16:42
1
$begingroup$
"One does not need to know that $B=P/G$ is a manifold, only that it is Hausdorff" How do we even know that it Hausdorff... Is it that we are assuming? You are also saying $B$ is homeomorphic to a smooth manifold... This is something I did not expect :) Not diffeomorphic but homeomorphic... Can you please give a reference where one can see this explanation...
$endgroup$
– Praphulla Koushik
Feb 4 at 18:41
1
$begingroup$
@PraphullaKoushik: Manifolds are normally assumed to be Hausdorff (and 2nd countable). In fact, I just realized that one does not need local triviality to conclude properness of the action, all you need to assume that $P/G$ is Hausdorff.
$endgroup$
– Moishe Kohan
Feb 5 at 16:16
|
show 8 more comments
$begingroup$
If you read closely what they write on page 50, you will notice that they make four requirements on the action of $G$ on $P$:
(0) You are given [ahead of time, as a part of a principal fiber bundle data] a smooth manifold $M$. [This part precedes their definition.]
(1) You are given a smooth free action $Gtimes Pto P$ of a Lie group $G$ on a smooth manifold $P$.
(2) The quotient space $P/G$, equipped with quotient topology [they did not specify this but this is the default assumption when dealing with quotient spaces of topological spaces] is homeomorphic to $M$. [Note that at this point it is meaningless to say "diffeomorphic" since $P/G$ is just a topological space.]
(3) The projection map $f: Pto P/Gto M$ [where the latter map is the homeomorphism in (2)] is locally a product in the smooth category, i.e. for every $xin M$ there is a neighborhood $U$ of $x$ in $M$ such that $f^{-1}(U)$ is $G$-equivariantly diffeomorphic to $Gtimes U$ where $G$ acts on itself via left multiplication and acts trivially on $U$.
I will refer to this as the KN-definition (of a principal fiber bundle).
This is a definition and you do not need to prove anything about it, so the answer to your question 1 is negative; the answer to your question 2 is that a smooth structure on $P/G$ is given by the requirements 0 and 2; the answer to your question 3 is also negative. In particular, the QMT (or the "slice theorem") is never relevant.
What I will prove now is that one can modify their definition so that the connection to the QMT/slice theorem is apparent.
Lemma A. KN definition implies that the action of $G$ on $P$ is proper.
This lemma will be a corollary of a stronger statement where we weaken the KN definition and still get properness of the action:
Lemma B. Suppose that $P$ is a Hausdorff locally compact topological space, $Gtimes Pto P$ is a continuous action of a topological group such that the quotient space $B=P/G$ is Hausdorff and the quotient map $q: Pto B$ is locally trivial in the topological sense: For every $xin B$ there is a neighborhood $U$ of $x$ in $B$ such that $q^{-1}(U)$ is $G$-equivariantly diffeomorphic to $Gtimes U$ where $G$ acts on itself via left multiplication and acts trivially on $U$. Then the action of $G$ on $P$ is proper. [Note that we have no manifold assumptions anywhere in this theorem, in particular, $G$ is not assumed to be a Lie group.]
Proof. Since $P$ is locally compact, it suffices to check that any two distinct points $x, yin P$ have neighborhoods $U_x, U_y$ in $P$ such that
$$
G_{U_x,U_y}:= {gin G: gU_xcap U_yne emptyset}
$$
is relatively compact in $G$.
Consider first the case when $q(x)ne q(y)$. Since $B$ is Hausdorff and $q(x)ne q(y)$, these points of $B$ have disjoint neighborhoods $V_x, V_y$. Therefore, $U_x:= q^{-1}(V_x)$, $U_y:= q^{-1}(V_y)$ are also disjoint. Since $G U_x=U_x$, $G U_y=U_y$,
$$
G_{U_x,U_y}= emptyset.
$$Suppose now that $q(x)=q(y)$, i.e. $Gx=Gy$. Thus, without loss of generality, we may assume that $x=y$. Local triviality of the projection
$$
q: Pto B
$$
means that there exists a neighborhood $V$ of $q(x)$ in $B$ such that $q^{-1}(V)$ is $G$-equivariantly homeomorphic to the product $Gtimes V$ (via a homeomorphism $f$). Then $f(x)= (g,q(x))$ for some $gin G$. Since $G$ acts on itself properly by left multiplication (I will leave this as an exercise), we can take as the required neighborhood of $x$ the subset $f^{-1}(Ktimes V)$, where $K$ is a relatively compact neighborhood of $gin G$. qed
It is clear that Lemma B implies Lemma A since manifolds are Hausdorff (at least, according to the standard definition).
