Koszul complexes and the shift formulas

Written by Claude Opus 5.5 (Anthropic) on 20 October 2026. Self-checked by the writing AI. Public domain (CC0).

Shifting a complex by one degree turns symmetric powers into exterior powers and exterior powers into divided powers: $\begin{gathered}LS^n(E[1])\cong(L\Lambda^nE)[n],\\ L\Lambda^n(E[1])\cong(L\Gamma^nE)[n].\end{gathered}$ These formulas convert questions about $L\Lambda^n$ of the cotangent complex, which appear in its Chern classes and its de Rham theory, into questions about symmetric powers, and back. They come from Koszul complexes. An exact sequence $0\to E\to F\to G\to0$ of flat modules gives exact sequences $0\to\Lambda^nE\to\cdots\to S^nF\to S^nG\to0$. These are applied to the sequence $0\to E\to C(E)\to E[1]\to0$ whose middle term, the cone of the identity, is contractible.

This lesson constructs the two Koszul complexes of a map of modules, proves their exactness for exact sequences of flat modules over a ring of a topos, and proves the shift formulas with their compatibility with products.

Prerequisites. Non-additive functors of complexes in a ringed topos, Sections 2–5; Homotopy sheaves and the free abelian Whitehead theorem, Section 5; Complexes, cones and localization, Lemma 1.2 and Proposition 5.2. From the AI-integrated Stacks project: spectral sequences of double complexes with finitely many nonzero terms on each antidiagonal [Stacks, Tag 0132] and bounded above complexes of flat modules are K-flat [Tag 06YQ].

Conventions

$A$ is a sheaf of commutative rings on a site $\mathcal C$. Tensor products are over $A$. For an $A$-module $M$, $S(M)$, $\Lambda(M)$ and $\Gamma(M)$ are the symmetric, exterior and divided power algebras: the sheaves associated with the constructions on sections. $\Gamma(M)$ is generated by symbols $\gamma_k(x)$, $x\in M$, $k\geq0$, subject to $\gamma_0(x)=1$, $\gamma_1(x)=x$, $\gamma_k(ax)=a^k\gamma_k(x)$, $\gamma_k(x+y)=\sum_{i+j=k}\gamma_i(x)\gamma_j(y)$ and $\gamma_i(x)\gamma_j(x)=\binom{i+j}{i}\gamma_{i+j}(x)$. For a complex $X$, $X[m]^k=X^{k+m}$ with differential $(-1)^md$.

1. The two Koszul complexes

A bigraded algebra $R=\bigoplus R^{p,q}$ is anticommutative in the first degree if $x'x=(-1)^{pp'}xx'$ for $x\in R^{p,q}$, $x'\in R^{p',q'}$, and similarly in the second degree. A derivation of bidegree $(-1,1)$ of an algebra anticommutative in the first degree is an $A$-linear map $d$ of that bidegree with $d(xx')=(dx)x'+(-1)^px(dx')$ for $x\in R^{p,q}$.

Proposition 1.1. Let $u\colon E\to F$ be a map of $A$-modules.

On $\Lambda E\otimes SF$, with $\Lambda^pE\otimes S^qF$ in bidegree $(p,q)$, there is a unique derivation $d$ of bidegree $(-1,1)$ with $d(x\otimes1)=1\otimes u(x)$ and $d(1\otimes y)=0$. It satisfies $d\circ d=0$.
On $\Gamma E\otimes\Lambda F$, with $\Gamma^pE\otimes\Lambda^qF$ in bidegree $(p,q)$, there is a unique derivation $d$ of bidegree $(-1,1)$, for the sign rule of the second degree, with $d(\gamma_k(x)\otimes1)=\gamma_{k-1}(x)\otimes u(x)$ and $d(1\otimes z)=0$. It satisfies $d\circ d=0$.

Proof. Uniqueness holds because the elements named generate the algebras. We prove existence on sections; the constructions are compatible with restriction and pass to the associated sheaves.

(1) For $x_1,\ldots,x_p\in E$ and $y\in S^qF$, put $\begin{aligned}&d(x_1\wedge\cdots\wedge x_p\otimes y)\\&\quad=\sum_{i=1}^p(-1)^{i-1}x_1\wedge\cdots\widehat{x_i}\cdots\wedge x_p\\&\qquad\qquad\otimes u(x_i)y.\end{aligned}$ The right side is multilinear and alternating in the $x_i$: exchanging $x_i$ and $x_{i+1}$ exchanges two terms and changes all signs, and if $x_i=x_{i+1}$ the two terms cancel and the others contain $x_i\wedge x_{i+1}=0$. So $d$ is well defined. The derivation rule holds by the same formula. The composite $d\circ d$ is a derivation of even degree that vanishes on $x\otimes1$ and on $1\otimes y$, so it is zero. (In $d^2$ of a monomial, the terms removing $x_i$ and then $x_j$ and those removing $x_j$ and then $x_i$ carry opposite signs and equal values, because $S F$ is commutative.)

