Is $GL(r,mathbb{C})$ diffeomorphic to direct product of $U(n)$ and euclidean space
$begingroup$
Can some one give a reference/proof of the following (possibly positive) result?
$GL(r,mathbb{C})$ is diffeomorphic to $U(n)$ and Euclidean space.
This is a (possible) special case of Iwasawa theorem
If $G$ is a connected Lie group, then $G$ is diffeomorphic to direct product of a maximal compact subgroup $H$ and Eucldiean space.
Here $Gl(r,mathbb{C})$ is connected Lie group, $U(n)$ is maximal compact subgroup. So, it seems $GL(r,mathbb{C})$ is diffeomorphic to $U(n)$ and Euclidean space. Can an easy proof/reference be given for this?
differential-geometry lie-groups
asked Dec 24 '18 at 16:04
Praphulla KoushikPraphulla Koushik
203119
$endgroup$
add a comment |
$begingroup$
Can some one give a reference/proof of the following (possibly positive) result?
$GL(r,mathbb{C})$ is diffeomorphic to $U(n)$ and Euclidean space.
This is a (possible) special case of Iwasawa theorem
If $G$ is a connected Lie group, then $G$ is diffeomorphic to direct product of a maximal compact subgroup $H$ and Eucldiean space.
Here $Gl(r,mathbb{C})$ is connected Lie group, $U(n)$ is maximal compact subgroup. So, it seems $GL(r,mathbb{C})$ is diffeomorphic to $U(n)$ and Euclidean space. Can an easy proof/reference be given for this?
differential-geometry lie-groups
asked Dec 24 '18 at 16:04
Praphulla KoushikPraphulla Koushik
203119
$endgroup$
$begingroup$
I think the theorem should say "direct product of a maximal compact subgroup and Euclidean space" (and then it holds trivially for a compact group with the Euclidean space being zero-dimensional).
$endgroup$
– Rob Arthan
Dec 24 '18 at 16:36
$begingroup$
See the discussion under "Topology" on en.wikipedia.org/wiki/Maximal_compact_subgroup.
$endgroup$
– Rob Arthan
Dec 24 '18 at 16:47
$begingroup$
@RobArthan that was my mistake,, corrected it.. thank you :)
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 3:20
add a comment |
$begingroup$
Can some one give a reference/proof of the following (possibly positive) result?
$GL(r,mathbb{C})$ is diffeomorphic to $U(n)$ and Euclidean space.
This is a (possible) special case of Iwasawa theorem
If $G$ is a connected Lie group, then $G$ is diffeomorphic to direct product of a maximal compact subgroup $H$ and Eucldiean space.
Here $Gl(r,mathbb{C})$ is connected Lie group, $U(n)$ is maximal compact subgroup. So, it seems $GL(r,mathbb{C})$ is diffeomorphic to $U(n)$ and Euclidean space. Can an easy proof/reference be given for this?
differential-geometry lie-groups
asked Dec 24 '18 at 16:04
Praphulla KoushikPraphulla Koushik
203119
$endgroup$
Can some one give a reference/proof of the following (possibly positive) result?
$GL(r,mathbb{C})$ is diffeomorphic to $U(n)$ and Euclidean space.
This is a (possible) special case of Iwasawa theorem
If $G$ is a connected Lie group, then $G$ is diffeomorphic to direct product of a maximal compact subgroup $H$ and Eucldiean space.
Here $Gl(r,mathbb{C})$ is connected Lie group, $U(n)$ is maximal compact subgroup. So, it seems $GL(r,mathbb{C})$ is diffeomorphic to $U(n)$ and Euclidean space. Can an easy proof/reference be given for this?
differential-geometry lie-groups
differential-geometry lie-groups
asked Dec 24 '18 at 16:04
Praphulla KoushikPraphulla Koushik
203119
asked Dec 24 '18 at 16:04
Praphulla KoushikPraphulla Koushik
203119
edited Dec 25 '18 at 3:20
Praphulla Koushik
asked Dec 24 '18 at 16:04
Praphulla KoushikPraphulla Koushik
203119
asked Dec 24 '18 at 16:04
Praphulla KoushikPraphulla Koushik
203119
asked Dec 24 '18 at 16:04
Praphulla KoushikPraphulla Koushik
203119
203119
$begingroup$
I think the theorem should say "direct product of a maximal compact subgroup and Euclidean space" (and then it holds trivially for a compact group with the Euclidean space being zero-dimensional).
