Isotropy/little group of $O(n)$
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I'm trying to prove that the little group of $O(n)$ acting on a $k$-dimensional subspace of $mathbb{R}^n$, call it $V$, is $O(k)times O(n - k)$ due to the Grassmann manifold is isomorphic to $O(n)/(O(k)times O(n - k))$.
I tried following the next steps: for elements in the little group, call them $g_l$, $q in V$ has to be invariant, that is $g_lq = q$ and therefore:
$$g_l = begin{pmatrix}1_k & 0 \0 & A_{n - k} end{pmatrix} tag1$$
Where $1_k$ is the $ktimes k$ identity matrix and $A_{n-k}$ a $(n - k)times (n - k)$ matrix. A similar process can be seen in Isotropy group of $SO(n)$. I presume that $A_{n - k} in O(n - k)$ because we are working with orthonormal transformations.
Therefore, we conclude that the little group is $1_ktimes O(n - k)$ which is different from the deduction by the isomorphism for Grassmann manifold.
I'm doing something bad but I don't know what. Can you show me the way?
group-theory differential-geometry lie-groups vector-space-isomorphism grassmannian
asked Nov 16 at 15:38
Vicky
1387
add a comment |
up vote
1
down vote
favorite
I'm trying to prove that the little group of $O(n)$ acting on a $k$-dimensional subspace of $mathbb{R}^n$, call it $V$, is $O(k)times O(n - k)$ due to the Grassmann manifold is isomorphic to $O(n)/(O(k)times O(n - k))$.
I tried following the next steps: for elements in the little group, call them $g_l$, $q in V$ has to be invariant, that is $g_lq = q$ and therefore:
$$g_l = begin{pmatrix}1_k & 0 \0 & A_{n - k} end{pmatrix} tag1$$
Where $1_k$ is the $ktimes k$ identity matrix and $A_{n-k}$ a $(n - k)times (n - k)$ matrix. A similar process can be seen in Isotropy group of $SO(n)$. I presume that $A_{n - k} in O(n - k)$ because we are working with orthonormal transformations.
Therefore, we conclude that the little group is $1_ktimes O(n - k)$ which is different from the deduction by the isomorphism for Grassmann manifold.
I'm doing something bad but I don't know what. Can you show me the way?
group-theory differential-geometry lie-groups vector-space-isomorphism grassmannian
asked Nov 16 at 15:38
Vicky
1387
1
First, in English, we say "subgroup" rather than "little group." So your mistake is that $1_k$ can be any orthogonal transformation preserving $q$. This is isomorphic to $O(k)$, not just the identity. (Take $q$, in particular, to be the subspace $Bbb R^ktimes{0}$.) And, of course, $A_{n-k}$ must be orthogonal, as well.
– Ted Shifrin
Nov 16 at 23:02
First, for 'little group' I didn't mean 'subgroup' but 'isotropic' which are synonymous. And second, my mistake is quite related to what you say. Instead of thinking in whole spaces ($q = mathbb{R}^ktimes{0}$), I was thinking in particular vectors ($q in mathbb{R}^ktimes{0}$). So thanks for your contribution
– Vicky
Nov 16 at 23:08
"Isotropy subgroup" is what you mean :)
– Ted Shifrin
Nov 16 at 23:11
That's it. The slash between 'isotropic' and 'little' was just to put together both names. And again thanks for your contribution, it enlightened me
– Vicky
Nov 16 at 23:12
You're most welcome.
– Ted Shifrin
Nov 16 at 23:15
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm trying to prove that the little group of $O(n)$ acting on a $k$-dimensional subspace of $mathbb{R}^n$, call it $V$, is $O(k)times O(n - k)$ due to the Grassmann manifold is isomorphic to $O(n)/(O(k)times O(n - k))$.
