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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 $3\times3$ matrix is not conjugate to a matrix with all zero values in the bottom right $3\times3$ matrix.

Calimero
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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 Willse Nov 12 '18 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 - on hiatus Nov 12 '18 at 18:32
  • Thanks for your comments. It is the symplectic group, which obeys $S^T\Omega S=\Omega$ where $\Omega =\begin{Bmatrix} 0 & I_n \ -I_n &0 \end{Bmatrix}$, and is indeed compact. – Calimero Nov 12 '18 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? – Calimero Nov 16 '18 at 11:31

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