Koszul duality

Remarks

The construction is an involution: $(A^!)^! \cong A$, because $(R^\perp)^\perp = R$ in finite dimension. If $A$ is Koszul, then $A^!$ is Koszul, and the algebra $\operatorname{Ext}^\bullet_A(k, k)$ is isomorphic to $A^!$ (some conventions give the opposite algebra). For Koszul $A$ the Hilbert series satisfy $h_A(t) \, h_{A^!}(-t) = 1$. Example: let $A = S(V)$ be the symmetric algebra, so $R$ is spanned by the tensors $x \otimes y - y \otimes x$. A form on $V \otimes V$ vanishes on $R$ exactly when it is symmetric, so $R^\perp$ is the space of symmetric tensors, spanned by the tensors $\xi \otimes \xi$ and $\xi \otimes \eta + \eta \otimes \xi$. Hence $A^! = \Lambda(V^*)$ is the exterior algebra, and $h_A(t) = (1 - t)^{-n}$, $h_{A^!}(-t) = (1 - t)^n$ for $n = \dim V$.