Quotient group as a manifold
$begingroup$
Let $V_k(n) subset prod _{i=1} ^k S^n$
the real Stiefel space endowed with subspace topology and defined via
$$V_k(n) := {(v_1, v_2, ..., v_k) vert text{ } v_i bot v_j text{ for } i neq j text{ and } left| v_i right|=1 }$$
I know that there are a lot of ways to show that $V_k(n)$ is a manifold but I intend to prove it using the group action by $O(k)$ and to show that $V_k(n) cong O(n)/O(n-k)$.
The problem is to show that $O(k)/O(n-k)$ is a manifold. The concrete point I'm struggle now is to verify the totally discontinuity of the action.
Therefore I don't know how to show that for every $(v_1, v_2, ..., v_k) in V_k(n)$ there exist an open neightbourhood $U$ such that for all $f neq g in O(n)$ we have $fU cap gU = emptyset$.
The problem is that $O(k)$ is not discrete so I suppose that I need a clever choice of $U$, but I don't find the correct choice.
group-theory manifolds quotient-group
edited Dec 5 '18 at 4:17
Andrews
3831317
asked Dec 5 '18 at 2:27
KarlPeterKarlPeter
5931315
$endgroup$
add a comment |
$begingroup$
Let $V_k(n) subset prod _{i=1} ^k S^n$
the real Stiefel space endowed with subspace topology and defined via
$$V_k(n) := {(v_1, v_2, ..., v_k) vert text{ } v_i bot v_j text{ for } i neq j text{ and } left| v_i right|=1 }$$
I know that there are a lot of ways to show that $V_k(n)$ is a manifold but I intend to prove it using the group action by $O(k)$ and to show that $V_k(n) cong O(n)/O(n-k)$.
The problem is to show that $O(k)/O(n-k)$ is a manifold. The concrete point I'm struggle now is to verify the totally discontinuity of the action.
Therefore I don't know how to show that for every $(v_1, v_2, ..., v_k) in V_k(n)$ there exist an open neightbourhood $U$ such that for all $f neq g in O(n)$ we have $fU cap gU = emptyset$.
The problem is that $O(k)$ is not discrete so I suppose that I need a clever choice of $U$, but I don't find the correct choice.
group-theory manifolds quotient-group
edited Dec 5 '18 at 4:17
Andrews
3831317
asked Dec 5 '18 at 2:27
KarlPeterKarlPeter
5931315
$endgroup$
1
$begingroup$
Since $O(k)$ is not discrete, I don't think any choice of $U$ will work. There are other conditions which guarantee the quotient is a manifold.
$endgroup$
– Jason DeVito
Dec 5 '18 at 3:36
$begingroup$
You should see a proof that if G acts freely on X then X/G is a smooth manifold (here G is assumed to be a Lie group, X a manifold and the action should be smooth). Also that's a really useful result.
$endgroup$
– Nicolas Hemelsoet
Dec 5 '18 at 6:52
add a comment |
$begingroup$
Let $V_k(n) subset prod _{i=1} ^k S^n$
the real Stiefel space endowed with subspace topology and defined via
$$V_k(n) := {(v_1, v_2, ..., v_k) vert text{ } v_i bot v_j text{ for } i neq j text{ and } left| v_i right|=1 }$$
I know that there are a lot of ways to show that $V_k(n)$ is a manifold but I intend to prove it using the group action by $O(k)$ and to show that $V_k(n) cong O(n)/O(n-k)$.
The problem is to show that $O(k)/O(n-k)$ is a manifold. The concrete point I'm struggle now is to verify the totally discontinuity of the action.
Therefore I don't know how to show that for every $(v_1, v_2, ..., v_k) in V_k(n)$ there exist an open neightbourhood $U$ such that for all $f neq g in O(n)$ we have $fU cap gU = emptyset$.
The problem is that $O(k)$ is not discrete so I suppose that I need a clever choice of $U$, but I don't find the correct choice.
group-theory manifolds quotient-group
edited Dec 5 '18 at 4:17
Andrews
3831317
asked Dec 5 '18 at 2:27
KarlPeterKarlPeter
5931315
$endgroup$
Let $V_k(n) subset prod _{i=1} ^k S^n$
the real Stiefel space endowed with subspace topology and defined via
$$V_k(n) := {(v_1, v_2, ..., v_k) vert text{ } v_i bot v_j text{ for } i neq j text{ and } left| v_i right|=1 }$$
I know that there are a lot of ways to show that $V_k(n)$ is a manifold but I intend to prove it using the group action by $O(k)$ and to show that $V_k(n) cong O(n)/O(n-k)$.
