TITLE: What, exactly, is the fundamental group of a free loop space?
QUESTION [20 upvotes]: 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?

REPLY [2 votes]: The action $\rho$ is indeed the standard action. This can be seen from your picture for the action of $\gamma$ on $\alpha \in \pi_2(X)$: the "horizontal central strip" of that picture can be regarded as the based loop $\rho(\gamma)(\alpha) \colon [0,1] \to LX$. The whole picture can then be regarded as a path homotopy between that loop and $\gamma \ast \alpha$ (regarded as a loop on $\Omega X \subset LX$) - see the picture below. 
Since the picture for $\gamma \ast \alpha$ is symmetric with respect the median horizontal line the picture below does indeed describe a path homotopy in LX.

The vertical edges of the rectangles are constant equal to the base point. In the middle rectangle, along the top and bottom horizontal edges we have the first half of $\gamma$, followed by a constant path equal to $\gamma(\frac 1 2)$, followed by the first half of $\gamma$ in reverse.