Real Grassmann manifold and orthonormal groups
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I'm trying to prove that the Grassmann manifold
$$G_k(mathbb{R}^n) = {E = {rm {it k} - dimensional subspace of } mathbb{R}^n}$$
is equivalent to:
$$G_k(mathbb{R}^n) = frac{O(n)}{O(k)times O(n - k)} tag1$$
Where $O(n)$ is the orthonormal group of $ntimes n$ matrices.
From my research I've seen that Eq. (1) is due to the idea of splitting the original $n$-dimensional subspace into a $k$-dimensional one and its orthonormal complement of $n - k$ dimension; but I don't get how this Grassmann manifold, which is made of vectors ($(ktimes 1)$-dimensional matrices - column vectors), is related to $ntimes n$ matrices since the quotient in Eq. (1) is the following set, as usual:
$$frac{O(n)}{O(k)times O(n - k)} = {M_ncdot (O(k)times O(n - k)) | M_n in O(n)} tag2$$
Can anyone explain me this relation and the prove of Eq. (1)? Thanks in advace! ;)
group-theory differential-geometry lie-groups quotient-spaces grassmannian
asked Nov 16 at 4:14
Vicky
1387
|
show 4 more comments
up vote
2
down vote
favorite
I'm trying to prove that the Grassmann manifold
$$G_k(mathbb{R}^n) = {E = {rm {it k} - dimensional subspace of } mathbb{R}^n}$$
is equivalent to:
$$G_k(mathbb{R}^n) = frac{O(n)}{O(k)times O(n - k)} tag1$$
Where $O(n)$ is the orthonormal group of $ntimes n$ matrices.
From my research I've seen that Eq. (1) is due to the idea of splitting the original $n$-dimensional subspace into a $k$-dimensional one and its orthonormal complement of $n - k$ dimension; but I don't get how this Grassmann manifold, which is made of vectors ($(ktimes 1)$-dimensional matrices - column vectors), is related to $ntimes n$ matrices since the quotient in Eq. (1) is the following set, as usual:
$$frac{O(n)}{O(k)times O(n - k)} = {M_ncdot (O(k)times O(n - k)) | M_n in O(n)} tag2$$
Can anyone explain me this relation and the prove of Eq. (1)? Thanks in advace! ;)
group-theory differential-geometry lie-groups quotient-spaces grassmannian
asked Nov 16 at 4:14
Vicky
1387
What is your definition of Grassmann manifold, i.e., in which sense is it made of "(1 times k)-dimensional matrices"?
– Travis
Nov 16 at 4:16
In any case, are you familiar with the characterization/construction of homogeneous spaces? See, e.g., Theorem 9.22 (Homogeneous Construction Space Theorem) in Lee's Introduction to Smooth Manifolds.
– Travis
Nov 16 at 4:18
It is a vector space, isn't it? A $k$-dimensional subspace of $mathbb{R}^n$, so $1times k$ - matrices are vectors with $k$ components
– Vicky
Nov 16 at 4:20
1
That description isn't quite right. The Grassmannian $G_k(V)$ is the space of all $k$-dimensional subspaces of $V$. It is true that we can specify any such plane by giving a basis, i.e., $k$ linearly independent vectors in $V$, but that choice is not unique.
– Travis
Nov 16 at 4:42
1
The question isn't quite a duplicate, but by chance I explained exactly this in an answer to another question a few days ago: math.stackexchange.com/a/2993101/155629
– Travis
Nov 16 at 4:43
|
show 4 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I'm trying to prove that the Grassmann manifold
$$G_k(mathbb{R}^n) = {E = {rm {it k} - dimensional subspace of } mathbb{R}^n}$$
is equivalent to:
$$G_k(mathbb{R}^n) = frac{O(n)}{O(k)times O(n - k)} tag1$$
Where $O(n)$ is the orthonormal group of $ntimes n$ matrices.