Now, let us connect this discussion to the QMT and slice theorem proven in
R. Palais, On the existence of slices for actions of non-compact Lie groups. Ann. of Math. (2) 73 (1961) 295-323
Proposition C. Suppose that $P$ is a smooth manifold, $Gtimes Pto P$ is a smooth free action of a Lie group which satisfies the assumptions of Lemma B (i.e. the quotient $B=P/G$ is Hausdorff and the projection $q: Pto B$ is locally trivial). Then $B$ has structure of a smooth manifold $M$ such that the quotient map $Pto B=M$ is a principal fiber bundle in the KN sense. Moreover such a smooth structure on $M$ is unique. [Answering the question that you asked in a comment.]
Proof. By Lemma B, the action of $G$ on $P$ is proper. Therefore, the QMT (or slice theorem) applies and, therefore, every $G$-orbit $Gysubset P$ has a $G$-invariant neighborhood $Vsubset P$ which is $G$-equivariantly diffeomorphic to the product $Gtimes U$, where $U$ is an open subset in $R^n$ and $n=dim(P)-dim(G)$. Equivariance of this diffeomorphism yields a homeomorphism $h: V/Gto U$. Thus, we obtain an atlas on $B$ given by the maps $h: V/Gto Usubset R^n$. The fact that transition maps are smooth follows comes from the fact that the equivariant maps $tilde{h}: Vto Gtimes U$ were diffeomorphisms. Thus, $B$ is now equipped with the structure of a smooth manifold $M$. (Hausdorfness of $B$ was an assumption and 2nd countability of $B$ follows from 2nd countability of $P$.)
The projection map $q: Pto B=M$ is smooth by the construction. Since the maps $tilde{h}$ are diffeomorphisms, we also obtain that the local trivializations appearing in Part 3 of KN-definition are diffeomorphisms. This proves the existence part of the proposition.
Let us prove the uniqueness part. Suppose that $Wsubset B$ is an open subset, $h: Wto Usubset R^n, h': Wto U'subset R^n$ are charts of two smooth atlases on $B$ satisfying the KN-requirements, i.e. there exist $G$-equivariant diffeomorphisms
$$
tilde{h}: q^{-1}(W)to Gtimes U, tilde{h}': q^{-1}(W)to Gtimes U'
$$
which project to the maps $h, h'$. I claim that the transition maps $h'circ h^{-1}$ is smooth, hence, atlases are equivalent. Indeed, the composition
$$
tilde{h}'circ tilde{h}^{-1}: Gtimes Uto Gtimes U'
$$
is smooth (since $tilde{h}, tilde{h}'$ are diffeomorphisms) and, due to its $G$-equivariance, has the form
$$
(g,u)mapsto (phi(g), h'circ h^{-1}(u)).
$$
This implies smoothness of $h'circ h^{-1}$. qed
One last thing, I initially thought that in Lemma B it suffices to assume that the quotient space $B$ is Hausdorff to conclude properness of the action (i.e. that the local triviality part is redundant). I now realized that I was wrong:
Example. There is an example (derived from the Reeb foliation) of a smooth free action of $G={mathbb R}$ on $P={mathbb R}^2- {(0,0)}$ such that the quotient space $P/G$ is homeomorphic to $[0,infty)$ (and, hence, is Hausdorff) but the action of $G$ on $P$ is not proper.
I do not know if there are similar examples (a nonproper smooth free action of a Lie group on a smooth manifold, $Gtimes Xto X$) where $X/G$ is a topological manifold (without boundary). Here is what I do know:
Theorem. Suppose that $G$ is a Lie group with finitely many connected components acting smoothly and freely on a smooth manifold $X$ such that $B=X/G$ is a topological manifold $M$ which admits a smooth structure such that the quotient map $Xto M$ is a submersion. Then the action of $G$ on $X$ is proper and, hence, $Gtimes Xto B$ is a principal fiber bundle.
answered Feb 3 at 23:34
Moishe KohanMoishe Kohan
48.5k344110
$endgroup$
1
$begingroup$
Moishe Cohen@ Thanks a lot! ...I did not think before that local triviality will imply properness of the action. I will try to prove this implication !
$endgroup$
– Wandereradi
Feb 4 at 0:39
1
$begingroup$
@ Moishe Cohen Sir, one thing is confusing me. Local triviality (3rd requirement) assumes the existence of $pi$ ,a differentiable map from $P$ to $M$ which is taken to be the canonical projection when we identify $M$ with $P/G$(which is the 2nd requirement). So the paradox is if I assume local triviality imply properness and hence $P/G$ has the differentiable structure by QMT then while identifying $M$ and $P/G$ we just used set bijection (not diffeomorphism) as while identifying we dont have the privilege of local triviality.. So I couldn't understand how local trivial will imply proper?