(2) A derivation of $\Gamma E$ into a $\Gamma E$-module is determined by its values on the generators $\gamma_k(x)$, and exists if and only if these values respect the defining relations. With the values $D\gamma_k(x)=\gamma_{k-1}(x)\otimes u(x)\in\Gamma E\otimes F$, the relations hold: $\begin{aligned} &D\gamma_k(ax)=a^{k-1}\gamma_{k-1}(x)\otimes au(x)\\&\quad=a^kD\gamma_k(x),\\ &D\gamma_k(x+y)\\&\quad=\sum_{i+j=k-1}\gamma_i(x)\gamma_j(y)\otimes(u(x)+u(y))\\&\quad=D\Bigl(\sum_{i+j=k}\gamma_i(x)\gamma_j(y)\Bigr),\\ &D\bigl(\gamma_i(x)\gamma_j(x)\bigr)\\&\quad=\Bigl(\tbinom{i+j-1}{i-1}+\tbinom{i+j-1}{i}\Bigr)\\&\qquad\quad\cdot\gamma_{i+j-1}(x)\otimes u(x)\\&\quad=\tbinom{i+j}{i}D\gamma_{i+j}(x). \end{aligned}$ Extend $D$ to $\Gamma E\otimes\Lambda F$ by $d(a\otimes z)=D(a)\wedge z$, where the $F$-component of $D(a)$ is multiplied into $\Lambda F$ on the left. This is a derivation for the sign rule of the second degree. The composite $d\circ d$ is a derivation that kills the generators: $d^2\gamma_k(x)=\gamma_{k-2}(x)\otimes u(x)\wedge u(x)=0$. Hence $d\circ d=0$. $\square$

We write $\operatorname{Kos}_\bullet(u)=(\Lambda E\otimes SF,d)$ and $\operatorname{Kos}^\bullet(u)=(\Gamma E\otimes\Lambda F,d)$, and $k^n$ for the part of total degree $p+q=n$. The first is a chain complex with the exterior degree as homological degree, the second a cochain complex with the exterior degree as cohomological degree. A commutative square from $u$ to $u'$ induces maps of differential bigraded algebras $\operatorname{Kos}(u)\to\operatorname{Kos}(u')$.

Lemma 1.2. For $u\colon E\to F$ and $u'\colon E'\to F'$, the isomorphisms $\begin{aligned}&\Lambda(E\oplus E')\otimes S(F\oplus F')\\&\quad\cong(\Lambda E\otimes SF)\otimes(\Lambda E'\otimes SF'),\\&\Gamma(E\oplus E')\otimes\Lambda(F\oplus F')\\&\quad\cong(\Gamma E\otimes\Lambda F)\otimes(\Gamma E'\otimes\Lambda F')\end{aligned}$ carry the Koszul differential of $u\oplus u'$ to the tensor-product differential.

Proof. Both sides are derivations of the same bidegree, and they agree on the generators $x\otimes1$, $x'\otimes1$, $\gamma_k(x)\otimes1$ and $\gamma_k(x')\otimes1$, and on $F$, $F'$. $\square$

Construction 1.3. Let $E\xrightarrow{u}F\xrightarrow{v}G$ be maps with $vu=0$.

$\operatorname{Kos}_\bullet(u)\to SG$, zero on $\Lambda^{\ge1}E\otimes SF$ and $S(v)$ on $SF$, is a map of differential bigraded algebras: $d(x\otimes1)=u(x)$ is sent to $vu(x)=0$.
$\Gamma E\to\operatorname{Kos}^\bullet(v)$, $\gamma_k(x)\mapsto\gamma_k(u(x))\otimes1$, is one as well: $d\gamma_k(u(x))$ equals $\gamma_{k-1}(u(x))\otimes vu(x)$, which is $0$.

2. Exactness

Theorem 2.1. Let $0\to E\xrightarrow{u}F\xrightarrow{v}G\to0$ be an exact sequence of flat $A$-modules. Then the maps of Construction 1.3 are quasi-isomorphisms. In total degree $n\geq1$ they give the exact Koszul sequences $\begin{gathered} 0\to\Lambda^nE\to\Lambda^{n-1}E\otimes F\to\cdots\\\cdots\to E\otimes S^{n-1}F\to S^nF\to S^nG\to0,\\[4pt] 0\to\Gamma^nE\to\Gamma^nF\to\Gamma^{n-1}F\otimes G\to\cdots\\\cdots\to F\otimes\Lambda^{n-1}G\to\Lambda^nG\to0. \end{gathered}$

