Lecture 4: Ext Functors, Sheaves, Double complexes

Teacher: Benjamin Hennion


$Ext$ functors

Fix $𝒞$ an abelian category (for $Tor$: we need a category of modules, but not for the $Ext$ functor).

Y ∈ 𝒞, \; \Hom_𝒞(-, Y): 𝒞^{op} ⟶ Ab

This functor is left exact.

For all

0 ⟶ X_1 ⟶ X_2 ⟶ X_3 ⟶ 0

an exact sequence in $𝒞$,

0 ⟶ \Hom_𝒞(X_3, Y) ⟶ \Hom_𝒞(X_2, Y) ⟶ \Hom_𝒞(X_1, Y) ⟶ \Hom_𝒞(X, Y)

Let’s derive it on the right.

Take an injective resolution in $𝒞^{op}$, i.e. a projective resolution in $𝒞$.

NB: it’s a bit tricky: $\Hom_𝒞(-, Y): 𝒞 ⟶ Ab^{op}$ is right exact, so we could derive it on the left. To avoid confusion, we always consider opposite categories as the domain.

Def: Assume $𝒞$ has enough projections (hence $𝒞^{op}$ has enough injectives).

Ext^n_𝒞(X,Y) ≝ R^n(\Hom_𝒞(-, Y))(X)

In practice: Take a projective resolution $P^\bullet$ of $X$ in $𝒞$, compute $H^n(\Hom_𝒞(P^\bullet, Y))$

Example: Bar construction

$G$ group.

𝒞 = Rep_ℤ(G) = \lbrace \text{ Abelian groups with a } G\text{-action}\rbrace = Mod_{ℤ[G]}

where $ℤ[G]$ is the group ring of $G$.

$ℤ ∈ Mod_{ℤ[G]}$: trivial representation. $G$ acts trivially: $g \cdot n = n$

Find a projective resolution of $ℤ$ in $𝒞$

Bar construction: Set

P^{-n} ≝ \text{ Free } ℤ[G] \text{-module generated by } G^n \\ = ℤ[G]^{(G^n)} = \bigoplus\limits_{g_1,…, g_n ∈ G} (g_1, ⋯, g_n) ℤ[G]

Consider the $ℤ[G]$-module:

\underbrace{P^0}_{= ℤ[G]} \overset{\text{projection } g \, ↦ \, 1}{⟶} ℤ ⟶ 0

Differential:

d^n: P^{-n-1} ⟶ P^{-n}
d ⟺ \text{ a map of sets }: \\ \begin{cases} G^{n+1} &⟶ P{-n} = ℤ[G]^{(G^n)}\\ g_0, …, g_n &⟼ \sum\limits_{ i=0 }^{n-1} (-1)^{i+1} (g_0, ⋯, g_i, g_{i+1}, …, g_n) + g_0 \cdot (g_1, …, g_n) + (-1)^n (g_0, …, g_{n-1}) \end{cases}

$P^\bullet$ is a resolution of $ℤ$ as a $ℤ[G]$-module

1. $d^2 = 0$: cf. exercise on group cohomology
2. The complex $⋯ ⟶ P^{-2} ⟶ P^{-1} ⟶ P^0 ⟶ ℤ ⟶ 0$ is exact

To prove (2), we actually build a homotopy from $0$ to the identity of this complex.

h: \begin{cases} P^{-n} &⟶ P^{-n-1} \\ G^n \ni (g_0, …, g_{n-1}) &⟼ (1, g_0, …, g_{n-1}) \end{cases}

Exercise: check that this is indeed a homotopy.

Corollary: Fix $M ∈ Mod_{ℤ[G]}$

H^\bullet(G, M) ≃ Ext^\bullet(ℤ, M)

Proof: $C^\bullet(G, M) ≅ \Hom_{ℤ[G]}(P^\bullet, M)$

Question: $\Hom_{ℤ[G]}(ℤ, M) ≅ M^G = H^0(G, M)$ : can we compute $H^\bullet(G, M)$ as a derived functor of $M ⟼ M^G$? (i.e. can we resolve $M$ instead of resolving $ℤ$?)

