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Is this a submanifold?
When is a conjugacy class of matrices an embedded submanifold?Finding covariant derivative of a riemanian submanifoldTotally geodesic submanifold of round spherepositive sectional curvature of submanifold in $R^n$?Distance function to a submanifoldTotally geodesic submanifold of a hyperbolic 3-manifoldSmoothness of the closest point on a submanifoldFundamental group of compact manifolds with non-negative Ricci curvature and Bieberbach theoremClosest point of one SO(3) submanifold onto another submanifoldCodimension reduction for developable Euclidean submanifold
$begingroup$
Let $(M,g)$ be a compact Riemannian manifold with an isometric action $rho : G to mathrmIso(M)$ by a compact Lie group $G$. There is a natural extension of $rho$ to $TM$ given by:
$$psi : G times TM to TM$$
$$psi(g,p,X) := (rho(g)p,drho(g)X).$$
I would like to know if the set:
$$TM supset tilde S := (p,X) in TM : G_X = G_p$$ is a submanifold of $TM$, and further, if $pi : TM to M$ is the natural projection, then $pi(tilde S)$ is a submanifold of $M$.
I tried the following approach:
For each $gin G$ consider $eta_g(p,X) equiv eta(g,p,X) := d^2_TMleft((p,X),psi(g,p,X)right).$ Then, one has:
$$eta_g^-1(0) = (p,X) :(rho(g)p,drho(g)X) = (p,X).$$
So,
$$tilde S = bigcup_gin Geta_g^-1(0). $$
But according to my calculation $0$ is not a regular value of $eta_g$.
I appreciate any help.
EDIT
Thanks to all the answers and comments. I shall change a little the candidate to manifold to another that will be more helpful to me.
I would like to know if one denotes by $cal H_p$ the orthogonal complement to $T_pGcdot p$ on the $g$-metric, then the set
$$S := p in M : exists X in mathcal H_p : G_X = G_p$$
is a submanifold of $M$. In fact, this is what I was trying to prove at first.
dg.differential-geometry riemannian-geometry differential-topology group-actions
asked 6 hours ago
L.F. CavenaghiL.F. Cavenaghi
598213
$endgroup$
add a comment |
$begingroup$
Let $(M,g)$ be a compact Riemannian manifold with an isometric action $rho : G to mathrmIso(M)$ by a compact Lie group $G$. There is a natural extension of $rho$ to $TM$ given by:
$$psi : G times TM to TM$$
$$psi(g,p,X) := (rho(g)p,drho(g)X).$$
I would like to know if the set:
$$TM supset tilde S := (p,X) in TM : G_X = G_p$$ is a submanifold of $TM$, and further, if $pi : TM to M$ is the natural projection, then $pi(tilde S)$ is a submanifold of $M$.
I tried the following approach:
For each $gin G$ consider $eta_g(p,X) equiv eta(g,p,X) := d^2_TMleft((p,X),psi(g,p,X)right).$ Then, one has:
$$eta_g^-1(0) = (p,X) :(rho(g)p,drho(g)X) = (p,X).$$
So,
$$tilde S = bigcup_gin Geta_g^-1(0). $$
But according to my calculation $0$ is not a regular value of $eta_g$.
I appreciate any help.
EDIT
Thanks to all the answers and comments. I shall change a little the candidate to manifold to another that will be more helpful to me.
I would like to know if one denotes by $cal H_p$ the orthogonal complement to $T_pGcdot p$ on the $g$-metric, then the set
$$S := p in M : exists X in mathcal H_p : G_X = G_p$$
is a submanifold of $M$. In fact, this is what I was trying to prove at first.
dg.differential-geometry riemannian-geometry differential-topology group-actions
asked 6 hours ago
L.F. CavenaghiL.F. Cavenaghi
598213
$endgroup$
$begingroup$
Is this a pigeon? ;-)
$endgroup$
– David Roberts
3 hours ago
$begingroup$
@DavidRoberts, I am sorry, what are you asking if it is a pigeon?
$endgroup$
– L.F. Cavenaghi
2 hours ago
1
$begingroup$
It's a meme, meant in good humour.
$endgroup$
– David Roberts
1 hour ago
$begingroup$
By $G_X$ and $G_p$ you are referring to the vector and point stabilizers, respectively? I would imagine that generally this is not a manifold without some pretty restrictive assumptions. I'm imagining the action of $G$ degenerating along a stratum, with the stabilized vector space changing. What are you hoping to do with a result like this?
