# Homology of a loop-suspension space and action of $\mathcal{D}_1$-operad

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.

• Yes. Take a look at May's "homology of iterated loop spaces". – user43326 Aug 22 '19 at 16:28
• @user43326 I couldn't really find such a statement in May's paper, which one are you referring to? – Najib Idrissi Aug 22 '19 at 16:44
• 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 $\mathbb{F}_p$. – Najib Idrissi Aug 22 '19 at 16:47
• 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. – Igor Sikora Aug 22 '19 at 16:52
• 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. – 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.

• Wooooooooooooo! This explained me a lot more than expected! On the side, are there any reference for your descirption of Dyer-Lashof operations? – Igor Sikora Aug 22 '19 at 16:56
• 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. – skd Aug 23 '19 at 1:26
• 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.) – Nicholas Kuhn Aug 29 '19 at 21:22
• @NicholasKuhn I agree, but that's what I interpreted the OP as asking. I might have misunderstood, though. – 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.