Ext-groups and extensions

$𝒞$ abelian with enough projectives.

Def: an extension of $X ∈ 𝒞$ by $Y ∈ 𝒞$ is an exact sequence $0 ⟶ Y ⟶ Z ⟶ X ⟶ 0$

Two extensions are isomorphic if we have a morphism $f: Z ⟶ Z’$ in $𝒞$ such that

\begin{xy} \xymatrix{ 0 \ar[r] & Y \ar[d]_{id_Y} \ar[r] & Z \ar[r] \ar[d]_{f} & X \ar[d]_{id_X} \ar[r] & 0 \\ 0 \ar[r] & Y \ar[r] & I^1 \ar[r] & X \ar[r] & 0 } \end{xy}

commutes (in which case $f$ is an isomorphism in $𝒞$)

Denote by $E(X,Y)$ the set of isom. classes of extensions of $X$ by $Y$.

*Goal**: Compare $E(X,Y)$ with some $Ext^n(X,Y)$

$n=1$

Fix an extension $0 ⟶ Y ⟶ Z ⟶ X ⟶ 0$

Apply $\Hom_𝒞(-, Y)$: it’s not exact, but we can derive it, and get a long exact sequence, of which we extract:

\Hom_𝒞(Y, Y) ⟶ \Hom_𝒞(Z, Y) ⟶ \Hom_𝒞(X, Y) \overset{\partial_z}{⟶} Ext^1_𝒞(X,Y) ⟶ Ext^1_𝒞(Z,Y)

Def:

φ: \begin{cases} E(X,Y) &⟶ Ext^1_𝒞(X,Y) \\ (0 → Y → Z → X → 0) &⟼ \partial_z(id_Y) \end{cases}

It is well defined: take $Z ≅ Z’$, get:

\begin{xy} \xymatrix{ \Hom_𝒞(Y,Y) \ar[r]^{\partial_z} \ar[d] & Ext^1_𝒞(X,Y) \ar[d]^{ id } \\ \Hom_𝒞(Y,Y) \ar[r]_{\partial_{z'}} & Ext^1_𝒞(X,Y) } \end{xy}

by functoriality of the boundary maps $\partial$.

Theorem: $φ$ is a bijection.

Proof:

Surjectivity: Start with $α ∈ Ext^1(X,Y)$. Let $P$ be a projective object with $f: P ⟶ X$ epimorphism.

Denote by $K$ the kernel of $f$.

0 ⟶ K ⟶ P ⟶ X ⟶ 0

is exact.

\Hom(P, Y) ⟶ \Hom(K, Y) ⟶ Ext^1(X,Y) \underbrace{\overbrace{⟶}^{∃ \, β: K → Y \, ⟼ \; α}}_{\text{surjective}} Ext^1(P,Y) = \underbrace{0}_{P \text{ projective}}

Fix such a morphism $β: K ⟶ Y$.

\begin{xy} \xymatrix{ 0 \ar[r] & K \ar[d]_{β} \ar[r] & P \ar[r]^f \ar[d] & X \ar[d]_{id_X} \ar[r] & 0 \\ 0 \ar[r] & Y \ar[r] & Z_β \ar[r] & X \ar[r] & 0 } \end{xy}

Let

Z_β ≝ Y \coprod_K P \underbrace{ \overset{(0, f)}{⟶}}_{\text{same as a morphism } f: P → X + 0: Y → X \text{ that coincides on } K} X

Check that

0 ⟶ Y ⟶ Z_β ⟶ X ⟶ 0

is exact (exercise)

