Is this a submanifold?












1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is this a pigeon? ;-)
    $endgroup$
    – David Roberts
    2 hours 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
    58 mins ago
















1












$begingroup$


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







share|cite|improve this question











$endgroup$












  • $begingroup$
    Is this a pigeon? ;-)
    $endgroup$
    – David Roberts
    2 hours 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
    58 mins ago














1












1








1


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 hours ago









Ali Taghavi

9152084




9152084










asked 5 hours ago









L.F. CavenaghiL.F. Cavenaghi

598213




598213












  • $begingroup$
    Is this a pigeon? ;-)
    $endgroup$
    – David Roberts
    2 hours 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
    58 mins ago


















  • $begingroup$
    Is this a pigeon? ;-)
    $endgroup$
    – David Roberts
    2 hours 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
    58 mins ago
















$begingroup$
Is this a pigeon? ;-)
$endgroup$
– David Roberts
2 hours ago




$begingroup$
Is this a pigeon? ;-)
$endgroup$
– David Roberts
2 hours 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
58 mins ago




$begingroup$
It's a meme, meant in good humour.
$endgroup$
– David Roberts
58 mins ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 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
    2 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
    2 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
    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
    1 hour 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
    40 mins ago



















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
    42 mins ago













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 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
    2 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
    2 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
    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
    1 hour 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
    40 mins ago
















2












$begingroup$

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.






share|cite|improve this answer











$endgroup$









  • 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
    2 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
    2 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
    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
    1 hour 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
    40 mins ago














2












2








2





$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited 2 hours ago

























answered 3 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
    2 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
    2 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
    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
    1 hour 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
    40 mins ago














  • 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
    2 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
    2 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
    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
    1 hour 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
    40 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
2 hours 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
2 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
2 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
2 hours 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
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
1 hour 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
40 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
40 mins ago











1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
    42 mins ago


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
    42 mins ago
















1












1








1





$begingroup$

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.






share|cite|improve this answer









$endgroup$



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.







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answered 54 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
    42 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
    42 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
42 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
42 mins ago




















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