“…(One exception is the proof of Proposition 5.2, where the
‐action plays an important role.) However, there is no loss in doing so: as explained in [
5, Section 3], the graded Lusztig–Vogan bijection is completely determined by the ordinary (ungraded) Lusztig–Vogan bijection. In particular, the main theorem of this paper implies that the graded Lusztig–Vogan bijection is also independent of
.…”