Is the de Rham complex a free (commutative?) differential graded algebra?

A differential graded algebra (dg-algebra) is a monoid object in the category of chain complexes with respect to the usual tensor product of complexes. A (graded) commutative dg-algebra is simply a commutative monoid object.

The de Rham complex for a smooth manifold is easily seen to be a commutative dg-algebra with respect to the wedge product.

I could have sworn that I saw somewhere that the de Rham complex is the free dg-algebra (or free commutative dg-algebra?) on some appropriate base category, i.e. left adjoint to a forgetful functor. I can't seem to find a reference for this, however.

With regards to my background I'm familiar with category theory and differential geometry, but my algebra is a bit weak.

• Perhaps you're thinking of the cotangent complex in algebraic geometry. – Zhen Lin Sep 15 '15 at 9:50
• Wedge product is not commutative. How does that fit into the story? – Amitai Yuval Sep 15 '15 at 10:01
• The first statement that comes to mind is that the de Rham complex should be freely generated as a commutative dg-algebra by its degree zero algebra (the algebra of smooth functions), but this is not actually true (see this answer on MO, for instance). – Eric Wofsey Sep 15 '15 at 10:04
• @AmitaiYuval It's graded commutative: $\alpha \wedge \beta = (-1)^{|\alpha| |\beta|} \beta \wedge \alpha$ for homogeneous elements. – Najib Idrissi Sep 15 '15 at 10:17
• I think, but I haven't thought it out much, that the de Rham dga is the free skew-commutative graded $C^\infty(M)$-dga with $\Omega^1(M)$ in degree 1. Doesn't one deduce this from proposition 2.6 (page 57) in "Elements of noncommutative geometry" by Varilly et al.? One avoids problems of the kind pointed by Eric Wofsey by actually starting from the 1-forms. – Bruno Stonek Apr 27 '16 at 8:28