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.
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.
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 reader 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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