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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?

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One can make both algebras quadratic. Add element "z^2" which is central and modify the relations like changing 1-> z^2 i.e. [x,d]=z^2 in Weyl case and xd+dx = z^2 in Clifford case. And may ask whether the two modified algebras Koszul dual in the standard sense. I am not sure this is something good trick, just a comment... – Alexander Chervov Nov 11 '11 at 6:28
I begin to remember that such trick is used somewhere in Feigin Tsygan works. They convert universal enveloping algebra into quadratic algebra - like this: adding central variable t, and modifying relations to [xi, xj] = c_ij^k xk*t - so algebra becomes quadratic... – Alexander Chervov Nov 11 '11 at 7:50
Surely, you can homogenize nonhomogeneous quadratic relations, but the homogenizations of the Clifford and Weyl algebras wouldn't be quadratic dual to each other. Generally, I don't think there can be a positive answer to the original question, if only because there is no natural correspondence between symmetric and skew-symmetric forms $g$ and $h$ on a given vector space, or indeed, on any two different vector spaces. A theory of nonhomogeneous quadratic duality exists, but it does not relate these two algebras to each other. – Leonid Positselski Nov 11 '11 at 8:31
up vote 19 down vote accepted

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.

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.

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