# Gromov-Hausdorff limits of flat Riemannian surfaces

Posted on January 27, 2019 by Dima

tags: degenerations, weight function, Gromov-Hausdorff limits

Let $$X$$ be a Riemannian surface of genus $$g \geq 1$$. A holomorphic 1-form $$\Omega$$ on $$X$$ has $$2g-2$$ zeroes and putting $$\omega=\dfrac{i}{2}(\Omega \wedge \bar \Omega)$$ we obtain a Kähler metric on the complement of the zeroes of $$\Omega$$ (it is pseudo-Kähler viewed as a metric on the entire $$X$$, since $$\omega$$ is degenerate at at the points where $$\Omega$$ has zeroes). This metric is flat since picking local coordinates $$\operatorname{Re}z$$, $$\operatorname{Im}z$$, where $$z$$ is a local holomorphic coordinate, one observes that $$X$$ is locally isometric (away from zeroes of $$\Omega$$) to $$\mathbb{C}$$ with the flat metric.

Assume we now have a family of such surfaces $$X_t$$ where $$t \in B$$, and that we endow each $$X_t$$ with a pseudo-Kähler metric of the form $$\dfrac{i}{2}{\Omega_t \wedge \bar\Omega_t}$$ where $$\Omega$$ is a relative holomorphic 1-form on $$X \to B$$. As $$t$$ tends towards a point $$O \in B$$, the shape of the Riemannian manifold $$(X_t, \omega_t)$$ changes and we would like to understand if it tends towards some limit shape. More precisely, we will consider the Gromov-Hausdorff limit of $$X_t$$ as $$t \to O$$.

Let $$X, Y$$ be two subsets of a metric space $$Z$$, the the Hausdorff distance between $$X$$ and $$Y$$ is the infimum of positive numbers $$\epsilon$$ such that $$X$$ is contained in the $$\epsilon$$-neighbourhood of $$Y$$ (the union of open balls of raidius $$\epsilon$$ with the center in $$Y$$) and vice versa. The Gromov-Hausdorff distance between two metric spaces $$X$$ and $$Y$$ is the infimum of Hausdorff distances between $$X$$ and $$Y$$ over all possible isometric embeddings of $$X$$ and $$Y$$ into a third metric space $$Z$$.

## Ultrapowers and Los theorem

An ultrafilter $$U$$ on a set $$A$$ is a collection of subsets closed under intersections and supersets such that for any subset $$I \subset A$$ either $$I \in U$$ or $$A \setminus I \in U$$. Consider a countable non-trivial ultra-power of $$\mathbb{R}$$, $${}^* \mathbb{R}$$. What it means is that we choose some ultrafilter $$U$$ on $$\mathbb{N}$$ that contains all cofinite sets (such an ultrafilter exists by the axiom of choice) and we consder the factor $$\prod_{i \in \mathbb{N}} \mathbb{R}/\sim$$ by the equivalence relation $(x_i) \sim (y_i) \textrm{ iff } \{ i \in \mathbb{N}\mid x_i=y_i \} \in U$ The quotient is a ring: we can apply the ring operations coordinate-wise and one checks that the equivalence class of the result does not depend on the representatives picked using the definition of the ultrafilter. Moreover, $$\mathbb{R}$$ admits a diagonal embedding into $${}^* \mathbb{R}$$; its image is called standard reals. Now one can observe that $${}^* \mathbb{R}$$ is actually a field. Indeed, let $$[(x_i)]$$ be some element of $${}^*\mathbb{R}$$ that is not equal to 0. Then there must be $$I \in U$$ such that for all $$i \in I$$, $$x_i \neq 0$$. Let $$y_i=x_i^{-1}$$ for $$i \in I$$ and let $$y_i$$ be anything for $$i \notin I$$. One then checks that $$(x_i \cdot y_i) \sim (1)$$ since $$\{ i \in \mathbb{N}\mid x_i y_i = 1\} \supset I$$ and ultrafilter is closed under supersets.

There is a more streamlined way of checking various properties of ultraproducts:

Theorem (Loś). Let $$\varphi(x_1, \ldots, x_n)$$ be a formula of first-order logic with free variables $$x_1, \ldots, x_n$$ (with $$n$$ possibly 0). Then $$\prod_U \mathbb{R}\models \varphi([(x^1_i)], \ldots, [(x^n_i)])$$ if and only if $$\{ i \in \mathbb{N}\mid \mathbb{R}\models \varphi(x^1_i, \ldots, x^n_i)\} \in U$$.