$endgroup$
– Ryan Budney
13 mins ago
$begingroup$
Dear @RyanBudney, please, consider seeing my edit, as It was suggested on the comments $S$ possibly is a manifold, and that was my original problem. I am working on positive curvature and I need on some step of my argument to be sure that $S$ is a submanifold.
$endgroup$
– L.F. Cavenaghi
11 mins ago
add a comment |
$begingroup$
Let $(M,g)$ be a compact Riemannian manifold with an isometric action $rho : G to mathrmIso(M)$ by a compact Lie group $G$. There is a natural extension of $rho$ to $TM$ given by:
$$psi : G times TM to TM$$
$$psi(g,p,X) := (rho(g)p,drho(g)X).$$
I would like to know if the set:
$$TM supset tilde S := (p,X) in TM : G_X = G_p$$ is a submanifold of $TM$, and further, if $pi : TM to M$ is the natural projection, then $pi(tilde S)$ is a submanifold of $M$.
I tried the following approach:
For each $gin G$ consider $eta_g(p,X) equiv eta(g,p,X) := d^2_TMleft((p,X),psi(g,p,X)right).$ Then, one has:
$$eta_g^-1(0) = (p,X) :(rho(g)p,drho(g)X) = (p,X).$$
So,
$$tilde S = bigcup_gin Geta_g^-1(0). $$
But according to my calculation $0$ is not a regular value of $eta_g$.
I appreciate any help.
EDIT
Thanks to all the answers and comments. I shall change a little the candidate to manifold to another that will be more helpful to me.
I would like to know if one denotes by $cal H_p$ the orthogonal complement to $T_pGcdot p$ on the $g$-metric, then the set
$$S := p in M : exists X in mathcal H_p : G_X = G_p$$
is a submanifold of $M$. In fact, this is what I was trying to prove at first.
dg.differential-geometry riemannian-geometry differential-topology group-actions
asked 6 hours ago
L.F. CavenaghiL.F. Cavenaghi
598213
$endgroup$
Let $(M,g)$ be a compact Riemannian manifold with an isometric action $rho : G to mathrmIso(M)$ by a compact Lie group $G$. There is a natural extension of $rho$ to $TM$ given by:
$$psi : G times TM to TM$$
$$psi(g,p,X) := (rho(g)p,drho(g)X).$$
I would like to know if the set:
$$TM supset tilde S := (p,X) in TM : G_X = G_p$$ is a submanifold of $TM$, and further, if $pi : TM to M$ is the natural projection, then $pi(tilde S)$ is a submanifold of $M$.
I tried the following approach:
For each $gin G$ consider $eta_g(p,X) equiv eta(g,p,X) := d^2_TMleft((p,X),psi(g,p,X)right).$ Then, one has:
$$eta_g^-1(0) = (p,X) :(rho(g)p,drho(g)X) = (p,X).$$
So,
$$tilde S = bigcup_gin Geta_g^-1(0). $$
But according to my calculation $0$ is not a regular value of $eta_g$.
I appreciate any help.
EDIT
Thanks to all the answers and comments. I shall change a little the candidate to manifold to another that will be more helpful to me.
I would like to know if one denotes by $cal H_p$ the orthogonal complement to $T_pGcdot p$ on the $g$-metric, then the set
$$S := p in M : exists X in mathcal H_p : G_X = G_p$$
is a submanifold of $M$. In fact, this is what I was trying to prove at first.
dg.differential-geometry riemannian-geometry differential-topology group-actions
dg.differential-geometry riemannian-geometry differential-topology group-actions
asked 6 hours ago
L.F. CavenaghiL.F. Cavenaghi
598213
asked 6 hours ago
L.F. CavenaghiL.F. Cavenaghi
598213
edited 12 mins ago
L.F. Cavenaghi
asked 6 hours ago
L.F. CavenaghiL.F. Cavenaghi
598213
asked 6 hours ago
L.F. CavenaghiL.F. Cavenaghi
598213
asked 6 hours ago
L.F. CavenaghiL.F. Cavenaghi
598213
598213
$begingroup$
Is this a pigeon? ;-)
$endgroup$
– David Roberts
3 hours ago
$begingroup$
@DavidRoberts, I am sorry, what are you asking if it is a pigeon?
$endgroup$
– L.F. Cavenaghi
2 hours ago
1
$begingroup$
It's a meme, meant in good humour.
