## 22.13 Hom complexes and differential graded modules

We urge the reader to skip this section.

Let $R$ be a ring and let $M^\bullet$ be a complex of $R$-modules. Consider the complex of $R$-modules

$E^\bullet = \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , M^\bullet )$

introduced in More on Algebra, Section 15.71. By More on Algebra, Lemma 15.71.3 there is a canonical composition law

$\text{Tot}(E^\bullet \otimes _ R E^\bullet ) \to E^\bullet$

which is a map of complexes. Thus we see that $E^\bullet$ with this multiplication is a differential graded $R$-algebra which we will denote $(E, \text{d})$. Moreover, viewing $M^\bullet$ as $\mathop{\mathrm{Hom}}\nolimits ^\bullet (R, M^\bullet )$ we see that composition defines a multiplication

$\text{Tot}(E^\bullet \otimes _ R M^\bullet ) \to M^\bullet$

which turns $M^\bullet$ into a left differential graded $E$-module which we will denote $M$.

Lemma 22.13.1. In the situation above, let $A$ be a differential graded $R$-algebra. To give a left $A$-module structure on $M$ is the same thing as giving a homomorphism $A \to E$ of differential graded $R$-algebras.

Proof. Proof omitted. Observe that no signs intervene in this correspondence. $\square$

We continue with the discussion above and we assume given another complex $N^\bullet$ of $R$-modules. Consider the complex of $R$-modules $\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , N^\bullet )$ introduced in More on Algebra, Section 15.71. As above we see that composition

$\text{Tot}(\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , N^\bullet ) \otimes _ R E^\bullet ) \to \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , N^\bullet )$

defines a multiplication which turns $\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , N^\bullet )$ into a right differential graded $E$-module. Using Lemma 22.13.1 we conclude that given a left differential graded $A$-module $M$ and a complex of $R$-modules $N^\bullet$ there is a canonical right differential graded $A$-module whose underlying complex of $R$-modules is $\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , N^\bullet )$ and where multiplication

$\mathop{\mathrm{Hom}}\nolimits ^ n(M^\bullet , N^\bullet ) \times A^ m \longrightarrow \mathop{\mathrm{Hom}}\nolimits ^{n + m}(M^\bullet , N^\bullet )$

sends $f = (f_{p, q})_{p + q = n}$ with $f_{p, q} \in \mathop{\mathrm{Hom}}\nolimits (M^{-q}, N^ p)$ and $a \in A^ m$ to the element $f \cdot a = (f_{p, q} \circ a)$ where $f_{p, q} \circ a$ is the map

$M^{-q - m} \xrightarrow {a} M^{-q} \xrightarrow {f_{p, q}} N^ p, \quad x \longmapsto f_{p, q}(ax)$

without the intervention of signs. Let us use the notation $\mathop{\mathrm{Hom}}\nolimits (M, N^\bullet )$ to denote this right differential graded $A$-module.

Lemma 22.13.2. Let $R$ be a ring. Let $(A, \text{d})$ be a differential graded $R$-algebra. Let $M'$ be a right differential graded $A$-module and let $M$ be a left differential graded $A$-module. Let $N^\bullet$ be a complex of $R$-modules. Then we have

$\mathop{\mathrm{Hom}}\nolimits _{\text{Mod}_{(A, d)}}(M', \mathop{\mathrm{Hom}}\nolimits (M, N^\bullet )) = \mathop{\mathrm{Hom}}\nolimits _{\text{Comp}(R)}(M' \otimes _ A M, N^\bullet )$

where $M \otimes _ A M$ is viewed as a complex of $R$-modules as in Section 22.12.

Proof. Let us show that both sides correspond to graded $A$-bilinear maps

$M' \times M \longrightarrow N^\bullet$

compatible with differentials. We have seen this is true for the right hand side in Section 22.12. Given an element $g$ of the left hand side, the equality of More on Algebra, Lemma 15.71.1 determines a map of complexes of $R$-modules $g' : \text{Tot}(M' \otimes _ R M) \to N^\bullet$. In other words, we obtain a graded $R$-bilinear map $g'' : M' \times M \to N^\bullet$ compatible with differentials. The $A$-linearity of $g$ translates immediately into $A$-bilinarity of $g''$. $\square$

Let $R$, $M^\bullet$, $E^\bullet$, $E$, and $M$ be as above. However, now suppose given a differential graded $R$-algebra $A$ and a right differential graded $A$-module structure on $M$. Then we can consider the map

$\text{Tot}(A^\bullet \otimes _ R M^\bullet ) \xrightarrow {\psi } \text{Tot}(A^\bullet \otimes _ R M^\bullet ) \to M^\bullet$

where the first arrow is the commutativity constraint on the differential graded category of complexes of $R$-modules. This corresponds to a map

