# What, exactly, is the fundamental group of a free loop space?

In these two questions from a few years ago, there was some discussion of the homotopy groups of the free loop space $LX$ of a topological space $X$, which, mysteriously, is the only reference I can find on the issue, although there has been a lot of work on (co)homology of free loop spaces.

The long exact sequence of the fibration $\Omega X\to LX\to X$ helps a lot in computing $\pi_i(LX,c(x))$, where $x\in X$ and $c:X\to LX$ is the embedding of constant loops. Since $c$ admits a retraction, all the boundary maps in the long exact sequence vanish. Then the higher homotopy groups are easy: $\pi_i(LX)\cong \pi_i X\oplus \pi_{i+1} X$, when $i>1$.

For the fundamental group, you get $\pi_1(LX,c(x))\cong \pi_2(X,x)\rtimes_\rho \pi_1(X,x)$ where $\rho$ is some action of $\pi_1$ on $\pi_2$. It's a natural guess that $\rho$ should be the usual action of $\pi_1$ on $\pi_2$, but I'm not so sure. Given $\gamma\in\pi_1(X)$ represented by $g$ and $\alpha\in\pi_2(X)$ represented by $a:S^2\to X$, the usual action makes $\gamma\cdot\alpha$ the class of some map $a'$ which is homotopic to $a$ by a homotopy sending the basepoint around $g$.

But $\rho$ has to be the conjugation action in $\pi_1(LX,c(x))$, which we can describe concretely as the group of pointed homotopy classes of pinched tori $TP\to X$ based at $x$. Then to calculate $\rho(\gamma)(\alpha)$, we view $g$ and $a$ as degenerate pinched tori, take the concatenation $g^{-1}* a * g$, and then contract the path $g^{-1}*g$ the base point follows in the result. This doesn't look the same as the usual action to me.

For illustration, here's a possibly-incomprehensible sketch. These pictures are describing maps $S^2\to X$ as maps from the square $I^2$. Darkened lines are mapped to $x$, arrows are copies of $g$, and diagonally filled regions are copies of $a$. I don't see that any two of the three 2-spheres $a,\rho(\gamma)(\alpha),$ and $\gamma * \alpha$ are equivalent in $\pi_2(X)$, although surely $\rho(\gamma)(\alpha)$ must be one or the other. Anybody have any idea which?

This question is discussed on p.293, Example 9.3.8, of the book partially titled Nonabelian Algebraic Topology. It is sufficient to assume that $X$ is the classifying space of a crossed module $\mathcal M$ of the form $\mu : M \to P$ over a group $P$. Then the aim is to extend $\mathcal M$ trivially to a crossed complex and to describe explicitly the crossed complex

$$L \mathcal M = CRS(\mathbb K(\mathbb Z,1), \mathcal M)$$ where $CRS$ is the internal hom for the monoidal closed structure on the category $Crs$ of crossed complexes, and $\mathbb K(\mathbb Z,1)$ is the crossed complex model of the circle. Since $CRS$ has an explicit description, given in that Chapter of the book, you can write things down.

This was being worked over in more detail after arXiv:1003.5617 with some GAP calculations by a colleague, but the final version did not get completed, I fear, so just the sketch appeared in the book. I'll look at it again.

So I am glad to see the question asked again, and hope this helps.

In view of the equivalence between the monoidal closed structure of crossed complexes and of their cubical analogue, it could be possible to relate this approach to Kevin's pictures.

• Thanks, Ronnie! The classifying space of a crossed module is an arbitrary 2-type, right? So, shouldn't this give the answer for an arbitrary $X$ as $\pi_1 LX$ only depends on the 2-type? – Kevin Carlson Jan 2 '18 at 23:49
• Exactly. I'll edit the answer to say "It is sufficient to assume that ..." – Ronnie Brown Jan 3 '18 at 10:34

In an answer to this question, I learned the following from Tyrone, citing G. Lang's The evaluation map and the EHP spectral sequence.

Let $$(X,x)$$ be a pointed connected space, and let $$\lambda \in LX$$ be a lift of the basepoint -- equivalently, $$\lambda \in \Omega X$$. Then we have a pointed fibration sequence

$$(\Omega X, \lambda) \to (LX, \lambda) \to (X,x)$$

and an induced long exact sequence in homotopy. To a large extent, then, your question boils down to determining what the boundary map $$\pi_{n+1}(X) \to \pi_{n}(\Omega X)$$ is in the long exact sequence. As you've anticipated, this map is not the canonical isomorphism between these groups. According to Lang, this boundary map is the Whitehead product with $$\lambda$$, i.e. it's $$[-,\lambda]: \pi_{n+1}(X) \to \pi_{n+1}(X)$$.

When $$n=0$$, this comes to saying that the map is $$\alpha \mapsto \alpha \lambda \alpha^{-1}$$, and we recover the fact that $$\pi_0(LX)$$ is in canonical bijection with the set of conjugacy classes in $$\pi_1(X)$$. The kernel of the map with basepoint $$\lambda$$ is the centralizer $$Z_{\pi_1(X)}(\lambda)$$.

So to partially answer your question, we have that $$\pi_1(LX,\lambda)$$ sits in a short exact sequence

$$0 \to \operatorname{coker}(\pi_2(X) \xrightarrow{[-,\lambda]} \pi_2(X)) \to \pi_1(LX,\lambda) \to Z_{\pi_1(X)}(\lambda) \to 0$$

and more generally, we have a short exact sequence

$$0 \to \operatorname{coker}(\pi_{n+1}(X) \xrightarrow{[-,\lambda]} \pi_{n+1}(X)) \to \pi_n(LX,\lambda) \to \operatorname{ker}(\pi_{n}(X) \xrightarrow{[-,\lambda]} \pi_{n}(X)) \to 0$$

But I don't know anything about describing the relevant extension. I think that the relevant action of $$Z_{\pi_1(X)}(\lambda)$$ on $$\pi_2(X)/[-,\lambda]$$ is at least the usual action of $$\pi_1$$ on $$\pi_2$$, but I'm not really sure.

• Ah, very cool, thanks! That's the first thing I've learned about the case of a noncontractible basepoint. I'm still very interested to pin down the extensions, but I think I also came up with nothing last time I thought about this. – Kevin Carlson Apr 19 at 15:06