$endgroup$
– David Roberts
1 hour ago
$begingroup$
By $G_X$ and $G_p$ you are referring to the vector and point stabilizers, respectively? I would imagine that generally this is not a manifold without some pretty restrictive assumptions. I'm imagining the action of $G$ degenerating along a stratum, with the stabilized vector space changing. What are you hoping to do with a result like this?
$endgroup$
– Ryan Budney
13 mins ago
$begingroup$
Dear @RyanBudney, please, consider seeing my edit, as It was suggested on the comments $S$ possibly is a manifold, and that was my original problem. I am working on positive curvature and I need on some step of my argument to be sure that $S$ is a submanifold.
$endgroup$
– L.F. Cavenaghi
11 mins ago
add a comment |
$begingroup$
Is this a pigeon? ;-)
$endgroup$
– David Roberts
3 hours ago
$begingroup$
@DavidRoberts, I am sorry, what are you asking if it is a pigeon?
$endgroup$
– L.F. Cavenaghi
2 hours ago
1
$begingroup$
It's a meme, meant in good humour.
$endgroup$
– David Roberts
1 hour ago
$begingroup$
By $G_X$ and $G_p$ you are referring to the vector and point stabilizers, respectively? I would imagine that generally this is not a manifold without some pretty restrictive assumptions. I'm imagining the action of $G$ degenerating along a stratum, with the stabilized vector space changing. What are you hoping to do with a result like this?
$endgroup$
– Ryan Budney
13 mins ago
$begingroup$
Dear @RyanBudney, please, consider seeing my edit, as It was suggested on the comments $S$ possibly is a manifold, and that was my original problem. I am working on positive curvature and I need on some step of my argument to be sure that $S$ is a submanifold.
$endgroup$
– L.F. Cavenaghi
11 mins ago
$begingroup$
Is this a pigeon? ;-)
$endgroup$
– David Roberts
3 hours ago
$begingroup$
Is this a pigeon? ;-)
$endgroup$
– David Roberts
3 hours ago
$begingroup$
@DavidRoberts, I am sorry, what are you asking if it is a pigeon?
$endgroup$
– L.F. Cavenaghi
2 hours ago
$begingroup$
@DavidRoberts, I am sorry, what are you asking if it is a pigeon?
$endgroup$
– L.F. Cavenaghi
2 hours ago
1
1
$begingroup$
It's a meme, meant in good humour.
$endgroup$
– David Roberts
1 hour ago
$begingroup$
It's a meme, meant in good humour.
$endgroup$
– David Roberts
1 hour ago
$begingroup$
By $G_X$ and $G_p$ you are referring to the vector and point stabilizers, respectively? I would imagine that generally this is not a manifold without some pretty restrictive assumptions. I'm imagining the action of $G$ degenerating along a stratum, with the stabilized vector space changing. What are you hoping to do with a result like this?
$endgroup$
– Ryan Budney
13 mins ago
$begingroup$
By $G_X$ and $G_p$ you are referring to the vector and point stabilizers, respectively? I would imagine that generally this is not a manifold without some pretty restrictive assumptions. I'm imagining the action of $G$ degenerating along a stratum, with the stabilized vector space changing. What are you hoping to do with a result like this?
$endgroup$
– Ryan Budney
13 mins ago
$begingroup$
Dear @RyanBudney, please, consider seeing my edit, as It was suggested on the comments $S$ possibly is a manifold, and that was my original problem. I am working on positive curvature and I need on some step of my argument to be sure that $S$ is a submanifold.
$endgroup$
– L.F. Cavenaghi
11 mins ago
$begingroup$
Dear @RyanBudney, please, consider seeing my edit, as It was suggested on the comments $S$ possibly is a manifold, and that was my original problem. I am working on positive curvature and I need on some step of my argument to be sure that $S$ is a submanifold.
$endgroup$
– L.F. Cavenaghi
11 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For the obvious reflection action of $mathbbZ/2mathbbZ$ on $S^1$ this set $tilde S $ has two singular points at the zero vectors tangent at points $p=(1,0)$ and $q=(-1,0)$.
Then $tilde S$ is $T(S^1setminus p,q) cup p_0,q_0$ where $p_0, q_0$ are zero vectors at $p,q$. This is a connected set and after removing $p_0,q_0$ we obtain a disconnected set. So obviously it is not a manifold because every connected manifold of dimension at least $2$, remains connected after removing a finite set.
answered 4 hours ago
Ali TaghaviAli Taghavi
9152084
$endgroup$
1
$begingroup$
"every connected manifold remains connected after removing a finite set" -- this is not true for 1-dimensional manifolds. Nonetheless, your example is valid: if $x$ is one of the fixed points of the reflection action, then a neighborhood of $(x,0)intilde S$ is homeomorphic to $mathbb R^2$ minus the coordinate axes but including the origin, so it cannot be a manifold.
