3
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Yes. Take a look at May's "homology of iterated loop spaces". $\endgroup$ – user43326 Aug 22 '19 at 16:28
  • 1
    $\begingroup$ @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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

$\newcommand{\E}{\mathbf{E}} \newcommand{\co}{\mathcal{O}} \newcommand{\free}{\mathrm{Free}} \newcommand{\H}{\mathrm{H}}$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.

$\endgroup$
  • 4
    $\begingroup$ 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
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.