Moreover:

\begin{xy} \xymatrix{ \overbrace{\Hom_𝒞(K,Y)}^{\ni \, β} \ar[r]^{\partial_P} \ar@{<-}[d]_{\text{composition with } β} & \overbrace{Ext^1_𝒞(X,Y)}^{\ni \, α} \ar@{<-}[d]^{ id } \\ \underbrace{\Hom_𝒞(Y,Y)}_{\ni \, id_Y} \ar[r]_{\partial_{z'}} & \underbrace{Ext^1_𝒞(X,Y)}_{\ni \, \partial_{Z_β} (id_Y) = φ(Z_β)} } \end{xy}
φ(Z_β) = α ⟹ φ \text{ surjective}

Injectivity:

Lemma: The construction $α ⟼ \text{ isom. class of } Z_β$ defines a map:

ψ: Ext^1(X,Y) ⟶ E(X,Y)

Proof: Show that given two morphisms $β, β’: K ⟶ Y$ such that $\partial_P (β) = \partial_P (β’) = α$, the extension $Z_β$ and $Z_{β’}$ are isomorphic.

\Hom(P, Y) \overset{∃ \, γ \, ↦ \, β - β'}{⟶} \underbrace{\Hom(K, Y)}_{β, β'} ⟶ \underbrace{Ext^1(X, Y)}_{α} ⟶ 0
\begin{xy} \xymatrix{ K \ar[r]^{ i } \ar[d]_{ β } & P \ar[d]^{ π_β } \ar@/^/[rdd]^{ π_{β'} + y_{β'} γ } \\ Y \ar[r]_{ y_β } \ar@/_/[rrd]_{ y_{β'} \text{ (no choice)}} & Z_β \\ & & Z_{β'} } \end{xy}
\begin{xy} \xymatrix{ 0 \ar[r] & Y \ar[d]_{id} \ar[r]^{y_β} & Z_β \ar[r] \ar[d] & X \ar[d]^{id_X} \ar[r] & 0 \\ 0 \ar[r] & Y \ar[r]_{y_{β'}} & Z_{β'} \ar[r] & X \ar[r] & 0 } \end{xy}

Need to check:

\underbrace{(π_{β'} + y_{β'} γ) i}_{= \underbrace{π_{β'} i}_{= y_{β'} β'} + y_{β'} (β-β') = y_{β'} β } = y_{β'} β

By the universal property of the pushout $Z_{β’}$, we get

Z_β \overset{g_γ}{⟶} Z_{β'}

Still have to check

\begin{xy} \xymatrix{ Z_β \ar[r]^{ f } \ar[d]_{ g_γ } & X \ar[d]^{ id } \\ Z_{β'} \ar[r]_{ g } & X } \end{xy}

Exercise: use the UP of $Z_β$ again

To finish the proof of the theorem,

E(X,Y) \overset{φ}{⟶} Ext'(X,Y) \overset{ψ}{⟶} E(X,Y)
• φ \circ ψ = id
• enough to check that $ψ$ is surjective. Start with $0 ⟶ Y ⟶ Z ⟶ X ⟶ 0$ and $α = φ(Z)$.

\begin{xy} \xymatrix{ 0 \ar[r] & K \ar@{.>}[d]_{β \text{ (induced on the kernels)}} \ar[r] & P \ar[rr] \ar@{.>}[d]^{∃ \text{ (}P \text{ projective)}} && X \ar[d]^{id_X} \ar[r] & 0 \\ 0 \ar[r] & Y \ar[r]_{y_{β'}} & Z \ar[rr] && X \ar[r] & 0 } \end{xy}

By construction, $Z ≅ P \coprod_K Y$

Rule: Baer sum

Take

0 ⟶ Y ⟶ Z_1 ⟶ X ⟶ 0 \\ 0 ⟶ Y ⟶ Z_2 ⟶ X ⟶ 0

two extensions.

\begin{xy} \xymatrix{ 0 \ar[r] & Y ⊕ Y \ar@{<-}[d]_{=} \ar[r] & Z_1 ⊕ Z_2 \ar[r]^f \ar@{<-}[d] & X ⊕ X \ar@{<-}[d] \ar[r] & 0 \\ 0 \ar[r] & Y ⊕ Y \ar[d] \ar[r] & Z_1 ×_X Z_2 \ar[r] \ar[d] & X \ar[r] \ar[d]^{=} & 0\\ 0 \ar[r] & Y \ar[r] & Z_{12} ≝ Y \coprod_{Y ⊕ Y} (Z_1 ×_X Z_2) \ar[r] & X \ar[r] & 0\\ } \end{xy}

Claim: $φ(Z_{12}) = φ(Z_1) + φ(Z_2)$

Sheaves

Definition

Fix $X$ a topological space.

