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Let f_1, …, f_c be elements of a ring R. We view the Koszul algebra B = K(R, f_1, …, f_c) as a differential graded R-algebra sitting in cohomological degrees -c, …, 0. So we have R = B^0 and B^{-1} is free over R with a basis x_1, …, x_c such that d(x_i) = f_i.

Of course, if f_1, …, f_c is a regular sequence, then B —> R/(f_1, …, f_c) is a quasi-isomorphism. But we are interested in the general case too.

Denote i_* : D(B) —> D(R) the pushforward and denote i* : D(R) —> D(B) the pullback. Let M be an object of D(B). We have a counit map

e : i*i_*M —> M

We want to find an element z in R such that for **every** M in D(B) there is a map s : M —> i*i_*M whose composition with e is multiplication by z on M. In other words, we want to split e up to multiplication by z.

Example: suppose that f_1, …, f_r is a regular sequence and that the map R —> R/(f_1, …, f_r) has a section. Then R —> B has a section too and we get what we want with z = 1.

Let M be a (right) dg module over B which is free as a graded module. Any dg module over B is quasi-isomorphic to one of these, so there is no loss of generality. The map e is the map

multiplication : M ⊗ B —> M

where the tensor product is over R. We are going to construct a map s : M —> M ⊗ B using derivations. Before we continue, observe that we do have a map of dg B-modules

ξ : M —> (M ⊗ B)[-c]

Namely, we can send m in M to

(sign)m ⊗ 1 (x_1 ⊗ 1 – 1 ⊗ x_1) … (x_c ⊗ 1 – 1 ⊗ x_c)

where the sign is (-1) to the power cdeg(m). The reason this works is that x_i ⊗ 1 – 1 ⊗ x_i is killed by the differential. There are some sign rules for the multiplication on B ⊗ B: we have (x_i ⊗ 1)(1 ⊗ x_j) = x_i ⊗ x_j and we have (1 ⊗ x_j)(x_i ⊗ 1) = – x_i ⊗ x_j.

Let D : R —> R be a derivation. Then we can extend D to a degree zero map D’ : B —> B which satisfies the Leibniz rule by setting D'(x_i) = 0. Of course D’ does not commute with d in general.

Suppose that M is a (right) dg module over B which is free as a graded module. (Any dg module over B is quasi-isomorphic to one of these.) Then we can similarly find a degree zero map D’ : M —> M which satisfies the Leibniz rule over D’.

In both cases consider the map θ : B —> B and θ : M —> M defined by the formula &theta = D’ o d – d o D’. Then θ has degree 1, defines a map B —> B[1] and M —> M[1] of complexes, θ : B —> B is a derivation, and θ : M —> M satisfies the Leibniz rule θ(mb) = θ(m)b + (sign) m θ(b) where the sign is (-1) to the power the degree of m.

A simple calculation shows that θ(x_i) = D(f_i).

Next, suppose we have c derivations D_1, …, D_c. Then we get c maps θ_1, …, θ_c : M —> M[1]. Then we can consider the composition

M — ξ –> (M ⊗ B)[-c] — θ_1 … θ_c ⊗ 1 –> M ⊗ B

Unless I made a calculation error (which is very possible) the composition of these maps with the map e : M ⊗ B —> M is equal to multiplication by

z = det(D_i(f_j))

Thus we conclude what we want with z as above.

The conclusion of this is a precise version of something we all already know: if we have a closed embedding i of codimension c and we have c tangent fields spanning the normal bundle, then we can split the counit map i*i_*M —> M using those tangent fields.