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

Question

If we consider any differential graded algebra $A^\bullet$, then its homology is a graded algebra, since the tensor product interacts well with homology.

A sufficient condidtion for the homology to be a commutative graded algebra is that $A^\bullet$ is graded commutative, i.e. that we have $ab=(-1)^{\deg a \cdot \deg b} ba$. The terminology “graded commutative dg-algebra” really makes it sound like this is also a necessary condition for $H^*(A^\bullet)$ to be commutative. Yet I've not found any comment on this, so I'm asking here (assuming I've missed the right place to look). Hence my question:

Is $H^*(A^\bullet)$ being commutative equivalent to $A^\bullet$ being graded commutative?

Answer 1

Let $A^\bullet=C^\bullet(X;R)$, the singular cochains on a topological space $X$ with coefficients in a commutative ring $R$, endowed with the cup product of cochains. Given cochains $\varphi\in C^k(X;R)$ and $\psi\in C^\ell(X;R)$ their cup product $\varphi\cup\psi\in C^{k+\ell}(X;R)$ is the cochain whose value on a singular simplex $\sigma:\Delta^{k+\ell}\to X$ is given by $$ (\varphi\cup\psi)(\sigma)=\varphi(\sigma|_{[v_0,\ldots, v_k]})\cdot\psi(\sigma|_{[v_k,\ldots , v_{k+\ell}]}). $$ That is, $\varphi$ restricted to the front $k$-face multiplied by $\psi$ restricted to the back $\ell$-face.

This is usually not graded commutative, yet when you pass to homology you get the singular cohomology $H^\bullet(X;R)$, which is. See Hatcher Section 3.2, especially Theorem 3.14.

Answer 2

I think that Mark Grant's answer is great, but if you want a simpler algebra, consider $$ k\langle a_2, b_2, c_3 \rangle / (ac-ca, bc-cb, [a,[a,b]], [b, [a,b]]). $$ (Subscripts indicate the degrees of the elements.) That is, this is the algebra is freely generated by $a$, $b$, and $c$, and then we impose relations that make $c$ central. We also want $a$ and $b$ to commute with $[a,b]$, explanation below. Note that $a$ and $b$ do not commute, even in the graded sense: the relations are generated by homogeneous elements of degree 5, so there is no relation between the degree 4 elements $ab$ and $ba$. Now put a differential on this: $$ c \mapsto ab-ba. $$ (The last two relations above are there so that this differential on $k\langle a,b,c \rangle$ will induce a differential on the quotient.) In cohomology, the classes induced by $a$ and $b$ will now commute, and I think that the cohomology algebra should be commutative. In fact I think it should just be polynomial on $[a]$ and $[b]$, if I use brackets to indicate cohomology classes. (Since the other answer is so good, I have not felt compelled to check the details.)