Fiber at infinity of an arithmetic surface $X$ as an element of $\widehat{\operatorname{Div}(X)}$

Question

Introduction:

Let $M$ be a Riemann surface, then a Green function on $M$ is an element $g\in C^\infty(V)$ where $V=M\setminus\{x_1,\ldots,x_r\}$ and around each point $p\in M$ we have:

$$g=a\log\left|\phi\right|^2+u$$ where $\phi$ is a complex chart centered in $p$, $u$ is a $C^\infty$ function and $a\in \mathbb R$ is independent from the chosen chart. We denote $\operatorname{ord}_p(g):=a$.

The Green functions on $M$ form a vector space $G(M)$and we have a group homomorphism $$\operatorname{div^G}:G(M)\to\operatorname{Div}(M)\otimes_\mathbb Z\mathbb R$$ $$g\mapsto\sum_p\operatorname{ord}_p(g)\{p\}$$

So each Green function defines a real divisor on $M$. The components of $\operatorname{div^G}(g)$ are exactly the points where $g$ is not defined. Now let $K$ be a number field with ring of integers $O_K$ and let $\pi:X\to\operatorname{Spec} O_K$ be a a projective, regular arithmetic surface. For every field embedding $\sigma:K\to \mathbb C$ we obtain a Riemann Surface $X_\sigma:X\times_\sigma\operatorname {Spec}\mathbb C$, and for every divisor $D\subset X$, $D_\sigma$ is the pullback divisor of $D$ through the canonical map $X_\sigma\to X$.

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Question:

The group of arithmetic divisors of $X$ can be defined in the following way: $$\widehat{\operatorname{Div}(X)}:=\left\{\left(D,\sum_\sigma g_\sigma\sigma\right)\in \operatorname{Div(X)\times\bigoplus_\sigma G(X_\sigma)\sigma}\,:\, \operatorname{div^G}(g_\sigma)=D_\sigma,\;\forall\sigma\right\}$$

The meaning of this is quite obvious: an element $\left(D,\sum_\sigma g_\sigma\sigma\right)$ can be "physically" seen as the divisor $D$ plus the finite set of points where $g_\sigma$ is not defined for each Riemann surface $X_\sigma$.

I don't understand why the arithmetic divisor $F_\sigma:=(0,2\sigma)\in \widehat{\operatorname{Div}(X)}$ is usually identified with the "fiber at infinity" $X_\sigma$. In particular what is the meaning of the constant $2$? How can I see $F_\sigma$ as the Riemann surface $X_\sigma$?

Answer 1

Here is another low brow way to look at it by tracing Arakelov's ideas in his paper.

Let $X$ be a curve over $K$, where $K$ is a number field. Let $\infty$ denoting the archimedean valuations of $K$ to $\mathbb{C}$. Let $\sigma\in \infty$, then $X_{\sigma}=X_{K}\otimes_{\sigma}\mathbb{C}$. This is a one dimensional algebraic curve over $\mathbb{C}$, therefore it is a Riemann Surface. In particular, every embedding of $K$ into $\mathbb{C}$ came in conjugate pairs. So it makes sense to group them together and identify $\sigma$, $\overline{\sigma}$ together. Clearly $X_{\sigma}=\overline{X_{\overline{\sigma}}}$ and they are the same as Riemann Surfaces.

In later work people usually take the $v$-adic distance completion of $K$ directly. Then we have two cases: either $\overline{K_{\sigma}}=\mathbb{R}$, or $\overline{K_{\sigma}}=\mathbb{C}$. The intersection pairing acquired a specific meaning like the finite part: Recall in the finite part we have $$ (D,E)_{v}=\log(q_v)*\textrm{ordinary intersection index} $$ where $q_v$ is the degree of residue field extension at $v$. Then it is natural to define analgous coefficient at infinity using the degree of functional field extension by $\epsilon_v=1$ for $K_v=\mathbb{R}$, and $\epsilon_v=2$ for $K_v=\mathbb{C}$. This corresponds to the classical intuition that the archimeadean primes should be viewed as unramified with degree of extension 1 or 2. The intersection index thus taken the following form: $$ (D,E)_{v}=\epsilon_v\cdot G(D,E) $$

It might be helpful to recall why we have the $\log(|x|)$ blow-up over the diagonal for Green functions in 2D. The Laplacian in polar coordinates can be written as $$ \Delta=\frac{\partial}{\partial^2 r}+\frac{1}{r}\frac{\partial}{\partial r}+\frac{1}{r^2}\frac{\partial}{\partial^2 \theta} $$ So if you assume the solution is rotation symmetric, then the last term drops out and you solve a first order ODE in terms of $r$, which gives you the desired expression for the Green's function. If you know $\Psi DO$ theory, the blow-up can be seen very precisely by $$ \int \frac{1}{|\xi|^2}e^{i(x-y)\cdot \xi}f(y)dyd\xi=2\pi\int \log(|r|)dr e^{i(x-y)\cdot \xi}dy $$ So from the perspective of Arakelov theory Green's function entered quite naturally as they are unique, symmetric and have the required decay behaviour at the diagonal. From the analysis perspective, there is indeed nothing mysterious about the constant 2; you can replace it with 1 if you replace Green's function with Neron's function. But it does make the theory cleaner.