$endgroup$
– Wandereradi
Feb 4 at 15:01
1
$begingroup$
"There is some smooth structure on $P/G$ and a diffeomorphism $Mrightarrow P/G$.. Then, local triviality condition implies that the action $G$ on $P$ is proper... Because the action of $G$ on $P$ is proper (it was free to start with), the quotient space $P/G$ has a unique smooth structure and all that... By uniqueness (unique smooth structure on $P/G$ such that $Prightarrow P/G$ is a submerison) in Quotient manifold theorem, it happens that, the some structure that we have started with is same to the structre coming from quotient manifold theorem".. Is this what you are saying Sir?
$endgroup$
– Praphulla Koushik
Feb 4 at 16:42
1
$begingroup$
"One does not need to know that $B=P/G$ is a manifold, only that it is Hausdorff" How do we even know that it Hausdorff... Is it that we are assuming? You are also saying $B$ is homeomorphic to a smooth manifold... This is something I did not expect :) Not diffeomorphic but homeomorphic... Can you please give a reference where one can see this explanation...
$endgroup$
– Praphulla Koushik
Feb 4 at 18:41
1
$begingroup$
@PraphullaKoushik: Manifolds are normally assumed to be Hausdorff (and 2nd countable). In fact, I just realized that one does not need local triviality to conclude properness of the action, all you need to assume that $P/G$ is Hausdorff.
$endgroup$
– Moishe Kohan
Feb 5 at 16:16
|
show 8 more comments
$begingroup$
If you read closely what they write on page 50, you will notice that they make four requirements on the action of $G$ on $P$:
(0) You are given [ahead of time, as a part of a principal fiber bundle data] a smooth manifold $M$. [This part precedes their definition.]
(1) You are given a smooth free action $Gtimes Pto P$ of a Lie group $G$ on a smooth manifold $P$.
(2) The quotient space $P/G$, equipped with quotient topology [they did not specify this but this is the default assumption when dealing with quotient spaces of topological spaces] is homeomorphic to $M$. [Note that at this point it is meaningless to say "diffeomorphic" since $P/G$ is just a topological space.]
(3) The projection map $f: Pto P/Gto M$ [where the latter map is the homeomorphism in (2)] is locally a product in the smooth category, i.e. for every $xin M$ there is a neighborhood $U$ of $x$ in $M$ such that $f^{-1}(U)$ is $G$-equivariantly diffeomorphic to $Gtimes U$ where $G$ acts on itself via left multiplication and acts trivially on $U$.
I will refer to this as the KN-definition (of a principal fiber bundle).
This is a definition and you do not need to prove anything about it, so the answer to your question 1 is negative; the answer to your question 2 is that a smooth structure on $P/G$ is given by the requirements 0 and 2; the answer to your question 3 is also negative. In particular, the QMT (or the "slice theorem") is never relevant.
What I will prove now is that one can modify their definition so that the connection to the QMT/slice theorem is apparent.
Lemma A. KN definition implies that the action of $G$ on $P$ is proper.
This lemma will be a corollary of a stronger statement where we weaken the KN definition and still get properness of the action:
Lemma B. Suppose that $P$ is a Hausdorff locally compact topological space, $Gtimes Pto P$ is a continuous action of a topological group such that the quotient space $B=P/G$ is Hausdorff and the quotient map $q: Pto B$ is locally trivial in the topological sense: For every $xin B$ there is a neighborhood $U$ of $x$ in $B$ such that $q^{-1}(U)$ is $G$-equivariantly diffeomorphic to $Gtimes U$ where $G$ acts on itself via left multiplication and acts trivially on $U$. Then the action of $G$ on $P$ is proper. [Note that we have no manifold assumptions anywhere in this theorem, in particular, $G$ is not assumed to be a Lie group.]
Proof. Since $P$ is locally compact, it suffices to check that any two distinct points $x, yin P$ have neighborhoods $U_x, U_y$ in $P$ such that
$$
G_{U_x,U_y}:= {gin G: gU_xcap U_yne emptyset}
$$
is relatively compact in $G$.
Consider first the case when $q(x)ne q(y)$. Since $B$ is Hausdorff and $q(x)ne q(y)$, these points of $B$ have disjoint neighborhoods $V_x, V_y$. Therefore, $U_x:= q^{-1}(V_x)$, $U_y:= q^{-1}(V_y)$ are also disjoint. Since $G U_x=U_x$, $G U_y=U_y$,
$$
G_{U_x,U_y}= emptyset.
$$Suppose now that $q(x)=q(y)$, i.e. $Gx=Gy$. Thus, without loss of generality, we may assume that $x=y$. Local triviality of the projection
$$
q: Pto B
$$
means that there exists a neighborhood $V$ of $q(x)$ in $B$ such that $q^{-1}(V)$ is $G$-equivariantly homeomorphic to the product $Gtimes V$ (via a homeomorphism $f$). Then $f(x)= (g,q(x))$ for some $gin G$. Since $G$ acts on itself properly by left multiplication (I will leave this as an exercise), we can take as the required neighborhood of $x$ the subset $f^{-1}(Ktimes V)$, where $K$ is a relatively compact neighborhood of $gin G$. qed
It is clear that Lemma B implies Lemma A since manifolds are Hausdorff (at least, according to the standard definition).
