What is the rational homology of $\Omega^2 ( \mathbb CP^n \vee S^d)$? Here $\Omega$ denotes based loop space.
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1$\begingroup$ The space you are looping has a very simple rational model (and is also formal). $\mathbb{C}P^n$ has a Sullivan model $\Lambda \{x,y\}$ with $|x|=2$ and $|y|=2n+1$ and $\partial(y)=x^{n+1}$ and $S^d$ has a Sullivan model $\Lambda \{z\}$ if $d$ is odd or $\Lambda \{z,w\}$ with $\partial w =z^2$ if $d$ is even. A model for the wedge is given by the product of the chain complexes in positive degrees. From here an explicit model of the based loop space can be written down (so do this twice). $\endgroup$– Connor MalinCommented Jan 31, 2021 at 15:27
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$\begingroup$ @ConnorMalin can you elaborate on this statement that a model for the wedge is given by a product? You’ll have to introduce new generators in the tensor product of the two cdgas you wrote in order for the result to have the correct cohomology (I’m assuming you want a minimal model, since from a minimal model you easily get one for the based loop space). For example, xz must vanish in cohomology so we need a generator in degree d+1. Then this forces new generators to be introduced, etc; it seems pretty cumbersome to me to get a full description without invoking Gregory Arone’s last 2 paragraphs $\endgroup$– Aleksandar MilivojevićCommented Feb 1, 2021 at 7:11
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$\begingroup$ @Aleksandar_Milivojevic Yes, that is right. Just an alternative POV. Perhaps the analysis will end up being similar to Cohen’s proof of his result. $\endgroup$– Connor MalinCommented Feb 1, 2021 at 13:06
1 Answer
Let us first consider the homotopy fiber of the map $f\colon\mathbb CP^n\vee S^d \to \mathbb CP^\infty$ which is the inclusion on $\mathbb CP^n$ and is trivial on $S^d$. The homotopy fiber of the map $\mathbb CP^n\to \mathbb CP^\infty$ is $S^{2n+1}$. The homotopy fiber of the map $*\to \mathbb CP^\infty$ is $S^1$, and the homotopy fiber of the trivial map $S^d\to \mathbb CP^\infty$ is $S^1\times S^d$.
It is well-known that taking homotopy pushout commutes with homotopy fibers, therefore the homotopy fiber of $f$ is equivalent to the homotopy pushout of the following diagram. $$ S^{2n+1}\leftarrow S^1 \to S^1\times S^d. $$ I will assume that $n>0$. In this case the map $S^1\to S^{2n+1}$ can only be null homotopic. It follows that the homotopy fiber is equivalent to $$S^{2n+1}\vee S^1_+ \wedge S^d\simeq S^{2n+1}\vee S^d \vee S^{d+1}$$ (for the second equivalence, assume $d\ge 1$).
Applying $\Omega^2$ we get a homotopy fibration sequence $$\Omega^2(S^{2n+1}\vee S^d\vee S^{d+1})\to \Omega^2 (\mathbb CP^n \vee S^d)\to \Omega^2 \mathbb CP^\infty\simeq \mathbb Z.$$ This is a fibration sequence over a homotopically discrete space $\mathbb Z$. Moreover, it is a fibration of (double) loop spaces, so all the homotopy fibers are homotopy equivalent to each other. It follows that there is a homotopy equivalence of spaces $$ \Omega^2 (\mathbb CP^n \vee S^d)\simeq \mathbb Z \times \Omega^2(S^{2n+1}\vee S^d\vee S^{d+1}). $$ So in order to calculate the homology of $\Omega^2 (\mathbb CP^n \vee S^d)$ (with any coefficients), it is enough to calculate the homology of $\Omega^2(S^{2n+1}\vee S^d\vee S^{d+1})$ (with same coefficients).
Let me assume for simplicity that $n\ge 1$ and $d\ge 2$. In this case we can write $$\Omega^2(S^{2n+1}\vee S^d\vee S^{d+1})\simeq \Omega^2\Sigma^2(S^{2n-1}\vee S^{d-2}\vee S^{d-1}).$$
The rational homology of spaces of the form $\Omega^2\Sigma^2 X$ is well-understood. If $X$ is path connected, then it is isomorphic to a certain type of free algebra on $X$. I think some people will call it the free Gerstenhaber algebra and some will say it is the free Poisson algebra. I am not sure what is the most accepted terminology. To get a description of $H_*(\Omega^2 \Sigma^2 X;\mathbb Q)$ as a vector space, first take the free Lie algebra on $H_*(X;\mathbb Q)$, where the Lie bracket has degree $+1$, and then take the free graded commutative algebra on that.
I believe the result about $H_*(\Omega^2\Sigma^2 X)$ is due to Fred Cohen ("The homology of $C_{n+1}$-Spaces", LNM 533). You can find a perhaps somewhat more modern explanation and proof in this paper of Alexander Berglund (Corollary 8 in particular).
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$\begingroup$ Thanks to this great answer. I just like to add that a general result of Gray, generalising Hilton-Milnor decomposition, seems to be applicable to the above question. A first application describes $\Omega(\mathbb{C}P^n\vee S^{d})$ and then from his result one can get something for the double loop space. Indeed, a very useful assumption, perhaps implicit, is that $d>0$ so one can apply refined version of Gray's result (see the comments after Theorem). The paper I refer to is titled ``A note on the Hilton–Milnor Theorem''. $\endgroup$ Commented Feb 2, 2021 at 4:16