Consider a connected reductive group G over the complex numbers. Is there a `simple' formula for the number of conjugacy classes of unipotent elements in G?
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1$\begingroup$ The question is a bit unfocused, since "unipotent" is not directly related to Lie group structure (instead it depends on the algebraic group structure, and the count of unipotent classes is then the same in all good characteristics). As Peter Crooks indicates, there is no "simple" formula even in the general or special linear groups. $\endgroup$– Jim HumphreysCommented Jul 15, 2014 at 14:52
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1$\begingroup$ Carter's book "Finite groups of Lie type" is a good source for this stuff. If you want an e-copy, email me. $\endgroup$– Nick GillCommented Jul 15, 2014 at 15:04
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$\begingroup$ Thanks for the reference! Chapter 13.1 in Carter's book seems to give precisely the kind of answer that I was looking for $\endgroup$– mnrCommented Jul 15, 2014 at 15:14
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$\begingroup$ Great! That book is amazing - I'm glad it helped. $\endgroup$– Nick GillCommented Jul 15, 2014 at 15:32
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1 Answer
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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.
answered Jul 15, 2014 at 14:32