How to show that Symplectic group $Sp(n,mathbb C)$ is not compact?
$begingroup$
I wanted to prove above just using basic facts .I had only given that Sympletic group preserve following bilinear form :
$B[x,y]=sum_{i=1}^kx_iy_{n+i}-x_{n+i}y_i$
I had defination of compactness as
Matrix Lie group is compact
1)If $A_m$ converges to A then A is in G
2) There exist constant C such that $|a_{ij}|<C,forall i,j$
How to prove using this thing $Sp(n,mathbb C)$ is not
compact?
Please, I had not done manifold, This is a question from introductory Lie Algebra.
ANy Help will be appreciated
general-topology lie-groups lie-algebras compactness
asked Jan 16 at 15:17
SRJSRJ
1,6651520
$endgroup$
add a comment |
$begingroup$
I wanted to prove above just using basic facts .I had only given that Sympletic group preserve following bilinear form :
$B[x,y]=sum_{i=1}^kx_iy_{n+i}-x_{n+i}y_i$
I had defination of compactness as
Matrix Lie group is compact
1)If $A_m$ converges to A then A is in G
2) There exist constant C such that $|a_{ij}|<C,forall i,j$
How to prove using this thing $Sp(n,mathbb C)$ is not
compact?
Please, I had not done manifold, This is a question from introductory Lie Algebra.
ANy Help will be appreciated
general-topology lie-groups lie-algebras compactness
asked Jan 16 at 15:17
SRJSRJ
1,6651520
$endgroup$
1
$begingroup$
A good first idea is to write down explicitly which matrices are in $Sp(n, Bbb C)$. Then you should see that your condition 2 is not satisfied, i.e. among those matrices you can make some entries arbitrarily large.
$endgroup$
– Torsten Schoeneberg
Jan 16 at 18:36
add a comment |
$begingroup$
I wanted to prove above just using basic facts .I had only given that Sympletic group preserve following bilinear form :
$B[x,y]=sum_{i=1}^kx_iy_{n+i}-x_{n+i}y_i$
I had defination of compactness as
Matrix Lie group is compact
1)If $A_m$ converges to A then A is in G
2) There exist constant C such that $|a_{ij}|<C,forall i,j$
How to prove using this thing $Sp(n,mathbb C)$ is not
compact?
Please, I had not done manifold, This is a question from introductory Lie Algebra.
ANy Help will be appreciated
general-topology lie-groups lie-algebras compactness
asked Jan 16 at 15:17
SRJSRJ
1,6651520
$endgroup$
I wanted to prove above just using basic facts .I had only given that Sympletic group preserve following bilinear form :
$B[x,y]=sum_{i=1}^kx_iy_{n+i}-x_{n+i}y_i$
I had defination of compactness as
Matrix Lie group is compact
1)If $A_m$ converges to A then A is in G
2) There exist constant C such that $|a_{ij}|<C,forall i,j$
How to prove using this thing $Sp(n,mathbb C)$ is not
compact?
Please, I had not done manifold, This is a question from introductory Lie Algebra.
ANy Help will be appreciated
general-topology lie-groups lie-algebras compactness
general-topology lie-groups lie-algebras compactness
asked Jan 16 at 15:17
SRJSRJ
1,6651520
asked Jan 16 at 15:17
SRJSRJ
1,6651520
edited Jan 16 at 15:51
SRJ
asked Jan 16 at 15:17
SRJSRJ
1,6651520
asked Jan 16 at 15:17
SRJSRJ
1,6651520
asked Jan 16 at 15:17
SRJSRJ
1,6651520
1,6651520
1
$begingroup$
A good first idea is to write down explicitly which matrices are in $Sp(n, Bbb C)$. Then you should see that your condition 2 is not satisfied, i.e. among those matrices you can make some entries arbitrarily large.
$endgroup$
– Torsten Schoeneberg
Jan 16 at 18:36
add a comment |
1
$begingroup$
A good first idea is to write down explicitly which matrices are in $Sp(n, Bbb C)$. Then you should see that your condition 2 is not satisfied, i.e. among those matrices you can make some entries arbitrarily large.
$endgroup$
– Torsten Schoeneberg
Jan 16 at 18:36
1
1
$begingroup$
A good first idea is to write down explicitly which matrices are in $Sp(n, Bbb C)$. Then you should see that your condition 2 is not satisfied, i.e. among those matrices you can make some entries arbitrarily large.
$endgroup$
– Torsten Schoeneberg
Jan 16 at 18:36
$begingroup$
A good first idea is to write down explicitly which matrices are in $Sp(n, Bbb C)$. Then you should see that your condition 2 is not satisfied, i.e. among those matrices you can make some entries arbitrarily large.