$endgroup$
– Arun Debray
3 hours ago
$begingroup$
@ArunDebray yes thank you but $tilde S $ is 2 dimensional at generic points. I revise the answer. Thanks again for your correction!
$endgroup$
– Ali Taghavi
3 hours ago
1
$begingroup$
@AliTaghavi, thank you very much. Do you know if one can restrict some hypothesis in order to obtain a manifold? Or, what kind of structure this set has?
$endgroup$
– L.F. Cavenaghi
2 hours ago
$begingroup$
You are well come. What about if we assume $G$ is connected? I am not sure what is the answer in this case.
$endgroup$
– Ali Taghavi
2 hours ago
$begingroup$
@AliTaghavi, take a look to the other answer and possible on the comments. I changed a little the problem.
$endgroup$
– L.F. Cavenaghi
1 hour ago
add a comment |
$begingroup$
Your definition implies that
$$ tilde S = bigcup_pin M TM_p^G_p. $$
In particular, $pi(tilde S) = M$, and $tilde S$ will be a submanifold of $TM$ iff the dimension of $TM_p^G_p$ is the same for all $pin M$.
$G$ being connected won't necessarily make this happen: Another counterexample is $S^1$ acting by rotation on $S^2$, fixing the north and south poles, $n$ and $s$. $tilde S$ is then $T(S^2setminusn,s)cup n,s$, and the points $n$ and $s$ do not have neighborhoods homeomorphic to open balls.
answered 1 hour ago
Steve CostenobleSteve Costenoble
9451514
$endgroup$
$begingroup$
what if we change a little bit my definition by asking the following: $S = p in M : exists 0 neq X in mathcal H_p subset T_pM : G_X = G_p$? Now I ask if such $S$ is a submanifold of $M$, where $cal H_p$ is the space $g$-orthogonal to $T_pGcdot p$.
$endgroup$
– L.F. Cavenaghi
1 hour ago
1
$begingroup$
My first reaction is that, yes, $S$ will be a submanifold. In fact, an open submanifold: If $pin S$ then all points in an open neighborhood of $p$ should also satisfy your new condition. Essentially, a neighborhood of $p$ will look like $T_p M$ with the action of $G_p$.
$endgroup$
– Steve Costenoble
43 mins ago
$begingroup$
I have edited the question in order to encompass $S$ to it, it would be extremely helpful if you explain to me your thoughts on why $S$ has a chance to be a submanifold.
$endgroup$
– L.F. Cavenaghi
9 mins ago
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For the obvious reflection action of $mathbbZ/2mathbbZ$ on $S^1$ this set $tilde S $ has two singular points at the zero vectors tangent at points $p=(1,0)$ and $q=(-1,0)$.
Then $tilde S$ is $T(S^1setminus p,q) cup p_0,q_0$ where $p_0, q_0$ are zero vectors at $p,q$. This is a connected set and after removing $p_0,q_0$ we obtain a disconnected set. So obviously it is not a manifold because every connected manifold of dimension at least $2$, remains connected after removing a finite set.
answered 4 hours ago
Ali TaghaviAli Taghavi
9152084
$endgroup$
1
$begingroup$
"every connected manifold remains connected after removing a finite set" -- this is not true for 1-dimensional manifolds. Nonetheless, your example is valid: if $x$ is one of the fixed points of the reflection action, then a neighborhood of $(x,0)intilde S$ is homeomorphic to $mathbb R^2$ minus the coordinate axes but including the origin, so it cannot be a manifold.
$endgroup$
– Arun Debray
3 hours ago
$begingroup$
@ArunDebray yes thank you but $tilde S $ is 2 dimensional at generic points. I revise the answer. Thanks again for your correction!
$endgroup$
– Ali Taghavi
3 hours ago
1
$begingroup$
@AliTaghavi, thank you very much. Do you know if one can restrict some hypothesis in order to obtain a manifold? Or, what kind of structure this set has?
$endgroup$
– L.F. Cavenaghi
2 hours ago
$begingroup$
You are well come. What about if we assume $G$ is connected? I am not sure what is the answer in this case.
$endgroup$
– Ali Taghavi
2 hours ago
$begingroup$
@AliTaghavi, take a look to the other answer and possible on the comments. I changed a little the problem.
