Conjugacy classes of the Lie group $Sp(6,mathbb{R})$
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I am trying to find the number of conjugacy classes of the Lie group $Sp(6,mathbb{R})$, and identify which elements each of them contains (in matrix form).
When searching literature on the topic, I find lots of examples where the conjugacy classes are calculated for discrete groups but I can't seem to find something similar for Lie groups. I understand that because it is not discrete there are an infinite number of possible transformation matrices, but we shouldn't we be able to define a general form those matrices should have? E.g. a matrix with all zero values in the bottom left $3times3$ matrix is not conjugate to a matrix with all zero values in the bottom right $3times3$ matrix.
lie-groups
asked Nov 12 at 16:39
Th.Phy
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I am trying to find the number of conjugacy classes of the Lie group $Sp(6,mathbb{R})$, and identify which elements each of them contains (in matrix form).
When searching literature on the topic, I find lots of examples where the conjugacy classes are calculated for discrete groups but I can't seem to find something similar for Lie groups. I understand that because it is not discrete there are an infinite number of possible transformation matrices, but we shouldn't we be able to define a general form those matrices should have? E.g. a matrix with all zero values in the bottom left $3times3$ matrix is not conjugate to a matrix with all zero values in the bottom right $3times3$ matrix.
lie-groups
asked Nov 12 at 16:39
Th.Phy
11
New contributor
Th.Phy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
One can certainly still ask about the conjugacy classes of an infinite group. NB the spectrum of a matrix is invariant under conjugacy.
– Travis
Nov 12 at 17:41
What group is $Sp(6,mathbb{R})$ again? If it's a compact Lie group, then every element is conjugate to one on a maximal torus $T^n$. The full set of conjugacy classes is given by $T^n/W$ where $W$ is the Weyl group of $G$.
– Jason DeVito
Nov 12 at 18:32
Thanks for your comments. It is the symplectic group, which obeys $S^TOmega S=Omega$ where $Omega =begin{Bmatrix} 0 & I_n \ -I_n &0 end{Bmatrix}$, and is indeed compact.
– Th.Phy
Nov 12 at 20:31
Just a quick question. Since two elements of a group are conjugate when $hg_1h^{-1}=g_2$, and taking the trace on both sides gives us $Tr g_1=Tr g_2$, doesn't this mean that any two matrices in our group which have different different trace are in a different conjugacy class? Meaning that there is an infinite number of conjugacy classes. Then would determining the full set of conjugacy classes provide any insight into the different properties of matrices beloning to different classes?
– Th.Phy
yesterday
add a comment |
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up vote
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down vote
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I am trying to find the number of conjugacy classes of the Lie group $Sp(6,mathbb{R})$, and identify which elements each of them contains (in matrix form).
When searching literature on the topic, I find lots of examples where the conjugacy classes are calculated for discrete groups but I can't seem to find something similar for Lie groups. I understand that because it is not discrete there are an infinite number of possible transformation matrices, but we shouldn't we be able to define a general form those matrices should have? E.g. a matrix with all zero values in the bottom left $3times3$ matrix is not conjugate to a matrix with all zero values in the bottom right $3times3$ matrix.
lie-groups
asked Nov 12 at 16:39
Th.Phy
11
New contributor
Th.Phy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I am trying to find the number of conjugacy classes of the Lie group $Sp(6,mathbb{R})$, and identify which elements each of them contains (in matrix form).
When searching literature on the topic, I find lots of examples where the conjugacy classes are calculated for discrete groups but I can't seem to find something similar for Lie groups. I understand that because it is not discrete there are an infinite number of possible transformation matrices, but we shouldn't we be able to define a general form those matrices should have? E.g. a matrix with all zero values in the bottom left $3times3$ matrix is not conjugate to a matrix with all zero values in the bottom right $3times3$ matrix.
lie-groups
lie-groups
asked Nov 12 at 16:39
Th.Phy
11
New contributor
Th.Phy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 12 at 16:39
Th.Phy
11
New contributor
Th.Phy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 12 at 16:39
Th.Phy
11
New contributor
Th.Phy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 12 at 16:39
Th.Phy
11
asked Nov 12 at 16:39
Th.Phy
11
11
New contributor
Th.Phy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Th.Phy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Th.Phy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
One can certainly still ask about the conjugacy classes of an infinite group. NB the spectrum of a matrix is invariant under conjugacy.
– Travis
Nov 12 at 17:41
What group is $Sp(6,mathbb{R})$ again? If it's a compact Lie group, then every element is conjugate to one on a maximal torus $T^n$. The full set of conjugacy classes is given by $T^n/W$ where $W$ is the Weyl group of $G$.
– Jason DeVito
Nov 12 at 18:32
Thanks for your comments. It is the symplectic group, which obeys $S^TOmega S=Omega$ where $Omega =begin{Bmatrix} 0 & I_n \ -I_n &0 end{Bmatrix}$, and is indeed compact.