A sheaf on $X$ is a functor:

F: Open(X)^{op} ⟶ Set

where $Open(X)$ si the category whose objects are open subsets of $X$ and $% $

Let $U ∈ Open(X)$ and $\bigcup\limits_{i} U_i$ be a covering (i.e. $U_i ⊆{open} U$ and $U = \bigcup\limits{i} U_i$)

\begin{cases} F(U) &⟶ F(U_i) \\ \underbrace{γ}_{\text{called "section of } F \text{"}} &⟼ γ_{| U_i} \end{cases}

$F$ is a sheaf if

• ∀ U, ∀ γ_1, γ_2 ∈ F(U), ∀ (U_i) \text{ covering}: \\ ∀i, {γ_1}_{| U_i} = {γ_2}_{| U_i} ⟹ γ_1 = γ_2
• ∀ U, ∀ U_i, ∀ (γ_i) ∈ \prod\limits_{ i } F(U_i), \\ \text{ if } ∀ i, j, {γ_i}_{| U_i ∩ U_j} = {γ_j}_{| U_i ∩ U_j} \text{ then } ∃ γ ∈ F(U) \text{ such that } γ_i = γ_{| U_i} \quad ∀i

Equivalently:

\begin{xy} \xymatrix{ F(U) \ar[r] & \prod\limits_{ i } F(U_i) \ar@<+2pt>[r] \ar@<-2pt>[r] & \prod\limits_{ i, j } F(U_i ∩ U_j) } \end{xy}

$F$ is a sheaf iff $∀ U, ∀ (U_i)$ covering of $U$, $F(U)$ is the limit (equalizer of the two maps) of $% [r] \ar@<-2pt>[r] & \prod\limits_{ i, j } F(U_i ∩ U_j) } \end{xy} %]]>$

Def: Elements of $F(U)$ are called sections of $F$ on $U$.

Define similarly sheaves on $X$ with values in any reasonable category $𝒞$. Denote by $Sh(X, 𝒞)$ the category of sheaves.

Sh(X, 𝒞) ⊆ \underbrace{Fun(Open(X)^{op}, 𝒞)}_{\text{abelian: everything computed pointwise}}

Exercise: Show that $Sheaves(X, Mod_A)$ is an abelian category for any ring $A$.

Why does it work?: The embedding functor $Sh(X, Ab) \hookrightarrow Fun(Open(X)^{op}, Ab)$ has a left adjoint: the sheafification functor. So it preserves comilits: cokernels are computed in $Fun(Open(X)^{op}, Ab)$, and then sheafified.

Example of derived functor

Consider the global section functor:

Γ(X, -): \begin{cases} Sh(X, Ab) &⟶ Ab \\ F &⟼ F(X) \end{cases}

Claim: $Γ(X, -)$ is not right exact, but left exact.

If

0 ⟶ F_1 ⟶ F_2 ⟶ F_3 ⟶ 0

in $Sh(X, Ab)$, then

0 ⟶ Γ(X, F_1) ⟶ Γ(X, F_2) \overset{\text{not epimorphism}}{⟶} Γ(X, F_3)

is exact.

