2
$\begingroup$

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?

$\endgroup$
4
  • 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. $\endgroup$ Jul 15, 2014 at 14:52
  • 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 Gill
    Jul 15, 2014 at 15:04
  • $\begingroup$ Thanks for the reference! Chapter 13.1 in Carter's book seems to give precisely the kind of answer that I was looking for $\endgroup$
    – mnr
    Jul 15, 2014 at 15:14
  • $\begingroup$ Great! That book is amazing - I'm glad it helped. $\endgroup$
    – Nick Gill
    Jul 15, 2014 at 15:32

1 Answer 1

4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.