Is this a submanifold?When is a conjugacy class of matrices an embedded submanifold?Finding covariant...
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 $rho : G to mathrm{Iso}(M)$ 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 $eta_g(p,X) equiv eta(g,p,X) := d^2_{TM}left((p,X),psi(g,p,X)right).$ Then, one has:
$$eta_g^{-1}(0) = {(p,X) :(rho(g)p,drho(g)X) = (p,X)}.$$
So,
$$tilde S = bigcup_{gin G}eta_g^{-1}(0). $$
But according to my calculation $0$ is not a regular value of $eta_g$.
I appreciate any help.
dg.differential-geometry riemannian-geometry differential-topology group-actions
edited 2 hours ago
Ali Taghavi
9152084
asked 4 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 $rho : G to mathrm{Iso}(M)$ 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 $eta_g(p,X) equiv eta(g,p,X) := d^2_{TM}left((p,X),psi(g,p,X)right).$ Then, one has:
$$eta_g^{-1}(0) = {(p,X) :(rho(g)p,drho(g)X) = (p,X)}.$$
So,
$$tilde S = bigcup_{gin G}eta_g^{-1}(0). $$
But according to my calculation $0$ is not a regular value of $eta_g$.
I appreciate any help.
dg.differential-geometry riemannian-geometry differential-topology group-actions
edited 2 hours ago
Ali Taghavi
9152084
asked 4 hours ago
L.F. CavenaghiL.F. Cavenaghi
598213
$endgroup$
$begingroup$
Is this a pigeon? ;-)
$endgroup$
– David Roberts
1 hour ago
$begingroup$
@DavidRoberts, I am sorry, what are you asking if it is a pigeon?
$endgroup$
– L.F. Cavenaghi
1 hour ago
1
$begingroup$
It's a meme, meant in good humour.
$endgroup$
– David Roberts
25 mins ago
add a comment |
$begingroup$
Let $(M,g)$ be a compact Riemannian manifold with an isometric action $rho : G to mathrm{Iso}(M)$ 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 $eta_g(p,X) equiv eta(g,p,X) := d^2_{TM}left((p,X),psi(g,p,X)right).$ Then, one has:
$$eta_g^{-1}(0) = {(p,X) :(rho(g)p,drho(g)X) = (p,X)}.$$
So,
$$tilde S = bigcup_{gin G}eta_g^{-1}(0). $$
But according to my calculation $0$ is not a regular value of $eta_g$.
I appreciate any help.
dg.differential-geometry riemannian-geometry differential-topology group-actions
edited 2 hours ago
Ali Taghavi
9152084
asked 4 hours ago
L.F. CavenaghiL.F. Cavenaghi
598213
$endgroup$
Let $(M,g)$ be a compact Riemannian manifold with an isometric action $rho : G to mathrm{Iso}(M)$ 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 $eta_g(p,X) equiv eta(g,p,X) := d^2_{TM}left((p,X),psi(g,p,X)right).$ Then, one has:
$$eta_g^{-1}(0) = {(p,X) :(rho(g)p,drho(g)X) = (p,X)}.$$
So,
$$tilde S = bigcup_{gin G}eta_g^{-1}(0). $$
But according to my calculation $0$ is not a regular value of $eta_g$.
I appreciate any help.
dg.differential-geometry riemannian-geometry differential-topology group-actions
dg.differential-geometry riemannian-geometry differential-topology group-actions
edited 2 hours ago
Ali Taghavi
9152084
asked 4 hours ago
L.F. CavenaghiL.F. Cavenaghi
598213
edited 2 hours ago
Ali Taghavi
9152084
asked 4 hours ago
L.F. CavenaghiL.F. Cavenaghi
598213
edited 2 hours ago
Ali Taghavi
9152084
edited 2 hours ago
Ali Taghavi
9152084
edited 2 hours ago
Ali Taghavi
9152084
9152084
asked 4 hours ago
L.F. CavenaghiL.F. Cavenaghi
598213
asked 4 hours ago
L.F. CavenaghiL.F. Cavenaghi
598213
asked 4 hours ago
L.F. CavenaghiL.F. Cavenaghi
598213
598213
$begingroup$
Is this a pigeon? ;-)
$endgroup$
– David Roberts
1 hour ago
$begingroup$
@DavidRoberts, I am sorry, what are you asking if it is a pigeon?
