Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Suppose $A$ is a commutative algebra over a field $k$.

It is well known that there is a module that generalizes the notion of differential $1$-forms. It is denoted $\Omega^1_{k}(A)$ and is called the module of Kahler differentials. By definition, it is a module over $A$ generated by symbols $da,a\in A$ satisfying

  • $dc=0$ if $c$ is "constant", i.e. $c\in k$ viewed as a subset of $A$.

  • $d(a+b)=da+db$

  • $d(ab)=(da)b+a(db)$

  • $(da)b=b(da)$

There is also a map $d\colon A\to \Omega^1_{k}(A)$, $d(a):=da$ called the de Rham differential.

It is well-known that $\Omega^1_{k}(A)$ and $d$ can be defined by a universal property. Recall first that a map $\phi\colon A\to M$ for some $A$-module $M$ is called a derivation of $A$ with values in $M$ if $\phi(ab)=\phi(a)b+\phi(b)a$. Then $\Omega^1_{k}(A)$ and $d$ are characterized by the property that for any derivation $X\colon A\to M$, there exists unique morphism $\mu_X\colon \Omega^1_{k}(A)\to M$ such that $X=\mu_x\circ d$. You can read all this in details here.

My question is the following. Is there a similar universal description of the de Rham differential $d^1\colon \Omega^1_{k}(A)\to \Omega^2_{k}(A)$? What about $d^n\colon \Omega^n_{k}(A)\to \Omega^{n+1}_{k}(A)$? I would like to see a description like this:

$d^1\colon \Omega^1_{k}(A)\to \Omega^2_{k}(A)$ is a map satisfying some properties, such that any other map $\Omega^1_{k}(A)\to M$ satisfying these properties factors through $d^1$.

Thank you very much for your help!

share|improve this question

1 Answer 1

up vote 8 down vote accepted

You can describe the universal property of all of the de Rham differentials at once as follows. The direct sum $\Omega(A) = \bigoplus_i \Omega^i(A)$ together with the differential has the structure of a graded-commutative dg-algebra. There is a forgetful functor from graded-commutative dg-algebras to commutative algebras sending a dg-algebra to its degree-$0$ subalgebra, and the functor $A \mapsto \Omega(A)$ is its left adjoint.

share|improve this answer
Can you, please, be a bit more precise? The adjunction you are talking about means $Hom(\Omega(A),B)=Hom(A,B_0)$ for any commutative algebra $A$ and any dg-algebra $B$. First of all, why there is such an adjunction? Why don't we have $B$ on the right instead of $B_0$? I might be missing something but I thought we should have a canonical map $A\to\Omega(A)$. But this adjunction does not seem to give such a map. Can you, please, explain this part a bit more carefully? –  Sasha Patotski Aug 11 '13 at 4:42
The unit of the adjunction is a map $A \to \Omega(A)_0$ (actually an isomorphism), which may be composed with the inclusion to $\Omega(A)$. Qiaochu's answer essentially contains the following statement, which you probably already know, because it is needed in the construction: The differential $d$ on $\Omega(A)$ is the unique one which satisfies the graded Leibniz rule. –  Martin Brandenburg Aug 11 '13 at 15:51
@Sasha: a map $\Omega(A) \to B$ is first of all a map of chain complexes, so it sends elements of $A$ to elements of $B_0$. Second of all it is completely determined by what it does to $A$ since it is also a map of dg-algebras. Finally it is freely determined by what it does to $A$ by the universal properties of $\Omega^1$ and exterior powers. So such a thing can be identified with the corresponding map $A \to B_0$. Martin's comment explains the map $A \mapsto \Omega(A)$ (as a map of algebras). –  Qiaochu Yuan Aug 11 '13 at 16:06
@Sasha: I'm not entirely sure you can universally describe the map $d : \Omega^1 \to \Omega^2$ without talking about the rest of the structure of $\Omega$, the main problem being that you're ignoring the multiplication $\Omega^1 \otimes \Omega^1 \to \Omega^2$, but this map interacts with $d : \Omega^2 \to \Omega^3$, so... –  Qiaochu Yuan Aug 11 '13 at 16:07
@QiaochuYuan, MartinBrandenburg: Thank you very much for the explanation! –  Sasha Patotski Aug 11 '13 at 16:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.