Now, let us connect this discussion to the QMT and slice theorem proven in
R. Palais, On the existence of slices for actions of non-compact Lie groups. Ann. of Math. (2) 73 (1961) 295-323
Proposition C. Suppose that $P$ is a smooth manifold, $Gtimes Pto P$ is a smooth free action of a Lie group which satisfies the assumptions of Lemma B (i.e. the quotient $B=P/G$ is Hausdorff and the projection $q: Pto B$ is locally trivial). Then $B$ has structure of a smooth manifold $M$ such that the quotient map $Pto B=M$ is a principal fiber bundle in the KN sense. Moreover such a smooth structure on $M$ is unique. [Answering the question that you asked in a comment.]
Proof. By Lemma B, the action of $G$ on $P$ is proper. Therefore, the QMT (or slice theorem) applies and, therefore, every $G$-orbit $Gysubset P$ has a $G$-invariant neighborhood $Vsubset P$ which is $G$-equivariantly diffeomorphic to the product $Gtimes U$, where $U$ is an open subset in $R^n$ and $n=dim(P)-dim(G)$. Equivariance of this diffeomorphism yields a homeomorphism $h: V/Gto U$. Thus, we obtain an atlas on $B$ given by the maps $h: V/Gto Usubset R^n$. The fact that transition maps are smooth follows comes from the fact that the equivariant maps $tilde{h}: Vto Gtimes U$ were diffeomorphisms. Thus, $B$ is now equipped with the structure of a smooth manifold $M$. (Hausdorfness of $B$ was an assumption and 2nd countability of $B$ follows from 2nd countability of $P$.)
The projection map $q: Pto B=M$ is smooth by the construction. Since the maps $tilde{h}$ are diffeomorphisms, we also obtain that the local trivializations appearing in Part 3 of KN-definition are diffeomorphisms. This proves the existence part of the proposition.
Let us prove the uniqueness part. Suppose that $Wsubset B$ is an open subset, $h: Wto Usubset R^n, h': Wto U'subset R^n$ are charts of two smooth atlases on $B$ satisfying the KN-requirements, i.e. there exist $G$-equivariant diffeomorphisms
$$
tilde{h}: q^{-1}(W)to Gtimes U, tilde{h}': q^{-1}(W)to Gtimes U'
$$
which project to the maps $h, h'$. I claim that the transition maps $h'circ h^{-1}$ is smooth, hence, atlases are equivalent. Indeed, the composition
$$
tilde{h}'circ tilde{h}^{-1}: Gtimes Uto Gtimes U'
$$
is smooth (since $tilde{h}, tilde{h}'$ are diffeomorphisms) and, due to its $G$-equivariance, has the form
$$
(g,u)mapsto (phi(g), h'circ h^{-1}(u)).
$$
This implies smoothness of $h'circ h^{-1}$. qed
One last thing, I initially thought that in Lemma B it suffices to assume that the quotient space $B$ is Hausdorff to conclude properness of the action (i.e. that the local triviality part is redundant). I now realized that I was wrong:
Example. There is an example (derived from the Reeb foliation) of a smooth free action of $G={mathbb R}$ on $P={mathbb R}^2- {(0,0)}$ such that the quotient space $P/G$ is homeomorphic to $[0,infty)$ (and, hence, is Hausdorff) but the action of $G$ on $P$ is not proper.
I do not know if there are similar examples (a nonproper smooth free action of a Lie group on a smooth manifold, $Gtimes Xto X$) where $X/G$ is a topological manifold (without boundary). Here is what I do know:
Theorem. Suppose that $G$ is a Lie group with finitely many connected components acting smoothly and freely on a smooth manifold $X$ such that $B=X/G$ is a topological manifold $M$ which admits a smooth structure such that the quotient map $Xto M$ is a submersion. Then the action of $G$ on $X$ is proper and, hence, $Gtimes Xto B$ is a principal fiber bundle.
answered Feb 3 at 23:34
Moishe KohanMoishe Kohan
48.5k344110
$endgroup$
If you read closely what they write on page 50, you will notice that they make four requirements on the action of $G$ on $P$:
(0) You are given [ahead of time, as a part of a principal fiber bundle data] a smooth manifold $M$. [This part precedes their definition.]
(1) You are given a smooth free action $Gtimes Pto P$ of a Lie group $G$ on a smooth manifold $P$.
(2) The quotient space $P/G$, equipped with quotient topology [they did not specify this but this is the default assumption when dealing with quotient spaces of topological spaces] is homeomorphic to $M$. [Note that at this point it is meaningless to say "diffeomorphic" since $P/G$ is just a topological space.]