– Th.Phy
Nov 12 at 20:31
Just a quick question. Since two elements of a group are conjugate when $hg_1h^{-1}=g_2$, and taking the trace on both sides gives us $Tr g_1=Tr g_2$, doesn't this mean that any two matrices in our group which have different different trace are in a different conjugacy class? Meaning that there is an infinite number of conjugacy classes. Then would determining the full set of conjugacy classes provide any insight into the different properties of matrices beloning to different classes?
– Th.Phy
yesterday
add a comment |
One can certainly still ask about the conjugacy classes of an infinite group. NB the spectrum of a matrix is invariant under conjugacy.
– Travis
Nov 12 at 17:41
What group is $Sp(6,mathbb{R})$ again? If it's a compact Lie group, then every element is conjugate to one on a maximal torus $T^n$. The full set of conjugacy classes is given by $T^n/W$ where $W$ is the Weyl group of $G$.
– Jason DeVito
Nov 12 at 18:32
Thanks for your comments. It is the symplectic group, which obeys $S^TOmega S=Omega$ where $Omega =begin{Bmatrix} 0 & I_n \ -I_n &0 end{Bmatrix}$, and is indeed compact.
– Th.Phy
Nov 12 at 20:31
Just a quick question. Since two elements of a group are conjugate when $hg_1h^{-1}=g_2$, and taking the trace on both sides gives us $Tr g_1=Tr g_2$, doesn't this mean that any two matrices in our group which have different different trace are in a different conjugacy class? Meaning that there is an infinite number of conjugacy classes. Then would determining the full set of conjugacy classes provide any insight into the different properties of matrices beloning to different classes?
– Th.Phy
yesterday
One can certainly still ask about the conjugacy classes of an infinite group. NB the spectrum of a matrix is invariant under conjugacy.
– Travis
Nov 12 at 17:41
One can certainly still ask about the conjugacy classes of an infinite group. NB the spectrum of a matrix is invariant under conjugacy.
– Travis
Nov 12 at 17:41
What group is $Sp(6,mathbb{R})$ again? If it's a compact Lie group, then every element is conjugate to one on a maximal torus $T^n$. The full set of conjugacy classes is given by $T^n/W$ where $W$ is the Weyl group of $G$.
– Jason DeVito
Nov 12 at 18:32
What group is $Sp(6,mathbb{R})$ again? If it's a compact Lie group, then every element is conjugate to one on a maximal torus $T^n$. The full set of conjugacy classes is given by $T^n/W$ where $W$ is the Weyl group of $G$.
– Jason DeVito
Nov 12 at 18:32
Thanks for your comments. It is the symplectic group, which obeys $S^TOmega S=Omega$ where $Omega =begin{Bmatrix} 0 & I_n \ -I_n &0 end{Bmatrix}$, and is indeed compact.
– Th.Phy
Nov 12 at 20:31
Thanks for your comments. It is the symplectic group, which obeys $S^TOmega S=Omega$ where $Omega =begin{Bmatrix} 0 & I_n \ -I_n &0 end{Bmatrix}$, and is indeed compact.
– Th.Phy
Nov 12 at 20:31
Just a quick question. Since two elements of a group are conjugate when $hg_1h^{-1}=g_2$, and taking the trace on both sides gives us $Tr g_1=Tr g_2$, doesn't this mean that any two matrices in our group which have different different trace are in a different conjugacy class? Meaning that there is an infinite number of conjugacy classes. Then would determining the full set of conjugacy classes provide any insight into the different properties of matrices beloning to different classes?
– Th.Phy
yesterday
Just a quick question. Since two elements of a group are conjugate when $hg_1h^{-1}=g_2$, and taking the trace on both sides gives us $Tr g_1=Tr g_2$, doesn't this mean that any two matrices in our group which have different different trace are in a different conjugacy class? Meaning that there is an infinite number of conjugacy classes. Then would determining the full set of conjugacy classes provide any insight into the different properties of matrices beloning to different classes?
– Th.Phy
yesterday
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One can certainly still ask about the conjugacy classes of an infinite group. NB the spectrum of a matrix is invariant under conjugacy.
– Travis
Nov 12 at 17:41
What group is $Sp(6,mathbb{R})$ again? If it's a compact Lie group, then every element is conjugate to one on a maximal torus $T^n$. The full set of conjugacy classes is given by $T^n/W$ where $W$ is the Weyl group of $G$.
– Jason DeVito
Nov 12 at 18:32
Thanks for your comments. It is the symplectic group, which obeys $S^TOmega S=Omega$ where $Omega =begin{Bmatrix} 0 & I_n \ -I_n &0 end{Bmatrix}$, and is indeed compact.
– Th.Phy
Nov 12 at 20:31
Just a quick question. Since two elements of a group are conjugate when $hg_1h^{-1}=g_2$, and taking the trace on both sides gives us $Tr g_1=Tr g_2$, doesn't this mean that any two matrices in our group which have different different trace are in a different conjugacy class? Meaning that there is an infinite number of conjugacy classes. Then would determining the full set of conjugacy classes provide any insight into the different properties of matrices beloning to different classes?
– Th.Phy
yesterday