This is a follow-up to a recent question asked by Peter Crooks here. The answer by Ben Webster includes a helpful link to the corrected arXiv version of Baohua Fu's 2003 Invent. Math. paper Symplectic resolutions for nilpotent orbits.

Most of this literature is unfamiliar to me, so I may be overlooking something. I've encountered nilpotent orbits mainly in connection with various types of representation theory (sometimes in good prime characteristic, where most properties of the orbits are the same as over $\mathbb{C}$). At this point I'm still confused about some details, such as:

Are there Richardson orbits whose closures fail to have a symplectic resolution (and if so, what is the lowest rank Lie algebra in which an example appears)?

EDIT: Here I'm using shorthand to avoid normality questions: read "for which the normalizations of their closures fail to have ...? (Apparently the sorting out of normal orbits isn't complete yet for some exceptional types.)

As Fu notes in Prop. 3.16, it follows from the main theorem of the paper that a nilpotent orbit whose closure admits a symplectic resolution must be Richardson (intersecting the nilradical of some parabolic subalgebra in a dense orbit). In turn a reviewer states: "But the converse is not always true." I don't see direct evidence of that in Fu's paper. Here the simple Lie algebras are studied case-by-case: all orbits in type $A_n$ are Richardson, with trivial component groups, forcing their closures to have symplectic resolutions. In types $G_2, F_4, E_6$, all Richardson orbits also have trivial component groups, whereas a few such orbits in types $E_7, E_8$ have component groups of order 2 and are left unsettled in the paper. (These cases were later treated geometrically here.)

The discussion of types $B_n, C_n, D_n$ leaves me somewhat confused, since the explicit examples mentioned between Prop. 3.21 and Prop. 3.22 aren't Richardson orbits. This prompts the question above.

ADDED: Fu reduces the problem (for a Richardson orbit) to the question of whether or not there exists a parabolic $P$ defining the orbit for which $N(P)=1$. This is the index in the full component group in $G$ (topologically, fundamental group) of the component group in $P$ of an orbit element $X$. (Here "component group" means the group $C_G(X)/C_G(X)^\circ$.) It's not clear how to compute $N(P)$ in all cases, which may be why Fu gave up on the leftover cases in types $E_7,E_8$. I'm wondering about the naive (presumably false?) statement that an orbit closure has a symplectic resolution iff the orbit is Richardson. What bothers me is that Fu seems to mention only examples of non-Richardson orbits such as minimal orbits in types other than $A_n$, etc.


2 Answers 2


The relevant information is in the article of Hesselink: Polarizations in the classical groups (Wayback Machine).

Fix a parabolic $P$ and a Richardson element $u$. Let $N(u,P)$ be the the number of conjugates of $P$ that contain $u$ (the number of polarizations of $u$). In Hesselink's notation, this is $N_1(P)$.

EDIT: I realized I had misread Hesselink; $N(P)$ is not the same for all polarizations.

Theorem (Fu): A nilpotent orbit closure $\bar{O}$ has a symplectic resolution if and only if $O$ is normal and Richardson with a polarization such that $N(P)=1$.

I misunderstood what was going on in Hesselink's tables. You should look for entries that only have one conjugacy class of polarizations which have $N_1=2$ (this is stronger than what you need, but all of Hesselink's examples have this form). It looks as though the first bad examples in each series are the Richardson orbits for the stabilizer of a line in $Sp(6)$ (denoted $C_2$), the stabilizer of a 4-space in $SO(9)$ (denoted $A_3$), and the stabilizer of a 5-space in $SO(12)$ (denoted $A_4$; that one I'm less confident I got right). If I understand correctly, inducing these up should give bad examples of higher rank in these series.

  • $\begingroup$ Thanks for the reference, which I'll check out further. I don't think the examples with $N_1 > 1$ are the issue here, but Hesselink's notation is complicated. (Your last remark is not helpful here, since only for type $A_n$ are all nilpotent orbits of Richardson type.) $\endgroup$ Commented Jun 13, 2014 at 15:25
  • $\begingroup$ @JimHumphreys I meant to write "Richardson" in that last sentence. Also, I realized I was wrong about how I was reading Hesselink (in particular, that statement was wrong for $Sp(4)$ and $SO(7)$). Corrected above. $\endgroup$
    – Ben Webster
    Commented Jun 13, 2014 at 23:27
  • $\begingroup$ Hesselink's tables are helpful but as you say not the whole story here; I never got far into that paper but suspect his tables are reliable. In some cases non-conjugate parabolics determine the same Richardson orbit, e.g., the subregular orbit, so $N(P)$ has to be checked for each such $P$. But for $Sp(6)$ the minimal special orbit (involving short root vectors) $[2^2, 1^2]$ comes from only one class of parabolics, with $N(P)=2$; so this would be the unique counterexample of least rank. Fu's examples are out of focus. $\endgroup$ Commented Jun 14, 2014 at 12:34
  • $\begingroup$ P.S. I tried to bypass the tricky issues about "normal", which I think are separate. I've edited my post (and corrected your spelling a bit). $\endgroup$ Commented Jun 14, 2014 at 12:41

Just adding a reference to the literature where this question seems to have cropped up :

"Calculating canonical distinguished involutions in the affine Weyl groups" - Chmutova, Ostrik (pdf)

They phrase the question as : "Are all Richardson orbits strongly Richardson ?" ('strongly Richardson' being their term for the existence of a resolution satisfying the necessary properties). The published version of the paper (pdf link above) also includes a family of counter examples(== one of the examples alluded to in Ben's answer).


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