From my research I've seen that Eq. (1) is due to the idea of splitting the original $n$-dimensional subspace into a $k$-dimensional one and its orthonormal complement of $n - k$ dimension; but I don't get how this Grassmann manifold, which is made of vectors ($(ktimes 1)$-dimensional matrices - column vectors), is related to $ntimes n$ matrices since the quotient in Eq. (1) is the following set, as usual:
$$frac{O(n)}{O(k)times O(n - k)} = {M_ncdot (O(k)times O(n - k)) | M_n in O(n)} tag2$$
Can anyone explain me this relation and the prove of Eq. (1)? Thanks in advace! ;)
group-theory differential-geometry lie-groups quotient-spaces grassmannian
asked Nov 16 at 4:14
Vicky
1387
I'm trying to prove that the Grassmann manifold
$$G_k(mathbb{R}^n) = {E = {rm {it k} - dimensional subspace of } mathbb{R}^n}$$
is equivalent to:
$$G_k(mathbb{R}^n) = frac{O(n)}{O(k)times O(n - k)} tag1$$
Where $O(n)$ is the orthonormal group of $ntimes n$ matrices.
From my research I've seen that Eq. (1) is due to the idea of splitting the original $n$-dimensional subspace into a $k$-dimensional one and its orthonormal complement of $n - k$ dimension; but I don't get how this Grassmann manifold, which is made of vectors ($(ktimes 1)$-dimensional matrices - column vectors), is related to $ntimes n$ matrices since the quotient in Eq. (1) is the following set, as usual:
$$frac{O(n)}{O(k)times O(n - k)} = {M_ncdot (O(k)times O(n - k)) | M_n in O(n)} tag2$$
Can anyone explain me this relation and the prove of Eq. (1)? Thanks in advace! ;)
group-theory differential-geometry lie-groups quotient-spaces grassmannian
group-theory differential-geometry lie-groups quotient-spaces grassmannian
asked Nov 16 at 4:14
Vicky
1387
asked Nov 16 at 4:14
Vicky
1387
edited Nov 16 at 13:52
asked Nov 16 at 4:14
Vicky
1387
asked Nov 16 at 4:14
Vicky
1387
asked Nov 16 at 4:14
Vicky
1387
1387
What is your definition of Grassmann manifold, i.e., in which sense is it made of "(1 times k)-dimensional matrices"?
– Travis
Nov 16 at 4:16
In any case, are you familiar with the characterization/construction of homogeneous spaces? See, e.g., Theorem 9.22 (Homogeneous Construction Space Theorem) in Lee's Introduction to Smooth Manifolds.
– Travis
Nov 16 at 4:18
It is a vector space, isn't it? A $k$-dimensional subspace of $mathbb{R}^n$, so $1times k$ - matrices are vectors with $k$ components
– Vicky
Nov 16 at 4:20
1
That description isn't quite right. The Grassmannian $G_k(V)$ is the space of all $k$-dimensional subspaces of $V$. It is true that we can specify any such plane by giving a basis, i.e., $k$ linearly independent vectors in $V$, but that choice is not unique.
– Travis
Nov 16 at 4:42
1
The question isn't quite a duplicate, but by chance I explained exactly this in an answer to another question a few days ago: math.stackexchange.com/a/2993101/155629
– Travis
Nov 16 at 4:43
|
show 4 more comments
What is your definition of Grassmann manifold, i.e., in which sense is it made of "(1 times k)-dimensional matrices"?
– Travis
Nov 16 at 4:16
In any case, are you familiar with the characterization/construction of homogeneous spaces? See, e.g., Theorem 9.22 (Homogeneous Construction Space Theorem) in Lee's Introduction to Smooth Manifolds.