$endgroup$
– Rob Arthan
Dec 24 '18 at 16:36
$begingroup$
See the discussion under "Topology" on en.wikipedia.org/wiki/Maximal_compact_subgroup.
$endgroup$
– Rob Arthan
Dec 24 '18 at 16:47
$begingroup$
@RobArthan that was my mistake,, corrected it.. thank you :)
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 3:20
add a comment |
$begingroup$
I think the theorem should say "direct product of a maximal compact subgroup and Euclidean space" (and then it holds trivially for a compact group with the Euclidean space being zero-dimensional).
$endgroup$
– Rob Arthan
Dec 24 '18 at 16:36
$begingroup$
See the discussion under "Topology" on en.wikipedia.org/wiki/Maximal_compact_subgroup.
$endgroup$
– Rob Arthan
Dec 24 '18 at 16:47
$begingroup$
@RobArthan that was my mistake,, corrected it.. thank you :)
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 3:20
$begingroup$
I think the theorem should say "direct product of a maximal compact subgroup and Euclidean space" (and then it holds trivially for a compact group with the Euclidean space being zero-dimensional).
$endgroup$
– Rob Arthan
Dec 24 '18 at 16:36
$begingroup$
I think the theorem should say "direct product of a maximal compact subgroup and Euclidean space" (and then it holds trivially for a compact group with the Euclidean space being zero-dimensional).
$endgroup$
– Rob Arthan
Dec 24 '18 at 16:36
$begingroup$
See the discussion under "Topology" on en.wikipedia.org/wiki/Maximal_compact_subgroup.
$endgroup$
– Rob Arthan
Dec 24 '18 at 16:47
$begingroup$
See the discussion under "Topology" on en.wikipedia.org/wiki/Maximal_compact_subgroup.
$endgroup$
– Rob Arthan
Dec 24 '18 at 16:47
$begingroup$
@RobArthan that was my mistake,, corrected it.. thank you :)
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 3:20
$begingroup$
@RobArthan that was my mistake,, corrected it.. thank you :)
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 3:20
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The Iwasawa theorem is not terribly easy, to my recollection. If you change your statement as in Rob Arthan's comment, it is correct. But for most of the classical groups you are probably interested in, there are simple proofs, as in the following (your specific case).
Let $mathcal H_n$ be the space of positive Hermitian forms on $Bbb C^n$; this is a convex open subset of the space of $n times n$ Hermitian (that is, complex self-adjoint) matrices. This space has real dimension $n^2$; any convex open subset of a real vector space of dimension $k$ is diffeomorphic to $Bbb R^k$ (see here for a more complicated proof of a more general fact).
The group $GL_n(Bbb C)$ acts smoothly on $mathcal H_n$ by pullback of forms (or if you're thinking of these as matrices, it acts by conjugation). It acts transitively because every $n$-dimensional complex Hilbert space is isomorphic, so given $H, H' in mathcal H_n$, there must be some isomorphism $A: (Bbb C^n, H) to (Bbb C^n, H')$; this isomorphism is $A in GL_n(Bbb C)$. If we write $g$ for the standard inner product corresponding to the diagonal matrix, the stabilizer of $g$ is $U(n)$ (by definition!). So $mathcal H_n$ is a homogeneous space, and hence sits into a principal bundle $$U(n) to GL_n(Bbb C) to GL_n(Bbb C)/U(n) cong mathcal H_n cong Bbb R^{n^2}.$$ Because every bundle over a contractible space is smoothly trivial, we can find a ($U(n)$-equivariant!) diffeomorphism $GL_n(Bbb C) cong U(n) times Bbb R^{n^2}$.
