If $X$ is a based connected topological space, it is well-known what the homology of $\Omega\Sigma X$ is: according to the Bott-Samelson theorem, it is a tensor algebra over reduced homology of $X$. (Some assumptions about coefficients should be put here).

However, it is also known that $\Omega\Sigma X$ is weakly equivalent to $D_1X$, where $D_1$ is a monad associated to little intervals operad. Passing to the homology, I can see that:

  1. $H_*(\mathcal{D}_1(*);R)$ forms an operad in graded $R$-modules;
  2. Homology of $D_1X$ is an algebra over this operad.

All of the proofs of the Bott-Samelson theorem which I know use rather geometric description of $\Omega\Sigma X$, without using any information coming from the operad action. So my (probably quite vague) question is: is it possible to prove B-S theorem using operadic data? The result being something like "free $\mathcal{D}_1$ algebra in $R$-modules with base $H_*(X;R)$"?

Maybe this is a question about reference, but any help would be appreciated.

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    $\begingroup$ Yes. Take a look at May's "homology of iterated loop spaces". $\endgroup$ – user43326 Aug 22 '19 at 16:28
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    $\begingroup$ @user43326 I couldn't really find such a statement in May's paper, which one are you referring to? $\endgroup$ – Najib Idrissi Aug 22 '19 at 16:44
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    $\begingroup$ I think a relevant theorem would be Theorem 3.1 in Cohen's paper The homology of $C_{n+1}$ spaces. But this is only for $n > 0$ (so it doesn't cover $C_1 = D_1$ in your notation). And skimming the proof, it seems to take as granted that $H_*(\Omega \Sigma X)$ is the free associative algebra on $\tilde{H}_*(X)$ (he calls them $W_0$-algebras). But I guess my point is that it's not automatic that $H_*(P(X)) = H_*(P)(H_*(X))$ in general, you have to work for it – Cohen's proof is an example of that: the homology of the free $D_n$-algebra is not the free $H_*(D_n)$-algebra over $\mathbb{F}_p$. $\endgroup$ – Najib Idrissi Aug 22 '19 at 16:47
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    $\begingroup$ That's correct, I also know that book (that doesn't mean I understand it). Anyway, homology of $\mathcal{D}_1(n)$ is rather simple - it is $R[\Sigma_n]$ concentrated in degree 0 - so I am not sure whether this case is simple, or operad action carries too few information to compute the homology of a monad. $\endgroup$ – Igor Sikora Aug 22 '19 at 16:52
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    $\begingroup$ It's more that homology behaves badly with two of the operations necessary to produce a monad from an operad: the cross product (problem unless over a field) and coinvariants with respect to the symmetric group action (problem unless in characteristic zero). The operad $D_1$ is particularly simple with respect to these two operations: it is equivariantly homotopy equivalent to a discrete free $\Sigma_n$ space, so not much can go wrong. $\endgroup$ – Najib Idrissi Aug 22 '19 at 17:55

$\newcommand{\E}{\mathbf{E}} \newcommand{\co}{\mathcal{O}} \newcommand{\free}{\mathrm{Free}} \newcommand{\H}{\mathrm{H}}$Here's one way of seeing the Bott-Samelson theorem. The James splitting gives an equivalence $$\Sigma \Omega \Sigma X \simeq \bigvee_{n>0} \Sigma X^{\wedge n},$$ so you find that if $k$ is a field, then the reduced homology $\H_\ast(\Omega \Sigma X;k)$ (which is the homology of the suspension spectrum of $\Omega \Sigma X$) is isomorphic to $\bigoplus_{n\geq 0} \H_\ast(X^{\wedge n}; k) \cong \bigoplus_{n\geq 0} \H_\ast(X; k)^{\otimes n}$. This is the tensor algebra on $\H_\ast(X; k)$; one observes that this isomorphism is actually multiplicative, too (it comes from the James splitting for $\Sigma^\infty_+ \Omega \Sigma X$ being multiplicative), so $\H_\ast(\Omega \Sigma X;k)$ is multiplicatively isomorphic to the tensor algebra on $\H_\ast(X; k)$.

I'll write $\E_k$ for what you call $D_k$, so that the space $\Omega \Sigma X$ is the free $\E_1$-space on $X$. In general, if $\co$ is an operad in spaces, then the $E$-homology $E_\ast(\free_\co(X))$ of the free $\co$-space on $X$ is given by the homotopy of $E \wedge \free_\co(X) \simeq \free_\co(E\wedge X)$, where the right hand side is the free $\co$-algebra on $E\wedge X$ in $E$-modules. This final equivalence is eseentially a consequence of the fact that the free functor is defined as a homotopy colimit, and these commute with smash products. The homotopy of this spectrum will be (in nice cases, e.g., homology with field coefficients) the free $E_\ast$-module generated by $\co$-$E$-Dyer-Lashof operations on $E_\ast X$.

One way to understand these Dyer-Lashof operations is as follows. There is a stable Snaith splitting which works more generally for any operad in spaces as above (that specializes to the stable version of the James splitting for $\co = \E_1$): $$\Sigma^\infty_+ \free_\co(X) \simeq \bigvee_{n\geq 0} (\co(n)_+ \wedge \Sigma^\infty X^{\wedge n})_{h\Sigma_n},$$ where $\co(n)$ are the spaces in $\co$. One can prove this essentially formally. The $\co$-$E$-Dyer-Lashof operations are encoded in the $E$-homology of $\co(n)$ --- this is the non-formal component of computing the $E$-homology of free $\co$-spaces. If $\co = \E_k$, then $\co(n) = \mathrm{Conf}_n(\mathbf{R}^k)$; when $k=1$, you find that there are no interesting Dyer-Lashof operations (other than "take powers"), so you get the Bott-Samelson result from these general considerations. For higher $k$, these Dyer-Lashof operations get a lot more interesting.

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    $\begingroup$ Wooooooooooooo! This explained me a lot more than expected! On the side, are there any reference for your descirption of Dyer-Lashof operations? $\endgroup$ – Igor Sikora Aug 22 '19 at 16:56
  • $\begingroup$ You can find a good description of them in, e.g., math.uchicago.edu/~may/BOOKS/homo_iter.pdf and math.uchicago.edu/~may/BOOKS/h_infty.pdf. $\endgroup$ – skd Aug 23 '19 at 1:26
  • $\begingroup$ It seems to me that this answer boarders on the circular. It seems inconceivable to me that someone would know that `$\Omega \Sigma X$ is the free $E_1$ algebra on $X$' without having already figured out that they both have the same homology enroute. (Technically this is possible, but very unlikely.) $\endgroup$ – Nicholas Kuhn Aug 29 '19 at 21:22
  • $\begingroup$ @NicholasKuhn I agree, but that's what I interpreted the OP as asking. I might have misunderstood, though. $\endgroup$ – skd Aug 31 '19 at 2:26

Your space $D_1(X)$ is equivalent to Ioan James' space $JX$ investigated in the 1950's. It is quite easy to directly show that the homology of this is the tensor algebra on the reduced homology of $X$, in parallel to the same computation of $H_*(\Omega \Sigma X)$ (assuming $X$ is connected). Indeed, this can be then used to prove the equivalence of $JX$ and $\Omega \Sigma X$. One exposition of this is in G.W.Whitehead's textbook Elements of Homotopy Theory: see section VII2.

This well known story from the 1950's -- $JX \simeq \Omega \Sigma X$ and the homology of both `freely' generated by the homology of $X$ -- was the model for explorations of models for higher loopspaces 15 years later. It is worth looking at some of the old papers by James and others.


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