Proof. Finite free split case. Suppose $F=E\oplus G$ with $u$ and $v$ the inclusion and projection, and $E$, $G$ finite free. Then $u=1_E\oplus(0\to G)$ and $v=(E\to0)\oplus1_G$. By Lemma 1.2, $\begin{gathered}\operatorname{Kos}_\bullet(u)\cong\operatorname{Kos}_\bullet(1_E)\otimes SG,\\\operatorname{Kos}^\bullet(v)\cong\Gamma E\otimes\operatorname{Kos}^\bullet(1_G),\end{gathered}$ with zero differential on $SG$ and on $\Gamma E$, and the maps of Construction 1.3 are the projections onto, respectively the inclusions of, the parts of degree $0$ of $\operatorname{Kos}(1)$. It suffices to show that $k^n\operatorname{Kos}_\bullet(1_M)$ and $k^n\operatorname{Kos}^\bullet(1_M)$ are acyclic for $n\geq1$ and $M$ finite free. By Lemma 1.2, $\operatorname{Kos}(1_{M'\oplus M''})=\operatorname{Kos}(1_{M'})\otimes\operatorname{Kos}(1_{M''})$. Its part of degree $n$ is a sum of tensor products $k^a\otimes k^b$ with $a+b=n$, each with an acyclic factor if $a$ or $b$ is positive. These factors are bounded complexes of finite free modules, so locally contractible, and the tensor products are acyclic. So it suffices to treat $M=A$. There $k^n\operatorname{Kos}_\bullet(1_A)$ is $\Lambda^1A\otimes S^{n-1}A\to S^nA$, $e\otimes t^{n-1}\mapsto t^n$, an isomorphism. Likewise $k^n\operatorname{Kos}^\bullet(1_A)$ is $\Gamma^nA\to\Gamma^{n-1}A\otimes\Lambda^1A$, $\gamma_n(1)\mapsto\gamma_{n-1}(1)\otimes e$, an isomorphism.

General case. The sequence is a bounded acyclic complex of flat modules. By Proposition 3.3 of Non-additive functors of complexes in a ringed topos, it is the local inductive limit of contractible complexes $0\to E_i\to F_i\to G_i\to0$ of finite free modules over objects $U_i$. Such a complex is split exact, and locally its terms decompose as in the first case. The Koszul complexes and the maps of Construction 1.3 are built from $S$, $\Lambda$, $\Gamma$, tensor products and finite sums, which commute with local inductive limits. Local inductive limits are exact (Section 5 of Homotopy sheaves and the free abelian Whitehead theorem). The cones of the maps are local inductive limits of locally acyclic complexes, so they are acyclic. $\square$

Proposition 2.2 (duality). Let $E$, $F$ be finite free and $M^\vee=\underline{\operatorname{Hom}}(M,A)$. Pair $\Lambda^pE$ with $\Lambda^p(E^\vee)$ by $\begin{aligned}&\langle x_1\wedge\cdots\wedge x_p,\xi_1\wedge\cdots\wedge\xi_p\rangle\\&\quad=\det(\xi_j(x_i)),\end{aligned}$ and $S^qF$ with $\Gamma^q(F^\vee)$ by $\begin{aligned}&\langle y_1\cdots y_q,\gamma_{k_1}(\eta_1)\cdots\gamma_{k_r}(\eta_r)\rangle\\&\quad=\sum_{\varphi}\prod_{i=1}^q\eta_{\varphi(i)}(y_i),\end{aligned}$ the sum running over the maps $\varphi\colon\{1,\ldots,q\}\to\{1,\ldots,r\}$ with $|\varphi^{-1}(j)|=k_j$. Pair $\Lambda^pE\otimes S^qF$ with $\Gamma^q(F^\vee)\otimes\Lambda^p(E^\vee)$ by the product of the two pairings. These pairings are perfect, and for $u\colon E\to F$, $\langle d\alpha,\beta\rangle=\langle\alpha,d\beta\rangle$ for $\alpha\in\Lambda^pE\otimes S^qF$ and $\beta\in\Gamma^{q+1}(F^\vee)\otimes\Lambda^{p-1}(E^\vee)$, where the second $d$ is that of $\operatorname{Kos}^\bullet(u^\vee)$. So $k^n\operatorname{Kos}_\bullet(u)$ is the dual complex of $k^n\operatorname{Kos}^\bullet(u^\vee)$.

Proof. Take a basis $e_1,\ldots,e_s$ of $E$, a basis $f_1,\ldots,f_r$ of $F$ and the dual bases, and write $u(e_i)=\sum_ja_{ji}f_j$. The monomials $e_I=e_{i_1}\wedge\cdots\wedge e_{i_p}$ ($i_1<\cdots<i_p$) and $e_I^\vee$ pair to $\delta_{IJ}$. For multi-indices $\rho,\nu$ the monomial $f^\rho$ and the divided monomial $\gamma_\nu(f^\vee)=\prod_j\gamma_{\nu_j}(f_j^\vee)$ pair to $\delta_{\rho\nu}$: if $f^\rho$ is written as a product of $q$ basis vectors, exactly one $\varphi$ sends every factor to its own index when $\rho=\nu$, and none otherwise. Both pairings are therefore perfect.

Let $\alpha=e_I\otimes f^\mu$ and $\beta=\gamma_\nu(f^\vee)\otimes e_J^\vee$ with $|I|=p$, $|J|=p-1$, $|\nu|=|\mu|+1$. Both sides vanish unless $J=I\smallsetminus\{i_0\}$ and $\nu=\mu+\delta_j$ for some $i_0$ and $j$. Then $d\alpha=\sum_{t}(-1)^{t-1}e_{I\smallsetminus\{i_t\}}\otimes u(e_{i_t})f^\mu$ gives $\langle d\alpha,\beta\rangle=(-1)^{t_0-1}a_{ji_0}$, where $i_0=i_{t_0}$. On the other side $d\gamma_\nu(f^\vee)=\sum_k\gamma_{\nu-\delta_k}(f^\vee)\otimes u^\vee(f_k^\vee)$ with $u^\vee(f_k^\vee)=\sum_ia_{ki}e_i^\vee$, so $\langle\alpha,d\beta\rangle=a_{ji_0}\langle e_I,e_{i_0}^\vee\wedge e_J^\vee\rangle$. Moving $e_{i_0}^\vee$ past the $t_0-1$ smaller indices of $J$ gives the sign $(-1)^{t_0-1}$. $\square$