$endgroup$
– L.F. Cavenaghi
1 hour ago
add a comment |
$begingroup$
For the obvious reflection action of $mathbbZ/2mathbbZ$ on $S^1$ this set $tilde S $ has two singular points at the zero vectors tangent at points $p=(1,0)$ and $q=(-1,0)$.
Then $tilde S$ is $T(S^1setminus p,q) cup p_0,q_0$ where $p_0, q_0$ are zero vectors at $p,q$. This is a connected set and after removing $p_0,q_0$ we obtain a disconnected set. So obviously it is not a manifold because every connected manifold of dimension at least $2$, remains connected after removing a finite set.
answered 4 hours ago
Ali TaghaviAli Taghavi
9152084
$endgroup$
1
$begingroup$
"every connected manifold remains connected after removing a finite set" -- this is not true for 1-dimensional manifolds. Nonetheless, your example is valid: if $x$ is one of the fixed points of the reflection action, then a neighborhood of $(x,0)intilde S$ is homeomorphic to $mathbb R^2$ minus the coordinate axes but including the origin, so it cannot be a manifold.
$endgroup$
– Arun Debray
3 hours ago
$begingroup$
@ArunDebray yes thank you but $tilde S $ is 2 dimensional at generic points. I revise the answer. Thanks again for your correction!
$endgroup$
– Ali Taghavi
3 hours ago
1
$begingroup$
@AliTaghavi, thank you very much. Do you know if one can restrict some hypothesis in order to obtain a manifold? Or, what kind of structure this set has?
$endgroup$
– L.F. Cavenaghi
2 hours ago
$begingroup$
You are well come. What about if we assume $G$ is connected? I am not sure what is the answer in this case.
$endgroup$
– Ali Taghavi
2 hours ago
$begingroup$
@AliTaghavi, take a look to the other answer and possible on the comments. I changed a little the problem.
$endgroup$
– L.F. Cavenaghi
1 hour ago
add a comment |
$begingroup$
For the obvious reflection action of $mathbbZ/2mathbbZ$ on $S^1$ this set $tilde S $ has two singular points at the zero vectors tangent at points $p=(1,0)$ and $q=(-1,0)$.
Then $tilde S$ is $T(S^1setminus p,q) cup p_0,q_0$ where $p_0, q_0$ are zero vectors at $p,q$. This is a connected set and after removing $p_0,q_0$ we obtain a disconnected set. So obviously it is not a manifold because every connected manifold of dimension at least $2$, remains connected after removing a finite set.
answered 4 hours ago
Ali TaghaviAli Taghavi
9152084
$endgroup$
For the obvious reflection action of $mathbbZ/2mathbbZ$ on $S^1$ this set $tilde S $ has two singular points at the zero vectors tangent at points $p=(1,0)$ and $q=(-1,0)$.
Then $tilde S$ is $T(S^1setminus p,q) cup p_0,q_0$ where $p_0, q_0$ are zero vectors at $p,q$. This is a connected set and after removing $p_0,q_0$ we obtain a disconnected set. So obviously it is not a manifold because every connected manifold of dimension at least $2$, remains connected after removing a finite set.
answered 4 hours ago
Ali TaghaviAli Taghavi
9152084
edited 3 hours ago
answered 4 hours ago
Ali TaghaviAli Taghavi
9152084
answered 4 hours ago
Ali TaghaviAli Taghavi
9152084
answered 4 hours ago
Ali TaghaviAli Taghavi
9152084
9152084
1
$begingroup$
"every connected manifold remains connected after removing a finite set" -- this is not true for 1-dimensional manifolds. Nonetheless, your example is valid: if $x$ is one of the fixed points of the reflection action, then a neighborhood of $(x,0)intilde S$ is homeomorphic to $mathbb R^2$ minus the coordinate axes but including the origin, so it cannot be a manifold.
$endgroup$
– Arun Debray
3 hours ago
$begingroup$
@ArunDebray yes thank you but $tilde S $ is 2 dimensional at generic points. I revise the answer. Thanks again for your correction!
$endgroup$
– Ali Taghavi
3 hours ago
1
$begingroup$
@AliTaghavi, thank you very much. Do you know if one can restrict some hypothesis in order to obtain a manifold? Or, what kind of structure this set has?
$endgroup$
– L.F. Cavenaghi
2 hours ago
$begingroup$
You are well come. What about if we assume $G$ is connected? I am not sure what is the answer in this case.