Def (Sheaf cohomology):

H^n (X, F) ≝ R^n(Γ(X, -))(F)

(exists since $Sh(X, Ab)$ has enough injectives (cf Webeil, Example 2.3.12))

De Rham cohomology

Sh(X, ℝ) \hookrightarrow Fun(Open(X)^{op}, Vect_ℝ)

Consider the sheafification:

\begin{cases} Fun(Open(X)^{op}, Vect_ℝ) &⟶ Sh(X, ℝ) \\ const ∈ ℝ &⟼ \underbrace{\underline ℝ_X }_{\text{constant sheaf: locally constant functions}} \end{cases}

Now, cohomology of this constant sheaf is De Rham cohomology:

H^n (X, \underline ℝ_X) = H^n_{dR}(X)

VI. Double complexes

Goal: Show that $Tor^A_n(M, N) ≅ Tor^A_n(N, M)$

• $Ext^n_𝒞(-, -)$ can be derived on either side
• $\Hom_𝒞(-, -)$ ⟺ $Ext^n_𝒞(X, Y) ≅ Ext^n_{𝒞^{op}}(Y, X)$

Fix $𝒞$ an abelian category.

Def: a double complex in $𝒞$ is a complex in complexes in $𝒞$.

X^{\bullet, \bullet} ∈ C^2 (𝒞) = C(C(𝒞))
\begin{xy} \xymatrix{ X^{p,q} \ar[r]^{ d_H } \ar[d]_{ d_v } & X^{p+1,q} \ar[d]^{ d_v } \\ X^{p,q+1} \ar[r]_{ d_H } & X^{p+1,q+1} } \end{xy}
d_H^2 = d_v^2 = 0 \\ d_v d_H = d_H d_v

Examples:

• $𝒞 = Mod_A \ni M, N$, $A$ commutative.

$P^\bullet_M, P^\bullet_N$ projective resolutions of $M$ and $N$ respectively.

X^{\bullet, \bullet} = P^\bullet_M ⊗_A P^\bullet_N \\ X^{p, q} = P^p_M ⊗_A P^q_N
• $X, Y ∈ 𝒞$. $P^\bullet$ projective resolution of $X$, $I^\bullet$ injective resolution of $Y$.

Y^{p,q} = \Hom_𝒞(P^{-p}, I^q)

(cf. picture)

Theorem: Let $X^{\bullet, \bullet}$ be a double complex such that

• $∀ p < 0, X^{\bullet, p} = 0 = X^{p, \bullet}$

• $∀ p > 0$,

• $X^{p, \bullet}$ is exact
• $X^{\bullet, p}$ is exact

then $∀ p ∈ ℤ$,

H^p(X^{\bullet, 0}) = H^p(X^{0, \bullet})

Proof: cf. pictures

Corollary: Let $𝒞, 𝒟, ℰ$ be abelian categories. Let $F: 𝒞 × 𝒟 ⟶ ℰ$ be a functor st:

• $∀ c ∈ 𝒞, \; F(c, -): 𝒟 ⟶ ℰ$ is additive and left exact.
• $∀ d ∈ 𝒟, \; F(-, d): 𝒞 ⟶ ℰ$ is additive and left exact.
• $∀I ∈ 𝒞$ injective, $F(I, -)$ is exact
• $∀J ∈ 𝒟$ injective, $F(-, J)$ is exact

(bilinear functor)

Then $∀n, ∀c ∈ 𝒞, ∀ d ∈ 𝒟$:

R^n (F(c, -))(d) ≅ R^n (F(-, d))(c)

You can derive whichever side is easier to compute!

Proof: Fix $(c,d) ∈ 𝒞 × 𝒟$, $I$ and $J$ injective resolutions of $c$ and $d$ respectively.