3. The shift formulas

For a complex $E$ of flat modules bounded above and $n\geq1$, put $\mathbf T(E)=\operatorname{Tot}N\mathbf T\Gamma\,iE$ for $\mathbf T=S^n,\Lambda^n,\Gamma^n$, as in Section 2.2 of Non-additive functors of complexes in a ringed topos. By Theorem 4.4 there, $L\mathbf T(E)=\mathbf T(E)$.

Lemma 3.1. Let $0\to P_n\xrightarrow{\alpha_n}I_{n-1}\to\cdots\to I_0\xrightarrow{\alpha_0}Q\to0$ be an exact sequence of complexes in which every $I_j$ is acyclic, $n\geq1$.

The sequence defines an isomorphism $\sigma\colon Q\to P_n[n]$ in the derived category, natural for morphisms of such sequences.
Let $X$ be the total complex of the double complex whose column of index $a$ is $P_n$ for $a=-n$ and $I_{-a}$ for $-n<a\leq0$, with the maps of the sequence as horizontal differentials and $(-1)^a$ times the differential of the column of index $a$ as vertical differential: $X^k=\bigoplus_a(\text{column }a)^{k-a}$, $d=d_h+(-1)^ad_v$. The map $e\colon X\to Q$ given by $\alpha_0$ on the column of index $0$ is a quasi-isomorphism, the projection $\pi\colon X\to P_n[n]$ onto the column of index $-n$ is a morphism of complexes, and $\sigma=(-1)^{n(n+1)/2}\,\pi\circ e^{-1}.$
Let $Y$ be the total complex, with the same conventions, of the double complex whose column of index $c$ is $I_{n-1-c}$ for $0\leq c<n$ and $Q$ for $c=n$. The map $e'\colon P_n\to Y$ given by $\alpha_n$ into the column of index $0$ is a quasi-isomorphism, the inclusion $\iota\colon Q[-n]\to Y$ of the column of index $n$ is a morphism of complexes, and $\sigma[-n]=(-1)^{n(n+1)/2}\,e'^{-1}\circ\iota.$

Proof. (1) Split the sequence into short exact sequences $0\to K_{j+1}\to I_j\to K_j\to0$, with $K_0=Q$ and $K_n=P_n$. Each gives a distinguished triangle whose third morphism is $\delta_j=(-p)r^{-1}\colon K_j\to K_{j+1}[1]$, where $r\colon C(K_{j+1}\to I_j)\to K_j$ is the quasi-isomorphism and $p$ the projection onto $K_{j+1}[1]$ (Proposition 5.2 of Complexes, cones and localization); on cohomology it is the boundary map of the long exact sequence. Since $I_j$ is acyclic, $\delta_j$ is an isomorphism. Put $\sigma=\delta_{n-1}[n-1]\circ\cdots\circ\delta_1[1]\circ\delta_0$, morphisms being shifted without sign. A morphism of sequences induces morphisms of the short exact sequences, and $r$ and $p$ are natural in them.

(2) The cone of $e$ is isomorphic to the shift by one of the total complex of the double complex obtained by adding $Q$ as a column of index $1$, through $(q,x)\mapsto(-q,x)$. That double complex has exact rows and finitely many columns, so its total complex is acyclic [Tag 0132], and $e$ is a quasi-isomorphism. The column of index $-n$ receives no horizontal differential and has the differential $(-1)^nd$ of $P_n[n]$, so $\pi$ is a morphism of complexes. Put $\rho_X=\pi\circ e^{-1}$ and $\varepsilon_n=(-1)^{n(n+1)/2}$.

For $n=1$, $X$ is the cone $C(\alpha_1)$, with $C(\alpha_1)^k=I_0^k\oplus P_1^{k+1}$ and $d(y,x)=(dy+\alpha_1x,-dx)$; $e$ is the map $r$ and $\pi$ the map $p$ of Proposition 5.2 there. So $\sigma=(-p)r^{-1}=-\rho_X$.