– Travis
Nov 16 at 4:18
It is a vector space, isn't it? A $k$-dimensional subspace of $mathbb{R}^n$, so $1times k$ - matrices are vectors with $k$ components
– Vicky
Nov 16 at 4:20
1
That description isn't quite right. The Grassmannian $G_k(V)$ is the space of all $k$-dimensional subspaces of $V$. It is true that we can specify any such plane by giving a basis, i.e., $k$ linearly independent vectors in $V$, but that choice is not unique.
– Travis
Nov 16 at 4:42
1
The question isn't quite a duplicate, but by chance I explained exactly this in an answer to another question a few days ago: math.stackexchange.com/a/2993101/155629
– Travis
Nov 16 at 4:43
What is your definition of Grassmann manifold, i.e., in which sense is it made of "(1 times k)-dimensional matrices"?
– Travis
Nov 16 at 4:16
What is your definition of Grassmann manifold, i.e., in which sense is it made of "(1 times k)-dimensional matrices"?
– Travis
Nov 16 at 4:16
In any case, are you familiar with the characterization/construction of homogeneous spaces? See, e.g., Theorem 9.22 (Homogeneous Construction Space Theorem) in Lee's Introduction to Smooth Manifolds.
– Travis
Nov 16 at 4:18
In any case, are you familiar with the characterization/construction of homogeneous spaces? See, e.g., Theorem 9.22 (Homogeneous Construction Space Theorem) in Lee's Introduction to Smooth Manifolds.
– Travis
Nov 16 at 4:18
It is a vector space, isn't it? A $k$-dimensional subspace of $mathbb{R}^n$, so $1times k$ - matrices are vectors with $k$ components
– Vicky
Nov 16 at 4:20
It is a vector space, isn't it? A $k$-dimensional subspace of $mathbb{R}^n$, so $1times k$ - matrices are vectors with $k$ components
– Vicky
Nov 16 at 4:20
1
1
That description isn't quite right. The Grassmannian $G_k(V)$ is the space of all $k$-dimensional subspaces of $V$. It is true that we can specify any such plane by giving a basis, i.e., $k$ linearly independent vectors in $V$, but that choice is not unique.
– Travis
Nov 16 at 4:42
That description isn't quite right. The Grassmannian $G_k(V)$ is the space of all $k$-dimensional subspaces of $V$. It is true that we can specify any such plane by giving a basis, i.e., $k$ linearly independent vectors in $V$, but that choice is not unique.
– Travis
Nov 16 at 4:42
1
1
The question isn't quite a duplicate, but by chance I explained exactly this in an answer to another question a few days ago: math.stackexchange.com/a/2993101/155629
– Travis
Nov 16 at 4:43
The question isn't quite a duplicate, but by chance I explained exactly this in an answer to another question a few days ago: math.stackexchange.com/a/2993101/155629
– Travis
Nov 16 at 4:43
|
show 4 more comments
1 Answer
1
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oldest
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up vote
1
down vote
Let $S_k(mathbb{R}^n)$ be the Steifel manifold of $k$-frames in $mathbb{R}^n$:
$$ S_k(mathbb{R}^n) = {(v_1, ldots, v_k) mid text{the $v_i$ are orthonormal} }$$
The group $O(n)$ acts transitively on $S_k(mathbb{R}^n)$ by acting on each vector: $g cdot (v_1, ldots, v_k) = (gv_1, ldots, gv_k)$. The stabiliser of a frame $(v_1, ldots, v_k)$ will be the subgroup of $O(n)$ which fixes the span of $(v_1, ldots, v_k)$ and acts nontrivially on its orthogonal complement. This subgroup is isomorphic to $O(n - k)$, and hence we have
$$ S_k(mathbb{R}^n) cong O(n) / O(n - k)$$
Next, there is a natural map $phi: S_k(mathbb{R}^n) to G_k(mathbb{R}^n)$ by taking the span of the frame. The fibre of this map over a point $E in G_k(mathbb{R}^n)$ will be the set of $k$-frames spanning $E$, which is a set isomorphic to $O(k)$ (an isomorphism of $O(k)$-sets). Hence $$G_k(mathbb{R}^n) cong S_k(mathbb{R}^n) / O(k)$$
These together give the equality you were after.
answered Nov 17 at 3:05
Joppy
5,503420
i don't get the last paragraph. Why do you talk about fibres and divide by $O(k)$?