Thus the formula $$\forall x \exists y x \cdot y = 1$$ in the language of rings $$L_{ring} = (+, \times, 0, 1)$$ is true in $$\mathbb{R}$$ since it is a field, and by Los’ theorem it is also true in $${}^* \mathbb{R}$$.

A real closed field $$R$$ can be characterized via one of the following equivalent statements (due to Artin and Schreier, not to confuse with the Artin-Schreier theory of cyclic extensions of degree $$p$$ in positive charectristic!):

• $$[R^{alg}:R] < \infty$$ (equivalently, $$R^{alg} = R(\sqrt{-1})$$);
• the relation “$$x \sim y$$ iff there exists $$z$$ such that $$x - y = z^2$$” is a total order

It is a fun exercise (for someone starting out in model theory at least) to show that that the first condition can be expressed by countably many first-order formulas, while the second one translates quite straightforwardly to the formula $$\forall x \forall y \exists z (xy=z^2) \lor (xy=-z^2)$$. Either way, we conclude, by Los theorem, that $${}^* \mathbb{R}$$ is a real closed field.

Let $$\mathcal{O}$$ be the convex hull of $$\mathbb{R}$$ in $${}^* \mathbb{R}$$, or in other words, the union of all intervals $$[a,b]$$ in $${}^* \mathbb{R}$$ such that $$a,b \in \mathbb{R}$$. One can easily check that $$\mathcal{O}$$ is a value ring, indeed, for any $$x \in {}^* \mathbb{R}$$ if $$x \notin \mathcal{O}$$ then $$|x| > n$$ for any $$n \in \mathbb{N}$$, so $$|x^{-1}| < 1/n$$, so clearly $$x \in \mathcal{O}$$. The standard part map $$st: \mathcal{O}\to \mathbb{R}$$ maps elements of the form $$a + \epsilon$$ to $$a$$, where $$a\in \mathbb{R}$$, $$|\epsilon| < 1/n$$ for all $$n \in \mathbb{N}$$, is obviously a homomorphism of rings, and its kernel is the maximal ideal in $$\mathcal{O}$$. While its well-definedness is immediate (clearly, it is not the case that $$|a-b| < 1/n$$ if $$a \neq b$$ and $$a,b \in \mathbb{R}$$), to see that it is defined on all of $$\mathcal{O}$$, observe that if $$st$$ is not defined on $$x$$ then for any $$a \in \mathbb{R}$$ there $$|x-a| \geq 1/n$$ for some $$n$$, or in other words $$|x|$$ is bigger than any standard real. It follows that cannot belong to $$\mathcal{O}$$ since any elemnt of $$\mathcal{O}$$ is bounded by some standard real.

Furthermore, the quotient $${}^*\mathbb{R}^\times/\mathcal{O}^\times = {}^*\mathbb{R}^\times/(\mathcal{O}\setminus\mathfrak{m})$$ is non-canonically isomorphic to $$(\mathbb{R},+)$$. Indeed, pick an element $$t \in \mathfrak{m}$$ and construct the following map: $$v_t([(x_i)] = st [(\log_{t_i} |x_i|)]|$$. It is well-defined: indeed, for any $$x \in \mathcal{O}^\times=\mathcal{O}\setminus\mathfrak{m}$$, one checks that $$|x|^n < 1/t$$, since $$1/t$$ is bigger than any standard real by our choice. So $$[(\log_{t_i} |x_i|)] < 1/n$$ and hence belongs . The choice of $$t$$ only changes the scaling: $$v_{t'}(x) = v_t(x) / v_t(t')$$ for $$t'$$ distinct from $$t$$.

Let $${}^*\mathbb{C}$$ be the algebraic closure of $${}^*\mathbb{R}$$, then one checks that there is a unique valuation ring $$\mathcal{O}_{{}^* \mathbb{C}} = \{ x \in {}^* \mathbb{C}\mid |x| \in \mathcal{O}\} \subset \mathbb{C}$$ that extends $$\mathcal{O}\subset \mathbb{R}$$.

