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I want to use the cohomology Eilenberg-Moore spectral sequence to calculate the cohomology of the fibre of the map $$ S^{n} \to \Omega S^{n+1}. $$

Question 1: Is anyone aware of any references for this?

My main reference for this spectral sequence is McCleary, John, User’s guide to spectral sequences, Mathematics Lecture Series, 12. Wilmington, Delaware: Publish or Perish, Inc. XIII, 423 p. {$} 40.00 (1985). ZBL0577.55001.

Here is what I have so far.

The ring $H^*(S^n)$ is given by the exterior algebra $\Lambda[x]$ where $x$ is in degree $n$.

The ring $H^*(\Omega S^{n+1})$ is either the divided polynomial algebra $\Gamma[x]$, again with $x$ in degree $n$ if $n$ is even, or $\Lambda[x_1] \otimes_\mathbb{Z} \Gamma[x_2]$ with $x_1$ in degree $n$ and $x_2$ in degree $2n$, if $n$ is odd.

The $E_2$-page of the spectral sequence is given by $$ E_2^{*,*} = \mathrm{Tor}^{*,*}_{H^*(\Omega S^{n+1})}(\mathbb{Z}, H^*(S^n)). $$

My thoughts are to take a projective resolution of $H^*(S^n)$ as a $H^*(\Omega S^{n+1})$-module and then tensor this projective resolution with $\mathbb{Z}$. This tensor product shouldn't be too hard to understand once I understand the projective resolution since as a graded ring $\mathbb{Z}$ consists of a $\mathbb{Z}$ in degree $0$ and zeros elsewhere.

Question 2: What is the projective resolution of $H^*(S^n)$ as a module over $H^*(\Omega S^{n+1})$.

I'd be grateful if someone could explain how to calculate a projective resolution for the $n$ even case. Every attempt I make at doing this ends up quite messy and involved.

Question 3: What does the $E_2$-page of this spectral sequence look like?

Due to the structure of $H^*(S^n)$ and $H^*(\Omega S^{n+1})$, the $E_2$-page should be contained in the columns which are a multiple of $n$ (I believe this is a second quadrant spectral sequence) and hence, should collapse at the $E_2$-page for degree reasons.

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  • $\begingroup$ Actually I think "the standard" technique is to use a projective resolution of $\mathbb Z$ instead. $\endgroup$
    – user43326
    Commented Mar 28, 2019 at 8:17
  • $\begingroup$ Another thing that could be useful is that if you work in homology with Cotor instead of cohomology, you won't have to separte the case according to the parity of $n$, but this is a minor point. $\endgroup$
    – user43326
    Commented Mar 28, 2019 at 8:30
  • $\begingroup$ My thought process behind taking a projective resolution of $\Lambda[x]$ rather than $\mathbb{Z}$ was to use the fact that $\mathbb{Z}$ as a graded module has zeros in all degrees except $0$, and as such then tensor product should be a lot simplier. (Although probably not a lot of difference since $\Lambda[x]$ only has another $\mathbb{Z}$ in degree $n$.) $\endgroup$
    – Niall Taggart
    Commented Mar 28, 2019 at 9:43
  • $\begingroup$ Well, the point is that by taking the resolution of $\mathbb Z$, you can easily compare the Eilenberg Moore spectral sequence for your fibration with that for the fibration $S^{n+1} \rightarrow point \rightarrow \Omega S^{n+1}$. $\endgroup$
    – user43326
    Commented Mar 28, 2019 at 21:47
  • $\begingroup$ Is your heart set on the EMSS? It may be that the Leray-Serre spectral sequence is easier to understand. $\endgroup$
    – Mark Grant
    Commented Apr 1, 2019 at 20:42

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