This blog post will be used in a later one. Please skip ahead to the next one.<\/p>\n
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.<\/p>\n
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.<\/p>\n
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<\/p>\n
e : i*i_*M —> M<\/p><\/blockquote>\n
We want to find an element z in R such that for every<\/strong> 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.<\/p>\n
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.<\/p>\n
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<\/p>\n
multiplication : M ⊗ B —> M<\/p><\/blockquote>\n
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<\/p>\n
ξ : M —> (M ⊗ B)[-c]<\/p><\/blockquote>\n
Namely, we can send m in M to<\/p>\n
(sign)m ⊗ 1 (x_1 ⊗ 1 – 1 ⊗ x_1) … (x_c ⊗ 1 – 1 ⊗ x_c)<\/p><\/blockquote>\n
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.<\/p>\n
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.<\/p>\n
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’.<\/p>\n
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.<\/p>\n
A simple calculation shows that θ(x_i) = D(f_i).<\/p>\n
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<\/p>\n
M — ξ –> (M ⊗ B)[-c] — θ_1 … θ_c ⊗ 1 –> M ⊗ B <\/p><\/blockquote>\n
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<\/p>\n
z = det(D_i(f_j)) <\/p><\/blockquote>\n
Thus we conclude what we want with z as above.<\/p>\n
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.<\/p>\n","protected":false},"excerpt":{"rendered":"
This blog post will be used in a later one. Please skip ahead to the next one. 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 … Continue reading