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In trying to understand the import of Akshay Venkatesh' most recent work I found myself wondering anew about that old gnawing mystery: why Langlands? Why should arithmetic of polynomial equations over the rationals (and their generalizations) have anything to do with modular forms (and their generalizations), let alone have such a precise relationship as conjectured by arithmetic Langlands correspondence?

Even after decades of its formulation and spectacularly successful resolution in some cases - even more so for its geometric and $p$-adic analogues - my understanding of the literature is that the essential mystery remains in that beyond the equality of L-functions we don't have much of an inkling, at least in print, of a more direct relationship between the two sides, e.g., an action of (entities on) one side on the other, a direct map from one to the other, or both sides springing from a common source.

Of course, my reading of the literature may be wrong. It is certainly very incomplete. The proofs of special cases may well have provided insights into why these two sides seemingly so far apart should be so intimately related. If so, are they available publicly in a manner that I can access them? I'd also appreciate if you'd share your own here.

Finally, to bring it back to where I had begun: is the recent work of Venkatesh and co-workers (e.g., Kartik Prasanna and Akshay Venkatesh, Automorphic cohomology, motivic cohomology, and the adjoint L-function, see appendix A esp.) a step in the direction of showing a more direct relationship?

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    $\begingroup$ Most $L$-functions are not motivic. They are just there, and this is good so. $\endgroup$
    – GH from MO
    Commented Aug 17, 2018 at 16:08
  • $\begingroup$ Regarding "The proofs of special cases may well have provided the practitioners with insights into why these two sides seemingly so far apart should be so intimately related," see this MO question: mathoverflow.net/questions/66500/… $\endgroup$
    – Stopple
    Commented Aug 17, 2018 at 17:56
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    $\begingroup$ "among the myriad (if not quite thirteen) ways of looking at a modular form...." A myriad is considerably greater than 13. $\endgroup$ Commented Aug 18, 2018 at 0:28
  • $\begingroup$ @Gerry "among the myriad ..." - too subtle an attempt at humor perhaps. I have removed the distraction. $\endgroup$
    – user127738
    Commented Aug 21, 2018 at 22:22
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    $\begingroup$ @GerryMyerson probably objects to people's usage of "decimate" as well ;) $\endgroup$ Commented Aug 21, 2018 at 22:39

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I apologize that this answer is well-known to any expert on Langlands, but I believe it responds to what the OP was asking:

Let $K$ be a number field and $S_K$ its set of places. The kind of Galois representations considered in the Langlands program have coefficients in $\overline{\mathbb{Q}_{\ell}}$ for a fixed prime $\ell$ and are unramified outside a finite set $S \subseteq S_K$, i.e. representations of $G_{K,S}$. For a finite place $\mathfrak{p} \in S_K \setminus S$, there is a conjugacy class of maps $G_{K_{\mathfrak{p}}} \to G_{K,S}$ factoring through the quotient $G_{K_{\mathfrak{p}}} \twoheadrightarrow \widehat{\mathbb{Z}}$ generated by $\mathrm{Frob}_{\mathfrak{p}}$. Thus for a representation $\rho \colon G_{K,S} \to \operatorname{GL}_n(\overline{\mathbb{Q}_{\ell}})$, we get a semisimple conjugacy class $\rho(\mathrm{Frob}_{\mathfrak{p}})^\text{ss}$ in $\operatorname{GL}_n(\overline{\mathbb{Q}_{\ell}})$. The Chebotarev Density Theorem tells us that this collection $\{\rho(\mathrm{Frob}_{\mathfrak{p}})^\text{ss}\}_{\mathfrak{p} \in S_K \setminus S}$ of conjugacy classes of $\operatorname{GL}_n(\overline{\mathbb{Q}_{\ell}})$ indexed by almost all places $\mathfrak{p}$ of $K$ determines $\rho$ up to semisimplification.

In other words, we have an injection from the collection of degree $n$ semisimple $\ell$-adic representations of $G_K$ unramified outside $S$ to $\operatorname{Conj}_\text{ss}(\operatorname{GL}_n(\overline{\mathbb{Q}_{\ell}}))^{S_K \setminus S}$, where $\operatorname{Conj}_\text{ss}$ denotes the set of semisimple conjugacy classes (in a given algebraic group).