## Gromov-Hausdroff limit via ultraproducts

Let $$f: (X,d) \to (Y,d'')$$ be a map between metric spaces, then the distortion of $$f$$ is $\mathrm{dist} f = \inf |d(x,y) - d'(f(x), f(y))|$ A subset $$Z$$ of a metric space $$X$$ is called an $$\epsilon$$-net if $$X$$ is equal to the $$\epsilon$$-neighbourhood of $$Z$$. A map $$f: X \to Y$$ is called $$\epsilon$$-isometry, if $$\mathrm{dist} f < \epsilon/2$$ and $$f(X)$$ is an $$\epsilon/2$$-net.

Lemma. Let $$f: X \to Y$$ be an $$\epsilon$$-isometry, then $$d_{GH}(X,Y) < \epsilon$$.

Let $$K_1:K$$ be a finite field extension; recall that for a variety $$X$$ over a field $$K_1$$ the Weil restriction of $$W_{K_1/K}(X)$$ is the variety $$Y$$ over $$K$$ that represents the functor $$Res_{K_1/K} X: K-Sch \to Sets$$: $$S \mapsto X(S \otimes_K K_1)$$, in particular, $$Y(K) \cong X(K_1)$$.

Let $$X \to B$$ be as above, and let $$X_{\mathbb{R}} \to B_{\mathbb{R}}$$ be the induced Weil restriction for the field extension $$[\mathbb{C}:\mathbb{R}]$$. Pick a generator $$t$$ of the maximal ideal of $$\mathcal{O}_{B,O}$$. Let $$\widehat{\mathcal{O}_{B,O}} \to \mathbb{C}((t))$$ be some isomorphism, and let $$\mathrm{Spec}\mathbb{C}((t)) \to B$$ be the induced morphism of schemes. Embed $$\mathbb{C}((t)) \hookrightarrow {}^* \mathbb{C}$$ so that $$\mathbb{C}[[t]] \subset \mathcal{O}_{{}^* \mathbb{C}}$$ and $$v(t)=1$$. For any $$\alpha \in \mathbb{C}$$ such that $$|\alpha|=1$$ consider the composition of isomorphism $$\widehat{\mathcal{O}_{B,O}} \xrightarrow{\sim} \mathbb{C}((t))$$ mentioned above with the automorphism of $$\mathbb{C}((t))$$ that sends $$t$$ to $$\alpha \cdot t$$, and with the fixed embedding $$\mathbb{C}((t)) \to {}^* \mathbb{C}$$, and call $$\eta^\alpha: \mathrm{Spec}{}^* \mathbb{C}\to B$$ the corresponding $${}^* \mathbb{C}$$-valued point of $$B$$,. Let $$\eta^\alpha_\mathbb{R}: {}^* \mathbb{R}\to B$$ be the $${}^* \mathbb{R}$$-valued points of $$B_\mathbb{R}$$ that correspond to $$\eta^\alpha$$ via the identification $$B({}^* \mathbb{C}) \cong B_{\mathbb{R}}({}^* \mathbb{R})$$. Denote respective fibres $$\overline{X}^\alpha$$ and $$\overline{X}^\alpha_{{}^* \mathbb{R}}$$ and note that $$\overline{X}^\alpha({}^* \mathbb{C})$$ is naturally identified with $$\overline{X}^\alpha_\mathbb{R}({}^* \mathbb{R})$$.

Recall that a semi-algebraic subset of a variety $$Z$$ defined over a real-closed field $$R$$ is a set of $$R$$-points of $$Z$$ that satisfy a boolean combination of polynomial equalities and inequalities (I refer to the Chapter 7 the book of Bochnak, Coste and Roy on real algebraic geometry for the necesseray foundations material). For our purposes it suffices to know that for a semi-algebraic subset $$O \subset Z$$ the notion of a set of $$R'$$-points of $$O$$ makes sense over any real-closed field extension $$R' \supset R$$, and that given a semi-algebaric set $$O \subset X_\mathbb{R}\to B_\mathbb{R}$$ and a map $$\mathrm{Spec}{}^* \mathbb{R}\to B_\mathbb{R}$$, one can define the semialgebraic subset $$\overline{O}= O \otimes_{\mathbb{R}} {}^* \mathbb{R}$$ (essentialy by substituting the corresponding variable in the polynomial equations and inequalities defining $$O$$ by a value in $${}^* \mathbb{R}$$, but in a more invariant way), a subset of $$\overline{X}_\mathbb{R}$$.