Let $n\geq2$. Let $K=K_1\subset I_0$, and let $R_1=(0\to K\to I_0\to Q\to0)$ and $R'=$ $(0\to P_n\to I_{n-1}\to\cdots\to I_1\to K\to0)$, of lengths $1$ and $n-1$, with total complexes $X_1$, $X'$ and maps $e_1,\pi_1$, $e_{X'},\pi_{X'}$ as in (2). By definition $\sigma=\sigma_{R'}[1]\circ\sigma_{R_1}$. Let $\Phi\colon X\to X_1$ be the identity of $I_0$ on the column of index $0$, the corestriction $I_1\to K$ of $\alpha_1$ on the column of index $-1$, and zero on the other columns; it is a morphism of complexes, because $\alpha_1\alpha_2=0$. Let $\Psi\colon X\to X'[1]$ be $(-1)^{a+1}$ times the identity from the column of index $a\leq-1$ of $X$ to the column of index $a+1$ of $X'$, and zero on the column of index $0$. The vertical differentials agree, $(-1)^a$ for $X$ and $-(-1)^{a+1}$ for $X'[1]$; the horizontal differentials of $X$ and $X'[1]$ differ by the sign $-1$, which the factors $(-1)^{a+1}$ compensate; and the horizontal differential of $X$ from the column of index $-1$ to that of index $0$ is killed by $\Psi$, while $X'$ has no column of index $1$. So $\Psi$ is a morphism of complexes. On the column of index $-1$, $\pi_1\Phi$ and $e_{X'}[1]\Psi$ are both the corestriction of $\alpha_1$, and both vanish on the other columns; and $\pi_{X'}[1]\Psi=(-1)^{n+1}\pi$. With $e_1\Phi=e$, this gives in the derived category $\begin{aligned} \rho_{X'}[1]\circ\rho_{X_1}\circ e&=\pi_{X'}[1]\,e_{X'}[1]^{-1}\,\pi_1\Phi\\ &=\pi_{X'}[1]\Psi=(-1)^{n+1}\pi, \end{aligned}$ so $\rho_{X'}[1]\circ\rho_{X_1}=(-1)^{n+1}\rho_X$. By the case $n=1$ and induction, $\sigma=$ $\varepsilon_{n-1}\rho_{X'}[1]\circ(-\rho_{X_1})=$ $(-1)^n\varepsilon_{n-1}\rho_X=$ $\varepsilon_n\rho_X$.

(3) As in (2), the cone of $e'$ is acyclic, being isomorphic to the total complex of the double complex obtained by adding $P_n$ as a column of index $-1$, and $\iota$ is a morphism of complexes, no horizontal differential leaving the column of index $n$. Put $\rho'_Y=e'^{-1}\circ\iota\colon Q[-n]\to P_n$.

For $n=1$, $Y$ has $Y^k=I_0^k\oplus Q^{k-1}$ and $d(j,q)=(dj,\alpha_0j-dq)$, and $\mu(j,q)=(-q,j)$ is an isomorphism of $Y$ onto $C(\alpha_0)[-1]$, where $C(\alpha_0)^k=Q^k\oplus I_0^{k+1}$. It carries $e'$ to $\kappa\colon P_1\to C(\alpha_0)[-1]$, $x\mapsto(0,\alpha_1x)$, and $\iota$ to $-i[-1]$, $i\colon Q\to C(\alpha_0)$ the inclusion. On $C(\alpha_1)$, with $r$, $p$ as in (2), the map $(y,x)\mapsto(0,y)$ to $C(\alpha_0)$, of degree $-1$, is a homotopy between $i\circ r$ and $-\kappa[1]\circ p$: its commutator with the differentials is $(y,x)\mapsto(\alpha_0y,\alpha_1x)$. So $i=\kappa[1]\circ(-p)r^{-1}=\kappa[1]\circ\sigma$ in the derived category, and $\rho'_Y=\kappa^{-1}\circ(-i[-1])=-\sigma[-1]$.

Let $n\geq2$. Let $K'=K_{n-1}\subset I_{n-2}$, and let $S_1=(0\to P_n\to I_{n-1}\to K'\to0)$ and $S'=$ $(0\to K'\to I_{n-2}\to\cdots\to I_0\to Q\to0)$, with total complexes $Y_1$, $Y'$ as in (3); by definition $\sigma=\sigma_{S_1}[n-1]\circ\sigma_{S'}$. Let $\lambda\colon Y_1\to Y$ be the identity of $I_{n-1}$ on the column of index $0$ and the inclusion $K'\to I_{n-2}$ on the column of index $1$, and $j\colon Y'[-1]\to Y$ be $(-1)^c$ times the identity from the column of index $c-1$ of $Y'$ to the column of index $c$ of $Y$. Both are morphisms of complexes, by the same comparison of signs as in (2), and $\lambda e'_1=e'$, $\lambda\iota_1=-j\circ e'_{Y'}[-1]$, $j\circ\iota_{Y'}[-1]=(-1)^n\iota$. Hence, in the derived category, $\begin{aligned}&\iota\\ &\quad =(-1)^nj\,e'_{Y'}[-1]\,\rho'_{Y'}[-1]\\ &\quad =(-1)^{n+1}\lambda\iota_1\rho'_{Y'}[-1]\\ &\quad =(-1)^{n+1}e'\circ\rho'_{Y_1}\circ\rho'_{Y'}[-1],\end{aligned}$ so $\rho'_Y=(-1)^{n+1}\rho'_{Y_1}\circ\rho'_{Y'}[-1]$. By the case $n=1$ and induction, $\sigma[-n]=$ $\sigma_{S_1}[-1]\circ\sigma_{S'}[-n]=$ $(-\rho'_{Y_1})\circ\varepsilon_{n-1}\rho'_{Y'}[-1]=$ $\varepsilon_n\rho'_Y$. $\square$

