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22.11 Left modules

Everything we have said sofar has an analogue in the setting of left differential graded modules, except that one has to take care with some sign rules.

Let $(A, \text{d})$ be a differential graded $R$-algebra. Exactly analogous to right modules, we define a left differential graded $A$-module $M$ as a left $A$-module $M$ which has a grading $M = \bigoplus M^ n$ and a differential $\text{d}$, such that $A^ n M^ m \subset M^{n + m}$, such that $\text{d}(M^ n) \subset M^{n + 1}$, and such that

\[ \text{d}(am) = \text{d}(a) m + (-1)^{\deg (a)}a \text{d}(m) \]

for homogeneous elements $a \in A$ and $m \in M$. As before this Leibniz rule exactly signifies that the multiplication defines a map of complexes

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

Here $A^\bullet $ and $M^\bullet $ denote the complexes of $R$-modules underlying $A$ and $M$.

Definition 22.11.1. Let $R$ be a ring. Let $(A, \text{d})$ be a differential graded algebra over $R$. The opposite differential graded algebra is the differential graded algebra $(A^{opp}, \text{d})$ over $R$ where $A^{opp} = A$ as a graded $R$-module, $\text{d} = \text{d}$, and multiplication is given by

\[ a \cdot _{opp} b = (-1)^{\deg (a)\deg (b)} b a \]

for homogeneous elements $a, b \in A$.

This makes sense because

\begin{align*} \text{d}(a \cdot _{opp} b) & = (-1)^{\deg (a)\deg (b)} \text{d}(b a) \\ & = (-1)^{\deg (a)\deg (b)} \text{d}(b) a + (-1)^{\deg (a)\deg (b) + \deg (b)}b\text{d}(a) \\ & = (-1)^{\deg (a)}a \cdot _{opp} \text{d}(b) + \text{d}(a) \cdot _{opp} b \end{align*}

as desired. In terms of underlying complexes of $R$-modules this means that the diagram

\[ \xymatrix{ \text{Tot}(A^\bullet \otimes _ R A^\bullet ) \ar[rrr]_-{\text{multiplication of }A^{opp}} \ar[d]_{\text{commutativity constraint}} & & & A^\bullet \ar[d]^{\text{id}} \\ \text{Tot}(A^\bullet \otimes _ R A^\bullet ) \ar[rrr]^-{\text{multiplication of }A} & & & A^\bullet } \]

commutes. Here the commutativity constraint on the symmetric monoidal category of complexes of $R$-modules is given in More on Algebra, Section 15.72.

Let $(A, \text{d})$ be a differential graded algebra over $R$. Let $M$ be a left differential graded $A$-module. We will denote $M^{opp}$ the module $M$ viewed as a right $A^{opp}$-module with multiplication $\cdot _{opp}$ defined by the rule

\[ m \cdot _{opp} a = (-1)^{\deg (a)\deg (m)} a m \]

for $a$ and $m$ homogeneous. This is compatible with differentials because we could have used the diagram

\[ \xymatrix{ \text{Tot}(M^\bullet \otimes _ R A^\bullet ) \ar[rrr]_-{\text{multiplication on }M^{opp}} \ar[d]_{\text{commutativity constraint}} & & & M^\bullet \ar[d]^{\text{id}} \\ \text{Tot}(A^\bullet \otimes _ R M^\bullet ) \ar[rrr]^-{\text{multiplication on }M} & & & M^\bullet } \]

to define the multiplication $\cdot _{opp}$ on $M^{opp}$. To see that it is an associative multiplication we compute for homogeneous elements $a, b \in A$ and $m \in M$ that

\begin{align*} m \cdot _{opp} (a \cdot _{opp} b) & = (-1)^{\deg (a)\deg (b)} m \cdot _{opp} (ba) \\ & = (-1)^{\deg (a)\deg (b) + \deg (ab)\deg (m)} bam \\ & = (-1)^{\deg (a)\deg (b) + \deg (ab)\deg (m) + \deg (b)\deg (am)} (am) \cdot _{opp} b \\ & = (-1)^{\deg (a)\deg (b) + \deg (ab)\deg (m) + \deg (b)\deg (am) + \deg (a)\deg (m)} (m \cdot _{opp} a) \cdot _{opp} b \\ & = (m \cdot _{opp} a) \cdot _{opp} b \end{align*}

Of course, we could have been shown this using the compatibility between the associativity and commutativity constraint on the symmetric monoidal category of complexes of $R$-modules as well.