$endgroup$
$begingroup$
Thank you. I corrected the statement. It was a typo :) I need some time to understand this. Thanks thanks
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 3:22
$begingroup$
Can you please give a reference where Iwasawa theorem is proved... Ofcourse original paper has the proof :D.. I am looking for some recent exposition... :)
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 6:23
$begingroup$
It should be in Hochschild's book on Lie groups. I mostly don't remember it. It's not hard at the level of Lie algebras, but the global result is interesting.
$endgroup$
– user98602
Dec 25 '18 at 12:31
$begingroup$
Thanks, I will have a look at it... Can you have a look at math.stackexchange.com/questions/3051220/…
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 12:33
add a comment |
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$begingroup$
The Iwasawa theorem is not terribly easy, to my recollection. If you change your statement as in Rob Arthan's comment, it is correct. But for most of the classical groups you are probably interested in, there are simple proofs, as in the following (your specific case).
Let $mathcal H_n$ be the space of positive Hermitian forms on $Bbb C^n$; this is a convex open subset of the space of $n times n$ Hermitian (that is, complex self-adjoint) matrices. This space has real dimension $n^2$; any convex open subset of a real vector space of dimension $k$ is diffeomorphic to $Bbb R^k$ (see here for a more complicated proof of a more general fact).
The group $GL_n(Bbb C)$ acts smoothly on $mathcal H_n$ by pullback of forms (or if you're thinking of these as matrices, it acts by conjugation). It acts transitively because every $n$-dimensional complex Hilbert space is isomorphic, so given $H, H' in mathcal H_n$, there must be some isomorphism $A: (Bbb C^n, H) to (Bbb C^n, H')$; this isomorphism is $A in GL_n(Bbb C)$. If we write $g$ for the standard inner product corresponding to the diagonal matrix, the stabilizer of $g$ is $U(n)$ (by definition!). So $mathcal H_n$ is a homogeneous space, and hence sits into a principal bundle $$U(n) to GL_n(Bbb C) to GL_n(Bbb C)/U(n) cong mathcal H_n cong Bbb R^{n^2}.$$ Because every bundle over a contractible space is smoothly trivial, we can find a ($U(n)$-equivariant!) diffeomorphism $GL_n(Bbb C) cong U(n) times Bbb R^{n^2}$.
$endgroup$
$begingroup$
Thank you. I corrected the statement. It was a typo :) I need some time to understand this. Thanks thanks
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 3:22
$begingroup$
Can you please give a reference where Iwasawa theorem is proved... Ofcourse original paper has the proof :D.. I am looking for some recent exposition... :)
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 6:23
$begingroup$
It should be in Hochschild's book on Lie groups. I mostly don't remember it. It's not hard at the level of Lie algebras, but the global result is interesting.
$endgroup$
– user98602
Dec 25 '18 at 12:31
$begingroup$
Thanks, I will have a look at it... Can you have a look at math.stackexchange.com/questions/3051220/…
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 12:33
add a comment |
$begingroup$
The Iwasawa theorem is not terribly easy, to my recollection. If you change your statement as in Rob Arthan's comment, it is correct. But for most of the classical groups you are probably interested in, there are simple proofs, as in the following (your specific case).
Let $mathcal H_n$ be the space of positive Hermitian forms on $Bbb C^n$; this is a convex open subset of the space of $n times n$ Hermitian (that is, complex self-adjoint) matrices. This space has real dimension $n^2$; any convex open subset of a real vector space of dimension $k$ is diffeomorphic to $Bbb R^k$ (see here for a more complicated proof of a more general fact).