Theorem 3.2. For $E\in D^-(A)$ and $n\ge1$ there are canonical isomorphisms $\begin{aligned}&\text{(i)}\quad LS^n(E[1])\cong(L\Lambda^nE)[n],\\&\text{(ii)}\quad L\Lambda^n(E[1])\cong(L\Gamma^nE)[n].\end{aligned}$

Proof. Represent $E$ by a bounded-above complex of flat modules. Let $C(E)$ be the cone of $1_E$; it is contractible. There is an exact sequence of complexes of flat modules $0\to E\to C(E)\to E[1]\to0,$ split in each degree. Apply the exact functor $\Gamma\,i$: this gives an exact sequence of mixed objects of flat modules, $0\to\Gamma\,iE\to\Gamma\,iC(E)\to\Gamma\,i(E[1])\to0$. Apply Theorem 2.1 term by term, then $\operatorname{Tot}N$, which is exact. For (i) this gives an exact sequence of complexes $\begin{gathered}0\to\Lambda^nE\to I_{n-1}\to\cdots\\\cdots\to I_0\to S^n(E[1])\to0,\end{gathered}$ with $I_{n-p}=\operatorname{Tot}N\bigl(\Lambda^{n-p}\Gamma\,iE\otimes S^p\Gamma\,iC(E)\bigr)$ for $1\le p\le n$. By (5.1) of the same lesson, $I_{n-p}$ is homotopy equivalent to $\Lambda^{n-p}E\otimes S^pC(E)$. Since $C(E)$ is homotopy equivalent to $0$ and $S^p(0)=0$, Proposition 2.3 there makes $S^pC(E)$ contractible, so $I_{n-p}$ is acyclic. Lemma 3.1 gives $S^n(E[1])\cong\Lambda^nE[n]$, which is (i). The second Koszul sequence gives (ii) in the same way, the intermediate terms being $\Gamma^{n-p}C(E)\otimes\Lambda^p(E[1])$ with $p<n$. Naturality in $E$ follows from the naturality of every step, and independence of the flat representative from Theorem 4.4 there. $\square$

Products. Let $\tau\colon X[m]\otimes Y[n]\to(X\otimes Y)[m+n]$, $x\otimes y\mapsto(-1)^{n\,|x|}x\otimes y$, where $|x|$ is the degree of $x$ in $X[m]$. This is an isomorphism of complexes, natural in $X$ and $Y$. For morphisms $f\colon Q[-m]\to P$ and $g\colon Q'[-n]\to P'$, a check of the signs on $q\otimes q'$ gives $\begin{aligned}&\bigl((f\otimes g)\circ\tau^{-1}\bigr)[m+n]\\&\quad =(-1)^{mn}\,\tau\\&\quad \circ(f[m]\otimes g[n]).\end{aligned} \tag{3.1}$

For double complexes $D$, $D'$ with the conventions of Lemma 3.1 (2), let $D\otimes D'$ be the double complex whose column of index $c$ is $\bigoplus_{a+a'=c}D_a\otimes D'_{a'}$, with the tensor product differential of complexes as vertical differential and $d_h(x\otimes y)=d_hx\otimes y+(-1)^ax\otimes d_hy$ for $x$ in the column of index $a$. Then $\begin{gathered}\varphi \colon\operatorname{Tot}D\otimes\operatorname{Tot}D'\\ \to\operatorname{Tot}(D\otimes D'),\\ x\otimes y\mapsto(-1)^{ba'}x\otimes y,\end{gathered}$ for $x$ of vertical degree $b$ and $y$ in the column of index $a'$, is an isomorphism of complexes: on each of the four components of the differential the two signs agree.

Lemma 3.3. Let $R\colon$ $0\to$ $P\to$ $I_{m-1}\to$ $\cdots\to$ $I_0\to$ $Q\to$ $0$ and $R'\colon$ $0\to$ $P'\to$ $I'_{n-1}\to$ $\cdots\to$ $I'_0\to$ $Q'\to$ $0$ be exact sequences of bounded above complexes of flat modules, with all $I_j$, $I'_j$ acyclic and $m,n\geq1$, and let $\sigma_R$, $\sigma_{R'}$ be their isomorphisms (Lemma 3.1).

Let $D$, $D'$ be the double complexes of Lemma 3.1 (2) (columns of index $-m,\ldots,0$ and $-n,\ldots,0$). The columns of $D\otimes D'$, followed by the map to $Q\otimes Q'$ induced by the two augmentations, form an exact sequence $R\otimes R'$ of length $m+n$, from $P\otimes P'$ to $Q\otimes Q'$, with acyclic intermediate terms, and its isomorphism is $\tau\circ(\sigma_R\otimes\sigma_{R'})$.
Let $D$, $D'$ be instead the double complexes of Lemma 3.1 (3) (columns of index $0,\ldots,m$ and $0,\ldots,n$). The map $P\otimes P'\to D_0\otimes D'_0$, followed by the columns of $D\otimes D'$, form an exact sequence $R\otimes'R'$ of length $m+n$, from $P\otimes P'$ to $Q\otimes Q'$, with acyclic intermediate terms, and its isomorphism is $(-1)^{mn}\tau\circ(\sigma_R\otimes\sigma_{R'})$.

