$\newcommand{\g}{\mathfrak{g}}$Setting: $\mathfrak{g}$ is a semisimple complex Lie algebra. Here $\chi_\lambda$ denotes the central character corresponding to the action of $Z(\g)$ on a highest weight module of weight $\lambda$, i.e. Harish-Chandra says $\chi_\lambda=\chi_\mu$ if and only if $\lambda$ and $\mu$ are related by a SHIFTED action of the Weyl group, i.e. $w(\lambda+\rho)-\rho=\mu$. I emphasize this because lots of people use another convention such that Harish-Chandra does not use the shifted action. This notational/conventional difference is the source of my question. I let $\operatorname{Mod}U(\g)/\chi$ denote the category of modules on which $Z(\g)$ acts by $\chi$, i.e. mod out by $z-\chi(z)$.

Question: Is the following true: that for $\lambda$ antidominant and regular (both in the unshifted sense), one has $\operatorname{Mod} U(\g)/\chi_{\lambda+\rho}\cong \operatorname{Mod} U(\g)/\chi_{\lambda-\rho}$?

The reason I ask this is because different sources state Beilinson–Bernstein localization using different conventions. I sat down and picked two sources (in case it matters, Yi Sun’s expository paper $\mathcal D$-Modules and Representation Theory and Miličić’s preprint Localization and Representation Theory of Reductive Lie Groups) and unwound the conventions and got that this must be true in order for both of them to be correct.

I think something like this is true for category $\cal O$ due to translation functors, but I’m not sure what I can say for all $U(\g)$ modules.

(I guess it’s also possible that this is false because I unwound the conventions incorrectly or because one of the sources I picked made a mistake in their conventions.)

  • $\begingroup$ Isn't this false even for category O if e.g. $\lambda=-2\rho?$ $\endgroup$
    – dhy
    Commented Mar 29, 2022 at 3:02
  • $\begingroup$ Oops you are absolutely correct, but is it possible this is true when you extend to $U\mathfrak g$ modules? (I guess the category $\cal O$ analogy can only go so far since the definition of action of $Z\frak g$ on blocks for $\cal O$ is slightly different than for the modules I’m considering here (nilpotent rather than zero on the nose)?) $\endgroup$
    – Fan Zhou
    Commented Mar 29, 2022 at 3:20
  • 1
    $\begingroup$ I can't immediately rule out that there may be some exotic category equivalence, but if there is it can't preserve $N$-integrality (for the same reasons why it doesn't exist on category O.) And any equivalence gotten via Beilinson-Bernstein should preserve $N$-integrality. $\endgroup$
    – dhy
    Commented Mar 29, 2022 at 3:54

1 Answer 1


False alarm —- upon revisiting my unwinding I have realized that I messed up a sign. So actually the question should read $\operatorname{Mod} U\mathfrak{g}/\chi_{\lambda-\rho}=\operatorname{Mod} U\mathfrak{g}/\chi_{\lambda-\rho}$, which is true. So I guess this is a warning to myself to be a little more careful in the future before posting questions on overflow.

In case anyone cares, let me record the different conventions of these sources:

(1) For Hotta et al’s book, $\lambda$ acts by $\lambda$ on $\cal L(\lambda)$. What they call “$\chi_\lambda$” is actually $\chi_{\lambda-\rho}$, which are in bijection with shifted Weyl-orbits (in the usual sense). Most confusingly their definition for the shifted Weyl action also differs from the usual one. Hotta does not discuss the construction of the twisted sheaves for nonintegral weights.

(2) For Sun, $\lambda$ acts by $-\lambda$ on $\cal L^\lambda$, and again what he calls “$\chi_\lambda$” is actually $\chi_{\lambda-\rho}$. In his notation, the differential operators on $\cal L^{\lambda-\rho}$ are isomorphic to the universal enveloping algebroid modded out by $\xi-(\rho-\lambda)\xi$ for $\xi$ in $\mathcal O_X\otimes\mathfrak b$.

(3) For Milicic, again “$\chi_\lambda$” means $\chi_{\lambda-\rho}$. His definition of $\mathcal{D}_X^\lambda$ is the universal enveloping algebroid modded out by $\xi-(\lambda+\rho)\xi$.

(4) For Gaitsgory’s notes, $\lambda$ acts by $\lambda$ on $\cal L(\lambda)$. His $\chi_\lambda$ actually means $\chi_\lambda$.

So a follow-up question: it seems like Milicic, Sun, and Hotta all agree. However Gaitsgory’s statement of localization seems to be (let’s say $\lambda$ integral) $\operatorname{Mod} U\mathfrak{g}/\chi_{-\lambda}=\operatorname{Mod} \mathcal{D}_{X,\mathcal{L}(\lambda)}$ (for $\lambda$ dominant), whereas the other three say $\operatorname{Mod} U\mathfrak{g}/\chi_{-\lambda-\rho}=\operatorname{Mod} \mathcal{D}_{X,\mathcal{L}(-\lambda+\rho)}$. These don’t seem to agree —- what gives?


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