– Vicky
Nov 17 at 13:08
@Vicky: For $E in G_k(mathbb{R^n}$, its preimage $phi^{-1}(E)$ is the set of $k$-frames in $E$. This set is isomorphic to $O(k)$, since $O(k)$ acts freely and transitively on it. This means that $S_k(mathbb{R}^n)$ looks like $G_k(mathbb{R}^n)$ with some copy of $O(k)$ glued to each point, so $S_k(mathbb{R}^n) cong O(k) times G_k(mathbb{R}^n)$ as sets.
– Joppy
Nov 18 at 1:43
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Let $S_k(mathbb{R}^n)$ be the Steifel manifold of $k$-frames in $mathbb{R}^n$:
$$ S_k(mathbb{R}^n) = {(v_1, ldots, v_k) mid text{the $v_i$ are orthonormal} }$$
The group $O(n)$ acts transitively on $S_k(mathbb{R}^n)$ by acting on each vector: $g cdot (v_1, ldots, v_k) = (gv_1, ldots, gv_k)$. The stabiliser of a frame $(v_1, ldots, v_k)$ will be the subgroup of $O(n)$ which fixes the span of $(v_1, ldots, v_k)$ and acts nontrivially on its orthogonal complement. This subgroup is isomorphic to $O(n - k)$, and hence we have
$$ S_k(mathbb{R}^n) cong O(n) / O(n - k)$$
Next, there is a natural map $phi: S_k(mathbb{R}^n) to G_k(mathbb{R}^n)$ by taking the span of the frame. The fibre of this map over a point $E in G_k(mathbb{R}^n)$ will be the set of $k$-frames spanning $E$, which is a set isomorphic to $O(k)$ (an isomorphism of $O(k)$-sets). Hence $$G_k(mathbb{R}^n) cong S_k(mathbb{R}^n) / O(k)$$
These together give the equality you were after.
answered Nov 17 at 3:05
Joppy
5,503420
i don't get the last paragraph. Why do you talk about fibres and divide by $O(k)$?
– Vicky
Nov 17 at 13:08
@Vicky: For $E in G_k(mathbb{R^n}$, its preimage $phi^{-1}(E)$ is the set of $k$-frames in $E$. This set is isomorphic to $O(k)$, since $O(k)$ acts freely and transitively on it. This means that $S_k(mathbb{R}^n)$ looks like $G_k(mathbb{R}^n)$ with some copy of $O(k)$ glued to each point, so $S_k(mathbb{R}^n) cong O(k) times G_k(mathbb{R}^n)$ as sets.
– Joppy
Nov 18 at 1:43
add a comment |
up vote
1
down vote
Let $S_k(mathbb{R}^n)$ be the Steifel manifold of $k$-frames in $mathbb{R}^n$:
$$ S_k(mathbb{R}^n) = {(v_1, ldots, v_k) mid text{the $v_i$ are orthonormal} }$$
The group $O(n)$ acts transitively on $S_k(mathbb{R}^n)$ by acting on each vector: $g cdot (v_1, ldots, v_k) = (gv_1, ldots, gv_k)$. The stabiliser of a frame $(v_1, ldots, v_k)$ will be the subgroup of $O(n)$ which fixes the span of $(v_1, ldots, v_k)$ and acts nontrivially on its orthogonal complement. This subgroup is isomorphic to $O(n - k)$, and hence we have
$$ S_k(mathbb{R}^n) cong O(n) / O(n - k)$$
Next, there is a natural map $phi: S_k(mathbb{R}^n) to G_k(mathbb{R}^n)$ by taking the span of the frame. The fibre of this map over a point $E in G_k(mathbb{R}^n)$ will be the set of $k$-frames spanning $E$, which is a set isomorphic to $O(k)$ (an isomorphism of $O(k)$-sets). Hence $$G_k(mathbb{R}^n) cong S_k(mathbb{R}^n) / O(k)$$
These together give the equality you were after.
answered Nov 17 at 3:05
Joppy
5,503420
i don't get the last paragraph. Why do you talk about fibres and divide by $O(k)$?