Proof. In each vertical degree the rows of $D$ and $D'$, augmented, are resolutions of flat modules by flat modules; their tensor product computes the Tor of these flat modules, so it is a resolution of their tensor product. Hence $R\otimes R'$ is exact, and likewise $R\otimes'R'$, whose rows are tensor products of coresolutions. An intermediate term of $R\otimes R'$ is a sum of tensor products $D_a\otimes D'_{a'}$ with $(a,a')\neq(-m,-n)$, each with an acyclic factor; the other factor is a bounded above complex of flat modules, hence K-flat [Tag 06YQ], and the tensor product is acyclic. The same holds for the terms of $R\otimes'R'$ other than $Q\otimes Q'$.

(1) With the maps $e$, $\pi$ of Lemma 3.1 (2), $e_{R\otimes R'}\circ\varphi=e_R\otimes e_{R'}$, and $\pi_{R\otimes R'}\circ\varphi=(-1)^{mn}\tau\circ(\pi_R\otimes\pi_{R'})$: for $x\in P^b$ in the column of index $-m$ and $y$ in the column of index $-n$, $\varphi$ gives $(-1)^{-bn}$, while $\tau$ gives $(-1)^{n(b-m)}$, $x$ having degree $b-m$ in $P[m]$. As $e_R\otimes e_{R'}$ is a quasi-isomorphism of K-flat complexes, $\pi_{R\otimes R'}e_{R\otimes R'}^{-1}=$ $(-1)^{mn}\tau\circ(\pi_Re_R^{-1}\otimes\pi_{R'}e_{R'}^{-1})$. By Lemma 3.1 (2) the isomorphism of $R\otimes R'$ is $\varepsilon_{m+n}(-1)^{mn}\varepsilon_m\varepsilon_n\,\tau\circ(\sigma_R\otimes\sigma_{R'})$, and $\varepsilon_{m+n}\varepsilon_m\varepsilon_n=(-1)^{mn}$.

(2) With the maps $e'$, $\iota$ of Lemma 3.1 (3), $\varphi\circ(e'_R\otimes e'_{R'})=e'_{R\otimes'R'}$ and $\varphi\circ(\iota_R\otimes\iota_{R'})=(-1)^{mn}\iota_{R\otimes'R'}\circ\tau$: for $x\in Q^b$ in the column of index $m$ and $y$ in the column of index $n$, $\varphi$ gives $(-1)^{bn}$ and $\tau$ gives $(-1)^{n(b+m)}$. Hence $\rho'_{R\otimes'R'}=(-1)^{mn}(\rho'_R\otimes\rho'_{R'})\circ\tau^{-1}$, and by Lemma 3.1 (3) and (3.1) the isomorphism of $R\otimes'R'$ is $\varepsilon_{m+n}(-1)^{mn}(-1)^{mn}\varepsilon_m\varepsilon_n\,\tau\circ$ $(\sigma_R\otimes\sigma_{R'})=$ $(-1)^{mn}\tau\circ$ $(\sigma_R\otimes\sigma_{R'})$. $\square$

Proposition 3.4. The isomorphisms of Theorem 3.2 are compatible with the products of Section 5 of Non-additive functors of complexes in a ringed topos, as follows. For $m,n\geq1$, the product $\begin{aligned}&LS^m(E[1])\otimes^{\mathbf L}LS^n(E[1])\\&\quad\to LS^{m+n}(E[1])\end{aligned}$ corresponds, through (i), to the product $\begin{aligned}&L\Lambda^mE[m]\otimes^{\mathbf L}L\Lambda^nE[n]\\&\quad\to L\Lambda^{m+n}E[m+n]\end{aligned}$ composed with $\tau$; and the product of $L\Lambda(E[1])$ corresponds, through (ii), to $(-1)^{mn}$ times the product of $L\Gamma(E)$, shifted, composed with $\tau$. That is, $\sigma\circ(\text{product})=$ $(\text{product})[m+n]\circ\tau\circ(\sigma\otimes\sigma)$ for (i), and the same with the factor $(-1)^{mn}$ for (ii).

Proof. (i) Let $R_m$ be the sequence of the proof of Theorem 3.2 for $S^m(E[1])$: its terms are $\Lambda^mE$ and the complexes $I_{m-p}$, in the columns of index minus their exterior degree, and its isomorphism is that of (i). These are the parts of total degree $m$ of the Koszul algebra of $\Gamma\,iE\to\Gamma\,iC(E)$, followed by Construction 1.3. The multiplication of the Koszul algebra, composed with (5.1), maps $D_a\otimes D_{a'}$ to the column of index $a+a'$, since exterior degrees add. It commutes with the horizontal differentials of Lemma 3.3, because the Koszul differential is a derivation with the sign $(-1)^p$ of the exterior degree $p=-a$; it commutes with the vertical differentials, because (5.1) and the termwise multiplication are morphisms of complexes; and Construction 1.3 is a homomorphism of algebras. So the multiplication is a morphism of sequences $R_m\otimes R_n\to R_{m+n}$ of length $m+n$, equal to the product $\Lambda^mE\otimes\Lambda^nE\to\Lambda^{m+n}E$ on the first terms and to the product $S^m(E[1])\otimes S^n(E[1])\to S^{m+n}(E[1])$ on the last ones. By the naturality in Lemma 3.1 (1) and Lemma 3.3 (1), $\sigma_{R_{m+n}}\circ(\text{product})=$ $(\text{product})[m+n]\circ\tau\circ(\sigma_{R_m}\otimes\sigma_{R_n})$. The complexes are bounded above with flat terms, so the tensor products compute the derived tensor products.

