I'm trying to understand K\"ahler differentials for CDGAs (commutative differential graded algebras). A few minor things have been confusing me all afternoon. Pointers and references are welcome.

Let $A$ be a CDGA over a field of characteristic zero with differential of degree $1$ and with no cohomology in positive degrees. (If we made the differential of degree $-1$, then we might call these connective CDGAs), let $M$ be a left dg-module over $A$, and let $I=ker(A \otimes A \rightarrow A)$. Then I'd like to verify that $\delta:A \rightarrow I/I^{2}, a \mapsto 1 \otimes a - a \otimes 1$ is a derivation, just like for classical (not dg) commutative algebras.

For this to work, I seem to be forced to take the following convention for what I mean by derivation: $\delta(a \cdot b)=b \cdot \delta(a)+a\cdot \delta(b)$. On the other hand, some informal references I read give $\delta(a \cdot b)=\delta(a)\cdot b+a\cdot \delta(b)$, and use the convention that an action on the right $m \cdot x$ is defined by $(-1)^{|m||x|}x \cdot m$, where $x,m$ are homogeneous. Thus altogether we would require $\delta(a \cdot b)=(-1)^{|a||b|+|b|}b \cdot \delta(a)+a\cdot \delta(b)$ for homogeneous $a,b$. But this seems to mess up the signs to make the given $\delta$ a derivation.

So the question is: what is the right notion of derivation $\delta: A \rightarrow M$ of degree $0$, and more generally of degree $k$?

My apologies for the triviality of the question.

  • $\begingroup$ Your informal references maybe work with $A$-bimodules which are symmetric $A$-modules, in that case both notions coincide. In general, your first definition is OK if $|\delta|=0$, more generally the derivation condition is $\delta(a\cdot b)=\delta(a)\cdot b+(-1)^{|\delta||a|}a\cdot\delta (b)$. Just the usual convention together with Koszul's sign convention. $\endgroup$ May 2, 2012 at 20:45
  • $\begingroup$ I meant symmetric $A$-bimodules $\endgroup$ May 2, 2012 at 21:10
  • $\begingroup$ Thank you. I agree that $\delta(a \cdot b)=\delta(a)\cdot b + (-1)^{|\delta||a|}a \cdot \delta(b)$ is what you expect. However, $A=M$ is not in general a symmetric $A$-bimodule, unless symmetric for dg bimodules means graded symmetric. In which case that is what the reference seems to assume: $m \cdot x=(-1)^{|x||m|}x \cdot m$. But I don't think that works to make $a \mapsto 1 \otimes a - a \otimes 1$ a degree zero derivation. $\endgroup$
    – dhagbert
    May 2, 2012 at 21:41
  • $\begingroup$ It does work. Bear in mind that $I/I^2$ is getting its $A$-module structure from an $(A\otimes A)/I$-module structure (and the ring isomorphism $A=(A\otimes A)/I$). $\endgroup$ May 2, 2012 at 22:15
  • $\begingroup$ And when you view an ideal as both a left module and a right module, these module structures are related in just the way (with signs) that we are talking about. $\endgroup$ May 2, 2012 at 22:15

2 Answers 2


Yes, left modules for a graded-commutative ring can be routinely converted to right modules by defining $ma=(-1)^{|m||a|}am$, and then the "right" definition of degree zero derivation is


More generally


There really isn't a sign problem if you follow the Koszul rule faithfully.

Note that $ab\otimes 1-1\otimes ab=a(b\otimes 1-1\otimes b)+(a\otimes 1-1\otimes a)b$.

