There is an idea (I think originally due to Tate) that class field theory is fundamentally a consequence of Pontrjagin duality and Hilbert Theorem 90. I'm curious whether this can phrased using modern (slightly motivic) language in something like the following 3-line proof sketch of global class field theory (the set-up is more than 3 lines :) ). The nub of the argument is a global compactly-supported Pontryagin duality statement #(5) below, which is standard in the function field case and which I wish (but don't know) to be true for global fields (and which is the real reason for this question).

Let $K$ be a number field with ring of integers $O$. For any collection of places $S$ of $K$ *including all infinite ones*, let's make believe that there is a "Galois cohomology" dg category (or maybe a stable infinity category) $\mathcal{A}$ which has as objects (at least) "complexes of abelian group schemes" over $Spec(O_S)$. Let's assume that this category satisfies the following properties.

(1) For any commutative algebraic group $A$ over $O_S$ there is an object, $A\in \mathcal{A}$.

(2) For any (reasonable) map of schemes $T\to Spec(S)$ and any $A\in \mathcal{A}$, we have a well-behaved (derived) stalk functor $\Gamma(T, A)$, with value the etale (or something) derived sections $T\to A$. Further, say we can evaluate the stalk of $A$ at $K_v$ for any place and $O_v$ for any place not in $s$. When $v$ is an infinite place, we set the derived stalk functor at $K_v$ to be the complex of (conjugation invariants in) chains on the corresponding Lie group.

(3) There is a constant motive, $const$, with $\text{RHom}(const, A) = R\Gamma(O_S, A)$ (derived sections of $A$ over $O_S$ in a suitable topology).

(4) The object $\mathbb{G}_m\in \mathcal{A}$ satisfies Hilbert Theorem 90, i.e. $R\Gamma^1 (Spec(K), \mathbb{G}_m) = 0$.

(5) The category admits an inner $\underline{Hom}$ and a global Pontryagin-Serre duality statement as follows.

For any object $A\in \mathcal{A}$, define its compactly supported derived sections $R\Gamma_c(O_S, A)$ to be the fiber of the map of complexes $R\Gamma(O_S, A)\to \prod_{v\in S} R\Gamma(K_v, A)$. Then Pontryagin duality states that $R\Gamma(A)$ and $R\Gamma(\underline{Hom}(A, \mathbb{G}_m))$ are dual complexes. (Note that infinite places are important here).

With this setup, we would have the following three-line proof of class field theory. Let $CFT_S$ be the group of characters of the Galois group unramified outside of $S$. Now we have (assuming $S$ contains all infinite places):

$CFT_S = R\Gamma^0(\text{Spec}(O_K), B(\mathbb{Q}/\mathbb{Z}))$ (as we are classifying Galois characters into $\mathbb{Q}/\mathbb{Z}$). Using standard derived machinery, we can take the $B\mathbb{Q}/\mathbb{Z}$ outside (with a shift) to get $CFT_S = H^1(R\Gamma(const)\otimes \mathbb{Q}/\mathbb{Z})$ (where the tensor product is taken in a suitably derived sense).

But now by duality, we have $R\Gamma(const) = R\Gamma_c(\mathbb{G}_m)^\vee$, which means that we are looking for $RHom^1(R\Gamma_c(\mathbb{G}_m), \mathbb{Q}/\mathbb{Z})$. Since $\mathbb{Q}/\mathbb{Z}$ is divisible (hence injective), we can take it outside, and write $CFT_S = Hom(R\Gamma_c^1(\text{Spec}(O_S), const), \mathbb{Q}/\mathbb{Z}).$

The definition of compact support now tells us that $H^1R\Gamma_c(\mathbb{G}_m)$ fits into an exact sequence $$H^0R\Gamma(O_K,\mathbb{G}_m)\to \prod_{v\in S} R\Gamma(K_v, \mathbb{G}_m)\to {\bf H^1 R\Gamma_c}\to H^1(O_S, \mathbb{G}_m).$$ Passing to the limit as $S$ becomes all places and using Hilbert's Theorem 90 to get rid of the $H^1$ on the right then gives us the usual statement of class field theory. $\square$

The one piece of the setup that seems to not be standard is a global duality with compact supports statement (5) above. There is a well-known analogue in the function-field case, and Dustin Clausen's thesis seems to rely on similar ideas, but I've never seen a mixed-characteristic statement of this sort in the literature (as far as I know, it may be false in general, or require a more homotopical language). Does anyone know whether something like this is true?

1more comment