Timeline for Is being graded commutative a necessary condition on $A$ such that $H^*(A)$ is commutative?

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Sep 26 at 19:50	comment	added	Dan Petersen		You can make this example simpler by not taking the quotient by any relations at all. If $A=k\langle a, b, c\rangle$ with differential $c \mapsto ab-ba$, then $H(A) \cong k[a,b]$ and in fact $A$ is a cofibrant replacement of $k[a,b]$ in the category of dg algebras. The easiest way to see this is that a polynomial ring is Koszul, and $A$ is the Koszul resolution of $k[a,b]$ (the cobar construction on the Koszul dual coalgebra, which is the exterior coalgebra on two generators).
Sep 26 at 5:14	comment	added	John Palmieri		@mme: I think you're right.
Sep 26 at 5:14	history	edited	John Palmieri	CC BY-SA 4.0	added 233 characters in body
Sep 25 at 23:01	comment	added	mme		I think this does not quite work. I assume $d$ is extended freely by the Leibniz rule; but then $d(ac-ca) = a[a,b] - [a,b] a$, so I think $d$ does not descend to your quotient. I think it's OK if you also add the relations $[a, [a,b]] = [b, [a, b]] = 0$.
Sep 25 at 22:47	comment	added	John Palmieri		@DenisT simpler if the contractible complex is small enough, yes.
Sep 25 at 22:45	comment	added	John Palmieri		@mme: yes, thank you, fixed.
Sep 25 at 22:44	history	edited	John Palmieri	CC BY-SA 4.0	edited body
Sep 25 at 22:34	comment	added	Denis T		Even simpler example would be the tensor algebra of any contractible complex.
Sep 25 at 21:43	history	answered	John Palmieri	CC BY-SA 4.0