  • $\begingroup$ Thanks. It turns out I wasn't being consistent about the degree of $\delta$. Just a note. I think that your last equation is only true ${\rm mod} I^{2}$, which is what we need anyway. We use that we can turn right modules into left, we have $(a \otimes 1 - 1 \otimes a)b=(-1)^{|a|)|b|$, but the naive thing $(a \otimes 1 - 1\otimes a)b=a \otimes b-1\otimes ab$, which would give your equation, is not I think applicable for a derivation into a left module. Please point out if I'm confused, since the naive thing would be very useful. $\endgroup$
    – dhagbert
    May 3, 2012 at 9:46
  • $\begingroup$ Yes. The last equation is true with the obvious left and right $A$-module structure on $A\otimes A$, so with the obvious left and right $A$-module structure on $I$. By the time you get to $I/I^2$ the left and right $A$-module structures are "the same", i.e. related according to that convention we discussed. $\endgroup$ May 3, 2012 at 10:26

There are two steps to working smoothly with sign conventions. The first is to do as much as possible categorically, where all choices of sign conventions are relegated to single black box. The second step is only to unpack the black box when necessary, usually at the very end of a calculation.

Let $\mathcal C$ be a symmetric monoidal category. I denote the monoidal structure by $\otimes$, the monoidal unit by $\mathbf 1$, and the symmetry by $\operatorname{flip}$. I assume moreover that $\mathcal C$ is enriched in abelian groups (maybe because it is enriched in $k$-modules for some commutative ring $k$). I assume also that $\mathcal C$ has kernels.

A commutative algebra in $\mathcal C$ is an object $A\in\mathcal C$ along with maps $m: A \otimes A \to A$ and $1: \mathbf 1 \to A$, such that: (commutativity) $m = m\circ \operatorname{flip}$, (associativity) $m\circ (m \otimes \operatorname{id}) = m \circ (\operatorname{id}\otimes m)$, and (units) $m \circ (1 \otimes \operatorname{id}) = \operatorname{id}$. A (left) module for $A$ is an object $M \in \mathcal C$ and a map $m : A\otimes M \to M$ such that $m\circ (m \otimes \operatorname{id}) = m \circ (\operatorname{id}\otimes m)$. Note that every left module is also a right module and in fact a bimodule by composing judiciously with $\operatorname{flip}$.

A morphism $\delta : A \to M$ is a derivation if $\delta \circ m = m\circ (\delta \otimes \operatorname{id}) + m\circ (\operatorname{id}\otimes \delta)$. Then the set $\operatorname{der}(A,M)$ of derivations is naturally an abelian subgroup of $\hom(A,B)$.

Aside: Suppose that $(\mathcal C,\otimes)$ is closed in the sense that for any object $X$, the functor $\otimes X : \mathcal C \to \mathcal C$ has a right adjoint $\underline{\hom}(X,-)$. It follows that $\underline{\hom}(-,-)$ is contravariant in the first component. Then in particular there is an object $\underline{\hom}(A,M) \in \mathcal C$. It has a subobject $\underline{\operatorname{der}}(A,M) = \ker\bigl( \underline{\hom}(A,M) \rightrightarrows \underline{\hom}(A\otimes A,M)\bigr)$. One of these maps corresponds to $m : A\otimes A \to A$ in the first component. The second corresponds to $m : A\otimes M \to M$, and uses the adjunctions a few times in its definition.

Anyway, I believe that your claim that $I = \ker(m : A\otimes A \to A)$ comes equipped with a canonical derivation is satisfied in any symmetric monoidal linear category with kernels. At worst, you need to include some more adjectives like "abelian" or "presentable". But it is checkable without reaching inside to see how the objects are builts

Let me describe now a notation for doing such computations. For any category $\mathcal C$ and any object $A\in \mathcal C$, a generalized element $x\in A$ is a morphism $X \to A$ for some $X\in \mathcal C$. The generalized elements with domain $X$ are the $X$-points of $A$. The Yoneda lemma implies that an object is determined by its generalized elements. (In fact, in a sense $A$ is determined by its canonical $A$-point $\operatorname{id}:A \to A$.) If $\mathcal C$ is symmetric monoidal, global element of $A \in \mathcal C$ is a morphism $\mathbf 1 \to A$. Note that global elements do not determine the object. Generalized elements in a symmetric monoidal category enriched over abelian groups can be manipulated like regular elements of regular abelian groups. The requirement is that in any expression, there are no duplications or deletions: the expression should be homogeneous linear in each element.