$endgroup$
– Ali Taghavi
2 hours ago
$begingroup$
@AliTaghavi, take a look to the other answer and possible on the comments. I changed a little the problem.
$endgroup$
– L.F. Cavenaghi
1 hour ago
add a comment |
1
$begingroup$
"every connected manifold remains connected after removing a finite set" -- this is not true for 1-dimensional manifolds. Nonetheless, your example is valid: if $x$ is one of the fixed points of the reflection action, then a neighborhood of $(x,0)intilde S$ is homeomorphic to $mathbb R^2$ minus the coordinate axes but including the origin, so it cannot be a manifold.
$endgroup$
– Arun Debray
3 hours ago
$begingroup$
@ArunDebray yes thank you but $tilde S $ is 2 dimensional at generic points. I revise the answer. Thanks again for your correction!
$endgroup$
– Ali Taghavi
3 hours ago
1
$begingroup$
@AliTaghavi, thank you very much. Do you know if one can restrict some hypothesis in order to obtain a manifold? Or, what kind of structure this set has?
$endgroup$
– L.F. Cavenaghi
2 hours ago
$begingroup$
You are well come. What about if we assume $G$ is connected? I am not sure what is the answer in this case.
$endgroup$
– Ali Taghavi
2 hours ago
$begingroup$
@AliTaghavi, take a look to the other answer and possible on the comments. I changed a little the problem.
$endgroup$
– L.F. Cavenaghi
1 hour ago
1
1
$begingroup$
"every connected manifold remains connected after removing a finite set" -- this is not true for 1-dimensional manifolds. Nonetheless, your example is valid: if $x$ is one of the fixed points of the reflection action, then a neighborhood of $(x,0)intilde S$ is homeomorphic to $mathbb R^2$ minus the coordinate axes but including the origin, so it cannot be a manifold.
$endgroup$
– Arun Debray
3 hours ago
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"every connected manifold remains connected after removing a finite set" -- this is not true for 1-dimensional manifolds. Nonetheless, your example is valid: if $x$ is one of the fixed points of the reflection action, then a neighborhood of $(x,0)intilde S$ is homeomorphic to $mathbb R^2$ minus the coordinate axes but including the origin, so it cannot be a manifold.
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– Arun Debray
3 hours ago
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@ArunDebray yes thank you but $tilde S $ is 2 dimensional at generic points. I revise the answer. Thanks again for your correction!
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– Ali Taghavi
3 hours ago
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@ArunDebray yes thank you but $tilde S $ is 2 dimensional at generic points. I revise the answer. Thanks again for your correction!
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– Ali Taghavi
3 hours ago
1
1
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@AliTaghavi, thank you very much. Do you know if one can restrict some hypothesis in order to obtain a manifold? Or, what kind of structure this set has?
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– L.F. Cavenaghi
2 hours ago
$begingroup$
@AliTaghavi, thank you very much. Do you know if one can restrict some hypothesis in order to obtain a manifold? Or, what kind of structure this set has?
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– L.F. Cavenaghi
2 hours ago
$begingroup$
You are well come. What about if we assume $G$ is connected? I am not sure what is the answer in this case.
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– Ali Taghavi
2 hours ago
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You are well come. What about if we assume $G$ is connected? I am not sure what is the answer in this case.
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– Ali Taghavi
2 hours ago
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@AliTaghavi, take a look to the other answer and possible on the comments. I changed a little the problem.
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– L.F. Cavenaghi
1 hour ago
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@AliTaghavi, take a look to the other answer and possible on the comments. I changed a little the problem.
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– L.F. Cavenaghi
1 hour ago
add a comment |
$begingroup$
Your definition implies that
$$ tilde S = bigcup_pin M TM_p^G_p. $$
In particular, $pi(tilde S) = M$, and $tilde S$ will be a submanifold of $TM$ iff the dimension of $TM_p^G_p$ is the same for all $pin M$.
$G$ being connected won't necessarily make this happen: Another counterexample is $S^1$ acting by rotation on $S^2$, fixing the north and south poles, $n$ and $s$. $tilde S$ is then $T(S^2setminusn,s)cup n,s$, and the points $n$ and $s$ do not have neighborhoods homeomorphic to open balls.
answered 1 hour ago
Steve CostenobleSteve Costenoble
9451514
$endgroup$
$begingroup$
what if we change a little bit my definition by asking the following: $S = p in M : exists 0 neq X in mathcal H_p subset T_pM : G_X = G_p$? Now I ask if such $S$ is a submanifold of $M$, where $cal H_p$ is the space $g$-orthogonal to $T_pGcdot p$.
