Skip to content
Preprint

Deformations of the $\mathfrak{osp}(n|2)$-action on the superspace of symbols of differential operators on $\R^{1|n}$

Sep 2026 · 0 citations · 13 references
Mathematics

Abstract

We study formal deformations of the natural $\osp(n|2)$-action, $n\geq 3$, on the superspace $\Sc^n_d=\bigoplus_{k\geq 0}\Fc^n_{d-\frac{k}{2}}$ of symbols of linear differential operators on weighted densities over $\R^{1|n}$. Starting from the first cohomology space computed in \cite{10}, we compute the cup-product $\Hd^1\vee \Hd^1\to \Hd^2$ which carries the quadratic obstructions. The answer is governed by the $\osp(n|2)$-invariant operators $A_k=\eta_1\cdots\eta_n\partial_x^{k-1}$: the two cocycles $h_k$ and $\widetilde{h}_k$ spanning the off-diagonal part of $\Hd^1$ are exactly the two derivatives of the coboundary of $A_k$ with respect to the two weights. Consequently, all the products of two off-diagonal classes and all the products of two diagonal classes vanish, and the whole obstruction is carried, for each $k$, by a single non-trivial 2-cocycle $\Omega_k$. If $2d\notin\N$ the space $\Hd^1\vee\Hd^1$ is identically zero, so every infinitesimal deformation is integrable. If $2d=m\in\N$ we obtain exactly $m$ quadratic integrability conditions, $\tau_{2-n-k}(t_k-\widetilde{t}_k)+\tau_k\widetilde{t}_k=0$, $1\leq k\leq m$, and we prove that they are also sufficient: no condition of order $\geq 3$ occurs and the versal deformation is of degree one in the parameters. In particular every integrable formal deformation is equivalent to its infinitesimal part.

View source

We use cookies to run the site and, with your consent, for analytics and to show ads. See our Cookie Policy.