Koszul duality
--------------


Given a finite-dimensional $k$-vector space $V$ (I am happy taking $k = \mathbb{C}$ anywhere in the following if it makes a difference) and a subspace $R \subseteq V \otimes V$, we can form the *quadratic algebra*
$$A = A(V,R) = T(V)/ \langle R \rangle,$$
where $\langle R \rangle$ is the 2-sided ideal in the tensor algebra generated by $R$.


We can then form the quadratic algebra $A^! = A(V^\*, R^\perp)$, where
$$ R^\perp = \{ \phi \in V^\* \otimes V^\* \mid \phi(R) = 0 \}, $$
and we have identified $V^\* \otimes V^\*$ with $(V \otimes V)^\*$. This algebra $A^!$ is also quadratic by construction, and is known as the *Koszul dual* of $A$. It's pretty clear that $(A^!)^! \simeq A$.


One example of this is given by the symmetric and exterior algebras of a vector space and its dual, i.e. for a finite-dimensional vector space $V$, we have
$$ S(V)^! \simeq \Lambda(V^\*), \quad \Lambda(V)^! \simeq S(V^\*). $$


Clifford and Weyl algebras
--------------------------


Now suppose that $V$ is even-dimensional, say $\mathrm{dim}\_\mathbb{C}(V) = 2n$, and let $h: V \otimes V \to k$ be a nondegenerate symmetric bilinear form on $V$. The *Clifford algebra* is the algebra
$$ \mathrm{Cl}(V,h) = T(V)/\langle x - h (x) \mid x \in S^2(V) \rangle, $$
and this can be viewed as a deformation of the exterior algebra in the sense that the Clifford algebra is naturally filtered and the associated graded is $\Lambda(V)$. If $h$ is nondegenerate, then (over $\mathbb{C}$, at least) we can show that $\mathrm{Cl}(V,h) \simeq M\_{2^n}(\mathbb{C})$.


If we take instead a nondegenerate alternating (i.e. symplectic) form $g:V \otimes V \to k$, then we can form the *Weyl algebra*
$$ A\_n = A\_n(V,g) = T(V)/\langle x - g(x) \mid x \in \Lambda^2(V) \rangle. $$
This too has a natural filtration from the tensor algebra, and the associated graded is $S(V)$.


These two deformations share some features in common. For instance, the Weyl algebra is isomorphic to the algebra of polynomial differential operators on $\mathbb{C}[x\_1, \dots, x\_n]$, and one can think of the Clifford algebra as being a $\mathbb{Z}/2$-graded analogue of that via creation and annihilation operators on $\Lambda(V)$. Both algebras are simple.


Main question
-------------


Is there any sort of non-quadratic Koszul duality that relates the Clifford and Weyl algebras?



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](http://math.unice.fr/~brunov/Operads.pdf) 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](http://arxiv.org/abs/hep-th/9411113) ($\S 5.3$).


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.