$\tau : A^\bullet \longrightarrow E^\bullet$

of complexes of $R$-modules. Recall that $E^ n = \prod _{p + q = n} \mathop{\mathrm{Hom}}\nolimits _ R(M^{-q}, M^ p)$ and write $\tau (a) = (\tau _{p, q}(a))_{p + q = n}$ for $a \in A^ n$. Then we see

$\tau _{p, q}(a) : M^{-q} \longrightarrow M^ p,\quad x \longmapsto (-1)^{\deg (a)\deg (x)}x a = (-1)^{-nq}xa$

This is not compatible with the product on $A$ as the readed should expect from the discussion in Section 22.11. Namely, we have

$\tau (a a') = (-1)^{\deg (a)\deg (a')}\tau (a') \tau (a)$

We conclude the following lemma is true

Lemma 22.13.3. In the situation above, let $A$ be a differential graded $R$-algebra. To give a right $A$-module structure on $M$ is the same thing as giving a homomorphism $\tau : A \to E^{opp}$ of differential graded $R$-algebras.

Proof. See discussion above and note that the construction of $\tau$ from the multiplication map $M^ n \times A^ m \to M^{n + m}$ uses signs. $\square$

Let $R$, $M^\bullet$, $E^\bullet$, $E$, $A$ and $M$ be as above and let a right differential graded $A$-module structure on $M$ be given as in the lemma. In this case there is a canonical left differential graded $A$-module whose underlying complex of $R$-modules is $\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , N^\bullet )$. Namely, for multiplication we can use

\begin{align*} \text{Tot}(A^\bullet \otimes _ R \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , N^\bullet )) & \xrightarrow {\psi } \text{Tot}(\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , N^\bullet ) \otimes _ R A^\bullet ) \\ & \xrightarrow {\tau } \text{Tot}(\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , N^\bullet ) \otimes _ R \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , M^\bullet )) \\ & \to \text{Tot}(\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , N^\bullet ) \end{align*}

The first arrow uses the commutativity constraint on the category of complexes of $R$-modules, the second arrow is described above, and the third arrow is the composition law for the Hom complex. Each map is a map of complexes, hence the result is a map of complexes. In fact, this construction turns $\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , N^\bullet )$ into a left differential graded $A$-module (associativity of the multiplication can be shown using the symmetric monoidal structure or by a direct calculation using the formulae below). Let us explicate the multiplication

$A^ n \times \mathop{\mathrm{Hom}}\nolimits ^ m(M^\bullet , N^\bullet ) \longrightarrow \mathop{\mathrm{Hom}}\nolimits ^{n + m}(M^\bullet , N^\bullet )$

It sends $a \in A^ n$ and $f = (f_{p, q})_{p + q = m}$ with $f_{p, q} \in \mathop{\mathrm{Hom}}\nolimits (M^{-q}, N^ p)$ to the element $a \cdot f$ with constituents

$(-1)^{nm}f_{p, q} \circ \tau _{-q, q + n}(a) = (-1)^{nm - n(q + n)}f_{p, q} \circ a = (-1)^{np + n} f_{p, q} \circ a$

in $\mathop{\mathrm{Hom}}\nolimits _ R(M^{-q - n}, N^ p)$ where $f_{p, q} \circ a$ is the map

$M^{-q - n} \xrightarrow {a} M^{-q} \xrightarrow {f_{p, q}} N^ p,\quad x \longmapsto f_{p, q}(xa)$

Here a sign of $(-1)^{np + n}$ does intervene. Let us use the notation $\mathop{\mathrm{Hom}}\nolimits (M, N^\bullet )$ to denote this left differential graded $A$-module.

Lemma 22.13.4. Let $R$ be a ring. Let $(A, \text{d})$ be a differential graded $R$-algebra. Let $M$ be a right differential graded $A$-module and let $M'$ be a left differential graded $A$-module. Let $N^\bullet$ be a complex of $R$-modules. Then we have

$\mathop{\mathrm{Hom}}\nolimits _{\text{left diff graded }A\text{-modules}}(M', \mathop{\mathrm{Hom}}\nolimits (M, N^\bullet )) = \mathop{\mathrm{Hom}}\nolimits _{\text{Comp}(R)}(M \otimes _ A M', N^\bullet )$

where $M \otimes _ A M'$ is viewed as a complex of $R$-modules as in Section 22.12.

