The complex of Kahler differentials and de Rham complex

Let $A$ be a unital commutative algebra (say over complex numbers). Consider the multiplication map $m:A \otimes A \to A$ and put $\Omega^1_u(A)=\ker m$ to be the space of universal differential forms. We turn $A \otimes A$ into $A-A$ bimodule: left action is the natural one, the right is such that $d_u$ defined via $d_u(a)=1 \otimes a -a \otimes 1$ satisfies Leibniz rule. In more detail: $(x \otimes y) \cdot z=x \otimes yz-xy\otimes z$. We take $M:=(\Omega^1_u(A))^2$ be a subbimodule of $\Omega^1_u(A)$ generated by the terms of the form $d_u(a)d_u(b)$. If $A$ is commutative one then shows that the quotient $\Omega_{ab}^1(A):=\Omega^1_u(A)/M$ is symmetric $A$ bimodule therefore can be regarded as left $A$ module. Put $d_{ab}:=1 \otimes a -a \otimes 1 (\mod M)$. This is easily seen to be a derivation. The pair $(\Omega^1_{ab}(A),d_{ab})$ is universal in the following sense: given any pair consisting of $A$ module $E$ and a differential (derivation) $d:A \to E$ one can find unique $A$-module map $\varphi_d$ such that $d=\varphi_d \circ d_{ab}$. This establishes a bijection $Hom_A(\Omega^1_{ab}(A),A)=(\Omega^1_{ab}(A))' \cong Der(A,A)$ which can easily be seen to be an $A$ module map (here $Der$ stands for the space of all derivations). One can consider $\Omega^*_{ab}(A)$ defined as an exterior power (over $A$) with the unique extension of differential (still to be denoted by $d_{ab}$) via (graded) Leibniz rule and the requirment $d^2_{ab}=0$. Let us now specify $A=C^{\infty}(M)$ where $M$ is a smooth manifold. Then $Der(A,A)=\mathcal{X}(M)$ is a space of vector fields. I found the following remarkable theorem which states that in this case one can identify $\Omega^*_{ab}(A)$ with the complex of de Rham forms. Not everything in the proof is clear for me:

Question 1: As I explained we have $\mathcal{X}(M) \cong (\Omega_{ab}^1(A))'$ where $A=C^{\infty}(M)$. Is it obvious that this would imply $\Omega^1_{ab}(A) \cong \Omega^1_{dR}(M)$? The problem is that I don't know whether $\Omega^1_{ab}(A)$ satisfies $\Omega^1_{ab}(A)=(\Omega^1_{ab}(A))''$.

Suppose that the answer for the first question is affirmative: we get as far as I understood only the isomorphism of $A$-modules. So my second qestion is the following:

Question 2 Does the fact $\Omega^1_{ab}(A) \cong \Omega^1_{dR}(M)$ imply the isomorphism of $\Omega^*_{dR}(M)$ and $\Omega^*_{ab}(A)$ and if the anser is positive-this is an isomorphism of what structures? I would be very suprised if we have an isomorphism of differential graded algebras.

Q1, No: $\Omega^1_{ab}(A)$ is generated as an $A$ -bimodule by all $df$, $f\in A$. For $A=C^\infty(\mathbb R)$ (say) $d^{dR} f = f'.dx$ which equals $d^{ab} f$ if and only if $f$ and $x$ are algebraically dependent. For $f(x) = e^x$ this is not the case.
What you want, may have been spelled out as $\Omega_{Der}(A)$ here.
• So you claim that $\Omega^1_{ab}(A)$ is not isomorphic to $\Omega^1_{dR}(M)$? My question was inspired by Proposition 8.1 in J. Varilly, J.Garcia-Bondia and H. Figueroa's book "Elements of noncommutative geometry" where they claim (in the proof of this proposition) that in fact these two objects ARE isomorphic. – truebaran Apr 6 '15 at 12:33
• @truebaran For commutative unital algebras over a field, the Kaehler module of A is naturally isomorphic to the 1st Hochschild homology group $H_1(A,A)$. (I think this fact might be in e.g. the early parts of Loday's book on Cyclic Homology.) This isomorphism should also work if one defines versions of both objects for sufficiently nice topological algebras, e.g. $C^\infty(M)$ as a Frechet algebra. Now if I recall correctly, in his NCDG paper Connes calculates the topological-algebra version of Hochschild (co)homology for $C^\infty(M)$. – Yemon Choi Apr 9 '15 at 23:33