– Vicky
Nov 17 at 13:08
@Vicky: For $E in G_k(mathbb{R^n}$, its preimage $phi^{-1}(E)$ is the set of $k$-frames in $E$. This set is isomorphic to $O(k)$, since $O(k)$ acts freely and transitively on it. This means that $S_k(mathbb{R}^n)$ looks like $G_k(mathbb{R}^n)$ with some copy of $O(k)$ glued to each point, so $S_k(mathbb{R}^n) cong O(k) times G_k(mathbb{R}^n)$ as sets.
– Joppy
Nov 18 at 1:43
add a comment |
up vote
1
down vote
up vote
1
down vote
Let $S_k(mathbb{R}^n)$ be the Steifel manifold of $k$-frames in $mathbb{R}^n$:
$$ S_k(mathbb{R}^n) = {(v_1, ldots, v_k) mid text{the $v_i$ are orthonormal} }$$
The group $O(n)$ acts transitively on $S_k(mathbb{R}^n)$ by acting on each vector: $g cdot (v_1, ldots, v_k) = (gv_1, ldots, gv_k)$. The stabiliser of a frame $(v_1, ldots, v_k)$ will be the subgroup of $O(n)$ which fixes the span of $(v_1, ldots, v_k)$ and acts nontrivially on its orthogonal complement. This subgroup is isomorphic to $O(n - k)$, and hence we have
$$ S_k(mathbb{R}^n) cong O(n) / O(n - k)$$
Next, there is a natural map $phi: S_k(mathbb{R}^n) to G_k(mathbb{R}^n)$ by taking the span of the frame. The fibre of this map over a point $E in G_k(mathbb{R}^n)$ will be the set of $k$-frames spanning $E$, which is a set isomorphic to $O(k)$ (an isomorphism of $O(k)$-sets). Hence $$G_k(mathbb{R}^n) cong S_k(mathbb{R}^n) / O(k)$$
These together give the equality you were after.
answered Nov 17 at 3:05
Joppy
5,503420
Let $S_k(mathbb{R}^n)$ be the Steifel manifold of $k$-frames in $mathbb{R}^n$:
$$ S_k(mathbb{R}^n) = {(v_1, ldots, v_k) mid text{the $v_i$ are orthonormal} }$$
The group $O(n)$ acts transitively on $S_k(mathbb{R}^n)$ by acting on each vector: $g cdot (v_1, ldots, v_k) = (gv_1, ldots, gv_k)$. The stabiliser of a frame $(v_1, ldots, v_k)$ will be the subgroup of $O(n)$ which fixes the span of $(v_1, ldots, v_k)$ and acts nontrivially on its orthogonal complement. This subgroup is isomorphic to $O(n - k)$, and hence we have
$$ S_k(mathbb{R}^n) cong O(n) / O(n - k)$$
Next, there is a natural map $phi: S_k(mathbb{R}^n) to G_k(mathbb{R}^n)$ by taking the span of the frame. The fibre of this map over a point $E in G_k(mathbb{R}^n)$ will be the set of $k$-frames spanning $E$, which is a set isomorphic to $O(k)$ (an isomorphism of $O(k)$-sets). Hence $$G_k(mathbb{R}^n) cong S_k(mathbb{R}^n) / O(k)$$
These together give the equality you were after.
answered Nov 17 at 3:05
Joppy
5,503420
edited Nov 18 at 1:44
answered Nov 17 at 3:05
Joppy
5,503420
answered Nov 17 at 3:05
Joppy
5,503420
answered Nov 17 at 3:05
Joppy
5,503420
5,503420
i don't get the last paragraph. Why do you talk about fibres and divide by $O(k)$?