Double complex:

\begin{xy} \xymatrix{ & & 0 \ar[d] & 0 \ar[d] \\ & 0 \ar[d] \ar[r] & F(I^0, d) \ar[r] \ar[d] & F(I^1, d) \ar[d] \ar[r] & ⋯ \\ 0 \ar[r] & F(c, J^0) \ar[d] \ar[r] & F(I^0, J^0) \ar[r] \ar[d] & F(I^1, J^0) \ar[r] \ar[d] & ⋯ \\ 0 \ar[r] & F(c,J^1) \ar[d] \ar[r] & F(I^0,J^1) \ar[d] \ar[r] & F(I^1,J^1) \ar[d] \ar[r] & ⋯ \\ & & 0 & 0 \\ } \end{xy}
X^{q,p} = F(I^{p-1}, J^{q-1})

with the convention

• $I^{-1} = c$
• $J^{-1} = d$
• $I^{-n} = 0 \quad n ≥ 2$
• $J^{-n} = 0 \quad n ≥ 2$
R^n(F(c,-))(d) ≅ H^{n+1}(X^{\bullet, 0})\\ R^n(F(-, d))(c) ≅ H^{n+1}(X^{0, \bullet})

Note that

∀ p > 1, X^{\bullet, p} = F(I^{p-1}, 0 → d → J^0 → J^1 → ⋯ \text{ exact complex})

an $I^{p-1}$ is injective, hence by (2) the complex $X^{\bullet, p}$ is exact.

Similarly, the rows are exact too.

We can apply the theorem and conclude…

Applications

• $A$ commutative ring.

F: \begin{cases} Mod_A × Mod_A &⟶ Mod_A \\ M, N &⟼ M ⊗_A N \end{cases}

We can either resolve $M$ or $N$ to compute $Tor^A_n(M, N) ≅ Tor^A_n(N, M)$

• F: \begin{cases} 𝒞^{op} × 𝒞 &⟶ Ab \\ X, Y &⟼ \Hom_𝒞(X,Y) \end{cases}
R^n (\Hom(X, -))(Y) ≅ Ext^n_𝒞(X, Y) = R^n (\Hom(-, Y))(X)

To compute (or even to define) $Ext^n_𝒞(X,Y)$, we can take either a projective resolution of $X$ or an injective resolution of $Y$ in $𝒞$.

Ex: $H^n(G,M) = R^nF(M)$ where

F: \begin{cases} Rep_ℤ &⟶ Ab = Mod_ℤ \\ M &⟼ M^G \end{cases}

Exercise: Let $X$ be a topological space. $F ∈ Sh(X, Ab)$

H^n(X, F) ≅ Ext^n_{Sh(X, Ab)}(\underline ℤ_X, F)

VII. Finite resolutions

Goal: Compute explicitely $Tor^A_n(M,N)$, or at least show that it vanishes for $n$ big enough, in some cases…

We work in $Mod_A$, $A$ commutative ring.

Finite resolution of $k$:

0 ⟶ x k[x] ⟶ \underbrace{k[x]}_{⟶ k} ⟶ 0

What about $k[ε] ≝ k[x]/(x^2=0)$?

⋯ ⟶ k[ε] ⟶ k[ε] \overset{μ_ε = \text{ mult by } ε}{⟶} \underbrace{k[ε]}_{⟶ k} ⟶ 0

And actually, there are no finite resolutions of $k[ε]$:

∀n, \; Tor^{k[ε]}_n(k,k) = k

Proof of that: consider an exact sequence

0 ⟶ ε k ⟶ k[ε] ⟶ k ⟶ 0 \text{ in } Mod_{k[ε]}

cf. picture

Koszul complexes

What are they: explicit resolutions of $A/I$ where $I$ is an ideal, which are finite.

Construction: $p,n ∈ ℕ, \; A^n$ of basis $(e_1, …, e_n)$

Exterior power:

\bigwedge^p A^n = \text{ Free } A \text{-mod generated by symbols } e_{i_1} ∧ ⋯ ∧ e_{i_p} \text{ with } 1 ≤ i_1 < ⋯ < i_p ≤ n\\ = A^{⊗p} / x_1 ⊗ ⋯ ⊗ x_p - x_1 ⊗ ⋯ ⊗ x_{i-1} ⊗ x_{i+1} ⊗ ⋯ ⊗ x_p

${A^n}^\ast \qquad (e_1^\ast, …, e_n^\ast)$

$p$-linear skew-symmetric $A$ for on ${A^n}^\ast$

NB:

• $\bigwedge^0 A^n = A$
• $\bigwedge^0 A^n = A^n$
• $\bigwedge^n A^n ≅ A$
• $\bigwedge^{n-p} A^n ≅ \bigwedge^{p} A^n$
• $\bigwedge^p A^n = 0$ if $p > n$

Def: Let $M$ be an $A$-module.