Let $$O \subset X_\mathbb{R}$$ be a semi-algebraic set such that its projection on $$B_\mathbb{R}$$ contains a punctured neighbourhood of $$O$$. Denote by $$\overline{O}^\alpha$$ the semi-algebraic subset of $$\overline{X}^\alpha$$ which is the fibre of $$O$$ over $$\eta_\alpha$$.

Lemma. Let $$U \subset \overline{X}^1$$ be a non-Archimedean semi-algebraic subset of $$\overline{X}$$, and denote its Galois conjugates $$U^\alpha \subset \overline{X}^\alpha$$. Let $$O \subset X_\mathbb{R}$$ be a semi-algebraic set such that $$\overline{O}^\alpha({}^* \mathbb{R}) \subset U^\alpha({}^* \mathbb{C}) \subset \overline{X}^\alpha({}^* \mathbb{C})$$ for all $$\alpha \in \mathbb{C}$$, $$|\alpha| = 1$$. Let $$W$$ be a Zariski open neighbourhood of $$X_O$$ such that $$W_\eta \supset U$$ and let $$f$$ be a regular function on $$W$$. Assume one of the following

• $$\inf_{x \in U} v(f(x)) > 0$$
• $$\inf_{x \in U} v(f(x)) \geq 0$$

Then, respectively,

• $$\sup_{x \in O_s} |f(x,s)| \to 0$$ as $$s \to 0$$
• $$\exists C_1\ \sup_{x \in O_s} |f(x,s)| \leq C_1$$ for $$|s|$$ sufficiently small

Proof. It is clear that the premises of the lemma hold also for the conjugates $$U^\alpha$$.

Assume that $$\inf_{x \in U} v(f(x)) > 0$$ but there exists $$\epsilon > 0$$ such that for all $$\delta > 0$$ there exists $$s_\delta$$, $$|s_\delta| < \delta$$ such that $$|f(x,s)| > \epsilon$$.

The formula $\varphi_{\epsilon,n} (s) = \exists x \in O_\mathbb{R}|f(x,s)| > \epsilon \land |s| < 1/n$ then is satisfiable for all values of $$n$$ and therefore, if $${}^* \mathbb{R}$$ is saturated enough, there exists a point $$s^* \in B({}^* \mathbb{R})$$ such that $$\varphi_{\epsilon,n}(s^*)$$ for all $$n \in \mathbb{N}$$. Without loss of generality we may assume then that $$(O_\mathbb{R})_{s^*}({}^* \mathbb{R}) \cong \overline{O}^\alpha({}^* \mathbb{R})$$ for some $$\alpha$$, and so there exists $$x \in \overline{O}^\alpha({}^* \mathbb{R})$$ such that $$|f(x)| > \epsilon$$. But then it would mean that $$v(f(x)) \leq 0$$ for some $$x \in U^\alpha$$ which contradicts the premise.

The second claim is proved similarly.

$$\Box$$.

Assume now that $$\sim$$ be a definable equivalence relation on $$X({}^* \mathbb{C})$$ that is locally given by $x \sim y \textrm{ iff } v(f(x)) = v(f(y))$ for some holomorphic function $$f$$. Assume that for any semi-algebraic set $$O$$ such that for any $${}^* \mathbb{C}$$-valued points $$x,y \in O$$, $$d(x,y) \to 0$$, and assume that the diameter is bounded as $$s \to 0$$.

# Local cohomology and cohomology of line bundles on projective spaces

Posted on December 4, 2017 by Dima

tags: cohomology

all this is terribly unpolished, and arguments are ridden with holes, as is customary on this blog

### Flatness

Frist things first: Nakayama’s lemma.

Statement 1: Let $$R$$ be a ring, $$I$$ an ideal Let $$M$$ be a finitely generated module, and assume $$IM=M$$. Then there exists $$x \in 1 + I$$ such that $$xM=0$$.