Each automorphic representation $\pi$ of $\operatorname{GL}_n(\mathbb{A}_K)$ splits as a restricted tensor product $\prod_{\mathfrak{p} \in S_K} \pi_{\mathfrak{p}}$, where $\pi_{\mathfrak{p}}$ is an irreducible representation of $\operatorname{GL}_n(K_{\mathfrak{p}})$. The definition of an automorphic representation ensures that this representation has a $\operatorname{GL}_n(\mathcal{O}_{\mathfrak{p}})$-fixed vector for almost all $\mathfrak{p}$, i.e., is unramified. Such representations are classified by the Satake isomorphism; they are given by semisimple conjugacy classes in $\operatorname{GL}_n(\mathbb{C})$. Furthermore, if $\pi$ is cuspidal, the collection of these conjugacy classes for almost all $\mathfrak{p}$ determines $\pi$ by the strong multiplicity one theorem.

In other words, we have an injection from the collection of cuspidal automorphic representations of $\operatorname{GL}_n(\mathbb{A}_K)$ unramified outside $S$ to $\operatorname{Conj}_\text{ss}(\operatorname{GL}_n(\mathbb{C}))^{S_K \setminus S}$.

Choosing an isomorphism $\overline{\mathbb{Q}_{\ell}} \cong \mathbb{C}$, we get a bijection $$ \operatorname{Conj}_\text{ss}(\operatorname{GL}_n(\overline{\mathbb{Q}_{\ell}}))^{S_K \setminus S} \cong \operatorname{Conj}_\text{ss}(\operatorname{GL}_n(\mathbb{C}))^{S_K \setminus S}. $$

In other words, a Galois representation gives you a collection of local data that is exactly the kind that uniquely determines a cuspidal automorphic representation, and an automorphic representation gives you a collection of local data that is exactly the kind that uniquely determines a Galois representation. The Langlands conjectures equate two different notions of this local data fitting together into a global object.

More precisely, the condition of fitting together into a representation of $G_K$ is a global condition on the former, and the condition of fitting together into an automorphic representation is a global condition on the latter. The Langlands conjecture essentially says that these two conditions are equivalent. (More precisely, it says that irreducible Galois representations that are de Rham at all places above $\ell$ should correspond to cuspidal automorphic representations that are “algebraic” at the infinite places.)

The idea of two different globality conditions being the same already sounds nice, but why should we really believe they correspond? Well, class field theory already tells us they do for $n=1$. That's pretty strong. But they are already known to correspond for many cases of $n>1$. For a more classical example, Eichler–Shimura says that an automorphic representation for $n=2$ with a particular condition at the infinite place (corresponding to holomorphic weight $2$) gives rise to a Galois representation compatibly with the above procedure.

Here's another way to see how the global conditions are similar. On the one hand, an element of $\operatorname{Conj}_\text{ss}(\operatorname{GL}_n(\overline{\mathbb{Q}_{\ell}}))^{S_K \setminus S}$ gives rise to a representation of the free product $\bigstar_{\mathfrak{p} \in S_K \setminus S} G_{\mathfrak{p}}$. Choosing lifts of each $\mathfrak{p}$ to $\overline{K}$, we get an epimorphism of topological groups $\bigstar_{\mathfrak{p} \in S_K \setminus S} G_{\mathfrak{p}} \to G_K$. The global condition is something like invariance under the kernel of this epimorphism. On the other hand, an element of $\operatorname{Conj}_\text{ss}(\mathrm{GL}_n(\mathbb{C}))^{S_K \setminus S}$ gives rise to a representation of $\prod_{\mathfrak{p} \in S_K \setminus S} \operatorname{GL}_n(K_{\mathfrak{p}})$, which we can restrict to the restricted direct product. The condition of automorphicity is something like invariance under $\operatorname{GL}_n(K)$.

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  • $\begingroup$ I should note that the Satake isomorphism is more general, relating the two ideas for an arbitrary reductive group. However, the fact that almost all local components determine the automorphic representation does not hold for general groups. $\endgroup$ Commented Jun 25 at 10:36

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