Proof. Let us show that both sides correspond to graded $A$-bilinear maps

$M \times M' \longrightarrow N^\bullet$

compatible with differentials. We have seen this is true for the right hand side in Section 22.12. Given an element $g$ of the left hand side, the equality of More on Algebra, Lemma 15.71.1 determines a map of complexes $g' : \text{Tot}(M' \otimes _ R M) \to N^\bullet$. We precompose with the commutativity constraint to get

$\text{Tot}(M \otimes _ R M') \xrightarrow {\psi } \text{Tot}(M' \otimes _ R M) \xrightarrow {g'} N^\bullet$

which corresponds to a graded $R$-bilinear map $g'' : M \times M' \to N^\bullet$ compatible with differentials. The $A$-linearity of $g$ translates immediately into $A$-bilinarity of $g''$. Namely, say $x \in M^ e$ and $x' \in (M')^{e'}$ and $a \in A^ n$. Then on the one hand we have

\begin{align*} g''(x, ax') & = (-1)^{e(n + e')} g'(ax' \otimes x) \\ & = (-1)^{e(n + e')} g(ax')(x) \\ & = (-1)^{e(n + e')} (a \cdot g(x'))(x) \\ & = (-1)^{e(n + e') + n(n + e + e') + n} g(x')(xa) \end{align*}

and on the other hand we have

$g''(xa, x') = (-1)^{(e + n)e'} g'(x' \otimes xa) = (-1)^{(e + n)e'} g(x')(xa)$

which is the same thing by a trivial mod $2$ calculation of the exponents. $\square$

Remark 22.13.5. Let $R$ be a ring. Let $A$ be a differential graded $R$-algebra. Let $M$ be a left differential graded $A$-module. Let $N^\bullet$ be a complex of $R$-modules. The constructions above produce a right differential graded $A$-module $\mathop{\mathrm{Hom}}\nolimits (M, N^\bullet )$ and then a leftt differential graded $A$-module $\mathop{\mathrm{Hom}}\nolimits (\mathop{\mathrm{Hom}}\nolimits (M, N^\bullet ), N^\bullet )$. We claim there is an evaluation map

$ev : M \longrightarrow \mathop{\mathrm{Hom}}\nolimits (\mathop{\mathrm{Hom}}\nolimits (M, N^\bullet ), N^\bullet )$

in the category of left differential graded $A$-modules. To define it, by Lemma 22.13.2 it suffices to construct an $A$-bilinear pairing

$\mathop{\mathrm{Hom}}\nolimits (M, N^\bullet ) \times M \longrightarrow N^\bullet$

compatible with grading and differentials. For this we take

$(f, x) \longmapsto f(x)$

We leave it to the reader to verify this is compatible with grading, differentials, and $A$-bilinear. The map $ev$ on underlying complexes of $R$-modules is More on Algebra, Item (17).

Remark 22.13.6. Let $R$ be a ring. Let $A$ be a differential graded $R$-algebra. Let $M$ be a right differential graded $A$-module. Let $N^\bullet$ be a complex of $R$-modules. The constructions above produce a left differential graded $A$-module $\mathop{\mathrm{Hom}}\nolimits (M, N^\bullet )$ and then a right differential graded $A$-module $\mathop{\mathrm{Hom}}\nolimits (\mathop{\mathrm{Hom}}\nolimits (M, N^\bullet ), N^\bullet )$. We claim there is an evaluation map

$ev : M \longrightarrow \mathop{\mathrm{Hom}}\nolimits (\mathop{\mathrm{Hom}}\nolimits (M, N^\bullet ), N^\bullet )$

in the category of right differential graded $A$-modules. To define it, by Lemma 22.13.2 it suffices to construct an $A$-bilinear pairing

$M \times \mathop{\mathrm{Hom}}\nolimits (M, N^\bullet ) \longrightarrow N^\bullet$

compatible with grading and differentials. For this we take

$(x, f) \longmapsto (-1)^{\deg (x)\deg (f)}f(x)$

We leave it to the reader to verify this is compatible with grading, differentials, and $A$-bilinear. The map $ev$ on underlying complexes of $R$-modules is More on Algebra, Item (17).

Remark 22.13.7. Let $R$ be a ring. Let $A$ be a differential graded $R$-algebra. Let $M^\bullet$ and $N^\bullet$ be complexes of $R$-modules. Let $k \in \mathbf{Z}$ and consider the isomorphism

$\mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet , N^\bullet )[-k] \longrightarrow \mathop{\mathrm{Hom}}\nolimits ^\bullet (M^\bullet [k], N^\bullet )$

of complexes of $R$-modules defined in More on Algebra, Item (18). If $M^\bullet$ has the structure of a left, resp. right differential graded $A$-module, then this is a map of right, resp. left differential graded $A$-modules (with the module structures as defined in this section). We omit the verification; we warn the reader that the $A$-module structure on the shift of a left graded $A$-module is defined using a sign, see Definition 22.11.3.

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