Koszul duality between Weyl and Clifford algebras? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T06:56:37Z http://mathoverflow.net/feeds/question/80627 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/80627/koszul-duality-between-weyl-and-clifford-algebras Koszul duality between Weyl and Clifford algebras? MTS 2011-11-10T20:22:07Z 2011-11-11T09:08:08Z <h2>Koszul duality</h2> <p>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 <em>quadratic algebra</em> $$A = A(V,R) = T(V)/ \langle R \rangle,$$ where $\langle R \rangle$ is the 2-sided ideal in the tensor algebra generated by $R$.</p> <p>We can then form the quadratic algebra <code>$A^! = A(V^*, R^\perp)$</code>, where <code>$$R^\perp = \{ \phi \in V^* \otimes V^* \mid \phi(R) = 0 \},$$</code> and we have identified <code>$V^* \otimes V^*$</code> with <code>$(V \otimes V)^*$</code>. This algebra $A^!$ is also quadratic by construction, and is known as the <em>Koszul dual</em> of $A$. It's pretty clear that $(A^!)^! \simeq A$.</p> <p>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 <code>$$S(V)^! \simeq \Lambda(V^*), \quad \Lambda(V)^! \simeq S(V^*).$$</code></p> <h2>Clifford and Weyl algebras</h2> <p>Now suppose that $V$ is even-dimensional, say <code>$\mathrm{dim}_\mathbb{C}(V) = 2n$</code>, and let $h: V \otimes V \to k$ be a nondegenerate symmetric bilinear form on $V$. The <em>Clifford algebra</em> 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 $\mathrm{Cl}(V,h) \simeq M_{2^n}(\mathbb{C})$.</p> <p>If we take instead a nondegenerate alternating (i.e. symplectic) form $g:V \otimes V \to k$, then we can form the <em>Weyl algebra</em> $$A_n = A_n(V,g) = T(V)/\langle x - g(x) \mid x \in \Lambda^2(V) \rangle.$$ This too has a natural filtration from the tensor algebra, and the associated graded is $S(V)$.</p> <p>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.</p> <h2>Main question</h2> <p>Is there any sort of non-quadratic Koszul duality that relates the Clifford and Weyl algebras?</p> http://mathoverflow.net/questions/80627/koszul-duality-between-weyl-and-clifford-algebras/80666#80666 Answer by DamienC for Koszul duality between Weyl and Clifford algebras? DamienC 2011-11-11T09:02:50Z 2011-11-11T09:08:08Z <p>Non-homogeneous Koszul duality is now well-understood. Here are a few references: </p> <ul> <li>I guess the original reference is </li> </ul> <blockquote> <p>L. E. Positsel′ski˘ı. Nonhomogeneous quadratic duality and curvature. Funktsional. Anal. i Prilozhen., 27:57–66, 96, 1993.</p> </blockquote> <ul> <li>for a more systematic study you can have alook at </li> </ul> <blockquote> <p>A. Polishchuk and L. Positselski. Quadratic algebras, volume 37 of University Lecture Series. American Mathematical Society, Providence, RI, 2005.</p> </blockquote> <ul> <li><p>As far as I remember the <a href="http://math.unice.fr/~brunov/Operads.pdf" rel="nofollow">new book of Loday and Vallette</a> discusses this too (see $\S 3.6$). </p></li> <li><p>You can find the statement that Weyl and Clifford algebras are Koszul in the inhomogenous sens in <a href="http://arxiv.org/abs/hep-th/9411113" rel="nofollow">this paper of Braverman-Gaistgory</a> ($\S 5.3$). </p></li> </ul> <p>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!</p> <ul> <li><p>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). </p></li> <li><p>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. </p></li> </ul>