Let $X$ be a connected space. According to Getzler BV-algebras and two-dimensional topologcial field theories , page 271, we have and isomorphism

$ H_*(\Omega^2\Sigma^2X) \cong {\cal G}( \widetilde{H}_* X ) $

where ${\cal G}( V)$ means the free Gerstenhaber (Getzler calls it "braid") algebra over the graded space $V$ and $\widetilde{H}$ is the reduced homology.

Getzler credits Cohen's results in The homology of iterated loop spaces for this isomorphism. The closest thing I can find there is Cohen's theorem 3.2, in his chapter "The homology of $C_{n+1}$-spaces, $n\geq 0$", which sounds like it, but I'm having some problems to deduce Getzler's claim.

First of all, Getzler says to be working with complex coefficients, and Cohen with $\mathbb{Z}_p$ ones. Is it clear that the result should be true no matter which coefficients? Rational coefficients too?

Secondly, Cohen's result for $n=1$ would be, I guess, Getzler's case:

$ H_*(\Omega^2\Sigma^2X) \cong GW_1(H_*X) \ . $

But here the free algebra functor is this $GW_1$ which I'm having some troubles to identify with ${\cal G}$.

Any hints or other references will be greatly appreciated.

  • 1
    $\begingroup$ Perhaps a bit too late, but did you look at Berglund's "Koszul Spaces", Section 4? $\endgroup$
    – Pedro
    Commented Aug 21, 2020 at 21:57
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    $\begingroup$ @Pedro Tamaroff. Hey, this is great! Thank you, Pedro. $\endgroup$ Commented Aug 22, 2020 at 4:09

1 Answer 1


Over $\mathbb{Z}_p$ it is not true that $H_*(\Omega^2\Sigma^2X)$ is the free Gerstenhaber algebra. Instead, Cohen proves that $H_*(\Omega^n\Sigma^nX)$ is a free object in a more elaborate category involving some Dyer-Lashof operations. In the case $n=2$ there is only one Dyer-Lashof operation but it still creates a lot of complexity. However, the Dyer-Lashof operations are controlled by the homology of symmetric groups. If we use rational coefficients then the homology of any finite group is zero in positive degrees, and because of this, all the Dyer-Lashof operations are zero. You therefore expect to get a free Gerstenhaber algebra, but I do not know where that is spelled out. There will not be any interesting difference between $\mathbb{Q}$ and $\mathbb{C}$ here; there is a natural isomorphism $H_*(X;\mathbb{C})=H_*(X;\mathbb{Q})\otimes\mathbb{C}$ for all spaces $X$.


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