Koszul duality between Weyl and Clifford algebras?
Non-homogeneous Koszul duality is now well-understood. Here are a few references:
- I guess the original reference is
L. E. Positsel′ski˘ı. Nonhomogeneous quadratic duality and curvature. Funktsional. Anal. i Prilozhen., 27:57–66, 96, 1993.
- for a more systematic study you can have alook at
A. Polishchuk and L. Positselski. Quadratic algebras, volume 37 of University Lecture Series. American Mathematical Society, Providence, RI, 2005.
As far as I remember the new book of Loday and Vallette discusses this too (see $\S 3.6$).
You can find the statement that Weyl and Clifford algebras are Koszul in the inhomogenous sens in this paper of Braverman-Gaistgory ($\S 5.3$).
Nevertheless, as it is said in Leonid Positselski's comment, Weyl and Clifford algebras are not Koszul dual to each other. The reason is that inhomogeneous Koszul duality is inhomogeneous!
quadratic-linear algebras are dual to DG quadratic algebras (e.g. the universal envelopping algebra of a Lie algebra is Koszul dual its Chevalley-Eilenberg algebra).
quadratic--linear-constant algebra (e.g. Weyl or Clifford, for which there is even no linear part) are dual to curved quadratic DG algebras. E.g. for the Weyl algebra $\mathcal W_{(V,\omega)}$, its Kozsul dual is the pair $(\wedge(V^*),\omega)$ where the symplectic form $\omega$ is viewed as a curvature (a degree 2 element) in the exterior algebra.