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Peter Crooks
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I'm not sure you will find this answer to be satisfactory, as it addresses only a special case. Nevertheless, a unipotent conjugacy class in $SL_n(\mathbb{C})$ is the same as a conjugacy class of a nilpotent $n\times n$ matrix. The latter classes are indexed by Jordan canonical forms, and hence also by the partitions of $n$. So, there are as many unipotent conjugacy classes in $SL_n(\mathbb{C})$ as there are partitions of $n$. I do not know of a nice formula for the partition function, but I believe it has a nice generating function.

I'm not sure you will find this answer to be satisfactory. Nevertheless, a unipotent conjugacy class in $SL_n(\mathbb{C})$ is the same as a conjugacy class of a nilpotent $n\times n$ matrix. The latter classes are indexed by Jordan canonical forms, and hence also by the partitions of $n$. So, there are as many unipotent conjugacy classes in $SL_n(\mathbb{C})$ as there are partitions of $n$. I do not know of a nice formula for the partition function, but I believe it has a nice generating function.

I'm not sure you will find this answer to be satisfactory, as it addresses only a special case. Nevertheless, a unipotent conjugacy class in $SL_n(\mathbb{C})$ is the same as a conjugacy class of a nilpotent $n\times n$ matrix. The latter classes are indexed by Jordan canonical forms, and hence also by the partitions of $n$. So, there are as many unipotent conjugacy classes in $SL_n(\mathbb{C})$ as there are partitions of $n$. I do not know of a nice formula for the partition function, but I believe it has a nice generating function.

Source Link
Peter Crooks
  • 4.9k
  • 2
  • 22
  • 42

I'm not sure you will find this answer to be satisfactory. Nevertheless, a unipotent conjugacy class in $SL_n(\mathbb{C})$ is the same as a conjugacy class of a nilpotent $n\times n$ matrix. The latter classes are indexed by Jordan canonical forms, and hence also by the partitions of $n$. So, there are as many unipotent conjugacy classes in $SL_n(\mathbb{C})$ as there are partitions of $n$. I do not know of a nice formula for the partition function, but I believe it has a nice generating function.