I tried following the next steps: for elements in the little group, call them $g_l$, $q in V$ has to be invariant, that is $g_lq = q$ and therefore:
$$g_l = begin{pmatrix}1_k & 0 \0 & A_{n - k} end{pmatrix} tag1$$
Where $1_k$ is the $ktimes k$ identity matrix and $A_{n-k}$ a $(n - k)times (n - k)$ matrix. A similar process can be seen in Isotropy group of $SO(n)$. I presume that $A_{n - k} in O(n - k)$ because we are working with orthonormal transformations.
Therefore, we conclude that the little group is $1_ktimes O(n - k)$ which is different from the deduction by the isomorphism for Grassmann manifold.
I'm doing something bad but I don't know what. Can you show me the way?
group-theory differential-geometry lie-groups vector-space-isomorphism grassmannian
asked Nov 16 at 15:38
Vicky
1387
I'm trying to prove that the little group of $O(n)$ acting on a $k$-dimensional subspace of $mathbb{R}^n$, call it $V$, is $O(k)times O(n - k)$ due to the Grassmann manifold is isomorphic to $O(n)/(O(k)times O(n - k))$.
I tried following the next steps: for elements in the little group, call them $g_l$, $q in V$ has to be invariant, that is $g_lq = q$ and therefore:
$$g_l = begin{pmatrix}1_k & 0 \0 & A_{n - k} end{pmatrix} tag1$$
Where $1_k$ is the $ktimes k$ identity matrix and $A_{n-k}$ a $(n - k)times (n - k)$ matrix. A similar process can be seen in Isotropy group of $SO(n)$. I presume that $A_{n - k} in O(n - k)$ because we are working with orthonormal transformations.
Therefore, we conclude that the little group is $1_ktimes O(n - k)$ which is different from the deduction by the isomorphism for Grassmann manifold.
I'm doing something bad but I don't know what. Can you show me the way?
group-theory differential-geometry lie-groups vector-space-isomorphism grassmannian
group-theory differential-geometry lie-groups vector-space-isomorphism grassmannian
asked Nov 16 at 15:38
Vicky
1387
asked Nov 16 at 15:38
Vicky
1387
edited Nov 16 at 18:36
asked Nov 16 at 15:38
Vicky
1387
asked Nov 16 at 15:38
Vicky
1387
asked Nov 16 at 15:38
Vicky
1387
1387
1
First, in English, we say "subgroup" rather than "little group." So your mistake is that $1_k$ can be any orthogonal transformation preserving $q$. This is isomorphic to $O(k)$, not just the identity. (Take $q$, in particular, to be the subspace $Bbb R^ktimes{0}$.) And, of course, $A_{n-k}$ must be orthogonal, as well.
– Ted Shifrin
Nov 16 at 23:02
First, for 'little group' I didn't mean 'subgroup' but 'isotropic' which are synonymous. And second, my mistake is quite related to what you say. Instead of thinking in whole spaces ($q = mathbb{R}^ktimes{0}$), I was thinking in particular vectors ($q in mathbb{R}^ktimes{0}$). So thanks for your contribution
– Vicky
Nov 16 at 23:08
"Isotropy subgroup" is what you mean :)
– Ted Shifrin
Nov 16 at 23:11
That's it. The slash between 'isotropic' and 'little' was just to put together both names. And again thanks for your contribution, it enlightened me
– Vicky
Nov 16 at 23:12
You're most welcome.
– Ted Shifrin
Nov 16 at 23:15
add a comment |
1
First, in English, we say "subgroup" rather than "little group." So your mistake is that $1_k$ can be any orthogonal transformation preserving $q$. This is isomorphic to $O(k)$, not just the identity. (Take $q$, in particular, to be the subspace $Bbb R^ktimes{0}$.) And, of course, $A_{n-k}$ must be orthogonal, as well.
