Mixing
Differential of element in the stabiliser
$begingroup$
$G$ a compact Lie group, $W$ a smooth G-manifold, $x in W$.
Then as $G$ is compact $G cdot x$ is an embedded submanifold.
Now the text I'm reading claims that for $g in G_x$,
$dg_x : text{T}_x W to text{T}_{gcdot{x}}W=text{T}_x W$
sends the tangent space of the orbit to itself by the identity map. However, I'm having difficulty showing this is actually the identity map on $text{T}_{x}Gcdot{x}$. I've tried passing to $G/G_x$ but as $G$ is not necessarily Abelian I don't think this is any easier. Is there something obvious I'm missing?
lie-groups smooth-manifolds
asked Jan 23 at 18:58
O.S.O.S.
964
$endgroup$
add a comment |
$begingroup$
$G$ a compact Lie group, $W$ a smooth G-manifold, $x in W$.
Then as $G$ is compact $G cdot x$ is an embedded submanifold.
Now the text I'm reading claims that for $g in G_x$,
$dg_x : text{T}_x W to text{T}_{gcdot{x}}W=text{T}_x W$
sends the tangent space of the orbit to itself by the identity map. However, I'm having difficulty showing this is actually the identity map on $text{T}_{x}Gcdot{x}$. I've tried passing to $G/G_x$ but as $G$ is not necessarily Abelian I don't think this is any easier. Is there something obvious I'm missing?
lie-groups smooth-manifolds
asked Jan 23 at 18:58
O.S.O.S.
964
$endgroup$
1
$begingroup$
I don't understand the statement (what does $T_xG$ mean?). Consider the group of rotations acting on Euclidean space. The group fixes the origin but does not act trivially on the tangent space. Is that a counterexample?
$endgroup$
– Max
Jan 23 at 19:05
$begingroup$
Here $Gcdot{x}$ is the orbit $text{Orb}{x}$ as an embedded submanifold of W. I think your example is correct in the fact that it doesn't act trivially on the tangent space of the origin as an element of the whole Euclidean space. However the orbit of the origin is just itself and the action does act trivially on the tangent space of the orbit. *also I've just seen the confusing notation: $text{T}_{x}Gcdot{x}$ is meant to be $text{T}_{x}(Gcdot{x})$
$endgroup$
– O.S.
Jan 23 at 19:12
$begingroup$
Ok, I understand now.
$endgroup$
– Max
Jan 23 at 19:31
1
$begingroup$
What about taking $SO(3)$ acting on the unit sphere? The stabilizer of a point is a copy of $SO(2)$ which acts nontrivially on the tangent space (the action is transitive so the orbit is all of $S^2$)? I don't think the statement is true unless $G_x$ is normal, which would imply every point in the orbit of $x$ has the same stabilizer.
$endgroup$
– Max
Jan 23 at 23:55
add a comment |
$begingroup$
$G$ a compact Lie group, $W$ a smooth G-manifold, $x in W$.
Then as $G$ is compact $G cdot x$ is an embedded submanifold.
Now the text I'm reading claims that for $g in G_x$,
$dg_x : text{T}_x W to text{T}_{gcdot{x}}W=text{T}_x W$
sends the tangent space of the orbit to itself by the identity map. However, I'm having difficulty showing this is actually the identity map on $text{T}_{x}Gcdot{x}$. I've tried passing to $G/G_x$ but as $G$ is not necessarily Abelian I don't think this is any easier. Is there something obvious I'm missing?
lie-groups smooth-manifolds
asked Jan 23 at 18:58
O.S.O.S.
964
$endgroup$
$G$ a compact Lie group, $W$ a smooth G-manifold, $x in W$.
Then as $G$ is compact $G cdot x$ is an embedded submanifold.
Now the text I'm reading claims that for $g in G_x$,
$dg_x : text{T}_x W to text{T}_{gcdot{x}}W=text{T}_x W$
sends the tangent space of the orbit to itself by the identity map. However, I'm having difficulty showing this is actually the identity map on $text{T}_{x}Gcdot{x}$. I've tried passing to $G/G_x$ but as $G$ is not necessarily Abelian I don't think this is any easier. Is there something obvious I'm missing?
lie-groups smooth-manifolds
lie-groups smooth-manifolds
asked Jan 23 at 18:58
O.S.O.S.
964
asked Jan 23 at 18:58
O.S.O.S.
964
asked Jan 23 at 18:58
O.S.O.S.
964
asked Jan 23 at 18:58
O.S.O.S.
964
asked Jan 23 at 18:58
O.S.O.S.
964
964
1
$begingroup$
I don't understand the statement (what does $T_xG$ mean?). Consider the group of rotations acting on Euclidean space. The group fixes the origin but does not act trivially on the tangent space. Is that a counterexample?
$endgroup$
– Max
Jan 23 at 19:05
$begingroup$
Here $Gcdot{x}$ is the orbit $text{Orb}{x}$ as an embedded submanifold of W. I think your example is correct in the fact that it doesn't act trivially on the tangent space of the origin as an element of the whole Euclidean space. However the orbit of the origin is just itself and the action does act trivially on the tangent space of the orbit. *also I've just seen the confusing notation: $text{T}_{x}Gcdot{x}$ is meant to be $text{T}_{x}(Gcdot{x})$
$endgroup$
– O.S.
Jan 23 at 19:12
$begingroup$
Ok, I understand now.
