springer_table_D.at

Springer correspondence from nilpotent orbits to representations of W in type D
orbit: partition -> representation= pair of partitions
see Carter pg. 419

diagram:

ComplexOrbit O            (for SO(2n))
Partition P               (for orbit of SO(2n))
D_irrep                   (for representation of W(SO(2n)))

from combinatorics.at:

D_irrep = { from folding pairs of partitions under swapping }
  ( (Partition,Partition) unsplit_irr { unequal partitions, larger first }
  | (Partition,bool) split_irr { for pair of equal partitions, with sign }
  )

The Springer map from nilpotent orbits to irreducible W(D_n)-representations
is for the most part governed by the map |core_quotient_2| from Jordan types
(partitions of $2n$ for type $D_n$) to pairs of partitions of sum $n$ (like
for $C_n$ the core is always empty for valid nilpotent Jordan types). The
subtle point is "doubly even" Jordan types (only even parts, each as always in
type $D_n$ repeated an even number of times, so they only occur when $n$ is
even) for which the $O(2n)$ nilpotent orbit splits into two $SO(2n)$ orbits,
and for which the 2-quotient is $(\lambda,\lambda)$ where $\lambda$ is the
partition of $n/2$ whose Young diagram is that of the Jordan type shrunk by a
factor 2 both horizontally and vertically. That partition pair determines a
"split irrep" for $W(D_n)$ and we need to consider the mapping from the
Boolean distinguishing the nilpotent orbits to the one distinguishing the
split irreducibles. The mapping is bijective, and in the end easy to describe:
it is the identity when the rank $n$ is divisible by 4, and Boolean NOT
(swapping |false| end |true|) when $n$ is (even but) not divisible by 4.

The following appears to be the justification for this choice. On page 423 of
Carter's book, a description is given couched in savant language of Richardson
classes, parabolic subgroups of G, and Macdonald representations (obtained
from the $j$-operation defined on page 368). For the doubly even Jordan types
this appears to translate into truncated induction of the sign representation
from the Bala-Carter Levi subgroup of $W(D_n)$ associated to the nilpotent
orbit, tensored (it appears one never misses an occasion for a twist) by the
sign representation of $W(D_n)$. Since we can focus on getting the information
distinguishing the two split irreducibles the evaluation on a particular split
class. We can take the "simplest" class with very even cycle type, namely the
class of $w=s_0s_2s_4..s_{n-2}$, whose signed cycle type has $n/2$ cycles,
each length $2$ and of positive cycle sign, and whose discriminating Boolean
value is |false|; the character of any split irrep on it is nonzero, and it is
negative if and only if that irrep has Boolean |true| attached (this can be
deduced from the formula implemented in |hyperoctahedral_character|). The
question then boils down to finding the Boolean attached to the truncated
induction of the Bala-Carter Levi subgroup associated to |O|, and this appears
to be the same as the discriminating Boolean attached to the split nilpotent
orbit; tensorsing with the sign representation then flips this with the sign
$(-1)^{n/2}$ of $w$, anf at is this flip that is implemented below (where
$n/2$ appears under the form of |sum(lambda)|). It may be noted that the
initial discriminating Boolean for |O| records the equivalent conditions (by
definition:) (0) |b| for which |O| is |nilpotent_orbit_D(rd)(Jordan_type,b)|,
(1) whether the final entry of |O.H| (dominant) is negative, (2) whether the
final entry of |diagram(O)| is zero. It is noteworthy that the condition
whether set of simple roots for the associated Bala-Carter Levi contains the
final node of the diagram is opposite to this condition whenever $n/2$ is odd,
a switch that apparently gets compensated to disappear in the truncated
induction process. The switch arises because, taking $S$ to be a disjoint
union of odd rank type $A$ sub-diagrams separated by a single absent node and
not containing the final node of the $D_n$ diagram, the unique associated
distinguished $H$ has half its entries negative, which is $n/2$ in number, and
making it dominant leaves a final negative entry if and only if $n/2$ is odd.
At this time I (MvL) don't understand how truncated induction undoes the flip.

after the initial |assert| statement, |D_adjust| is identical to |B_adjust|