Skip to main content
+links
Source Link
Myshkin
  • 17.6k
  • 5
  • 71
  • 137

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

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

  • Robert Langlands, "Automorphic representations, Shimura varieties, and motives. Ein Märchen" (1977)

Some other relevant references are

  • James Milne, Kuang-yen Shih, "Langlands's Construction of the Taniyama Group" (1982)

  • Norbert Schappacher, "CM motives and the Taniyama group" (1994)

  • James Arthur, "A note on the automorphic Langlands group" (2002)

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

Source Link
Myshkin
  • 17.6k
  • 5
  • 71
  • 137

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

  • Robert Langlands, "Automorphic representations, Shimura varieties, and motives. Ein Märchen" (1977)

Some other relevant references are

  • James Milne, Kuang-yen Shih, "Langlands's Construction of the Taniyama Group" (1982)

  • Norbert Schappacher, "CM motives and the Taniyama group" (1994)

  • James Arthur, "A note on the automorphic Langlands group" (2002)