My question is simple: 



> 
> How do the little disks operad and $Gal (\bar {Q}/Q)$ relate?
> 
> 
> 


I realize that a huge amount of heavy-machinery can be brought into an answer to this, but I'm struggling with the basics. All papers I've found just seem to jump into the deep-end or involve musings that are more inspirational than precise; so I'm eager to read what people here say.



Little disks operad and $Gal (\bar {Q}/Q)$ 
A short answer would be: $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ acts faithfully on the profinite fundamental groupoïd of the operad of little discs.


If $X$ is an algebraic variety over $\mathbb{Q}$ we have an exact sequence 
$$
 1 \to\pi\_1(X\otimes \overline{\mathbb{Q}},p) \to 
 \pi\_1(X,p) \to Gal(\overline{\mathbb{Q}}/\mathbb{Q}) \to 1
$$
Here $\pi\_1(X\otimes \overline{\mathbb{Q}};p)$ is canonically identified with the profinite completion of the usual topological fundamental group $\pi\_1(X(\mathbb{C}),p)$. 
If the basepoint is defined over $\mathbb{Q}$, this split and we have an action 
$$
 Gal(\overline{\mathbb{Q}}/\mathbb{Q}) \to Aut(\widehat{\pi}\_1(X(\mathbb{C}),p)). 
$$ 


The (profinite completion of the) fundamental groupoïds of the $C\_2(n)$ inherit the operad structure. The trick is that all of it can be defined over $\mathbb{Q}$ as $C\_2(n)$ is homotopy equivalent to the configuration space of points on the affine line $F(\mathbb{A}^1\_{\mathbb{Q}},n)(\mathbb{C})$. One has to define rational "tangential base points" and check that the operad structure on the fundamental groupoïds is also defined over $\mathbb{Q}$. The resulting operad is described [here](https://mathoverflow.net/questions/39945/compatibility-of-braids-as-a-simplicial-set-and-as-a-braided-monoidal-category/39981#39981). One can explicitly compute its automorphism group. This is the Grothendieck-Teichmuller group $\widehat{GT}$. 


As everything is defined over $\mathbb{Q}$, $Gal(\overline{\mathbb{Q}}/\mathbb{Q})$ operates on the whole operad. So we have a morphism 
$$
 Gal(\overline{\mathbb{Q}}/\mathbb{Q}) \to \widehat{GT}
$$ 
It follows from a theorem of Belyi that it is injective.