(3) The projection map $f: Pto P/Gto M$ [where the latter map is the homeomorphism in (2)] is locally a product in the smooth category, i.e. for every $xin M$ there is a neighborhood $U$ of $x$ in $M$ such that $f^{-1}(U)$ is $G$-equivariantly diffeomorphic to $Gtimes U$ where $G$ acts on itself via left multiplication and acts trivially on $U$.
I will refer to this as the KN-definition (of a principal fiber bundle).
This is a definition and you do not need to prove anything about it, so the answer to your question 1 is negative; the answer to your question 2 is that a smooth structure on $P/G$ is given by the requirements 0 and 2; the answer to your question 3 is also negative. In particular, the QMT (or the "slice theorem") is never relevant.
What I will prove now is that one can modify their definition so that the connection to the QMT/slice theorem is apparent.
Lemma A. KN definition implies that the action of $G$ on $P$ is proper.
This lemma will be a corollary of a stronger statement where we weaken the KN definition and still get properness of the action:
Lemma B. Suppose that $P$ is a Hausdorff locally compact topological space, $Gtimes Pto P$ is a continuous action of a topological group such that the quotient space $B=P/G$ is Hausdorff and the quotient map $q: Pto B$ is locally trivial in the topological sense: For every $xin B$ there is a neighborhood $U$ of $x$ in $B$ such that $q^{-1}(U)$ is $G$-equivariantly diffeomorphic to $Gtimes U$ where $G$ acts on itself via left multiplication and acts trivially on $U$. Then the action of $G$ on $P$ is proper. [Note that we have no manifold assumptions anywhere in this theorem, in particular, $G$ is not assumed to be a Lie group.]
Proof. Since $P$ is locally compact, it suffices to check that any two distinct points $x, yin P$ have neighborhoods $U_x, U_y$ in $P$ such that
$$
G_{U_x,U_y}:= {gin G: gU_xcap U_yne emptyset}
$$
is relatively compact in $G$.
Consider first the case when $q(x)ne q(y)$. Since $B$ is Hausdorff and $q(x)ne q(y)$, these points of $B$ have disjoint neighborhoods $V_x, V_y$. Therefore, $U_x:= q^{-1}(V_x)$, $U_y:= q^{-1}(V_y)$ are also disjoint. Since $G U_x=U_x$, $G U_y=U_y$,
$$
G_{U_x,U_y}= emptyset.
$$Suppose now that $q(x)=q(y)$, i.e. $Gx=Gy$. Thus, without loss of generality, we may assume that $x=y$. Local triviality of the projection
$$
q: Pto B
$$
means that there exists a neighborhood $V$ of $q(x)$ in $B$ such that $q^{-1}(V)$ is $G$-equivariantly homeomorphic to the product $Gtimes V$ (via a homeomorphism $f$). Then $f(x)= (g,q(x))$ for some $gin G$. Since $G$ acts on itself properly by left multiplication (I will leave this as an exercise), we can take as the required neighborhood of $x$ the subset $f^{-1}(Ktimes V)$, where $K$ is a relatively compact neighborhood of $gin G$. qed
It is clear that Lemma B implies Lemma A since manifolds are Hausdorff (at least, according to the standard definition).
Now, let us connect this discussion to the QMT and slice theorem proven in
R. Palais, On the existence of slices for actions of non-compact Lie groups. Ann. of Math. (2) 73 (1961) 295-323
Proposition C. Suppose that $P$ is a smooth manifold, $Gtimes Pto P$ is a smooth free action of a Lie group which satisfies the assumptions of Lemma B (i.e. the quotient $B=P/G$ is Hausdorff and the projection $q: Pto B$ is locally trivial). Then $B$ has structure of a smooth manifold $M$ such that the quotient map $Pto B=M$ is a principal fiber bundle in the KN sense. Moreover such a smooth structure on $M$ is unique. [Answering the question that you asked in a comment.]
Proof. By Lemma B, the action of $G$ on $P$ is proper. Therefore, the QMT (or slice theorem) applies and, therefore, every $G$-orbit $Gysubset P$ has a $G$-invariant neighborhood $Vsubset P$ which is $G$-equivariantly diffeomorphic to the product $Gtimes U$, where $U$ is an open subset in $R^n$ and $n=dim(P)-dim(G)$. Equivariance of this diffeomorphism yields a homeomorphism $h: V/Gto U$. Thus, we obtain an atlas on $B$ given by the maps $h: V/Gto Usubset R^n$. The fact that transition maps are smooth follows comes from the fact that the equivariant maps $tilde{h}: Vto Gtimes U$ were diffeomorphisms. Thus, $B$ is now equipped with the structure of a smooth manifold $M$. (Hausdorfness of $B$ was an assumption and 2nd countability of $B$ follows from 2nd countability of $P$.)