$endgroup$
– L.F. Cavenaghi
1 hour ago
1
$begingroup$
It's a meme, meant in good humour.
$endgroup$
– David Roberts
25 mins ago
add a comment |
$begingroup$
Is this a pigeon? ;-)
$endgroup$
– David Roberts
1 hour ago
$begingroup$
@DavidRoberts, I am sorry, what are you asking if it is a pigeon?
$endgroup$
– L.F. Cavenaghi
1 hour ago
1
$begingroup$
It's a meme, meant in good humour.
$endgroup$
– David Roberts
25 mins ago
$begingroup$
Is this a pigeon? ;-)
$endgroup$
– David Roberts
1 hour ago
$begingroup$
Is this a pigeon? ;-)
$endgroup$
– David Roberts
1 hour ago
$begingroup$
@DavidRoberts, I am sorry, what are you asking if it is a pigeon?
$endgroup$
– L.F. Cavenaghi
1 hour ago
$begingroup$
@DavidRoberts, I am sorry, what are you asking if it is a pigeon?
$endgroup$
– L.F. Cavenaghi
1 hour ago
1
1
$begingroup$
It's a meme, meant in good humour.
$endgroup$
– David Roberts
25 mins ago
$begingroup$
It's a meme, meant in good humour.
$endgroup$
– David Roberts
25 mins ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For the obvious reflection action of $mathbb{Z}/2mathbb{Z}$ 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 2 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
1 hour 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
1 hour 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
1 hour 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
53 mins 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
7 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 $T(S^2setminus{n,s})cup {n,s}$, and the points $n$ and $s$ do not have neighborhoods homeomorphic to open balls.
answered 20 mins 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
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 $mathbb{Z}/2mathbb{Z}$ 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 2 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
1 hour 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
1 hour 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
1 hour 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
53 mins 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
7 mins ago
add a comment |
$begingroup$
For the obvious reflection action of $mathbb{Z}/2mathbb{Z}$ 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 2 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
1 hour 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
1 hour 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
1 hour 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
53 mins 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
7 mins ago
add a comment |
$begingroup$
For the obvious reflection action of $mathbb{Z}/2mathbb{Z}$ 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 2 hours ago
Ali TaghaviAli Taghavi
9152084
$endgroup$
For the obvious reflection action of $mathbb{Z}/2mathbb{Z}$ 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 2 hours ago
Ali TaghaviAli Taghavi
9152084
edited 1 hour ago
answered 2 hours ago
Ali TaghaviAli Taghavi
9152084
answered 2 hours ago
Ali TaghaviAli Taghavi
9152084
answered 2 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
1 hour 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
1 hour 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
1 hour 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
53 mins 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
7 mins 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
1 hour 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
1 hour 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
1 hour 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
53 mins 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
7 mins 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
1 hour ago
$begingroup$
"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
1 hour 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
1 hour 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
1 hour ago
1
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
1 hour 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?
$endgroup$
– L.F. Cavenaghi
1 hour 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
53 mins 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
53 mins 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
7 mins 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
7 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 $T(S^2setminus{n,s})cup {n,s}$, and the points $n$ and $s$ do not have neighborhoods homeomorphic to open balls.
answered 20 mins 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
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 $T(S^2setminus{n,s})cup {n,s}$, and the points $n$ and $s$ do not have neighborhoods homeomorphic to open balls.
answered 20 mins 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
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 $T(S^2setminus{n,s})cup {n,s}$, and the points $n$ and $s$ do not have neighborhoods homeomorphic to open balls.
answered 20 mins 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 $T(S^2setminus{n,s})cup {n,s}$, and the points $n$ and $s$ do not have neighborhoods homeomorphic to open balls.
answered 20 mins ago
Steve CostenobleSteve Costenoble
9451514
answered 20 mins ago
Steve CostenobleSteve Costenoble
9451514
answered 20 mins ago
Steve CostenobleSteve Costenoble
9451514
answered 20 mins 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
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
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
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
9 mins ago
add a comment |
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$begingroup$
Is this a pigeon? ;-)
$endgroup$
– David Roberts
1 hour ago
$begingroup$
@DavidRoberts, I am sorry, what are you asking if it is a pigeon?
$endgroup$
– L.F. Cavenaghi
1 hour ago
1
$begingroup$
It's a meme, meant in good humour.
$endgroup$
– David Roberts
25 mins ago