What are the maps in the long exact sequence of homotopy groups for the free loop space fibration?

Question

Let $(X,x)$ be a pointed connected CW complex, and let $\mathcal L X = Map(S^1, X)$ be its free loop space. We have a fibration $\mathcal L X \to X$ given by evaluating at the basepoint $0 \in S^1$, whose fiber is the pointed loop space $\Omega_{x} X$ based at $x$. Let $\lambda \in \mathcal L X$ be a basepoint; we may assume without loss of generality that $\lambda(0)= x$. Then properly speaking, we have a fibration sequence of pointed spaces:

$$(\Omega_{x} X, \lambda) \to (\mathcal L X, \lambda) \to (X,x)$$

We get a long exact sequence in homotopy. The end of it looks like this:

$$\dots \to \pi_1(X,x) \to \pi_0(\Omega_{x} X, \lambda) \to \pi_0(\mathcal L X, \lambda) \to 0$$

Now, I'm told that $\pi_0(\mathcal L X)$ is in bijection with the conjugacy classes of elements of $\pi_1(X)$. Because the middle term is in canonical bijection with $\pi_1(X, x)$,

• This leads me to guess that the map $\pi_1(X,x) \to \pi_0(\Omega_{x} X, \lambda)$ is $\gamma \mapsto \gamma \lambda \gamma^{-1}$.

If that's the case, then the kernel of this map is the centralizer $Z_{\pi_1(X,x)}(\lambda)$. So the next part of the long exact sequence is:

$$ \dots \to \pi_2(X,x) \to \pi_1(\Omega_x X, \lambda) \to \pi_1(\mathcal L X, \lambda) \to Z_{\pi_1(X)}(\lambda) \to 0$$

I don't know what the map $\pi_2(X,x) \to \pi_1(\Omega_x X, \lambda)$ is:

• The natural guess is that this map is $\beta \mapsto \beta^\lambda$ where this denotes the natural action of $\pi_1(X,x)$ on $\pi_2(X,x)$.

But this map is an isomorphism; if this pattern were to persist to higher homotopy groups, it would imply that $\mathcal L X$ is aspherical, and that can't be true -- for instance, the path components of constant loops contain $X$ as a retract, and $X$ need not be aspherical. So I don't think this guess can be correct.

Question: Are either of the bulleted statements above correct? If not, is there a good description of the map $\pi_{n+1}(X,x) \to \pi_n(\Omega_x X, \lambda)$ in the long exact sequence of the free loop space fibration?

Answer 1

The fibration sequence in question is the evaluation sequence

$$\dots\rightarrow\Omega X\xrightarrow{\Delta_\lambda}Map^*_\lambda(S^1,X)\rightarrow Map_\lambda(S^1,X)\xrightarrow{e} X$$

Here the subscript $\lambda$ denotes the path component of $\lambda$ in the mapping spaces, $e$ is the basepoint evaluation map and $\Delta_\lambda$ is the fibration connecting map, which depends very much on the particular choice of $\lambda$. In your notation this is exactly

$$\dots \rightarrow X\xrightarrow{\Delta_\lambda}\Omega_\lambda X\rightarrow\mathcal{L}_\lambda X\xrightarrow{e}X.$$

I'm afraid this notation is a little distinct from yours: here $\Omega_\lambda X$ denotes the path component of the loop $\lambda$ in $\Omega X$, and $\Omega_0 X$ denote the path component of the constant loop in $X$. We're carrying round the basepoint $x\in X$ implicitly here, and all loops are to be based at $X$. Mixing notations we have $\pi_n(\Omega_\lambda X)=\pi_n(\Omega_xX;\lambda)$ (I'll stick to the former).

Now note that the homomorphisms you are studying are exactly those induced by the connecting map $\Delta_\lambda$. And here you're in luck, since this situation has been studied by both Whitehead and Lang. I'll point you towards Lang's paper The Evaluation Map and EHP Sequences for further details than I include below, since I have this reference to hand.

Now $\Omega X$ is a grouplike H-space, so all of its path components have the same homotopy type. In particular the right translation map

$$\Omega_\lambda X\rightarrow \Omega_0X,\qquad \omega\mapsto \omega\lambda^{-1}$$

is a homotopy equivalence. If we insert this transformation into the evaluation fibration the connecting map becomes a map

$$P_\lambda:\Omega X\xrightarrow{\Delta_\lambda} \Omega_\lambda X\xrightarrow{\simeq}\Omega_0X.$$

The utility of this is that $\pi_n(\Omega_0X)=\pi_nX$ (I'm assuming wlog that $X$ is path connected), so with these identifications our long exact sequence is

$$\dots\pi_nX\xrightarrow{P_\lambda}\pi_{n}X\rightarrow \pi_{n-1}\mathcal{L}_\lambda X\xrightarrow{e_*} \pi_{n-1}X\rightarrow\dots$$

and Lang tells us

The homomorphism $P_\lambda$ is the Whitehead product $\alpha\mapsto [\alpha,\lambda]$.

In particular your first map is correct, since on $\pi_1$ the Whitehead product in question is $[\alpha,\lambda]=\alpha\lambda\alpha^{-1}\lambda^{-1}$ (recall our translation map).

So hopefully now you know what the map at $\pi_2$ is, and more generally what the map at $\pi_n$ is.