$endgroup$
– L.F. Cavenaghi
1 hour ago
1
$begingroup$
My first reaction is that, yes, $S$ will be a submanifold. In fact, an open submanifold: If $pin S$ then all points in an open neighborhood of $p$ should also satisfy your new condition. Essentially, a neighborhood of $p$ will look like $T_p M$ with the action of $G_p$.
$endgroup$
– Steve Costenoble
43 mins ago
$begingroup$
I have edited the question in order to encompass $S$ to it, it would be extremely helpful if you explain to me your thoughts on why $S$ has a chance to be a submanifold.
$endgroup$
– L.F. Cavenaghi
9 mins ago
add a comment |
$begingroup$
Your definition implies that
$$ tilde S = bigcup_pin M TM_p^G_p. $$
In particular, $pi(tilde S) = M$, and $tilde S$ will be a submanifold of $TM$ iff the dimension of $TM_p^G_p$ is the same for all $pin M$.
$G$ being connected won't necessarily make this happen: Another counterexample is $S^1$ acting by rotation on $S^2$, fixing the north and south poles, $n$ and $s$. $tilde S$ is then $T(S^2setminusn,s)cup n,s$, and the points $n$ and $s$ do not have neighborhoods homeomorphic to open balls.
answered 1 hour ago
Steve CostenobleSteve Costenoble
9451514
$endgroup$
$begingroup$
what if we change a little bit my definition by asking the following: $S = p in M : exists 0 neq X in mathcal H_p subset T_pM : G_X = G_p$? Now I ask if such $S$ is a submanifold of $M$, where $cal H_p$ is the space $g$-orthogonal to $T_pGcdot p$.
$endgroup$
– L.F. Cavenaghi
1 hour ago
1
$begingroup$
My first reaction is that, yes, $S$ will be a submanifold. In fact, an open submanifold: If $pin S$ then all points in an open neighborhood of $p$ should also satisfy your new condition. Essentially, a neighborhood of $p$ will look like $T_p M$ with the action of $G_p$.
$endgroup$
– Steve Costenoble
43 mins ago
$begingroup$
I have edited the question in order to encompass $S$ to it, it would be extremely helpful if you explain to me your thoughts on why $S$ has a chance to be a submanifold.
$endgroup$
– L.F. Cavenaghi
9 mins ago
add a comment |
$begingroup$
Your definition implies that
$$ tilde S = bigcup_pin M TM_p^G_p. $$
In particular, $pi(tilde S) = M$, and $tilde S$ will be a submanifold of $TM$ iff the dimension of $TM_p^G_p$ is the same for all $pin M$.
$G$ being connected won't necessarily make this happen: Another counterexample is $S^1$ acting by rotation on $S^2$, fixing the north and south poles, $n$ and $s$. $tilde S$ is then $T(S^2setminusn,s)cup n,s$, and the points $n$ and $s$ do not have neighborhoods homeomorphic to open balls.
answered 1 hour ago
Steve CostenobleSteve Costenoble
9451514
$endgroup$
Your definition implies that
$$ tilde S = bigcup_pin M TM_p^G_p. $$
In particular, $pi(tilde S) = M$, and $tilde S$ will be a submanifold of $TM$ iff the dimension of $TM_p^G_p$ is the same for all $pin M$.
$G$ being connected won't necessarily make this happen: Another counterexample is $S^1$ acting by rotation on $S^2$, fixing the north and south poles, $n$ and $s$. $tilde S$ is then $T(S^2setminusn,s)cup n,s$, and the points $n$ and $s$ do not have neighborhoods homeomorphic to open balls.
answered 1 hour ago
Steve CostenobleSteve Costenoble
9451514
answered 1 hour ago
Steve CostenobleSteve Costenoble
9451514
answered 1 hour ago
Steve CostenobleSteve Costenoble
9451514
answered 1 hour ago
Steve CostenobleSteve Costenoble
9451514
9451514
$begingroup$
what if we change a little bit my definition by asking the following: $S = p in M : exists 0 neq X in mathcal H_p subset T_pM : G_X = G_p$? Now I ask if such $S$ is a submanifold of $M$, where $cal H_p$ is the space $g$-orthogonal to $T_pGcdot p$.