For example, suppose that you have generalized elements $x: X \to A$ and $y : Y \to A$. Then $xy = m(x\otimes y)$ is a generalized element with domain $X\otimes Y$. On the other hand, $yx \in A$ has domain $Y \otimes X$. There is a unique canonical isomorphism $\operatorname{flip}:X\otimes Y \to Y \otimes X$. Then an equation like "$\forall x,y\in A$, $xy = yx$" has the interpretation: "for all pairs of generalized elements $x : X \to A$ and $y : Y \to A$, we have an equality of morphisms $m\circ (x\otimes y) = m\circ (y\otimes x)\circ \operatorname{flip} : X\otimes Y \to A$." With this convention, you will never see signs. For example, a morphism $\delta : A\to M$ is a derivation if for all generalized elements $a,b \in A$, we have $\delta(ab) = a\delta(b) + b\delta(a)$.

This is all well and good, but how do you unpack it? The category of graded vector spaces is a monoidal category using the "convolution product" — this is what you're used to. Its symmetric structure is determined by the requirement that $\operatorname{flip} : [1] \otimes [1] \to [1] \otimes [1]$ is minus the identity, where $[1]$ denotes the one-dimensional vector space in degree $1$ (or in degree $-1$, whichever is your convention); it and its inverse $\otimes$-generate the category of graded vector spaces. You can skeletalize your category to decide that every object is a direct sum of various $[n] = [1]\otimes \dots \otimes [1]$. With respect to this skeletalization, $[m+n] = [m]\otimes [n] \overset{\operatorname{flip}}\to [n]\otimes[m] = [n+m]$ is the map $(-1)^{mn}$, which is the usual Koszul rule. This category has a distinguished Lie algebra, which is the abelian Lie algebra on the object $[1]$. (By "abelian", I mean that the Lie bracket is $0$.) The category of dg vector spaces is the category of representations of this Lie algebra (the Lie algebra action records the differential). It is a symmetric monoidal category in the same way that the representation theory of any Lie algebra is a symmetric monoidal category. In particular, forgetting from a dg vector space to a graded vector space is faithful and preserves the symmetric monoidal structure. So a morphism in dgVect is a morphism in gVect that is equivariant for the Lie algebra action.

It's then important to remember that forgetting from a graded vector space to a regular vector space is monoidal, but does not preserve the symmetric structure. So when working with generalized elements in gVect, you can say that a derivation should satisfy $\delta(ab) = a\delta(b) + b\delta(a)$. But when working with elements of the underlying vector space, you must remember that there was an implicit $\operatorname{flip}$ when the $a$ and $b$ switched places; for that (worse) convention, we have $\delta(ab) = a\delta(b) + (-1)^{|a||b|}b\delta(a)$.

Finally, you asked about derivations of non-zero degree. As in the aside, suppose that $\mathcal C$ is closed. Then one always has $\hom(\mathbf 1,\underline{\hom}(A,M)) = \hom(A,M)$ as abelian groups, and also $\hom(\mathbf 1,\underline{\operatorname{der}}(A,M)) = \operatorname{der}(A,M)$. When $\mathcal C$ is the category of graded or dg vector spaces, the monoidal unit is $\mathbf 1 = [0] = $ the one-dimensional vector space in degree $0$. Up to a grading convention, a degree-$k$ differential is a morphism $[k] \to \underline{\operatorname{der}}(A,M)$. It is not hard to unpack this definition: it says that $\delta(ab) = (-1)^{|a|k}a\delta(b) + (-1)^{|b|(|a|+k)}b\delta(a)$. I think one of your formulas must have been about degree-odd derivations.

It is an unfortunate but common occurrence for writers to use worse conventions. For example, remember that the canonical isomorphism $[k] \otimes X \cong X\otimes [k]$ is not the identity on underlying vector spaces. In the conventions above, the adjunction implies that a map $[k] \to \underline{\operatorname{der}}(A,M)$ is a map $[k] \otimes A \to M$ satisfying some condition. It is the same data as a map $A[k] = A\otimes [k] \to M$, but on underlying vector spaces those two maps are different.

The moral of the story is that you should never cut-and-paste sign-full formulas from anyone else. Rather, pick a convention and stick to it, and as much as possible make no choices at all by using a categorical framework.


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