Universal property of de Rham differential.

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!

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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.
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