TITLE: Different way to view action of fundamental group on higher homotopy groups
QUESTION [31 upvotes]: There are a couple of ways to define an action of $\pi_1(X)$ on $\pi_n(X)$. When $n = 1$, there is the natural action via conjugation of loops. However, the picture seems to blur a bit when looking at the action on higher $\pi_n$. All of them have the flavor of the conjugation map, but are more geometric than algebraic, and in some cases work is needed to show the map is well defined. Here are a couple I have seen: 
There is a homotopy equivalence $f : S^n \to S^n \vee I$. taking the basepoint of $S^n$ to the endpoint of the unit interval "far away" from $S^n$. Given a path $\alpha$ from $x_0$ to $x_1$, one can get a basepoint changing homomorphism $\pi_n(X,x_0) \to \pi_n(X,x_1)$ by taking $g : S^n \to X$ and mapping it to $(g \vee \alpha) \circ f$. If $\alpha$ is a loop this gives an action of $\pi_1$
Another way to proceed may be to look at elements of $\pi_n(X,x_0)$ as homotopy classes of maps $I^n \to X$ that send $\partial I^n$ to $x_0$. Then a base change homomorphism could be obtained by using a path $\alpha$ to define a map $I^n \cup (\partial I^n \times I) \to X$, which can be filled in to a map $I^{n+1} \to X$. Then the action would be to take the face opposite the original $I^n \subset I^{n+1}$. 
These both define the same standard action of $\pi_1$ on $\pi_n$, but lose the algebraic flavor of the group action and instead have this stronger geometric feel, which can make working with the action a bit cumbersome. Are there other ways of looking at this action that are more algebraic? 
Perhaps, can something be done wherein $\pi_0(Y)$ acts on $\pi_n(Y)$, where $Y$ is some sufficiently nice space like $\Omega X$, and does this coincide with the above defined actions? Is this a useful way of viewing the action?

REPLY [6 votes]: I doubt this would be considered less geometric than the actions in your question, but if $(X,x_0)$ has a universal cover with covering map $p:(\tilde X, \tilde x_0)\rightarrow (X,X_0)$, then $p$ induces isomorphisms $p_*:\pi_n(\tilde X, \tilde x_0)\rightarrow \pi_n(X,x_0)$ and  $\tilde p: D\rightarrow \pi_1(X,x_0)$ where $D$ is the group of deck transformations (covering transformations) of $(\tilde X,\tilde x_0)$.  
Now for $\alpha\in \pi_1(X,x_0)$ and $\rho\in\pi_n(X,x_0)$ let $\tilde \alpha$ be the preimage of $\alpha$ under $\tilde p$ and $\tilde\rho$ be the preimage of $\rho$ under $p_*$.  
Then $\sigma=(\tilde\alpha)_*(\tilde\rho)$ is in $\pi_n(\tilde X, \tilde x_0)$ and we get
$\alpha\cdot \rho$ as $p_*(\sigma)$.
Now that I think about it, this is at least as geometric as the actions in your question, but I like the picture better.  In the picture, the universal cover not only unrolls the elements of $\pi_1$, but it also unrolls the action of the elements of $\pi_1$ on the maps of spheres.