– Vicky
Nov 17 at 13:08
@Vicky: For $E in G_k(mathbb{R^n}$, its preimage $phi^{-1}(E)$ is the set of $k$-frames in $E$. This set is isomorphic to $O(k)$, since $O(k)$ acts freely and transitively on it. This means that $S_k(mathbb{R}^n)$ looks like $G_k(mathbb{R}^n)$ with some copy of $O(k)$ glued to each point, so $S_k(mathbb{R}^n) cong O(k) times G_k(mathbb{R}^n)$ as sets.
– Joppy
Nov 18 at 1:43
add a comment |
i don't get the last paragraph. Why do you talk about fibres and divide by $O(k)$?
– Vicky
Nov 17 at 13:08
@Vicky: For $E in G_k(mathbb{R^n}$, its preimage $phi^{-1}(E)$ is the set of $k$-frames in $E$. This set is isomorphic to $O(k)$, since $O(k)$ acts freely and transitively on it. This means that $S_k(mathbb{R}^n)$ looks like $G_k(mathbb{R}^n)$ with some copy of $O(k)$ glued to each point, so $S_k(mathbb{R}^n) cong O(k) times G_k(mathbb{R}^n)$ as sets.
– Joppy
Nov 18 at 1:43
i don't get the last paragraph. Why do you talk about fibres and divide by $O(k)$?
– Vicky
Nov 17 at 13:08
i don't get the last paragraph. Why do you talk about fibres and divide by $O(k)$?
– Vicky
Nov 17 at 13:08
@Vicky: For $E in G_k(mathbb{R^n}$, its preimage $phi^{-1}(E)$ is the set of $k$-frames in $E$. This set is isomorphic to $O(k)$, since $O(k)$ acts freely and transitively on it. This means that $S_k(mathbb{R}^n)$ looks like $G_k(mathbb{R}^n)$ with some copy of $O(k)$ glued to each point, so $S_k(mathbb{R}^n) cong O(k) times G_k(mathbb{R}^n)$ as sets.
– Joppy
Nov 18 at 1:43
@Vicky: For $E in G_k(mathbb{R^n}$, its preimage $phi^{-1}(E)$ is the set of $k$-frames in $E$. This set is isomorphic to $O(k)$, since $O(k)$ acts freely and transitively on it. This means that $S_k(mathbb{R}^n)$ looks like $G_k(mathbb{R}^n)$ with some copy of $O(k)$ glued to each point, so $S_k(mathbb{R}^n) cong O(k) times G_k(mathbb{R}^n)$ as sets.
– Joppy
Nov 18 at 1:43
add a comment |
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What is your definition of Grassmann manifold, i.e., in which sense is it made of "(1 times k)-dimensional matrices"?
– Travis
Nov 16 at 4:16
In any case, are you familiar with the characterization/construction of homogeneous spaces? See, e.g., Theorem 9.22 (Homogeneous Construction Space Theorem) in Lee's Introduction to Smooth Manifolds.
– Travis
Nov 16 at 4:18
It is a vector space, isn't it? A $k$-dimensional subspace of $mathbb{R}^n$, so $1times k$ - matrices are vectors with $k$ components
– Vicky
Nov 16 at 4:20
1
That description isn't quite right. The Grassmannian $G_k(V)$ is the space of all $k$-dimensional subspaces of $V$. It is true that we can specify any such plane by giving a basis, i.e., $k$ linearly independent vectors in $V$, but that choice is not unique.
– Travis
Nov 16 at 4:42
1
The question isn't quite a duplicate, but by chance I explained exactly this in an answer to another question a few days ago: math.stackexchange.com/a/2993101/155629
– Travis
Nov 16 at 4:43