Cohomological induction equivalence conjecture for generalized Harish-Chandra modules

About 13 years old · traced to

Let g\mathfrak{g} be a finite-dimensional Lie algebra, let k≃sl(2)\mathfrak{k}\simeq\mathfrak{sl}(2) be a subalgebra, and let t\mathfrak{t} be a Cartan subalgebra of k\mathfrak{k}. For n≥0n\geq 0, let Cpˉ,t,n\mathcal{C}_{\bar{\mathfrak{p}},\mathfrak{t},n} be the full subcategory of g\mathfrak{g}-modules consisting of finite-length modules whose simple subquotients are pˉ\bar{\mathfrak{p}}-locally finite (g,t)(\mathfrak{g},\mathfrak{t})-modules with t\mathfrak{t}-weight spaces NβN^\beta, β∈Z\beta\in\mathbb{Z}, satisfying β≥n\beta\geq n. Let Ck,n\mathcal{C}_{\mathfrak{k},n} be the full subcategory of g\mathfrak{g}-modules consisting of finite-length modules whose simple subquotients are (g,k)(\mathfrak{g},\mathfrak{k})-modules with minimal k\mathfrak{k}-type V(μ)V(\mu) for μ≥n\mu\geq n. Cohomological induction equivalence conjecture. If n≥Λn\geq\Lambda, then

R1Γk,t:Cpˉ,t,n+2⟶Ck,nR^{1}\Gamma_{\mathfrak{k},\mathfrak{t}}:\mathcal{C}_{\bar{\mathfrak{p}},\mathfrak{t},n+2}\longrightarrow\mathcal{C}_{\mathfrak{k},n}

is an equivalence between the categories Cpˉ,t,n+2\mathcal{C}_{\bar{\mathfrak{p}},\mathfrak{t},n+2} and Ck,n\mathcal{C}_{\mathfrak{k},n}. Here Λ\Lambda is the bound for the genericity condition associated with the pair (g,k)(\mathfrak{g},\mathfrak{k}). The conjecture strengthens the known result that R1Γk,tR^{1}\Gamma_{\mathfrak{k},\mathfrak{t}} is fully faithful from Cpˉ,t,n+2\mathcal{C}_{\bar{\mathfrak{p}},\mathfrak{t},n+2} to Ck,n\mathcal{C}_{\mathfrak{k},n} for n≥0n\geq0; the source does not state whether essential surjectivity, and hence the equivalence, has been proved.

References

Primary source

Ivan Penkov and Gregg Zuckerman, “Algebraic methods in the theory of generalized Harish-Chandra modules”, arXiv:1310.8058 (2013).

Progress summary

Never refreshed

Nothing recorded yet. Refresh searches the literature and the public web for attempts on this problem, and writes the first summary here.

Solutions 0

No solutions have been posted yet.