In this section we review the sign rules used so far and we discuss some of their ramifications. It also seems appropriate to discuss these issues in the setting of the category of complexes of modules over a ring, as most interesting phenomena already occur in this case. We sincerely hope the reader will not need to use the more esoteric aspects of this section.

For the rest of this section, we fix a ring $R$ and we denote $M^\bullet $ a complex of $R$-modules with differentials $d^ n_ M : M^ n \to M^{n + 1}$.

The $k$th shifted complex $M^\bullet [k]$ has terms $(M^\bullet [k])^ n = M^{n + k}$ and differentials $d_{M[k]}^ n = (-1)^ kd^{n + k}_ M$, see Homology, Definition 12.14.7.

Given a map $f : M^\bullet \to N^\bullet $ of complexes, we define $f[k] : M^\bullet [k] \to N^\bullet [k]$ without the intervention of signs, see Homology, Definition 12.14.7.

We identify $H^ n(M^\bullet [k])$ with $H^{n + k}(M^\bullet )$ without the intervention of signs, see Homology, Definition 12.14.8.

The boundary map of a short exact sequence of complexes is defined as in the snake lemma without the intervention of signs, see Homology, Lemma 12.13.12.

The distinguished triangle associated to a termwise split short exact sequence $0 \to K^\bullet \to L^\bullet \to M^\bullet \to 0$ of complexes is given by

\[ K^\bullet \to L^\bullet \to M^\bullet \to K^\bullet [1] \]

where $M^ n \to K^{n + 1}$ is the map $\pi ^{n + 1} \circ d^ n_ L \circ s^ n$ if $s$ and $\pi $ are compatible termwise splittings. In other words, without the intervention of signs. See Derived Categories, Definitions 13.10.1 and 13.9.9.

The total complex $\text{Tot}(M^\bullet \otimes _ R N^\bullet )$ has differential $d$ satisfying the Leibniz rule $d(x \otimes y) = d(x) \otimes y + (-1)^{\deg (x)}x \otimes d(y)$. See Homology, Example 12.18.2 and Homology, Definition 12.18.3.

There is a canonical isomorphism

\[ \text{Tot}(M^\bullet \otimes _ R N^\bullet )[a + b] \to \text{Tot}(M^\bullet [a] \otimes _ R N^\bullet [b]) \]

which uses the sign $(-1)^{pb}$ on the summand $M^ p \otimes _ R N^ q$, see Homology, Remark 12.18.5. It is often more convenient to consider the corresponding shifted map $\text{Tot}(M^\bullet \otimes _ R N^\bullet ) \to \text{Tot}(M^\bullet [a] \otimes _ R N^\bullet [b])[-a - b]$.

There is a canonical isomorphism of complexes

\[ \text{Tot}( \text{Tot}(K^\bullet \otimes _ R L^\bullet ) \otimes _ R M^\bullet ) \to \text{Tot}(K^\bullet \otimes _ R \text{Tot}(L^\bullet \otimes _ R M^\bullet )) \]

defined without the intervention of signs. See Section 15.58.

There is a canonical isomorphism

\[ \text{Tot}(L^\bullet \otimes _ R M^\bullet ) \to \text{Tot}(M^\bullet \otimes _ R L^\bullet ) \]

which uses the sign $(-1)^{pq}$ on the summand $L^ p \otimes _ R M^ q$. See Section 15.58.

Before we get into a discussion of the sign conventions regarding Hom-complexes, we construct the dual of a complex with respect to the conventions above.

**Proof.**
The assumptions mean that

\[ M \xrightarrow {\eta \otimes 1} M \otimes _ R N \otimes _ R M \xrightarrow {1 \otimes \epsilon } M \quad \text{and}\quad N \xrightarrow {1 \otimes \eta } N \otimes _ R M \otimes _ R N \xrightarrow {\epsilon \otimes 1} N \]

are the identity map. We can choose a finite free module $F$, an $R$-module map $F \to M$, and a lift $\tilde\eta : R \to F \otimes _ R N$ of $\eta $. We obtain a commutative diagram

\[ \xymatrix{ M \ar[rr]_-{\eta \otimes 1} \ar[rrd]_-{\tilde\eta \otimes 1} & & M \otimes _ R N \otimes _ R M \ar[r]_-{1 \otimes \epsilon } & M \\ & & F \otimes _ R N \otimes _ R M \ar[u] \ar[r]^-{1 \otimes \epsilon } & F \ar[u] } \]