$endgroup$
– Torsten Schoeneberg
Jan 16 at 18:36
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Nobody can do that. since it is not compact. Perhaps that what you have in mind is the compact symplectic group, which is $Sp(n,mathbb{C})cap U(2n)$. This is a compact real Lie group and the complexification of its Lie algebra is isomorphic to $mathfrak{sp}(n,mathbb{C})$.
answered Jan 16 at 15:30
José Carlos SantosJosé Carlos Santos
160k22126232
$endgroup$
$begingroup$
Sorry Sir for huge typo. Actually I have to prove not compact. For Sp(n ) is compact that I had proved using U(n) is compact.
$endgroup$
– SRJ
Jan 16 at 15:39
$begingroup$
Sp(n)=sp(n,c)$cap$u(2n)
$endgroup$
– SRJ
Jan 16 at 15:52
add a comment |
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1 Answer
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$begingroup$
Nobody can do that. since it is not compact. Perhaps that what you have in mind is the compact symplectic group, which is $Sp(n,mathbb{C})cap U(2n)$. This is a compact real Lie group and the complexification of its Lie algebra is isomorphic to $mathfrak{sp}(n,mathbb{C})$.
answered Jan 16 at 15:30
José Carlos SantosJosé Carlos Santos
160k22126232
$endgroup$
$begingroup$
Sorry Sir for huge typo. Actually I have to prove not compact. For Sp(n ) is compact that I had proved using U(n) is compact.
$endgroup$
– SRJ
Jan 16 at 15:39
$begingroup$
Sp(n)=sp(n,c)$cap$u(2n)
$endgroup$
– SRJ
Jan 16 at 15:52
add a comment |
$begingroup$
Nobody can do that. since it is not compact. Perhaps that what you have in mind is the compact symplectic group, which is $Sp(n,mathbb{C})cap U(2n)$. This is a compact real Lie group and the complexification of its Lie algebra is isomorphic to $mathfrak{sp}(n,mathbb{C})$.
answered Jan 16 at 15:30
José Carlos SantosJosé Carlos Santos
160k22126232
$endgroup$
$begingroup$
Sorry Sir for huge typo. Actually I have to prove not compact. For Sp(n ) is compact that I had proved using U(n) is compact.
$endgroup$
– SRJ
Jan 16 at 15:39
$begingroup$
Sp(n)=sp(n,c)$cap$u(2n)
$endgroup$
– SRJ
Jan 16 at 15:52
add a comment |
$begingroup$
Nobody can do that. since it is not compact. Perhaps that what you have in mind is the compact symplectic group, which is $Sp(n,mathbb{C})cap U(2n)$. This is a compact real Lie group and the complexification of its Lie algebra is isomorphic to $mathfrak{sp}(n,mathbb{C})$.
answered Jan 16 at 15:30
José Carlos SantosJosé Carlos Santos
160k22126232
$endgroup$
Nobody can do that. since it is not compact. Perhaps that what you have in mind is the compact symplectic group, which is $Sp(n,mathbb{C})cap U(2n)$. This is a compact real Lie group and the complexification of its Lie algebra is isomorphic to $mathfrak{sp}(n,mathbb{C})$.
answered Jan 16 at 15:30
José Carlos SantosJosé Carlos Santos
160k22126232
answered Jan 16 at 15:30
José Carlos SantosJosé Carlos Santos
160k22126232
answered Jan 16 at 15:30
José Carlos SantosJosé Carlos Santos
160k22126232
answered Jan 16 at 15:30
José Carlos SantosJosé Carlos Santos
160k22126232
160k22126232
$begingroup$
Sorry Sir for huge typo. Actually I have to prove not compact. For Sp(n ) is compact that I had proved using U(n) is compact.
$endgroup$
– SRJ
Jan 16 at 15:39
$begingroup$
Sp(n)=sp(n,c)$cap$u(2n)
$endgroup$
– SRJ
Jan 16 at 15:52
add a comment |
$begingroup$
Sorry Sir for huge typo. Actually I have to prove not compact. For Sp(n ) is compact that I had proved using U(n) is compact.
$endgroup$
– SRJ
Jan 16 at 15:39
$begingroup$
Sp(n)=sp(n,c)$cap$u(2n)
$endgroup$
– SRJ
Jan 16 at 15:52
$begingroup$
Sorry Sir for huge typo. Actually I have to prove not compact. For Sp(n ) is compact that I had proved using U(n) is compact.
$endgroup$
– SRJ
Jan 16 at 15:39
$begingroup$
Sorry Sir for huge typo. Actually I have to prove not compact. For Sp(n ) is compact that I had proved using U(n) is compact.
$endgroup$
– SRJ
Jan 16 at 15:39
$begingroup$
Sp(n)=sp(n,c)$cap$u(2n)
$endgroup$
– SRJ
Jan 16 at 15:52
$begingroup$
Sp(n)=sp(n,c)$cap$u(2n)
$endgroup$
– SRJ
Jan 16 at 15:52
add a comment |
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A good first idea is to write down explicitly which matrices are in $Sp(n, Bbb C)$. Then you should see that your condition 2 is not satisfied, i.e. among those matrices you can make some entries arbitrarily large.
$endgroup$
– Torsten Schoeneberg
Jan 16 at 18:36