The projection map $q: Pto B=M$ is smooth by the construction. Since the maps $tilde{h}$ are diffeomorphisms, we also obtain that the local trivializations appearing in Part 3 of KN-definition are diffeomorphisms. This proves the existence part of the proposition.
Let us prove the uniqueness part. Suppose that $Wsubset B$ is an open subset, $h: Wto Usubset R^n, h': Wto U'subset R^n$ are charts of two smooth atlases on $B$ satisfying the KN-requirements, i.e. there exist $G$-equivariant diffeomorphisms
$$
tilde{h}: q^{-1}(W)to Gtimes U, tilde{h}': q^{-1}(W)to Gtimes U'
$$
which project to the maps $h, h'$. I claim that the transition maps $h'circ h^{-1}$ is smooth, hence, atlases are equivalent. Indeed, the composition
$$
tilde{h}'circ tilde{h}^{-1}: Gtimes Uto Gtimes U'
$$
is smooth (since $tilde{h}, tilde{h}'$ are diffeomorphisms) and, due to its $G$-equivariance, has the form
$$
(g,u)mapsto (phi(g), h'circ h^{-1}(u)).
$$
This implies smoothness of $h'circ h^{-1}$. qed
One last thing, I initially thought that in Lemma B it suffices to assume that the quotient space $B$ is Hausdorff to conclude properness of the action (i.e. that the local triviality part is redundant). I now realized that I was wrong:
Example. There is an example (derived from the Reeb foliation) of a smooth free action of $G={mathbb R}$ on $P={mathbb R}^2- {(0,0)}$ such that the quotient space $P/G$ is homeomorphic to $[0,infty)$ (and, hence, is Hausdorff) but the action of $G$ on $P$ is not proper.
I do not know if there are similar examples (a nonproper smooth free action of a Lie group on a smooth manifold, $Gtimes Xto X$) where $X/G$ is a topological manifold (without boundary). Here is what I do know:
Theorem. Suppose that $G$ is a Lie group with finitely many connected components acting smoothly and freely on a smooth manifold $X$ such that $B=X/G$ is a topological manifold $M$ which admits a smooth structure such that the quotient map $Xto M$ is a submersion. Then the action of $G$ on $X$ is proper and, hence, $Gtimes Xto B$ is a principal fiber bundle.
answered Feb 3 at 23:34
Moishe KohanMoishe Kohan
48.5k344110
edited Feb 6 at 17:22
answered Feb 3 at 23:34
Moishe KohanMoishe Kohan
48.5k344110
answered Feb 3 at 23:34
Moishe KohanMoishe Kohan
48.5k344110
answered Feb 3 at 23:34
Moishe KohanMoishe Kohan
48.5k344110
48.5k344110
1
$begingroup$
Moishe Cohen@ Thanks a lot! ...I did not think before that local triviality will imply properness of the action. I will try to prove this implication !
$endgroup$
– Wandereradi
Feb 4 at 0:39
1
$begingroup$
@ Moishe Cohen Sir, one thing is confusing me. Local triviality (3rd requirement) assumes the existence of $pi$ ,a differentiable map from $P$ to $M$ which is taken to be the canonical projection when we identify $M$ with $P/G$(which is the 2nd requirement). So the paradox is if I assume local triviality imply properness and hence $P/G$ has the differentiable structure by QMT then while identifying $M$ and $P/G$ we just used set bijection (not diffeomorphism) as while identifying we dont have the privilege of local triviality.. So I couldn't understand how local trivial will imply proper?
$endgroup$
– Wandereradi
Feb 4 at 15:01
1
$begingroup$
"There is some smooth structure on $P/G$ and a diffeomorphism $Mrightarrow P/G$.. Then, local triviality condition implies that the action $G$ on $P$ is proper... Because the action of $G$ on $P$ is proper (it was free to start with), the quotient space $P/G$ has a unique smooth structure and all that... By uniqueness (unique smooth structure on $P/G$ such that $Prightarrow P/G$ is a submerison) in Quotient manifold theorem, it happens that, the some structure that we have started with is same to the structre coming from quotient manifold theorem".. Is this what you are saying Sir?
$endgroup$
– Praphulla Koushik
Feb 4 at 16:42
1
$begingroup$
"One does not need to know that $B=P/G$ is a manifold, only that it is Hausdorff" How do we even know that it Hausdorff... Is it that we are assuming? You are also saying $B$ is homeomorphic to a smooth manifold... This is something I did not expect :) Not diffeomorphic but homeomorphic... Can you please give a reference where one can see this explanation...
$endgroup$
– Praphulla Koushik
Feb 4 at 18:41
1
$begingroup$
@PraphullaKoushik: Manifolds are normally assumed to be Hausdorff (and 2nd countable). In fact, I just realized that one does not need local triviality to conclude properness of the action, all you need to assume that $P/G$ is Hausdorff.