Proof: let $$x_1, \ldots, x_n$$ be the generators of $$M$$. Then there exists a matrix $$A=(a_{ij}) \in I^{n \times n}$$ such that $$x_i = \sum a_{ij} x_j$$, or, $$(I-a)x=0$$. Multiply by the matrix formed of minors on the left and get $$\det (Id-A) \cdot I \cdot x = 0$$ where $$x=(x_i)$$. Note that $$\det (Id-A) \in 1 + I$$

Statement 2: Let $$R$$ be a ring, $$I$$ an ideal that annihilates all simple $$R$$-modules (equivalently: if it lies in the intersection of all maximal ideals, i.e. in a local ring, in the maximal ideal), then $$M=0$$.

Proof. Follows from Statement 1, since $$x$$ is invertible. Indeed, otherwise $$(x)$$ would be contained in a maxmial ideal ideal that intersects trivially with $$I$$ or otherwise $$1 \in I$$, but $$I \cap \mathfrak{m}\neq 0$$ for any maxmial $$\mathfrak{m}$$.

But now there is also a more direct proof. Let $$x_1, \ldots, x_n$$ be the minimal set of generators of $$M$$ and assume $$\mathfrak{m}M = M$$. Then $(1-a_n)x_n = \sum_{i=1}^{n-1} a_i x_i,$ and since $$1-a_n$$ is invertible, $$x_1, \ldots, x_{n-1}$$ generate $$M$$, which contradicts the minimality of the set of generators.

A module $$M$$ is flat iff $$-- \otimes M$$ is an exact functor.

Lemma. $$M$$ is flat if for all ideals $$\mathfrak{p}\subset R$$ $0 \to \mathfrak{p}\otimes M \to M \to M/\mathfrak{p}M \to 0$ is exact (and for that it suffices to check that the first arrow is inclusion).

Proposition. Assume $$R$$ is a Dedekind domain (i.e. every ideal is a product of prime ideals; equivalently, 1-dimensional integrally closed). Then an (integral) $$R$$-algebra $$S$$ is flat iff $$R \hookrightarrow S$$ is injective.

Proposition. Let $$f: X=\mathrm{Spec}S \to Y=\mathrm{Spec}R$$ be a morphism. Then $$S$$ is a flat $$R$$-algebraic iff $$\mathcal{O}_{X,x}$$ is flat over $$\mathcal{O}_{Y,y}$$ for all $$x \in X$$, all $$y = f(x)$$.

Lemma. Let $$R$$ be a local ring withe the maximal ideal $$\mathfrak{m}$$. Let $$M$$ be a finitely generated $$R$$-module. Then $$M$$ is free iff $$\mathrm{Tor}_1(M,k) = 0$$ (and moreover both conditions are equivalent to being projective and flat).

Proof. Assume $$\mathrm{Tor}_1(M,k)$$ vanishes. Since $$M$$ is finitely generated, there is a surjective morphism $$f: R^n \to M$$ where $$n = \dim k$$. Tensoring the short exact sequence $$\mathrm{Ker}f \to R^n \to M$$ with $$k$$ we get that $$\mathrm{Ker}f \otimes k = 0$$ which by Nakayama lemma implies $$\mathrm{Ker}f = 0$$.

Lemma. Setting as above. $$M$$ has a projective resolution of length $$n$$ if and only if $$Tor^n(M,k) = 0$$.

Proof. The non-trivial direction is right-to-left. Assume $\ldots \to P_{n+1} \to P_n \to P_{n-1} \to \ldots \to P_0 \to M \to 0$ is a resolution. Let $$Z_i = \mathrm{Ker}d_i \hookrightarrow P_i$$. Then we have short exact sequences $0 Z_i \to P_i \to Z_{i-1} \to 0$ Let $$i=0$$ and tensor with $$k$$, then $\ldots\mathrm{Tor}_n(M,k)=0 \to \mathrm{Tor}_{n-1}(Z_0,k) \to \mathrm{Tor}_{n-1}(P_0, k)=0 \to \ldots$ So $$\mathrm{Tor}_{n-1}(Z_0,k)=0$$. Arguing inductively, we get $$\mathrm{Tor}_{n-i-1}(Z_i,k) = 0$$. In particular, $$\mathrm{Tor}_{1}(Z_{n-2}, k)=0$$ and by the previous lemma it’s free and hence projective. Then $Z_{n-2} \to P_{n_2}\to P_{n-1} \to \ldots$ is a length $$n$$ projective resolution.

