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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.

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1 Answer

up vote 19 down vote accepted

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. 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.

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So once you have an action of the operad on the Hochschild complex (as in Deligne's conjecture) you have many, each conjugated by an element of $\mathrm{Gal}$. Is there an explicit way of seeing the difference between two of this conjugated actions on the complex? –  Mariano Suárez-Alvarez Sep 28 '10 at 11:59
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The Galois action only exists on the profinite completion of the fundamental group. So I don't see how Gal would act in your situation. I don't think the absolute Galois group really appears in the context of Deligne's conjecture. –  YBL Sep 28 '10 at 12:15
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Instead, what appears naturally is the motivic Galois group of mixed Tate motives over $\mathbb{Z}$ (in part because the $F(\mathbb{A}^1,n)$ are mixed Tate over $\mathbb{Z}$). It plays the same role as Gal but for the motivic prounipotent version of the fundamental groups. The two are related by a Zarsiki dense morphism $G_{\mathbb{Q}} \to G^{mot}_{\mathbb{Z}}(\mathbb{Q}_\ell)$ corresponding to the $\ell$-adic realization. See Kontsevich's "Operads and motives in deformation quantization". –  YBL Sep 28 '10 at 12:18
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PS: I think that writing an explicit action of the operad on the Hochschild complex of $k$-vector spaces usually requires the choice of an associator over $k$. In this case the $k$-points of the motivic Galois group $G_{\mathbb{Z}}(k)$ will act through the prounipotent $GT(k)$ on the coefficients of the associator. If $\mathbb{Q}_\ell \subset k$ you will get an action of $G_{\mathbb{Q}}$ through the $\ell$-adic realization morphism. But the coefficient of this action are very complicated, the simplest are given by the cyclotomic character and Soulé's characters. –  YBL Sep 28 '10 at 12:34
    
Very excellent answer, thank you. –  Romeo Sep 28 '10 at 17:05
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