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If one wants to understand representations of $\mathfrak{g}$ (a finite dimensional semisimple Lie algebra) of weight $\lambda$, the happiest you could be is if $\lambda+\rho$ is (integral) regular dominant, i.e. it's an element of the weight lattice whose product $\langle \lambda+\rho,\check{\alpha}\rangle$ with every simple coroot $\check{\alpha}$ is a negative integer. In this case there's an equivalence between the category of these representations and twisted $\mathscr{D}$ modules on the flag variety $U(\mathfrak{g})_\lambda\text{-mod}\simeq \mathscr{D}_\lambda\text{-mod}$. See the book by Hotta, Toshiyuki, Tanisaki, section 11.2.

If $\lambda+\rho$ is just (integral) regular (nothing in $W$ stabilises it), there is an equivalence of derived categories. I've not worked with this, but presumably along with translation functors (tensoring with $\mathscr{O}(\mu)$ to make the twist of the $\mathscr{D}$ module dominant) this allows you to use standard Beilinson-Bernstein above to get a handle on such modules.

My question is: what is currently known when $\lambda+\rho$ is non-integral? If it splits up into rational and irrational case (where all $\langle \lambda+\rho,\check{\alpha}\rangle\in\mathbf{Q}$ or otherwise), I am more interested in the rational case.

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    $\begingroup$ IIRC the story is very similar to the integral case, except that instead of $W$ you need to consider the subgroup $W'$ generated by those simple reflections $s_{\alpha}$ that send $\lambda$ to $\lambda+n\alpha$ for some integer $n$. Then if $\lambda$ is $W'$-dominant then Beilinson-Bernstein holds and there is a theory of reflection functors that takes care of the other weights. But I might be misremembering/confusing with something else. $\endgroup$
    – dhy
    Commented Sep 18, 2020 at 1:17
  • $\begingroup$ (I suspect that your rational/irrational bifurcation is motivated by the case of affine Kac-Moody algebras, where rationality of the level controls what category O looks like. There it appears because you have the entire affine Weyl group to play with, but here you only have the finite Weyl group and so the possibilities are more limited.) $\endgroup$
    – dhy
    Commented Sep 18, 2020 at 1:23

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Integrality doesn't really seem to play a role in the statement of Beilinson-Bernstein as far as I can tell. For a general weight $\lambda \in \mathfrak h^\ast$ it makes sense to talk about the notions of dominant and regular.

If $\lambda$ is regular, then there is a derived localization:

$D(D_{G/B}^\lambda-mod) \simeq D(U_\lambda -mod)$

If $\lambda$ is both regular and dominant, there is an abelian category localization:

$D_{G/B}^\lambda-mod \simeq U_\lambda-mod$

There are no integrality assumptions in the statement of Theorem 3.3.1 in the paper "A Proof of the Jantzen Conjecture", for example.

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  • $\begingroup$ I suppose that defining the ring of twisted differential operators requires a bit more discussion in the non-integral case. $\endgroup$
    – Sam Gunningham
    Commented Sep 18, 2020 at 13:11
  • $\begingroup$ Thanks! I think my confusion came from the fact that in Hotta et al, integrality is assumed. $\endgroup$
    – Pulcinella
    Commented Sep 18, 2020 at 14:10

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