### Koszul complex

Let $$R$$ be a local ring with the maximal ideal $$\mathfrak{m}$$ and residue field $$k=R/\mathfrak{m}$$. Then there exists a particularly interesting projective (in fact, free) resolution of $$k$$. Let $$x_1, \ldots, x_n$$ be a set of generators of $$\mathfrak{m}$$. Let $$K\langle\xi_1, \ldots, xi_n \rangle$$ be the ring of grassamanian polynomials in $$\xi_1, \ldots, x_n$$ graded by degree, where $$n$$-th graded piece is isomorphic to $$\wedge^n R$$. Define the differential map to be $d(\xi_i) = x_i$ on the 1-homogeneaus piece, and extend it linearly and applying Leibniz’s rule $d(a \wedge b) = da \wedge b + (-1)^{|a||b|}a \wedge db$ The graded pieces of $$K\langle\xi_1, \ldots, \xi_n \rangle$$ with this differential are element of the Koszul complex.

# From abelianized inertia to divisors

Posted on December 1, 2017 by Dima

tags: curves, Abelian varieties, valuations

The goal of this note is to clarify some points from a paper of Bogomolov, Korotiaev and Tschinkel.

Bogomolov’s approach to birational anabelian geometry consists in looking at abelianisations of absolute Galois groups of function fields of varieties over a fixed algebraically closed field $$k$$ and discerning some extra structures that can be computed from the full absolute Galois group in order to reconstruct the function field.

If $$X$$ is a curve over $$k$$, then by results of Florian Pop $$\operatorname{Gal}(k(X)^{alg}/k(X))$$ is profinite free on $$|X(k)|$$ generators (result independently established by David Harbater: MR1352282 (97b:14035) Harbater, David. Fundamental groups and embedding problems in characteristic p.), and so one cannot distinguish between the curves by the absolute Galois group of their function fields alone.

The anabelian problrem for function fields of curves is reformulated in the BKT paper as follows: let $$(G_1, \mathcal{I}_1)$$ and $$(G_2, \mathcal{I})$$ be a two pairs of abelianizations of absolute Galois groups of functions fields $$k(X_1)$$ and $$k(X_2)$$ of curves of genuse $$\geq 2$$ together with collections of abelianized inertia groups of points.

Let us unpack the terminology.

Let $$L \supset K$$ be a finite extension of fields and $$v: K^\times \to \mathbb{Z}$$ a discrete valuation. Assume $$v': L^\times \to \mathbb{Z}$$ is a valuation such that $$v'|_K = v|_K$$. The Galois group $$\operatorname{Gal}(L/K)$$ acts transitively on all such extensions, and the decomposition group $$D_{v'}$$ of $$v'$$ is the subgroup of elements of $$\operatorname{Gal}(L/K)$$ that fixes $$v'$$: $$v(\sigma\ x) = v(x)$$ for all $$x \in L^\times$$. Decomposition groups of different extensions of $$v$$ to $$L$$ are conjugated. Therefore if the extension $$L/K$$ is Abelian, all decomposition groups coincide. Decomposition group naturally maps to the Galois group of the extension of residue fields: residue field $$\mathcal{O}_{v'}/\mathfrak{m}_{v'}$$ of $$v'$$ over the residue field of $$v$$, the kernel of this homomorphism is called the inertia group. In fact, everything said above applies to non-finitely generated extensions as well. We will denote the abelianized inertia group of any extension of $$v$$ as $$I_{v,a}$$ or simply $$I_v$$.

Here’s how to interpet the decomposition and inertia group in terms of Kummer theory. A cyclic Galois extension $$L/K$$ gives rise to the $$\mu_n$$-torsor $$\operatorname{Spec}L \otimes_K L$$ defined over $$K$$, and such things are classified by $$H^1$$ of the absolute Galois group: $$H^1(K, \mu_n)$$. Since $$\mu_n$$ carries trivial Galois action, cocycle condition degenerates into the requirement of being a homomorphism from the absolute Galois group of $$K$$ to $$\mu_n$$, and each such homomorphism factors through the maximal abelian factor, let us denote it $$\mathcal{G}_a$$ after BKT. Thus the $$H^1$$ in question is isomorphic to $$\operatorname{Hom}(\mathcal{G}_a, \mu_n)$$. On the other hand, applying group cohomology functor to the Kummer exact sequence gives $$K^\times/(K^\times)^n \cong H^1(K, \mu_n)$$. Now restricting attention to powers of a prime $$l$$, and taking limit, we finally get $\widehat{K^\times} \cong \operatorname{Hom}(\mathcal{G}_{a,l}, \mathbb{Z}_l)$ where $$\widehat{K^\times}$$ is the pro-$$l$$ completion, and $$\mathcal{G}_{a,l}$$ is the maximal Abelian pro-$$l$$ quotient of the absolute Galois group. This identification, when dualized, tells us that $$\operatorname{Hom}(K^\times, \mathbb{Z}_l) \cong \mathcal{G}_{a,l}$$.

