Let $G$ be a compact Lie group (or reductive algebraic group over $\mathbb{C}$), and let $BG$ be its classifying space. Fix a prime $p$. Let $\mathcal{A}$ denote the dg algebra of singular cochains on $BG$ with coefficients in a field or characteristic $p$ (or if you prefer the dg algebra of endomorphisms of the constant sheaf). My question is:

Is it known for which primes $p$ the dg algebra $\mathcal{A}$ is formal, that is, quasi-isomorphic to a dg algebra with trivial differential?

I assume / hope that the answer is that this is true if $p$ is not a torsion prime for $G$ (i.e. $p$ arbitrary in types $A$ and $C$, $p \ne 2$ in types $B$, $D$ and $G_2$, $p \ne 2, 3$ in types $F_4$, $E_6$ and $E_7$, and $p \ne 2,3,5$ in type $E_8$.)

Note that we know* that $\mathcal{A}$ is formal in characteristic 0.

Can one then conclude that it is formal in any characteristic in which the cohomology of $\mathcal{A}$ is torsion free?

If so I think this would give the above list of primes.

*) for example because $H(BG, \mathbb{Q})$ is a poynomial algebra, and $\mathcal{A}$ admits a graded commutative model using the de Rham complex -- see Bernstein-Lunts "Equivariant sheaves and functors".


I think that you outlined the proof. In more detail, let $W$ be the Weyl group of $G$, and $T$ its maximal torus. Pick $p$ coprime to $|W|$; this allows to ignore higher $W$ group cohomology in the computation

$$H^*(BG, \mathbb{F}_p) \cong H^*(BT, \mathbb{F}_p)^W$$

Since $W$ is a reflection group, $H^*(BT, \mathbb{F}_p)^W$ is a polynomial algebra, say on $d$ generators. Pick cocycle representatives $x_1, \dots, x_d \in C^*(BG, \mathbb{F}_p)$. Now let $R = \mathbb{F}_p[y_1, \dots, y_d]$ be the free graded commutative algebra on generators $y_i$ in the same degree as $x_i$, and equip $R$ with the $0$ differential. By freeness (and the fact that $d(x_i) = 0$), you get a map $R \to C^*(BG, \mathbb{F}_p)$ of DGA's which sends $y_i$ to $x_i$. You know (because you constructed it that way) that it induces an isomorphism in cohomology, and so $BG$ is formal at the prime $p$.

If you have some other mechanism for ensuring that $H^*(BG, \mathbb{F}_p)$ is a polynomial algebra (e.g., the statement is known integrally, as for $G = U(n)$, $Sp(n)$), the same argument works.

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    $\begingroup$ What worries me here is that cochains with coefficients in $\mathbb{F}_p$ is only an $E_\infty$-algebra, and so I'm not sure that you can make the generators honestly commute. I think you'd need to try something like mapping the generators in separately by maps of associative algebras (where a polynomial algebra on one generator is free) and then use the fact that the range is $E_\infty$ to construct a map from the tensor product. $\endgroup$ – Tyler Lawson Apr 5 '11 at 14:54
  • $\begingroup$ That's a good point. Can you say more about how to get the map from the tensor product into the range? $\endgroup$ – Craig Westerland Apr 7 '11 at 2:30
  • $\begingroup$ Yes, this was also my concern. For tori I think one can make this work, but I don't see a priori how it works in this case. (The reason that Bernstein and Lunts develop the whole machinery of the de Rham complex on the classifying space is in order to make a choice of representatives so that they commute.) $\endgroup$ – Geordie Williamson Apr 7 '11 at 6:27
  • $\begingroup$ @Craig: If we were talking honest commutativity, then I'd construct maps of associative algebras $F_p[y_i] \to R$ and then use a composite $\otimes F_p[y_i] \to \otimes R \to R$; this uses that associative algebras are closed under tensor, and for commutative objects the product map is a map of commutative algebras. The same is true in this case, you just have to $E_\infty$-up everything or work with some kind of honest commutative or associative replacements and deal with the model category issues. $\endgroup$ – Tyler Lawson Apr 8 '11 at 5:54
  • $\begingroup$ Let me do a bit of gravedigging here. Let $G$ be the compact lie group $\mathbb{Z}/3$, then $C^*(BG;\mathbb{F}_3)$ is not formal, and especially it is not even quasiisomorphic to a graded commutative dga. If $x$ is a generator of $H^1(BG;\mathbb{F}_3)$, then the Massey product $\langle x,x,x\rangle\in H^2(BG;\mathbb{F}_3)$ is non-zero. I believe for arbitrary $p>2$, the same works with $p$-fold Massey-products. $\endgroup$ – HenrikRüping May 25 '20 at 18:12

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