$endgroup$
– Moishe Kohan
Feb 5 at 16:16
|
show 8 more comments
1
$begingroup$
Moishe Cohen@ Thanks a lot! ...I did not think before that local triviality will imply properness of the action. I will try to prove this implication !
$endgroup$
– Wandereradi
Feb 4 at 0:39
1
$begingroup$
@ Moishe Cohen Sir, one thing is confusing me. Local triviality (3rd requirement) assumes the existence of $pi$ ,a differentiable map from $P$ to $M$ which is taken to be the canonical projection when we identify $M$ with $P/G$(which is the 2nd requirement). So the paradox is if I assume local triviality imply properness and hence $P/G$ has the differentiable structure by QMT then while identifying $M$ and $P/G$ we just used set bijection (not diffeomorphism) as while identifying we dont have the privilege of local triviality.. So I couldn't understand how local trivial will imply proper?
$endgroup$
– Wandereradi
Feb 4 at 15:01
1
$begingroup$
"There is some smooth structure on $P/G$ and a diffeomorphism $Mrightarrow P/G$.. Then, local triviality condition implies that the action $G$ on $P$ is proper... Because the action of $G$ on $P$ is proper (it was free to start with), the quotient space $P/G$ has a unique smooth structure and all that... By uniqueness (unique smooth structure on $P/G$ such that $Prightarrow P/G$ is a submerison) in Quotient manifold theorem, it happens that, the some structure that we have started with is same to the structre coming from quotient manifold theorem".. Is this what you are saying Sir?
$endgroup$
– Praphulla Koushik
Feb 4 at 16:42
1
$begingroup$
"One does not need to know that $B=P/G$ is a manifold, only that it is Hausdorff" How do we even know that it Hausdorff... Is it that we are assuming? You are also saying $B$ is homeomorphic to a smooth manifold... This is something I did not expect :) Not diffeomorphic but homeomorphic... Can you please give a reference where one can see this explanation...
$endgroup$
– Praphulla Koushik
Feb 4 at 18:41
1
$begingroup$
@PraphullaKoushik: Manifolds are normally assumed to be Hausdorff (and 2nd countable). In fact, I just realized that one does not need local triviality to conclude properness of the action, all you need to assume that $P/G$ is Hausdorff.
$endgroup$
– Moishe Kohan
Feb 5 at 16:16
1
1
$begingroup$
Moishe Cohen@ Thanks a lot! ...I did not think before that local triviality will imply properness of the action. I will try to prove this implication !
$endgroup$
– Wandereradi
Feb 4 at 0:39
$begingroup$
Moishe Cohen@ Thanks a lot! ...I did not think before that local triviality will imply properness of the action. I will try to prove this implication !
$endgroup$
– Wandereradi
Feb 4 at 0:39
1
1
$begingroup$
@ Moishe Cohen Sir, one thing is confusing me. Local triviality (3rd requirement) assumes the existence of $pi$ ,a differentiable map from $P$ to $M$ which is taken to be the canonical projection when we identify $M$ with $P/G$(which is the 2nd requirement). So the paradox is if I assume local triviality imply properness and hence $P/G$ has the differentiable structure by QMT then while identifying $M$ and $P/G$ we just used set bijection (not diffeomorphism) as while identifying we dont have the privilege of local triviality.. So I couldn't understand how local trivial will imply proper?
$endgroup$
– Wandereradi
Feb 4 at 15:01
$begingroup$
@ Moishe Cohen Sir, one thing is confusing me. Local triviality (3rd requirement) assumes the existence of $pi$ ,a differentiable map from $P$ to $M$ which is taken to be the canonical projection when we identify $M$ with $P/G$(which is the 2nd requirement). So the paradox is if I assume local triviality imply properness and hence $P/G$ has the differentiable structure by QMT then while identifying $M$ and $P/G$ we just used set bijection (not diffeomorphism) as while identifying we dont have the privilege of local triviality.. So I couldn't understand how local trivial will imply proper?
$endgroup$
– Wandereradi
Feb 4 at 15:01
1
1
$begingroup$
"There is some smooth structure on $P/G$ and a diffeomorphism $Mrightarrow P/G$.. Then, local triviality condition implies that the action $G$ on $P$ is proper... Because the action of $G$ on $P$ is proper (it was free to start with), the quotient space $P/G$ has a unique smooth structure and all that... By uniqueness (unique smooth structure on $P/G$ such that $Prightarrow P/G$ is a submerison) in Quotient manifold theorem, it happens that, the some structure that we have started with is same to the structre coming from quotient manifold theorem".. Is this what you are saying Sir?