The problem is to show that $O(k)/O(n-k)$ is a manifold. The concrete point I'm struggle now is to verify the totally discontinuity of the action.
Therefore I don't know how to show that for every $(v_1, v_2, ..., v_k) in V_k(n)$ there exist an open neightbourhood $U$ such that for all $f neq g in O(n)$ we have $fU cap gU = emptyset$.
The problem is that $O(k)$ is not discrete so I suppose that I need a clever choice of $U$, but I don't find the correct choice.
group-theory manifolds quotient-group
group-theory manifolds quotient-group
edited Dec 5 '18 at 4:17
Andrews
3831317
asked Dec 5 '18 at 2:27
KarlPeterKarlPeter
5931315
edited Dec 5 '18 at 4:17
Andrews
3831317
asked Dec 5 '18 at 2:27
KarlPeterKarlPeter
5931315
edited Dec 5 '18 at 4:17
Andrews
3831317
edited Dec 5 '18 at 4:17
Andrews
3831317
edited Dec 5 '18 at 4:17
Andrews
3831317
3831317
asked Dec 5 '18 at 2:27
KarlPeterKarlPeter
5931315
asked Dec 5 '18 at 2:27
KarlPeterKarlPeter
5931315
asked Dec 5 '18 at 2:27
KarlPeterKarlPeter
5931315
5931315
1
$begingroup$
Since $O(k)$ is not discrete, I don't think any choice of $U$ will work. There are other conditions which guarantee the quotient is a manifold.
$endgroup$
– Jason DeVito
Dec 5 '18 at 3:36
$begingroup$
You should see a proof that if G acts freely on X then X/G is a smooth manifold (here G is assumed to be a Lie group, X a manifold and the action should be smooth). Also that's a really useful result.
$endgroup$
– Nicolas Hemelsoet
Dec 5 '18 at 6:52
add a comment |
1
$begingroup$
Since $O(k)$ is not discrete, I don't think any choice of $U$ will work. There are other conditions which guarantee the quotient is a manifold.
$endgroup$
– Jason DeVito
Dec 5 '18 at 3:36
$begingroup$
You should see a proof that if G acts freely on X then X/G is a smooth manifold (here G is assumed to be a Lie group, X a manifold and the action should be smooth). Also that's a really useful result.
$endgroup$
– Nicolas Hemelsoet
Dec 5 '18 at 6:52
1
1
$begingroup$
Since $O(k)$ is not discrete, I don't think any choice of $U$ will work. There are other conditions which guarantee the quotient is a manifold.
$endgroup$
– Jason DeVito
Dec 5 '18 at 3:36
$begingroup$
Since $O(k)$ is not discrete, I don't think any choice of $U$ will work. There are other conditions which guarantee the quotient is a manifold.
$endgroup$
– Jason DeVito
Dec 5 '18 at 3:36
$begingroup$
You should see a proof that if G acts freely on X then X/G is a smooth manifold (here G is assumed to be a Lie group, X a manifold and the action should be smooth). Also that's a really useful result.
$endgroup$
– Nicolas Hemelsoet
Dec 5 '18 at 6:52
$begingroup$
You should see a proof that if G acts freely on X then X/G is a smooth manifold (here G is assumed to be a Lie group, X a manifold and the action should be smooth). Also that's a really useful result.
$endgroup$
– Nicolas Hemelsoet
Dec 5 '18 at 6:52
add a comment |
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$begingroup$
Since $O(k)$ is not discrete, I don't think any choice of $U$ will work. There are other conditions which guarantee the quotient is a manifold.
$endgroup$
– Jason DeVito
Dec 5 '18 at 3:36
$begingroup$
You should see a proof that if G acts freely on X then X/G is a smooth manifold (here G is assumed to be a Lie group, X a manifold and the action should be smooth). Also that's a really useful result.
$endgroup$
– Nicolas Hemelsoet
Dec 5 '18 at 6:52