# Components of a loop space, semidirect products, and multiplicativity

Let $(X, x_0)$ be a based topological space, and $\Omega X$ its based loop space. The group of path components of $\Omega X$ is $\pi_0(\Omega X) = \pi_1(X, x_0)$. For brevity, let's call this group $G$. Since $\Omega X$ is a loop space, all of its components are homotopy equivalent, and so if we define $\Omega_1 X$ to be the component associated to $1 \in G$ (i.e., the null-homotopic loops), then there is a homotopy equivalence

$$\varphi: G \times \Omega_1 X \to \Omega X.$$

Explicitly, $\varphi(g, f) = r(g)*f$, where $r: G \to \Omega X$ is a section of $\pi_0$, and $*$ is concatenation of loops. My question is about the extent to which this map is a loop map (or $A_\infty$ map).

We can equip the domain of $\varphi$ with the structure of an $H$-space with multiplication

$$(g_1, f_1) \cdot (g_2, f_2) := (g_1 g_2, r(g_2)^{-1}*f_1 * r(g_2) * f_2);$$

i.e., a sort of semidirect product structure where $G$ acts on $\Omega_1 X$ by conjugation of loops (via $r$). Then $\varphi$ is an $H$-map with respect to this multiplication: i.e., it is multiplicative up to homotopy. However, the domain is not a strictly associative monoid with respect to the multiplication $\cdot$, nor is $\varphi$ a strict morphism.

So, my question is: is there an $A_\infty$ structure on $G \times \Omega_1 X$ agreeing the multiplicative structure described above (up to homotopy) and a map $\Phi: G \times \Omega_1 X \to \Omega X$, homotopic $\varphi$, which is an $A_\infty$ map?

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Doesn't the $A_\infty$-structure given by transferring that of $\Omega X$ along $\varphi$ work? –  Oscar Randal-Williams Jul 23 '13 at 13:58
Wait, do you mean a homotopy inverse of $\varphi$? –  Craig Westerland Jul 23 '13 at 14:04
I guess you need both $\varphi$ and its homotopy inverse to transfer an $A_\infty$ structure. –  Oscar Randal-Williams Jul 23 '13 at 14:09
Right, of course. That sounds perfect; thank you, Oscar. –  Craig Westerland Jul 23 '13 at 14:11

I think the answer is ' yes' for trivial reasons. If the loop space were a topological group the map $\varphi$ would be a homeomorphisms and the group structure on the target could be transferred to the source. Now, any space has models for the loop space which are honest topological groups, so you can transfer the result along these homotopy equivalences.

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No in general. Yes if and only if the fibration $\tilde X \to X \to BG$ is trivializable, where $X\to BG$ classifies the universal cover $\tilde X$. (Let me assume here that $X$ is connected.)

For if there is such an $A_\infty$-map $G\times \Omega_1 X \to \Omega X$, it would deloop to a based map $BG \times B\Omega_1 X \to X$. Note that $\tilde X = B\Omega_1 X$.

Conversely, if the fibration is trivializable, we obtain a map $BG \times B\Omega_1 X \to X$ which we can loop to give an to get an $A_\infty$-map $G\times \Omega_1 X \to X$.

Here I am using the following fact: the space of $A_\infty$-maps $U\to V$ has the homotopy type of of the space of based maps $BU \to BV$ whenever $U,V$ are group-like.

Note: an obstruction to the fibration being trivial is that the first $k$-invariant $X \to BG$ should split off. In particular, $X = \Bbb RP^2$ gives a counterexample.

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I think you are answering a slightly different question than the one Craig asked. Namely you are tacitly assuming that the A-infinity structure is the product structure (which is not, even up to homotopy, the one Craig describes). Otherwise B does not commute with products. –  Jesper Grodal Jul 23 '13 at 22:13
Thanks Jesper. I should have read it more carefully. In that case, Oscar's comment (transfer the $A_\infty$-structure using $\varphi$) gives an affirmative answer. –  John Klein Jul 24 '13 at 4:44