$endgroup$
– Max
Jan 23 at 19:31
1
$begingroup$
What about taking $SO(3)$ acting on the unit sphere? The stabilizer of a point is a copy of $SO(2)$ which acts nontrivially on the tangent space (the action is transitive so the orbit is all of $S^2$)? I don't think the statement is true unless $G_x$ is normal, which would imply every point in the orbit of $x$ has the same stabilizer.
$endgroup$
– Max
Jan 23 at 23:55
add a comment |
1
$begingroup$
I don't understand the statement (what does $T_xG$ mean?). Consider the group of rotations acting on Euclidean space. The group fixes the origin but does not act trivially on the tangent space. Is that a counterexample?
$endgroup$
– Max
Jan 23 at 19:05
$begingroup$
Here $Gcdot{x}$ is the orbit $text{Orb}{x}$ as an embedded submanifold of W. I think your example is correct in the fact that it doesn't act trivially on the tangent space of the origin as an element of the whole Euclidean space. However the orbit of the origin is just itself and the action does act trivially on the tangent space of the orbit. *also I've just seen the confusing notation: $text{T}_{x}Gcdot{x}$ is meant to be $text{T}_{x}(Gcdot{x})$
$endgroup$
– O.S.
Jan 23 at 19:12
$begingroup$
Ok, I understand now.
$endgroup$
– Max
Jan 23 at 19:31
1
$begingroup$
What about taking $SO(3)$ acting on the unit sphere? The stabilizer of a point is a copy of $SO(2)$ which acts nontrivially on the tangent space (the action is transitive so the orbit is all of $S^2$)? I don't think the statement is true unless $G_x$ is normal, which would imply every point in the orbit of $x$ has the same stabilizer.
$endgroup$
– Max
Jan 23 at 23:55
1
1
$begingroup$
I don't understand the statement (what does $T_xG$ mean?). Consider the group of rotations acting on Euclidean space. The group fixes the origin but does not act trivially on the tangent space. Is that a counterexample?
$endgroup$
– Max
Jan 23 at 19:05
$begingroup$
I don't understand the statement (what does $T_xG$ mean?). Consider the group of rotations acting on Euclidean space. The group fixes the origin but does not act trivially on the tangent space. Is that a counterexample?
$endgroup$
– Max
Jan 23 at 19:05
$begingroup$
Here $Gcdot{x}$ is the orbit $text{Orb}{x}$ as an embedded submanifold of W. I think your example is correct in the fact that it doesn't act trivially on the tangent space of the origin as an element of the whole Euclidean space. However the orbit of the origin is just itself and the action does act trivially on the tangent space of the orbit. *also I've just seen the confusing notation: $text{T}_{x}Gcdot{x}$ is meant to be $text{T}_{x}(Gcdot{x})$
$endgroup$
– O.S.
Jan 23 at 19:12
$begingroup$
Here $Gcdot{x}$ is the orbit $text{Orb}{x}$ as an embedded submanifold of W. I think your example is correct in the fact that it doesn't act trivially on the tangent space of the origin as an element of the whole Euclidean space. However the orbit of the origin is just itself and the action does act trivially on the tangent space of the orbit. *also I've just seen the confusing notation: $text{T}_{x}Gcdot{x}$ is meant to be $text{T}_{x}(Gcdot{x})$
$endgroup$
– O.S.
Jan 23 at 19:12
$begingroup$
Ok, I understand now.
$endgroup$
– Max
Jan 23 at 19:31
$begingroup$
Ok, I understand now.
$endgroup$
– Max
Jan 23 at 19:31
1
1
$begingroup$
What about taking $SO(3)$ acting on the unit sphere? The stabilizer of a point is a copy of $SO(2)$ which acts nontrivially on the tangent space (the action is transitive so the orbit is all of $S^2$)? I don't think the statement is true unless $G_x$ is normal, which would imply every point in the orbit of $x$ has the same stabilizer.
$endgroup$
– Max
Jan 23 at 23:55
$begingroup$
What about taking $SO(3)$ acting on the unit sphere? The stabilizer of a point is a copy of $SO(2)$ which acts nontrivially on the tangent space (the action is transitive so the orbit is all of $S^2$)? I don't think the statement is true unless $G_x$ is normal, which would imply every point in the orbit of $x$ has the same stabilizer.
$endgroup$
– Max
Jan 23 at 23:55
add a comment |
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$begingroup$
I don't understand the statement (what does $T_xG$ mean?). Consider the group of rotations acting on Euclidean space. The group fixes the origin but does not act trivially on the tangent space. Is that a counterexample?
$endgroup$
– Max
Jan 23 at 19:05
$begingroup$
Here $Gcdot{x}$ is the orbit $text{Orb}{x}$ as an embedded submanifold of W. I think your example is correct in the fact that it doesn't act trivially on the tangent space of the origin as an element of the whole Euclidean space. However the orbit of the origin is just itself and the action does act trivially on the tangent space of the orbit. *also I've just seen the confusing notation: $text{T}_{x}Gcdot{x}$ is meant to be $text{T}_{x}(Gcdot{x})$
$endgroup$
– O.S.
Jan 23 at 19:12
$begingroup$
Ok, I understand now.
$endgroup$
– Max
Jan 23 at 19:31
1
$begingroup$
What about taking $SO(3)$ acting on the unit sphere? The stabilizer of a point is a copy of $SO(2)$ which acts nontrivially on the tangent space (the action is transitive so the orbit is all of $S^2$)? I don't think the statement is true unless $G_x$ is normal, which would imply every point in the orbit of $x$ has the same stabilizer.
$endgroup$
– Max
Jan 23 at 23:55