This shows that the identity on $M$ factors through a finite free module and hence $M$ is finite projective. By symmetry we see that $N$ is finite projective. This proves part (1). Part (2) follows from Categories, Lemma 4.43.6 and its proof. Part (3) follows from the first equality of the proof.
$\square$

Lemma 15.72.2. Let $R$ be a ring. Let $M^\bullet $ be a complex of $R$-modules. Let $N^\bullet , \eta , \epsilon $ be a left dual of $M^\bullet $ in the monoidal category of complexes of $R$-modules. Then

$M^\bullet $ and $N^\bullet $ are bounded,

$M^ n$ and $N^ n$ are finite projective $R$-modules,

writing $\epsilon = \sum \epsilon _ n$ with $\epsilon _ n : N^{-n} \otimes _ R M^ n \to R$ and $\eta = \sum \eta _ n$ with $\eta _ n : R \to M^ n \otimes _ R N^{-n}$ then $(N^{-n}, \eta _ n, \epsilon _ n)$ is the left dual of $M^ n$ as in Lemma 15.72.1,

the differential $d_ N^ n : N^ n \to N^{n + 1}$ is equal to $-(-1)^ n$ times the map

\[ N^ n = \mathop{\mathrm{Hom}}\nolimits _ R(M^{-n}, R) \xrightarrow {d_ M^{-n - 1}} \mathop{\mathrm{Hom}}\nolimits _ R(M^{-n - 1}, R) = N^{n + 1} \]

where the equality signs are the identifications from Lemma 15.72.1 part (2).

Conversely, given a bounded complex $M^\bullet $ of finite projective $R$-modules, setting $N^ n = \mathop{\mathrm{Hom}}\nolimits _ R(M^{-n}, R)$ with differentials as above, setting $\epsilon = \sum \epsilon _ n$ with $\epsilon _ n : N^{-n} \otimes _ R M^ n \to R$ given by evaluation, and setting $\eta = \sum \eta _ n$ with $\eta _ n : R \to M^ n \otimes _ R N^{-n}$ mapping $1$ to $\text{id}_{M_ n}$ we obtain a left dual of $M^\bullet $ in the monoidal category of complexes of $R$-modules.

**Proof.**
Since $(1 \otimes \epsilon ) \circ (\eta \otimes 1) = \text{id}_{M^\bullet }$ and $(\epsilon \otimes 1) \circ (1 \otimes \eta ) = \text{id}_{N^\bullet }$ by Categories, Definition 4.43.5 we see immediately that we have $(1 \otimes \epsilon _ n) \circ (\eta _ n \otimes 1) = \text{id}_{M^ n}$ and $(\epsilon _ n \otimes 1) \circ (1 \otimes \eta _ n) = \text{id}_{N^{-n}}$ which proves (3). By Lemma 15.72.1 we have (2). Since the sum $\eta = \sum \eta _ n$ is finite, we get (1). Since $\eta = \sum \eta _ n$ is a map of complexes $R \to \text{Tot}(M^\bullet \otimes _ R N^\bullet )$ we see that

\[ (d_ M^{-n - 1} \otimes 1) \circ \eta _{-n - 1} + (-1)^ n (1 \otimes d_ N^{-n}) \circ \eta _{-n} = 0 \]

by our choice of signs for the differential on $\text{Tot}(M^\bullet \otimes _ R N^\bullet )$. Unwinding definitions, this proves (4). To see the final statement of the lemma one reads the above backwards.
$\square$

Given complexes $K^\bullet $, $M^\bullet $ we let $\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , K^\bullet )$ be the complex with terms

\[ \mathop{\mathrm{Hom}}\nolimits ^ n(M^\bullet , K^\bullet ) = \prod \nolimits _{n = p + q} \mathop{\mathrm{Hom}}\nolimits _ R(M^{-q}, K^ p) \]

and differential given by the rule

\[ d(f) = d_ K \circ f - (-1)^ n f \circ d_ M \]

The choice above is such that if $M^\bullet $ has a left dual $N^\bullet $ as in Lemma 15.72.2, then we have a canonical isomorphism

\[ \text{Tot}(K^\bullet \otimes _ R N^\bullet ) \longrightarrow \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , K^\bullet ) \]

defined without the intervention of signs sending the summand $K^ p \otimes _ R N^ q$ to the summand $\mathop{\mathrm{Hom}}\nolimits _ R(M^{-q}, K^ p)$ via $N^ q = \mathop{\mathrm{Hom}}\nolimits _ R(M^{-q}, R)$ and the canonical map $K^ p \otimes _ R \mathop{\mathrm{Hom}}\nolimits _ R(M^{-q}, R) \to \mathop{\mathrm{Hom}}\nolimits _ R(M^{-q}, K^ p)$.