The group $GL_n(Bbb C)$ acts smoothly on $mathcal H_n$ by pullback of forms (or if you're thinking of these as matrices, it acts by conjugation). It acts transitively because every $n$-dimensional complex Hilbert space is isomorphic, so given $H, H' in mathcal H_n$, there must be some isomorphism $A: (Bbb C^n, H) to (Bbb C^n, H')$; this isomorphism is $A in GL_n(Bbb C)$. If we write $g$ for the standard inner product corresponding to the diagonal matrix, the stabilizer of $g$ is $U(n)$ (by definition!). So $mathcal H_n$ is a homogeneous space, and hence sits into a principal bundle $$U(n) to GL_n(Bbb C) to GL_n(Bbb C)/U(n) cong mathcal H_n cong Bbb R^{n^2}.$$ Because every bundle over a contractible space is smoothly trivial, we can find a ($U(n)$-equivariant!) diffeomorphism $GL_n(Bbb C) cong U(n) times Bbb R^{n^2}$.
$endgroup$
$begingroup$
Thank you. I corrected the statement. It was a typo :) I need some time to understand this. Thanks thanks
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 3:22
$begingroup$
Can you please give a reference where Iwasawa theorem is proved... Ofcourse original paper has the proof :D.. I am looking for some recent exposition... :)
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 6:23
$begingroup$
It should be in Hochschild's book on Lie groups. I mostly don't remember it. It's not hard at the level of Lie algebras, but the global result is interesting.
$endgroup$
– user98602
Dec 25 '18 at 12:31
$begingroup$
Thanks, I will have a look at it... Can you have a look at math.stackexchange.com/questions/3051220/…
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 12:33
add a comment |
$begingroup$
The Iwasawa theorem is not terribly easy, to my recollection. If you change your statement as in Rob Arthan's comment, it is correct. But for most of the classical groups you are probably interested in, there are simple proofs, as in the following (your specific case).
Let $mathcal H_n$ be the space of positive Hermitian forms on $Bbb C^n$; this is a convex open subset of the space of $n times n$ Hermitian (that is, complex self-adjoint) matrices. This space has real dimension $n^2$; any convex open subset of a real vector space of dimension $k$ is diffeomorphic to $Bbb R^k$ (see here for a more complicated proof of a more general fact).
The group $GL_n(Bbb C)$ acts smoothly on $mathcal H_n$ by pullback of forms (or if you're thinking of these as matrices, it acts by conjugation). It acts transitively because every $n$-dimensional complex Hilbert space is isomorphic, so given $H, H' in mathcal H_n$, there must be some isomorphism $A: (Bbb C^n, H) to (Bbb C^n, H')$; this isomorphism is $A in GL_n(Bbb C)$. If we write $g$ for the standard inner product corresponding to the diagonal matrix, the stabilizer of $g$ is $U(n)$ (by definition!). So $mathcal H_n$ is a homogeneous space, and hence sits into a principal bundle $$U(n) to GL_n(Bbb C) to GL_n(Bbb C)/U(n) cong mathcal H_n cong Bbb R^{n^2}.$$ Because every bundle over a contractible space is smoothly trivial, we can find a ($U(n)$-equivariant!) diffeomorphism $GL_n(Bbb C) cong U(n) times Bbb R^{n^2}$.
$endgroup$
The Iwasawa theorem is not terribly easy, to my recollection. If you change your statement as in Rob Arthan's comment, it is correct. But for most of the classical groups you are probably interested in, there are simple proofs, as in the following (your specific case).
Let $mathcal H_n$ be the space of positive Hermitian forms on $Bbb C^n$; this is a convex open subset of the space of $n times n$ Hermitian (that is, complex self-adjoint) matrices. This space has real dimension $n^2$; any convex open subset of a real vector space of dimension $k$ is diffeomorphic to $Bbb R^k$ (see here for a more complicated proof of a more general fact).
