Given a commutative algebra $A$ smooth over a field $k$ of characteristic zero, the module of K\"ahler differentials $\Omega^{1}$ is projective of finite rank and so the sum of all wedge powers $\Omega^{\bullet}=\oplus_{p} \Omega^{p}$ is again projective of finite rank and is a commutative differential graded algebra with respect to the de Rham differential $d_{dR}$. Given a derivation $X$ of $A$, we can define a `contraction' operator on $\Omega^{\bullet}$ by requiring $\iota_{X}a=0$ for $a \in A$, $\iota_{X}(d_{dR}(a)$, and extending $\iota_{X}$ to all of $\Omega^{\bullet}$ by requiring it to be a derivation of degree $-1$.

Now if $\dots \rightarrow A_{1} \rightarrow A_{0}=A$ is a cdga with differential $d$ of degree $-1$, then the K\"ahler differentials $\Omega^{1}$ are a dg-module over $A$ with the (internal) differential also denoted $d$. Given suitable finiteness hypotheses on $A$, $\Omega^{1}$ is projective of finite rank. Then $\Omega^{\bullet}$ is also a dg-module over $A$, with the differential $d$ of degree $-1$ with respect to the internal grading and of degree $0$ with respect to the grading by $p$. Again, it has a de Rham differential $d_{dR}$ of degree $1$ with respect to $p$ and degree $0$ with respect to the internal grading.


Given a derivation $X$ of $A$ of some degree $|X|$, I would like to consider the contraction $\iota_{X}$ acting on $\Omega^{\bullet}$. It should have the same definition on $a \in A$ and $d_{dR}a \in \Omega^{1}$, and then I would like it to be uniquely extending by making it a derivation of some (bi?)degree. I can imagine the degree to be various things. Roughly, the formula should be something like

$$\iota_{X}(\alpha \wedge \beta)=\iota_{X}(\alpha) \wedge \beta + (-1)^{|X||\alpha|}\alpha \wedge \iota_{X}(\beta),$$

except I don't know if $|\alpha|$ should take into account only the $p$ in $\alpha \in \Omega^{p}$ or also the internal grading $i$ in the dg-module $\Omega^{p}$.

What is the natural thing to do? Probably Theo Johnson-Freyd will give some nice monoidal answer, but I'll also appreciate something very concrete.


1 Answer 1


I don't know whether to take your last line as an invitation or a slight, but I'll do my best with an answer. As you probably already know, I am (often uselessly) verbose.

I like to think of $\Omega^\bullet$ as a bigraded commutative algebra. The monoidal category of bigraded vector spaces can be equipped with a symmetric structure which is determined up to unique equivalence of symmetric monoidal categories by the request that for the generating lines $L_{(0,1)}$ and $L_{(1,0)}$ in bidegrees $(0,1)$ and $(1,0)$ respectively, the braiding $L \otimes L \to L\otimes L$ is minus the identity. This forces the line $L_{(0,1)} \otimes L_{(1,0)}$, along with any isomorph of it, to braid with itself by $+1$. More generally, a line in bidegree $(m,n)$ is forced to braid with itself for a factor of $(-1)^{m+n}$.

Note that I never said how different lines are supposed to braid with each other. That's because the question is not well-posed. If you skeletalize your category and fix a monoidal structure, then you do get an isomorphism between, say, $L_{(0,1)} \otimes L_{(1,0)} \cong L_{(1,1)} \cong L_{(1,0)} \otimes L_{(0,1)}$, and you can ask by what factor the braiding acts relative to this isomorphism. But from the point of view of category theory (i.e. without choosing extra data, like a skeletalization), there is no distinguished isomorphism between $L_{(0,1)} \otimes L_{(1,0)}$ and $L_{(1,0)} \otimes L_{(0,1)}$ except for the braiding. I will come back to this point in a moment.

