This question is a repost from stack.exchange. It didn't get a lot of attention there. Perhaps it is badly written (or silly?). If so, I'd be happy to get comments/suggestions about that.

Let $X$ be a (nice) connected topological space. Let $LX=Map(S^1,X)$ be the free loop space and $\Omega X = Map_*(S^1,X)$ the subspace of based loops (with some choice of base point for X). Now, there is a fiber sequence $$ \Omega X \to LX\to X $$

where the map $LX\to X$ is evaluation at the base point. I am interested in the following question:

Under what general conditions this sequence splits and we have a (weak) homotopy equivalence $LX \simeq X\times \Omega X$ ?

This is of course not true in general. In the case of a classifying space $BG$ of a discrete group $G$, we have $LBG$ equivalent to a disjoint union of $BC(x)$ where $x$ runs over conjugacy classes of $G$ and $C(x)$ is its centralizer subgroup (as explained nicely in the answer to this MO question). Hence, the splitting holds if and only if $G$ is abelian.

Partial Results

I have managed to prove this in the case where $X$ itself is of the homotopy type of a based loop space. i.e. $X\simeq \Omega Y$ for some $Y$ and a little bit more generally, when $X\simeq Map_*(W,Y)$ for some (nice) connected pointed space $W$ and some pointed space $Y$. There are several ways to see this, but the slickest proof I got is just a sequence of standard adjunctions. Denote by $X_+$ the based space obtained from $X$ by adding a disjoint base point. On the one hand,

$$ LX = Map(S^1,Map_*(W,Y))=Map_*(S^1_+,Map_*(W,Y))=Map_*(S^1_+\wedge W,Y) $$ and on the other, $$ \Omega X\times X=Map_*(S^1 \vee S^0,Map_*(W,Y))=Map_*((S^1 \vee S^0)\wedge W,Y) $$

The only difference between $S^1_+$ and $S^1\vee S^0$ is in the selection of the base point, but since $W$ is connected, after smashing with it we get a connected space so the choice of the base point does not matter.

Generalizing the loop space case, it is natural (I think) to ask whether this is true for a general H-space. I also have a feeling that this has something to do with vanishing of whitehead products. So I'll add a more specific version of the question:

Is it true that if all whitehead products of $X$ vanish, then the above sequence splits?


1 Answer 1


Let $X$ be a H-space with multiplication $\mu$ and unit $e$, let us consider the map $$\Phi:\Omega_e X\times X\rightarrow \mathcal{L}X$$ given by $\Phi(\omega,x)=\mu(\omega(t),x)$. This continuous map satisfies: $$ev_0(\Phi(\omega(t),x))=\mu(\omega(0),x)=0=pr_2((\omega,x)).$$ Then it induces a morphism of fibrations between: $$\Omega_eX\times X\stackrel{pr_2}{\longrightarrow} X$$ and $$\mathcal{L}X\stackrel{ev_0}{\longrightarrow} X.$$ Playing with long exact sequences of homotopy groups, you deduce that $\Phi$ is a weak homotopy equivalence.

We know from a result of Stasheff and Arkowitz that a space $X$ has all its generalized Whitehead products trivial if and only if its loopspace $\Omega X$ is homotopy abelian. For example $\Omega \mathbb{C}P^3$ is homotopy abelian. Let us look at the fibration: $$\mathcal{L}\mathbb{C}P^3\rightarrow \mathbb{C}P^3$$ it does not split, this answers negatively your second question. Let me give some details:

  • $\Omega \mathbb{C}P^n\simeq S^1\times \Omega S^{2n+1}$,
  • thus the homology of $\Omega \mathbb{C}P^n\times \mathbb{C}P^n$ is torsion free,
  • whereas the homology $\mathcal{L}\mathbb{C}P^n$ has $n+1$ torsion in fact: $$H_{2nk}(\mathcal{L}\mathbb{C}P^n,\mathbb{Z})\cong \mathbb{Z}\oplus \mathbb{Z}_{n+1}, k>0.$$

Ref: W. Ziller, The Free Loop Space of Globally Symmetric Spaces, Inv. Math. 41, 1-22 (1977).

  • 1
    $\begingroup$ Great! The first part about H-spaces is easy enough for me to follow (and I actually feel bad for not thinking of it myself...). I'll need to think a little more about the second part though. Thanks. $\endgroup$
    – KotelKanim
    May 28, 2015 at 17:29
  • $\begingroup$ What if you suppose $X$ admits an $S^1$-action? maybe up to homotopy. I thought it also will guarantee some splitting! $\endgroup$
    – user51223
    May 28, 2015 at 20:57
  • $\begingroup$ Thanks for the extra details. Now I see it clearly (modulo the the reference that I'll check when I have access to it). I am accepting this answer as it answers completely the two concrete questions I asked. $\endgroup$
    – KotelKanim
    May 29, 2015 at 7:07

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.