– Ted Shifrin
Nov 16 at 23:02
First, for 'little group' I didn't mean 'subgroup' but 'isotropic' which are synonymous. And second, my mistake is quite related to what you say. Instead of thinking in whole spaces ($q = mathbb{R}^ktimes{0}$), I was thinking in particular vectors ($q in mathbb{R}^ktimes{0}$). So thanks for your contribution
– Vicky
Nov 16 at 23:08
"Isotropy subgroup" is what you mean :)
– Ted Shifrin
Nov 16 at 23:11
That's it. The slash between 'isotropic' and 'little' was just to put together both names. And again thanks for your contribution, it enlightened me
– Vicky
Nov 16 at 23:12
You're most welcome.
– Ted Shifrin
Nov 16 at 23:15
1
1
First, in English, we say "subgroup" rather than "little group." So your mistake is that $1_k$ can be any orthogonal transformation preserving $q$. This is isomorphic to $O(k)$, not just the identity. (Take $q$, in particular, to be the subspace $Bbb R^ktimes{0}$.) And, of course, $A_{n-k}$ must be orthogonal, as well.
– Ted Shifrin
Nov 16 at 23:02
First, in English, we say "subgroup" rather than "little group." So your mistake is that $1_k$ can be any orthogonal transformation preserving $q$. This is isomorphic to $O(k)$, not just the identity. (Take $q$, in particular, to be the subspace $Bbb R^ktimes{0}$.) And, of course, $A_{n-k}$ must be orthogonal, as well.
– Ted Shifrin
Nov 16 at 23:02
First, for 'little group' I didn't mean 'subgroup' but 'isotropic' which are synonymous. And second, my mistake is quite related to what you say. Instead of thinking in whole spaces ($q = mathbb{R}^ktimes{0}$), I was thinking in particular vectors ($q in mathbb{R}^ktimes{0}$). So thanks for your contribution
– Vicky
Nov 16 at 23:08
First, for 'little group' I didn't mean 'subgroup' but 'isotropic' which are synonymous. And second, my mistake is quite related to what you say. Instead of thinking in whole spaces ($q = mathbb{R}^ktimes{0}$), I was thinking in particular vectors ($q in mathbb{R}^ktimes{0}$). So thanks for your contribution
– Vicky
Nov 16 at 23:08
"Isotropy subgroup" is what you mean :)
– Ted Shifrin
Nov 16 at 23:11
"Isotropy subgroup" is what you mean :)
– Ted Shifrin
Nov 16 at 23:11
That's it. The slash between 'isotropic' and 'little' was just to put together both names. And again thanks for your contribution, it enlightened me
– Vicky
Nov 16 at 23:12
That's it. The slash between 'isotropic' and 'little' was just to put together both names. And again thanks for your contribution, it enlightened me
– Vicky
Nov 16 at 23:12
You're most welcome.
– Ted Shifrin
Nov 16 at 23:15
You're most welcome.
– Ted Shifrin
Nov 16 at 23:15
add a comment |
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1
First, in English, we say "subgroup" rather than "little group." So your mistake is that $1_k$ can be any orthogonal transformation preserving $q$. This is isomorphic to $O(k)$, not just the identity. (Take $q$, in particular, to be the subspace $Bbb R^ktimes{0}$.) And, of course, $A_{n-k}$ must be orthogonal, as well.
– Ted Shifrin
Nov 16 at 23:02
First, for 'little group' I didn't mean 'subgroup' but 'isotropic' which are synonymous. And second, my mistake is quite related to what you say. Instead of thinking in whole spaces ($q = mathbb{R}^ktimes{0}$), I was thinking in particular vectors ($q in mathbb{R}^ktimes{0}$). So thanks for your contribution
– Vicky
Nov 16 at 23:08
"Isotropy subgroup" is what you mean :)
– Ted Shifrin
Nov 16 at 23:11
That's it. The slash between 'isotropic' and 'little' was just to put together both names. And again thanks for your contribution, it enlightened me
– Vicky
Nov 16 at 23:12
You're most welcome.
– Ted Shifrin
Nov 16 at 23:15