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A topological space $X$ is $\mathbb Z$-formal, if the singular cochain complex $C^*(X,\mathbb Z)$ is quasi-isomorphic to $H^*(X, \mathbb Z)$ as an augmented differential graded ring.

It's quite simple to write down specific quasi-isomorphisms to show that the Spheres $S^n$ are $\mathbb Q$-formal spaces by fixing a volume form $v \in \Omega^n(S^n)$ and considering the maps $H^*(S^n)=\operatorname{span}(1,[v]) \to \Omega^*(S^n)$ sending $1$ to the $1$-form and $[v]$ to $v$ and the canonical map $C^*(S^n) \to \Omega^*$.

Is it also possible to show the $\mathbb Z$-formality of the Spheres $S^n$ by writing down specific quasi-isomorphisms, or is it easier to use another method for showing $\mathbb Z$-formality?

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    $\begingroup$ I am probably missing something here: if you map 1 to 1 and a generator $g$ of $H^n(S^n,\mathbb{Z})$ to a closed element of $C^n(S^n,\mathbb{Z})$ with cohomology class $g$, doesn't that give you a quasi-isomorphism over $\mathbb{Z}$? $\endgroup$
    – abx
    Commented Jun 13, 2021 at 15:11
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    $\begingroup$ @abx Your reasoning shows that $H^*(S^n)$ and $C^*(S^n)$ are quasi-isomorphic as cochain complexes. Formality is the much stronger condition that they are quasi-isomorphic as differential graded algebras. $\endgroup$
    – John Pardon
    Commented Jun 13, 2021 at 16:26
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    $\begingroup$ @A.S. (see previous comment) $\endgroup$
    – John Pardon
    Commented Jun 13, 2021 at 16:26
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    $\begingroup$ If we try to construct a map of $A_\infty$-algebras $F:H^*(S^n)\to C^*(S^n)$ by induction on $k$ (indexing the operations $F^k:H^*(S^n)^{\otimes k}\to C^*(S^n)$) whose $k=1$ component $F^1:H^*(S^n)\to C^*(S^n)$ induces the identity map on cohomology, then we run into potential obstructions when $k\equiv 2\mod n$. $\endgroup$
    – John Pardon
    Commented Jun 13, 2021 at 16:32
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    $\begingroup$ What is wrong with the following? Introduce a simplicial structure on $S^n$. We get a zig-zag of algebra quasi-isomorphisms $C^*(S^n; \Bbb Z) \to C^*_\Delta(S^n;\Bbb Z) \xleftarrow{} H^*(S^n;\Bbb Z)$, the first map given by restricting the domain of cochains to simplicial chains, the last map existing because $C^{2n}_\Delta(S^n;\Bbb Z)$ is zero on the nose. $\endgroup$
    – mme
    Commented Jun 13, 2021 at 17:14

1 Answer 1

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Consider the simplicial set $\def\S{{\bf S}} \def\Sing{\mathop{\rm Sing}} \S^n=Δ^n/∂Δ^n$, which has exactly two nondegenerate simplices: a 0-simplex and an $n$-simplex.

Consider the map $\S^n→\Sing S^n$ that sends the only vertex of $\S^n$ to the given basepoint of $S^n$ and the only nondegenerate $n$-simplex of $\S^n$ to some singular $n$-simplex $Δ^n→S^n$ that sends the boundary to the basepoint and induces a degree 1 map once we mod out the boundary.

The map $\S^n→\Sing S^n$ is a simplicial weak equivalence. Thus, the induced map on integral normalized simplicial cochains $\def\Z{{\bf Z}} \def\C{{\rm C}} \C(S^n,\Z)=\C(\Sing S^n,\Z)→\C(\S^n,\Z)$ is a quasi-isomorphism of augmented differential graded rings.

Observe that $\C(\S^n,\Z)$ is a differential graded ring with exactly two generators: one in degree 0 and another one in degree $n$. The differentials of both generators vanish. Furthermore, the degree $n$ generator squares to 0 for dimensional reasons. Thus, the normalized simplicial cochain ring $\C(\S^n,\Z)$ is precisely the graded cohomology ring $\def\H{{\rm H}} \H(S^n,\Z)$.

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