The Stacks project

Example 22.19.8 (Differential graded category of differential graded modules). Let $(A, \text{d})$ be a differential graded algebra over a ring $R$. We will construct a differential graded category $\text{Mod}^{dg}_{(A, \text{d})}$ over $R$ whose category of complexes is $\text{Mod}_{(A, \text{d})}$ and whose homotopy category is $K(\text{Mod}_{(A, \text{d})})$. As objects of $\text{Mod}^{dg}_{(A, \text{d})}$ we take the differential graded $A$-modules. Given differential graded $A$-modules $L$ and $M$ we set

\[ \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(A, \text{d})}}(L, M) = \mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{gr}_ A}(L, M) = \bigoplus \mathop{\mathrm{Hom}}\nolimits ^ n(L, M) \]

as a graded $R$-module where the right hand side is defined as in Example 22.18.6. In other words, the $n$th graded piece $\mathop{\mathrm{Hom}}\nolimits ^ n(L, M)$ is the $R$-module of right $A$-module maps homogeneous of degree $n$. For an element $f \in \mathop{\mathrm{Hom}}\nolimits ^ n(L, M)$ we set

\[ \text{d}(f) = \text{d}_ M \circ f - (-1)^ n f \circ \text{d}_ L \]

To make sense of this we think of $\text{d}_ M$ and $\text{d}_ L$ as graded $R$-module maps and we use composition of graded $R$-module maps. It is clear that $\text{d}(f)$ is homogeneous of degree $n + 1$ as a graded $R$-module map, and it is linear because

\begin{align*} \text{d}(f)(xa) & = \text{d}_ M(f(x) a) - (-1)^ n f (\text{d}_ L(xa)) \\ & = \text{d}_ M(f(x)) a + (-1)^{\deg (x) + n} f(x) \text{d}(a) - (-1)^ n f(\text{d}_ L(x)) a - (-1)^{n + \deg (x)} f(x) \text{d}(a) \\ & = \text{d}(f)(x) a \end{align*}

as desired (observe that this calculation would not work without the sign in the definition of our differential on $\mathop{\mathrm{Hom}}\nolimits $). Similar formulae to those of Example 22.19.6 hold for the differential of $f$ in terms of components. The reader checks (in the same way as in Example 22.19.6) that

  1. $\text{d}$ has square zero,

  2. an element $f$ in $\mathop{\mathrm{Hom}}\nolimits ^ n(L, M)$ has $\text{d}(f) = 0$ if and only if $f : L \to M[n]$ is a homomorphism of differential graded $A$-modules,

  3. in particular, the category of complexes of $\text{Mod}^{dg}_{(A, \text{d})}$ is $\text{Mod}_{(A, \text{d})}$,

  4. the homomorphism defined by $f$ as in (2) is homotopy equivalent to zero if and only if $f = \text{d}(g)$ for some $g \in \mathop{\mathrm{Hom}}\nolimits ^{n - 1}(L, M)$.

  5. in particular, we obtain a canonical isomorphism

    \[ \mathop{\mathrm{Hom}}\nolimits _{K(\text{Mod}_{(A, \text{d})})}(L, M) \longrightarrow H^0(\mathop{\mathrm{Hom}}\nolimits _{\text{Mod}^{dg}_{(A, \text{d})}}(L, M)) \]

    and the homotopy category of $\text{Mod}^{dg}_{(A, \text{d})}$ is $K(\text{Mod}_{(A, \text{d})})$.

Given differential graded $A$-modules $K$, $L$, $M$ we define composition

\[ \mathop{\mathrm{Hom}}\nolimits ^ m(L, M) \times \mathop{\mathrm{Hom}}\nolimits ^ n(K, L) \longrightarrow \mathop{\mathrm{Hom}}\nolimits ^{n + m}(K, M) \]

by composition of homogeneous right $A$-module maps $(g, f) \mapsto g \circ f$. This defines a map of differential graded modules as in Definition 22.19.1 because

\begin{align*} \text{d}(g \circ f) & = \text{d}_ M \circ g \circ f - (-1)^{n + m} g \circ f \circ \text{d}_ K \\ & = \left(\text{d}_ M \circ g - (-1)^ m g \circ \text{d}_ L\right) \circ f + (-1)^ m g \circ \left(\text{d}_ L \circ f - (-1)^ n f \circ \text{d}_ K\right) \\ & = \text{d}(g) \circ f + (-1)^ m g \circ \text{d}(f) \end{align*}

as desired.


Comments (1)

Comment #287 by arp on

Typo: When defining elements of , in the line the on the LHS should be replaced with an .


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