$endgroup$
– L.F. Cavenaghi
1 hour ago
1
$begingroup$
My first reaction is that, yes, $S$ will be a submanifold. In fact, an open submanifold: If $pin S$ then all points in an open neighborhood of $p$ should also satisfy your new condition. Essentially, a neighborhood of $p$ will look like $T_p M$ with the action of $G_p$.
$endgroup$
– Steve Costenoble
43 mins ago
$begingroup$
I have edited the question in order to encompass $S$ to it, it would be extremely helpful if you explain to me your thoughts on why $S$ has a chance to be a submanifold.
$endgroup$
– L.F. Cavenaghi
9 mins ago
add a comment |
$begingroup$
what if we change a little bit my definition by asking the following: $S = p in M : exists 0 neq X in mathcal H_p subset T_pM : G_X = G_p$? Now I ask if such $S$ is a submanifold of $M$, where $cal H_p$ is the space $g$-orthogonal to $T_pGcdot p$.
$endgroup$
– L.F. Cavenaghi
1 hour ago
1
$begingroup$
My first reaction is that, yes, $S$ will be a submanifold. In fact, an open submanifold: If $pin S$ then all points in an open neighborhood of $p$ should also satisfy your new condition. Essentially, a neighborhood of $p$ will look like $T_p M$ with the action of $G_p$.
$endgroup$
– Steve Costenoble
43 mins ago
$begingroup$
I have edited the question in order to encompass $S$ to it, it would be extremely helpful if you explain to me your thoughts on why $S$ has a chance to be a submanifold.
$endgroup$
– L.F. Cavenaghi
9 mins ago
$begingroup$
what if we change a little bit my definition by asking the following: $S = p in M : exists 0 neq X in mathcal H_p subset T_pM : G_X = G_p$? Now I ask if such $S$ is a submanifold of $M$, where $cal H_p$ is the space $g$-orthogonal to $T_pGcdot p$.
$endgroup$
– L.F. Cavenaghi
1 hour ago
$begingroup$
what if we change a little bit my definition by asking the following: $S = p in M : exists 0 neq X in mathcal H_p subset T_pM : G_X = G_p$? Now I ask if such $S$ is a submanifold of $M$, where $cal H_p$ is the space $g$-orthogonal to $T_pGcdot p$.
$endgroup$
– L.F. Cavenaghi
1 hour ago
1
1
$begingroup$
My first reaction is that, yes, $S$ will be a submanifold. In fact, an open submanifold: If $pin S$ then all points in an open neighborhood of $p$ should also satisfy your new condition. Essentially, a neighborhood of $p$ will look like $T_p M$ with the action of $G_p$.
$endgroup$
– Steve Costenoble
43 mins ago
$begingroup$
My first reaction is that, yes, $S$ will be a submanifold. In fact, an open submanifold: If $pin S$ then all points in an open neighborhood of $p$ should also satisfy your new condition. Essentially, a neighborhood of $p$ will look like $T_p M$ with the action of $G_p$.
$endgroup$
– Steve Costenoble
43 mins ago
$begingroup$
I have edited the question in order to encompass $S$ to it, it would be extremely helpful if you explain to me your thoughts on why $S$ has a chance to be a submanifold.
$endgroup$
– L.F. Cavenaghi
9 mins ago
$begingroup$
I have edited the question in order to encompass $S$ to it, it would be extremely helpful if you explain to me your thoughts on why $S$ has a chance to be a submanifold.
$endgroup$
– L.F. Cavenaghi
9 mins ago
add a comment |
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$begingroup$
Is this a pigeon? ;-)
$endgroup$
– David Roberts
3 hours ago
$begingroup$
@DavidRoberts, I am sorry, what are you asking if it is a pigeon?
$endgroup$
– L.F. Cavenaghi
2 hours ago
1
$begingroup$
It's a meme, meant in good humour.
$endgroup$
– David Roberts
1 hour ago
$begingroup$
By $G_X$ and $G_p$ you are referring to the vector and point stabilizers, respectively? I would imagine that generally this is not a manifold without some pretty restrictive assumptions. I'm imagining the action of $G$ degenerating along a stratum, with the stabilized vector space changing. What are you hoping to do with a result like this?
$endgroup$
– Ryan Budney
13 mins ago
$begingroup$
Dear @RyanBudney, please, consider seeing my edit, as It was suggested on the comments $S$ possibly is a manifold, and that was my original problem. I am working on positive curvature and I need on some step of my argument to be sure that $S$ is a submanifold.
$endgroup$
– L.F. Cavenaghi
11 mins ago