Let $φ_1, …, φ_n$ be endomorphisms of $M$ st

φ_i φ_j = φ_j φ_i \quad ∀i,j
K(M, φ)^{-p} = \bigwedge^p A^n ⊗_A M
⋯ ⟶ \bigwedge^2 A^n ⊗_A M ⟶ \bigwedge^1 A^n ⊗_A M ⟶ M ⟶ 0

Differential:

d: \begin{cases} \bigwedge^p A^n ⊗_A M &⟶ \bigwedge^{p-1} A^n ⊗_A M \\ (e_{i_1} ∧ ⋯ ∧ e_{i_p}) ⊗ m &⟼ \sum\limits_{ k=1 }^p (-1)^{k+1} (e_{i_1} ∧ ⋯ ∧ \widehat {e_{i_k}} ∧ ⋯ ∧ e_{i_p}) ⊗ φ_{i_k}(m) \end{cases}

Let’s show that $d^2 = 0$ (notation: $δ_{l < k} = 1$ if $l < k$ else $0$)

\begin{align*} d^2((e_{i_1} ∧ ⋯ ∧ e_{i_p}) ⊗ m) & = \sum\limits_{ k=1 }^p (-1)^{k+1} \sum\limits_{ \substack{l=1 \\ l ≠ k} } (-1)^{l+ δ_{l < k}} (e_{i_1} ∧ ⋯ ∧ \widehat {e_{i_l}} ∧ ⋯ ∧ \widehat {e_{i_k}} ∧ ⋯ ∧ e_{i_p}) ⊗ φ_{i_l} φ_{i_k}(m) \\ & = \sum\limits_{ 1 ≤ l < k ≤ p } (-1)^{k+l} ( ⋯ ) ⊗ φ_{i_l} φ_{i_k}(m) + \sum\limits_{ 1 ≤ k < l ≤ p } (-1)^{k+l+1} ( ⋯ ) ⊗ φ_{i_l} φ_{i_k}(m) \\ & = 0 && \text{ since } φ_i φ_j = φ_j φ_i \end{align*}

NB:

• in Shapira: it’s the dual definition
• in Weibel/Wikipedia: special case where $M=A$
• the complex is bounded, since $\bigwedge^p A^n = 0$ if $p > n$

Prop:

1. H^0(K(M, φ_\bullet)) = M/(\Im φ_1 + ⋯ + \Im φ_n)
2. H^{-n}(K(M, φ_\bullet)) = \ker φ_1 ∩ ⋯ ∩ \ker φ_n
3. H^{-p}(K(M, φ_\bullet)) = 0 \qquad \text{ if } p > n

Def: We say that $(φ_1, …, φ_n)$ is a regular sequence if for any $i=1, …, n$, the morphism

φ_i: M/(\Im φ_1 + ⋯ + \Im φ_{i-1}) ⟶ M/(\Im φ_1 + ⋯ + \Im φ_{i-1})

is injective.

Prop: If $(φ_1, …, φ_n)$ is a regular sequence, then $K(M, φ_\bullet)$ is a resolution of $M/(\Im φ_1 + ⋯ + \Im φ_{i-1})$ (which is equivalent to saying that $H^{-p}(K(M, φ_\bullet)) = 0$ if $p ≠ 0$)

NB: The notion of regular sequences depends on the order of the $φ_i$’s.

Bonus: Skyscrapers of a sheaf: injectives of sheaves. $x ∈ X, c ∈ 𝒞$:

Sh(X, 𝒞) \ni h_c: U ⟼ \begin{cases} c &&\text{ if } x ∈ U \\ 0 &&\text{ else} \end{cases}

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