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Let $G$ be a simple classical algebraic group with corresponding root system $\Phi$ and natural module $W$ of dimension $n$.

For a (closed) subsystem $\Psi$ of $\Phi$, let $G_\Psi = \{ U_\alpha \mid \alpha \in \Psi\}$, where $U_\alpha$ is the root subgroup corresponding to $\alpha \in \Psi$. Take a regular unipotent element $x_{\Psi} \in G_\Psi$.

It is well known that there is a function from the set of classes of unipotent elements in $G$ to the set of partitions $\lambda \vdash n$, which is injective most of the time (except some of the time in $D_n$).

How can I find the partition of $\mu \vdash n$ corresponding to $x_{\Psi}$?

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    $\begingroup$ Your statement about partitions is only correct when $G$ is of type $A$. For example, for a unipotent element of $Sp_{2n}$, each Jordan block of odd size must have even multiplicity. So for example in $Sp_4$ there is no unipotent element with corresponding partition $(1, 3)$. $\endgroup$
    – spin
    Commented Aug 30, 2019 at 12:28
  • $\begingroup$ Yes, you're right. I've updated the question to reflect this. $\endgroup$
    – ChockaBlock
    Commented Aug 30, 2019 at 12:33
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    $\begingroup$ It is still not completely correct. In type $D$ you have some pairs of unipotent classes corresponding to the same partition. Some unipotent classes of $O_{2n}$ split into two unipotent classes in $SO_{2n}$. $\endgroup$
    – spin
    Commented Aug 30, 2019 at 12:58
  • $\begingroup$ In any case, to find the corresponding partitions, try to identify the subsystem subgroup in terms of the natural module. For example in type A (say $SL_n$) you get subsystem subgroups which look like $SL_{n_1} \times \cdots \times SL_{n_t}$, so the corresponding partition is $(n_1, \ldots, n_t)$ plus some number of $1$'s. $\endgroup$
    – spin
    Commented Aug 30, 2019 at 13:02
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    $\begingroup$ I think that in that case the partition is $(2,3^2)$. In general for the Levi $A_{n-1}$ in $C_n$, the regular unipotent of $A_{n-1}$ has partition $(n,n)$ on the natural module of $C_n$. The embedding $SL_n < Sp_{2n}$ is given by a maximal totally singular decomposition. $\endgroup$
    – spin
    Commented Aug 30, 2019 at 14:14

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