$endgroup$
– Praphulla Koushik
Feb 4 at 16:42
$begingroup$
"There is some smooth structure on $P/G$ and a diffeomorphism $Mrightarrow P/G$.. Then, local triviality condition implies that the action $G$ on $P$ is proper... Because the action of $G$ on $P$ is proper (it was free to start with), the quotient space $P/G$ has a unique smooth structure and all that... By uniqueness (unique smooth structure on $P/G$ such that $Prightarrow P/G$ is a submerison) in Quotient manifold theorem, it happens that, the some structure that we have started with is same to the structre coming from quotient manifold theorem".. Is this what you are saying Sir?
$endgroup$
– Praphulla Koushik
Feb 4 at 16:42
1
1
$begingroup$
"One does not need to know that $B=P/G$ is a manifold, only that it is Hausdorff" How do we even know that it Hausdorff... Is it that we are assuming? You are also saying $B$ is homeomorphic to a smooth manifold... This is something I did not expect :) Not diffeomorphic but homeomorphic... Can you please give a reference where one can see this explanation...
$endgroup$
– Praphulla Koushik
Feb 4 at 18:41
$begingroup$
"One does not need to know that $B=P/G$ is a manifold, only that it is Hausdorff" How do we even know that it Hausdorff... Is it that we are assuming? You are also saying $B$ is homeomorphic to a smooth manifold... This is something I did not expect :) Not diffeomorphic but homeomorphic... Can you please give a reference where one can see this explanation...
$endgroup$
– Praphulla Koushik
Feb 4 at 18:41
1
1
$begingroup$
@PraphullaKoushik: Manifolds are normally assumed to be Hausdorff (and 2nd countable). In fact, I just realized that one does not need local triviality to conclude properness of the action, all you need to assume that $P/G$ is Hausdorff.
$endgroup$
– Moishe Kohan
Feb 5 at 16:16
$begingroup$
@PraphullaKoushik: Manifolds are normally assumed to be Hausdorff (and 2nd countable). In fact, I just realized that one does not need local triviality to conclude properness of the action, all you need to assume that $P/G$ is Hausdorff.
$endgroup$
– Moishe Kohan
Feb 5 at 16:16
|
show 8 more comments
$begingroup$
Prof. Jack Lee proved here that action of $G$ on $P$ is proper in a principal $G$ bunlde. The problem there is, the user who asked the question does not mention what definition of principal bundle he is using. I only see that he is using local trivialization. He is also assuming that the quotient has a smooth structure.
I am only giving link to the answer. I will make this answer self explanatory when I get time.
answered Feb 5 at 5:26
Praphulla KoushikPraphulla Koushik
203119
$endgroup$
add a comment |
$begingroup$
Prof. Jack Lee proved here that action of $G$ on $P$ is proper in a principal $G$ bunlde. The problem there is, the user who asked the question does not mention what definition of principal bundle he is using. I only see that he is using local trivialization. He is also assuming that the quotient has a smooth structure.
I am only giving link to the answer. I will make this answer self explanatory when I get time.
answered Feb 5 at 5:26
Praphulla KoushikPraphulla Koushik
203119
$endgroup$
add a comment |
$begingroup$
Prof. Jack Lee proved here that action of $G$ on $P$ is proper in a principal $G$ bunlde. The problem there is, the user who asked the question does not mention what definition of principal bundle he is using. I only see that he is using local trivialization. He is also assuming that the quotient has a smooth structure.
I am only giving link to the answer. I will make this answer self explanatory when I get time.
answered Feb 5 at 5:26
Praphulla KoushikPraphulla Koushik
203119
$endgroup$
Prof. Jack Lee proved here that action of $G$ on $P$ is proper in a principal $G$ bunlde. The problem there is, the user who asked the question does not mention what definition of principal bundle he is using. I only see that he is using local trivialization. He is also assuming that the quotient has a smooth structure.
I am only giving link to the answer. I will make this answer self explanatory when I get time.
answered Feb 5 at 5:26
Praphulla KoushikPraphulla Koushik
203119
answered Feb 5 at 5:26
Praphulla KoushikPraphulla Koushik
203119
answered Feb 5 at 5:26
Praphulla KoushikPraphulla Koushik
203119
answered Feb 5 at 5:26
Praphulla KoushikPraphulla Koushik
203119
203119
add a comment |
add a comment |
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$begingroup$
math.stackexchange.com/questions/560371/…
$endgroup$
– Moishe Kohan
Feb 3 at 4:43
1
$begingroup$
@MoisheCohen Can you please specify what in particular you want to specify in that linked question/answer...
$endgroup$
– Praphulla Koushik
Feb 3 at 16:04
$begingroup$
MoisheCohen@ Thanks... I saw the link but couldn’t understand the relevance to my question..please can you elaborate a little about it?..
$endgroup$
– Wandereradi
Feb 3 at 16:18