Lemma 22.11.2. Let $(A, \text{d})$ be a differential graded $R$-algebra. The functor $M \mapsto M^{opp}$ from the category of left differential graded $A$-modules to the category of right differential graded $A^{opp}$-modules is an equivalence.

Proof. Omitted. $\square$

Mext, we come to shifts. Let $(A, \text{d})$ be a differential graded algebra. Let $M$ be a left differential graded $A$-module whose underlying complex of $R$-modules is denoted $M^\bullet $. For any $k \in \mathbf{Z}$ we define the $k$-shifted module $M[k]$ as follows

  1. the underlying complex of $R$-modules of $M[k]$ is $M^\bullet [k]$

  2. as $A$-module the multiplication

    \[ A^ n \times (M[k])^ m \longrightarrow (M[k])^{n + m} \]

    is equal to $(-1)^{nk}$ times the given multiplication $A^ n \times M^{m + k} \to M^{n + m + k}$.

Let us check that with this choice the Leibniz rule is satisfied. Let $a \in A^ n$ and $x \in M[k]^ m = M^{m + k}$ and denoting $\cdot _{M[k]}$ the product in $M[k]$ then we see

\begin{align*} \text{d}_{M[k]}(a \cdot _{M[k]} x) & = (-1)^{k + nk} \text{d}_ M(ax) \\ & = (-1)^{k + nk} \text{d}(a) x + (-1)^{k + nk + n} a \text{d}_ M(x) \\ & = \text{d}(a) \cdot _{M[k]} x + (-1)^{nk + n} a \text{d}_{M[k]}(x) \\ & = \text{d}(a) \cdot _{M[k]} x + (-1)^ n a \cdot _{M[k]} \text{d}_{M[k]}(x) \end{align*}

This is what we want as $a$ has degree $n$ as a homogeneous element of $A$. We also observe that with these choices we may think of the multiplication map as the map of complexes

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

where the first arrow is More on Algebra, Section 15.72 (7) which in this case involves exactly the sign we chose above. (In fact, we could have deduced that the Liebniz rule holds from this observation.)

With the rule above we have canonical identifications

\[ (M[k])^{opp} = M^{opp}[k] \]

of right differential graded $A^{opp}$-modules defined without the intervention of signs, in other words, the equivalence of Lemma 22.11.2 is compatible with shift functors.

Our choice above necessitates the following definition.

Definition 22.11.3. Let $R$ be a ring. Let $A$ be a $\mathbf{Z}$-graded $R$-algebra.

  1. Given a right graded $A$-module $M$ we define the $k$th shifted $A$-module $M[k]$ as the same as a right $A$-module but with grading $(M[k])^ n = M^{n + k}$.

  2. Given a left graded $A$-module $M$ we define the $k$th shifted $A$-module $M[k]$ as the module with grading $(M[k])^ n = M^{n + k}$ and multiplication $A^ n \times (M[k])^ m \to (M[k])^{n + m}$ equal to $(-1)^{nk}$ times the given multiplication $A^ n \times M^{m + k} \to M^{n + m + k}$.

Let $(A, \text{d})$ be a differential graded algebra. Let $f, g : M \to N$ be homomorphisms of left differential graded $A$-modules. A homotopy between $f$ and $g$ is a graded $A$-module map $h : M \to N[-1]$ (observe the shift!) such that

\[ f(x) - g(x) = \text{d}_ N(h(x)) + h(\text{d}_ M(x)) \]

for all $x \in M$. If a homotopy exists, then we say $f$ and $g$ are homotopic. Thus $h$ is compatible with the $A$-module structure (with the shifted one on $N$) and the grading (with shifted grading on $N$) but not with the differential. If $f = g$ and $h$ is a homotopy, then $h$ defines a morphism $h : M \to N[-1]$ of left differential graded $A$-modules.

With the rule above we find that $f, g : M \to N$ are homotopic if and only if the induced morphisms $f^{opp}, g^{opp} : M^{opp} \to N^{opp}$ are homotopic as right differential graded $A^{opp}$-module homomorphisms (with the same homotopy).

The homotopy category, cones, admissible short exact sequences, distinguished triangles are all defined in exactly the same manner as for right differential graded modules (and everything agrees on underlying complexes of $R$-modules with the constructions for complexes of $R$-modules). In this manner we obtain the analogue of Proposition 22.10.3 for left modules as well, or we can deduce it by working with right modules over the opposite algebra.

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