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

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