Koszul duality between Weyl and Clifford algebras?

2011-11-10T20:22:07Z

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?

Answer by DamienC for Koszul duality between Weyl and Clifford algebras?

2011-11-11T09:02:50Z

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.

Koszul duality between Weyl and Clifford algebras? - MathOverflow

Koszul duality between Weyl and Clifford algebras?

Koszul duality

Clifford and Weyl algebras

Main question

Answer by DamienC for Koszul duality between Weyl and Clifford algebras?