(ii) The sequence $R'_m$ for $\Lambda^m(E[1])$ is formed by $\Gamma^mE$ and the parts $\Gamma^{m-c}C(E)\otimes\Lambda^c(E[1])$ of the second Koszul complex, $0\leq c\leq m$, after $\Gamma\,i$ and $\operatorname{Tot}N$; in the double complex of Lemma 3.1 (3) the part of exterior degree $c$ is the column of index $c$. The multiplication of the second Koszul algebra adds exterior degrees, its differential is a derivation with the sign $(-1)^c$ of the exterior degree, and Construction 1.3 is a homomorphism of algebras; so the multiplication is a morphism of sequences $R'_m\otimes'R'_n\to R'_{m+n}$, and Lemma 3.3 (2) gives the factor $(-1)^{mn}$. $\square$

4. Exercises

Exercise 4.1. Let $A$ be a ring and $f\in A$ a nonzerodivisor. Write down $\operatorname{Kos}_\bullet$ of $f\colon A\to A$ in total degree $n$, and compare it with the Koszul complex of the element $f$.

Exercise 4.2. Use Theorem 3.2 to compute $LS^2(A[1])$ and $L\Lambda^2(A[1])$ for $A$ a ring.

Exercise 4.3. Over $\mathbb F_2$, show that $\Lambda^2$ and $\Gamma^2$ differ, and explain why the second shift formula needs $\Gamma$ rather than $S$.

Exercise 4.4. Let $E=Ae_1\oplus Ae_2$, placed in degree $0$, and let $x_1,x_2$ be the classes of $e_1,e_2$ in $H^{-1}(LS^1(E[1]))=E$, and $y_1,y_2$ the same classes in $H^{-1}(L\Lambda^1(E[1]))$. Show that for $n=1$ both isomorphisms of Theorem 3.2 are the identity of $E[1]$, and that the isomorphisms for $n=2$ send the product $x_1x_2\in H^{-2}(LS^2(E[1]))$ to $-e_1\wedge e_2\in\Lambda^2E$ and the product $y_1y_2\in H^{-2}(L\Lambda^2(E[1]))$ to $e_1e_2\in\Gamma^2E$.

5. Complete solutions

4.1. In total degree $n$, $\operatorname{Kos}_\bullet(f)$ is $\Lambda^1A\otimes S^{n-1}A\to S^nA$, $e\otimes t^{n-1}\mapsto ft^n$. Identifying both terms with $A$, it is multiplication by $f$: the Koszul complex of $f$, which is a resolution of $A/f$ because $f$ is a nonzerodivisor.

4.2. By (i), $LS^2(A[1])\cong\Lambda^2(A)[2]=0$. By (ii), $L\Lambda^2(A[1])\cong\Gamma^2(A)[2]\cong A[2]$, free of rank one on $\gamma_2(1)$.

4.3. For $M=\mathbb F_2$, $\Lambda^2M=0$, while $\Gamma^2M=\mathbb F_2\gamma_2(1)$ and $S^2M=\mathbb F_2$. The canonical algebra map $S(M)\to\Gamma(M)$ sends $x^2$ to $x\cdot x=2\gamma_2(x)$, which is $0$ in characteristic $2$; so $S^2M\to\Gamma^2M$ is zero, and $S^2$ cannot replace $\Gamma^2$ in (ii). By Exercise 4.2, $L\Lambda^2(\mathbb F_2[1])\cong\mathbb F_2[2]$. The second Koszul sequence, which is exact over every ring, is the one that involves $\Gamma$.

4.4. For $n=1$ both sequences are $0\to E\to C(E)\to E[1]\to0$, with $C(E)^{-1}=E$, $C(E)^0=E$ and $d=1$. Its third morphism $(-p)r^{-1}$ is the identity: $s(x)=((0,x),-x)$ is a morphism of complexes $E[1]\to C(E\to C(E))$ with $r\circ s=1$, so $r^{-1}=s$ and $-p\circ s(x)=x$. By Proposition 3.4 (i), the image of $x_1x_2$ is the product of $\tau(x_1\otimes x_2)$; since $x_1$ has degree $-1$ in $E[1]$ and the second factor is shifted by $1$, $\tau(x_1\otimes x_2)=(-1)^{-1}x_1\otimes x_2$, with product $-e_1\wedge e_2$. For (ii) the same computation gives $-e_1e_2$, and the factor $(-1)^{1\cdot1}$ of Proposition 3.4 turns it into $e_1e_2$.

Sources

Koszul complexes are classical; Proposition 1.1, Theorem 2.1 for flat modules over a ring of a topos and the shift formulas with their product compatibility are as in Illusie. The shift formulas for derived functors of non-additive functors appear in modern form in L. Breen and R. Mikhailov, Derived functors of non-additive functors and homotopy theory, arXiv:0910.2817.