In terms of this correspondence $D_v = \operatorname{Hom}(K^\times/1+m_v, \mathbb{Z}_l) \qquad I_v = \operatorname{Hom}(K^\times/\mathcal{O}_v^\times, \mathbb{Z}_l)$ Indeed, there is a natural mapping $\operatorname{Hom}(K^\times/1+m_v, \mathbb{Z}_l) \to \operatorname{Hom}(\mathcal{O}^\times/1+\mathfrak{m}_v, \mathbb{Z}_l)$ where $$\mathcal{O}^\times/1+\mathfrak{m}_v$$ is the multiplicative group of the residue field, so the homomorphisms to $$\mathbb{Z}_l$$ that lie in the inertia must vanish on $$\mathcal{O}_v^\times$$.

Now let $$X$$ be a smooth projective curve over an algebraically closed field $$k$$, and pick a point $$x \in X(k)$$. In view of the identification above, $$I_x$$ corresponds to the homomorphism obtained as follows. Let $$v_x: K^\times \to \mathbb{Z}$$ be the valuation associated with the point $$x$$. Then $$I_{x,a}$$ is the subgroup of $$\operatorname{Hom}(K^\times, \mathbb{Z}_l)$$ topologically generated by $$v_x$$, or, equivalently, the subgroup of maps $$\varphi: K^\times \to \mathbb{Z}_l$$ such that $$\varphi(\mathcal{O}_{v_x}^\times) = 0$$.

Now let $$k=\mathbb{F}_p^{alg}$$, and let $$X$$ be a curve of genuse $$\geq 2$$, $$K=k(X)$$. Then $0 \to K^\times/k^\times \to \operatorname{Div}(X) \to \operatorname{Pic}(X) \to 0$ Dualize it: $0 \to \operatorname{Hom}(\operatorname{Pic}, \mathbb{Z}_l) \to \operatorname{Hom}(\operatorname{Div}, \mathbb{Z}_l) \to \operatorname{Hom}(K^\times, \mathbb{Z}_l) \to \operatorname{Ext}^1(\operatorname{Pic}, \mathbb{Z}_l) \to 0$ (and $$\operatorname{Ext}^1(\operatorname{Div}, \mathbb{Z}_l)$$ vanishes, since $$\operatorname{Div}$$ is projective as a direct sum of $$X(k)$$ copies of $$\mathbb{Z}$$). Some identifications:

• $$\operatorname{Hom}(K^\times, \mathbb{Z}_l) \cong \mathcal{G}_{a,l}$$ by Kummer theory (above), note that all morphisms here vanish on $$k^\times$$ since it is divisible;
• $$\operatorname{Hom}(\operatorname{Pic}(X), \mathbb{Z}_l) \cong \mathbb{Z}_l$$, since $$\operatorname{Pic}^0$$ is torsion;
• $$\operatorname{Hom}(\operatorname{Div}(X), \mathbb{Z}_l)$$ is the $$\mathbb{Z}_l$$-linear space of maps $$X(k) \to \mathbb{Z}_l$$, and $$\operatorname{Hom}(\operatorname{Pic}, \mathbb{Z}_l)$$ are the constant maps;

Consider the pairing $\operatorname{Maps}(X(k), \mathbb{Q}_l) \times K^\times/k^\times \to \mathbb{Q}_l, \qquad [\mu, f] = \sum_x v_x(f) \mu(x)$ Then $I_x := \{ [a \cdot \delta_x, -]\}_{a \in \mathbb{Z}_l}$ where $$\delta_x(x)=1$$, $$\delta_x(y)=0$$ for all $$y \neq x$$.