There is a composition

\[ \text{Tot}( \mathop{\mathrm{Hom}}\nolimits ^\bullet (L^\bullet , K^\bullet ) \otimes _ R \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , L^\bullet )) \longrightarrow \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , K^\bullet ) \]

defined without the intervention of signs, see Lemma 15.71.3.

There is a canonical isomorphism

\[ \mathop{\mathrm{Hom}}\nolimits ^\bullet (K^\bullet , \mathop{\mathrm{Hom}}\nolimits ^\bullet (L^\bullet , M^\bullet )) = \mathop{\mathrm{Hom}}\nolimits ^\bullet (\text{Tot}(K^\bullet \otimes _ R L^\bullet ), M^\bullet ) \]

defined without the intervention of signs, see Lemma 15.71.1.

There is a canonical map

\[ \text{Tot}(K^\bullet \otimes _ R \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , L^\bullet )) \longrightarrow \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , \text{Tot}(K^\bullet \otimes _ R L^\bullet )) \]

defined without the intervention of signs, see Lemma 15.71.4.

There is a canonical map

\[ K^\bullet \longrightarrow \mathop{\mathrm{Hom}}\nolimits ^\bullet (L^\bullet , \text{Tot}(K^\bullet \otimes _ R L^\bullet )) \]

defined without the intervention of signs, see Lemma 15.71.5.

By Lemma 15.71.6 is a canonical map

\[ \text{Tot}(\mathop{\mathrm{Hom}}\nolimits ^\bullet (L^\bullet , M^\bullet ) \otimes _ R K^\bullet ) \longrightarrow \mathop{\mathrm{Hom}}\nolimits ^\bullet (\mathop{\mathrm{Hom}}\nolimits ^\bullet (K^\bullet , L^\bullet ), M^\bullet ) \]

which uses a sign $(-1)^{r + qr}$ on the module $\mathop{\mathrm{Hom}}\nolimits _ R(L^{-q}, M^ p) \otimes _ R K^ r$ whose reason is explained in Remark 15.71.7.

Taking $L^\bullet = M^\bullet $ and using $R \to \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , M^\bullet )$ the map from the previous item becomes the evaluation map

\[ ev : K^\bullet \longrightarrow \mathop{\mathrm{Hom}}\nolimits ^\bullet (\mathop{\mathrm{Hom}}\nolimits ^\bullet (K^\bullet , M^\bullet ), M^\bullet ) \]

It sends $x \in K^ n$ to the map which sends $f \in \mathop{\mathrm{Hom}}\nolimits ^ m(K^\bullet , M^\bullet )$ to $(-1)^{nm}f(x)$.

There is a canonical identification

\[ \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , K^\bullet )[a - b] \to \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet [b], K^\bullet [a]) \]

which uses signs. It is defined as the map whose corresponding shifted map

\[ \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , K^\bullet ) \to \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet [b], K^\bullet [a])[b - a] \]

uses the sign $(-1)^{nb}$ on the module $\mathop{\mathrm{Hom}}\nolimits _ R(M^{-q}, K^ p)$ with $p + q = n$. Namely, if $f \in \mathop{\mathrm{Hom}}\nolimits ^ n(M^\bullet , K^\bullet )$ then

\[ d(f) = d_ K \circ f - (-1)^ n f \circ d_ M \]

on the source, whereas on the target $f$ lies in $\left(\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet [b], K^\bullet [a])[b - a]\right)^ n = \mathop{\mathrm{Hom}}\nolimits ^{n + b -a}(M^\bullet [b], K^\bullet [a])$ and hence we get

\begin{align*} d(f) & = (-1)^{b - a} \left(d_{K[a]} \circ f - (-1)^{n + b - a} f \circ d_{M[b]}\right) \\ & = (-1)^{b - a} \left((-1)^ a d_ K \circ f - (-1)^{n + b - a} f \circ (-1)^ b d_ M \right) \\ & = (-1)^ b d_ K \circ f - (-1)^{n + b} f \circ d_ M \end{align*}

and one sees that the chosen sign of $(-1)^{nb}$ in degree $n$ produces a map of complexes for these differentials.

## Comments (2)

Comment #6986 by David Roberts on

Comment #7219 by Johan on