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I would like to know if there is any relationship between the motivic Galois groups and the Langlands program.

Many thanks.

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  • $\begingroup$ I think the paper of Clozel, Motifs et formes automorphes has something to do with this, but I've not readed the paper... $\endgroup$ – tttbase Oct 19 '16 at 3:19
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    $\begingroup$ Related: mathoverflow.net/q/74698/6518 and mathoverflow.net/q/71743/6518 - have you tried looking at the papers mentioned in the latter? $\endgroup$ – Kimball Oct 19 '16 at 12:00
  • $\begingroup$ Langlands' Taniyama group is isomorphic to the Motivic galois group of the category of absolute hodge cycle CM motives over Q. Deligne, Hodge cycles, motives, and shimura varieties. $\endgroup$ – tttbase Nov 4 '16 at 14:19
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Take $K=\mathbb{Q}$ for simplicity, but this applies to any number field $K$, and let $L$ be the Langlands group and $\mathcal{G}$ the motivic Galois group of $\mathbb{Q}$.

Then the conjectural relation between automorphic forms and motives (the Langlands program) implies that there is an homomorphism (up to a certain conjugation) such that the diagram

$$\require{AMScd}\begin{CD} L @>{}>> \mathcal{G};\\ @VVV @VVV \\ \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) @>{}>>\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}); \end{CD}$$

is commutative.

To get a better idea of both sides of this picture it is useful to use some classic (and better understood) intermediate groups. Let $W$ be the Weil group, $\mathcal{S}$ the Serre group and $\mathcal{T}$ the Taniyama group, then we have something like this

$$L \longrightarrow W \longrightarrow \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{(automorphic side)}$$

$$\mathcal{G} \longrightarrow \mathcal{S} \longrightarrow \mathcal{T} \longrightarrow \mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q}) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{(motivic side)}$$

Most of this goes back to

Some other relevant references are

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