# Endomorphisms of the Koszul complex

Let k be a ring, for example a field. Let R be a k-algebra. Let f_1, …, f_r be a regular sequence in R such that k → R/(f_1, …, f_r) is an isomorphism. Let K be the Koszul complex over R on f_1, …, f_r viewed as a cochain complex sitting in degrees -r, …, 0. See Tag 0621. Then we are interested in the hom-complex

E = Hom_R(K, K)

constructed in Tag 0A8H which we may and do view as a differential graded R-algebra, see for example Tag 0FQ2. This algebra is interesting for many reasons; for example because there is an equivalence

D(E) = D_{QCoh, V(f_1, …, f_r)}(Spec(R))

Here the LHS is the derived category of dg E-modules and the RHS is the derived category of complexes of quasi-coherent modules on Spec(R) supported set theoretically on f_1 = … = f_r = 0. To prove this equivalence, use Tag 09IR to see that K gives a generator of the RHS and argue as in the proof of Tag 09M5.

Recall that the underlying R-module of K is the exterior algebra on the free module of rank r, say with basis e_1, …, e_r. For i = 1, .., r let v_i : K → K be the operator given by contraction by the dual basis element to e_i. Of course v_i has degree 1 and a computation shows that v_i : K → K[1] is a map of complexes. Similarly, the reader shows that v_i ∘ v_j = – v_j ∘ v_i and v_i ∘ v_i = 0. Thus we obtain a map of differential graded k-algebras

k ⟨ v_1, …, v_r ⟩ → E

with target E and source the exterior algebra over k on v_1, …, v_r in degree 1 and vanishing differential. A bit more work shows that this map is a quasi-isomorphism of differential graded k-algebras; this is where we use the assumption that f_1, …, f_r is a regular sequence so the koszul complex is a resolution of R/(f_1, …, f_r) by Tag 062F.

Applying Tag 09S6 we conclude that

D(k ⟨ v_1, …, v_r ⟩) = D_{QCoh, V(f_1, …, f_r)}(Spec(R))

Cheers!