This category-theoretic language is the cleanest way I know to set up the basic definitions. For example, given any bigraded commutative algebra $\Omega^\bullet$ with multiplication $m$, there is a bigraded vector Lie algebra $\operatorname{Der}(\Omega^\bullet)$ of derivations. It is the universal object with a map $\triangleright : \operatorname{Der}(\Omega^\bullet) \otimes \Omega^\bullet \to \Omega^\bullet$ such that for any $\phi : \Phi \to \operatorname{Der}(\Omega^\bullet)$ and $\alpha : A\to \Omega^\bullet$ and $\beta : B \to \Omega^\bullet$, the following is an equality of maps $\Phi \otimes A \otimes B \to \Omega^\bullet$: $$\triangleright \circ (\phi \otimes m) \circ (\alpha \otimes \beta) = m \circ ((\triangleright \circ(\phi \otimes \alpha))\otimes \beta) + m\circ(\alpha \otimes (\triangleright \circ(\phi \otimes \beta))) $$ You can clean up the notation a bit by inventing "generalized elements". What's important to mention is that the second summand on the RHS as written is not a map $\Phi \otimes A \otimes B \to \Omega^\bullet$, but rather a map $A \otimes \Phi \otimes B \to \Omega^\bullet$; to add them, I need them to have the same codomain, and so I insist that you identify $\Phi \otimes A \otimes B \cong A \otimes \Phi \otimes B$ by using the symmetry, and not any other way. Finally, you've distinguished a differential $d_{dR}$ on $\Omega^\bullet$ of bidegree $(0,1)$ say, by which I actually mean that you've distinguished a particular line $L_{(0,1)}$ and a map $d_{dR} : L_{(0,1)} \to \Omega^\bullet$; such a choice gives you a map $\iota: (L_{(0,1)})^{\otimes (-1)} \otimes \operatorname{Der}(\mathcal A) \to \operatorname{Der}(\Omega^\bullet)$, where $(L_{(0,1)})^{\otimes (-1)}$ comes equipped as the tensor inverse of $L_{(0,1)}$, and $\mathcal A$ is your original algebra.

Ok, so I said this was the cleanest way to set up the definitions, but by no means is it always the smoothest way to perform computations. Sometimes, generalized elements (which I have not defined, but I have used implicitly) are the best way to do some computations. But sometimes you do need to sit down and use homogeneous elements.

Talking about homogeneous elements is basically the same as skeletalizing the category of bigraded vector spaces, and thus requires you to make more choices than you would otherwise have to. There are two reasonable choices (and many unreasonable ones), and as I asserted above, the two choices give you equivalent categories, and hence equivalent theorems. But since there are two reasonable choices, you need to be careful that you make one and use it consistently, and don't trust formulas in other papers, because they probably used different choices.

The two choices are to declare, after skeletalizing, that the generators $L_{(0,1)}$ and $L_{(1,0)}$ braid past each other as $+1$ relative to the skeletalization, or as $-1$. The former of these implies that if $\alpha$ is of bidegree $(p,q)$ and $\beta$ is of bidegree $(m,n)$ then $\alpha$ and $\beta$ commute for a factor of $(-1)^{pm + qn}$. The latter choice would have them commute for a factor of $(-1)^{(p+q)(m+n)}$. This latter choice makes it easier to present the symmetric monoidal functor that collapses the gradings. The former choice makes it easier to see that a bigraded vector space is the same as a graded (graded vector space).

See, you never said your conventions even for how to think of $\Omega^\bullet$ as a commutative algebra. I recommend that you use the same convention there as for derivations.

Anyway, if your vector field $X$ on $\mathcal A$ was of degree $m$, then $\iota_X$ will be of bidegree $(m,-1)$. If $\alpha$ is of bidegree $(p,q)$, then you should use one of: $$ \iota_X(\alpha \wedge\beta) = \iota_X(\alpha)\wedge\beta + (-1)^{mp + q}\alpha \wedge \iota_X(\beta)$$ or: $$ \iota_X(\alpha \wedge\beta) = \iota_X(\alpha)\wedge\beta + (-1)^{(m+1)(p + q)}\alpha \wedge \iota_X(\beta)$$ Either rule will work if you apply it consistently. Probably other rules will work as well.

  • 1
    $\begingroup$ I'm sorry if it could be interpreted as a slight. It was meant as an invitation for a diversity of answers. Thank you for this very nice answer. One thing about which I'm unclear. You say that the de Rham derivation determines a line object $L_{(0,1)}$. But the source of the differential is the cdga $A$, which is a line in $A$-dg-modules, but not in bigraded vector spaces. $\endgroup$
    – dhagbert
    Mar 8, 2013 at 9:09
  • $\begingroup$ Inconsequential nit-pick. Shouldn't you have $\iota_{X}(\alpha \wedge \beta)=\iota_{X}(\alpha) \wedge \beta + (-1)^{mp-q}\alpha \wedge \iota_{X}(\beta)$ rather than Inconsequential nit-pick. Shouldn't you have $\iota_{X}(\alpha \wedge \beta)=\iota_{X}(\alpha) \wedge \beta + (-1)^{mp+q}\alpha \wedge \iota_{X}(\beta)$? Of course, it doesn't matter in the end. $\endgroup$
    – dhagbert
    Mar 8, 2013 at 15:31
  • $\begingroup$ Sorry, that was a typo. I meant that to speak about $d_{dR}$, you need to say how it's supposed to have bidegree $(0,1)$. What I should have written was $d_{dR} : L_{(0,1)} \otimes \Omega^\bullet \to \Omega^\bullet$. But if you and I disagree about which the generating line in degree $(0,1)$ is (and in particular don't choose an isomorphism between them) then we will each have a $d_{dR}$, and no way to compare them, except that in any fixed skeletalization they'll differ by some scalar. $\endgroup$ Mar 8, 2013 at 18:25

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