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Aug 30, 2019 at 14:14 comment added spin 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.
Aug 30, 2019 at 14:05 review Close votes
Sep 1, 2019 at 18:07
Aug 30, 2019 at 13:40 history edited ChockaBlock CC BY-SA 4.0
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Aug 30, 2019 at 13:37 comment added ChockaBlock Thanks for the suggestion! I actually find the other types more confusing and am mostly okay with type $A$. One example that's causing me some grief is $\Psi = A_2C_1 < C_4$, where it seems like the partition should be $(3,2,1^3)$ but odd parts need even multiplicity.
Aug 30, 2019 at 13:02 comment added spin 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.
Aug 30, 2019 at 12:58 comment added spin 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 $SO_{2n}$.
Aug 30, 2019 at 12:33 comment added ChockaBlock Yes, you're right. I've updated the question to reflect this.
Aug 30, 2019 at 12:33 history edited ChockaBlock CC BY-SA 4.0
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Aug 30, 2019 at 12:28 comment added spin 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 $(1, 3)$.
Aug 30, 2019 at 12:23 history asked ChockaBlock CC BY-SA 4.0