The group $GL_n(Bbb C)$ acts smoothly on $mathcal H_n$ by pullback of forms (or if you're thinking of these as matrices, it acts by conjugation). It acts transitively because every $n$-dimensional complex Hilbert space is isomorphic, so given $H, H' in mathcal H_n$, there must be some isomorphism $A: (Bbb C^n, H) to (Bbb C^n, H')$; this isomorphism is $A in GL_n(Bbb C)$. If we write $g$ for the standard inner product corresponding to the diagonal matrix, the stabilizer of $g$ is $U(n)$ (by definition!). So $mathcal H_n$ is a homogeneous space, and hence sits into a principal bundle $$U(n) to GL_n(Bbb C) to GL_n(Bbb C)/U(n) cong mathcal H_n cong Bbb R^{n^2}.$$ Because every bundle over a contractible space is smoothly trivial, we can find a ($U(n)$-equivariant!) diffeomorphism $GL_n(Bbb C) cong U(n) times Bbb R^{n^2}$.
answered Dec 24 '18 at 19:09
user98602
$begingroup$
Thank you. I corrected the statement. It was a typo :) I need some time to understand this. Thanks thanks
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 3:22
$begingroup$
Can you please give a reference where Iwasawa theorem is proved... Ofcourse original paper has the proof :D.. I am looking for some recent exposition... :)
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 6:23
$begingroup$
It should be in Hochschild's book on Lie groups. I mostly don't remember it. It's not hard at the level of Lie algebras, but the global result is interesting.
$endgroup$
– user98602
Dec 25 '18 at 12:31
$begingroup$
Thanks, I will have a look at it... Can you have a look at math.stackexchange.com/questions/3051220/…
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 12:33
add a comment |
$begingroup$
Thank you. I corrected the statement. It was a typo :) I need some time to understand this. Thanks thanks
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 3:22
$begingroup$
Can you please give a reference where Iwasawa theorem is proved... Ofcourse original paper has the proof :D.. I am looking for some recent exposition... :)
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 6:23
$begingroup$
It should be in Hochschild's book on Lie groups. I mostly don't remember it. It's not hard at the level of Lie algebras, but the global result is interesting.
$endgroup$
– user98602
Dec 25 '18 at 12:31
$begingroup$
Thanks, I will have a look at it... Can you have a look at math.stackexchange.com/questions/3051220/…
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 12:33
$begingroup$
Thank you. I corrected the statement. It was a typo :) I need some time to understand this. Thanks thanks
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 3:22
$begingroup$
Thank you. I corrected the statement. It was a typo :) I need some time to understand this. Thanks thanks
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 3:22
$begingroup$
Can you please give a reference where Iwasawa theorem is proved... Ofcourse original paper has the proof :D.. I am looking for some recent exposition... :)
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 6:23
$begingroup$
Can you please give a reference where Iwasawa theorem is proved... Ofcourse original paper has the proof :D.. I am looking for some recent exposition... :)
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 6:23
$begingroup$
It should be in Hochschild's book on Lie groups. I mostly don't remember it. It's not hard at the level of Lie algebras, but the global result is interesting.
$endgroup$
– user98602
Dec 25 '18 at 12:31
$begingroup$
It should be in Hochschild's book on Lie groups. I mostly don't remember it. It's not hard at the level of Lie algebras, but the global result is interesting.
$endgroup$
– user98602
Dec 25 '18 at 12:31
$begingroup$
Thanks, I will have a look at it... Can you have a look at math.stackexchange.com/questions/3051220/…
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 12:33
$begingroup$
Thanks, I will have a look at it... Can you have a look at math.stackexchange.com/questions/3051220/…
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 12:33
add a comment |
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$begingroup$
I think the theorem should say "direct product of a maximal compact subgroup and Euclidean space" (and then it holds trivially for a compact group with the Euclidean space being zero-dimensional).
$endgroup$
– Rob Arthan
Dec 24 '18 at 16:36
$begingroup$
See the discussion under "Topology" on en.wikipedia.org/wiki/Maximal_compact_subgroup.
$endgroup$
– Rob Arthan
Dec 24 '18 at 16:47
$begingroup$
@RobArthan that was my mistake,, corrected it.. thank you :)
$endgroup$
– Praphulla Koushik
Dec 25 '18 at 3:20