Let $F$ be a number field such that $[F:\mathbb{Q}]=n$ and with ring of integers $O_F$. Let's put $B=\operatorname{Spec } O_F$, then an Arakelov divisor is an element of: $$Div(X)\times \bigoplus_\sigma \mathbb R[\sigma]$$ namely it can be written as $$\bigg(\sum_{\text{$\mathfrak p$ prime $\neq 0$}}n_{\mathfrak p}[\mathfrak p],\sum_{\sigma}x_{\sigma}[\sigma]\bigg)\quad \text{with $x_\sigma\in\mathbb R$, $n_{\mathfrak p}\in\mathbb Z$}$$

Now: some references say that $\sigma$ runs among the archimedean places of $F$ other say that $\sigma$ runs among the complex embeddings of $F$.

There is a substantial difference in the two cases, because in the former the dimension of the vector space of "divisors at infinity" is $r_1+r_2$, whereas in the latter it is $n=r_1+2r_2$ (as usual $r_1$ and $2r_2$ are the number of real and complex embeddings of $F$ ).

Which is the most natural approach?

In the first case the degree of an Arakelov divisor is $\sum_{\mathfrak p} n_{\mathfrak p}\log(\#(O_F/\mathfrak p))+\sum_\sigma x_\sigma$ whereas in the second case I think that we should put $$\sum_{\mathfrak p} n_{\mathfrak p}\log(\#(O_F/\mathfrak p))+\sum_{\text{$\sigma$ real }} x_\sigma+\frac{1}{2}\sum_{\text{$\sigma$ complex }} x_\sigma$$

Is it correct?

Edit: books/papers using places: Van De Geer,Schoof- Effectivity of Arakelov divisors of a number field,Neukirch ANT.

books/papers using embeddings: Moriwaki's book, De Jong thesis.

Moreover I'd like to understand what happens when you consider Arakelov theory on arithmetic surfaces $X\to B$. Here I think that would be meaningful to consider all the embeddings.

  • 2
    $\begingroup$ Embeddings up to conjugation is the correct assertion. $\endgroup$
    – user40276
    Oct 7 '16 at 6:32
  • $\begingroup$ The point is: Should we consider equivalent two conjugate embeddings or not? $\endgroup$
    – Dubious
    Oct 7 '16 at 10:47
  • $\begingroup$ I've never seen anyone considering both conjugates. Could you tell me the reference? This would be like considering a bunch of $Aut (\overline{\mathbb{Q}_p}/\mathbb{Q}_p)$-conjugated non-archimedean places. Since we don't do this for primes why would we do this for the infinity? $\endgroup$
    – user40276
    Oct 7 '16 at 12:46
  • $\begingroup$ De Jong thesis page 2: Here he says that the set of $\sigma$'s has cardinality $[K:\mathbb Q]$. So he is considering distinct conjugates. Moriwaki- Arakelov geometry: pag 63 (6). By the way he considers a different degree map in order to use the product formula. $\endgroup$
    – Dubious
    Oct 7 '16 at 12:55

You would usually want the principal Arakelov divisors, i.e. those of the form $(\sum_{\mathfrak{p}}{\rm ord}_{\mathfrak{p}}(a), \sum_\sigma -\log|\sigma(a)|)$ for $a\in F^\times$, to be cocompact in the group of degree $0$ divisors — the volume of the quotient should be the familiar product ${\rm Reg}(F)\cdot{\rm h}(F)$, where ${\rm h}$ denotes the class number. For that, you want $\sigma$ to only run over places, not over all embeddings. Incidentally, you also want to normalise your degree function carefully: the degree of a real place should be $1$, while the degree of a complex place should be $2$.


I think the natural definition is the one using archimedean places (or equivalently, all embeddings, with the pairs of complex conjugates considered as a single embedding), and I'm very curious of the context where you have seen the definition using complex embeddings only.

Consider for example the basic result that the kernel of the mapping $\mathrm{div}:K^*\to \mathcal{Div}(\overline{\mathcal{O}})$ is $\mu(K)$.

This follows directly from the exact sequence

$$1\longrightarrow \mu(K) \longrightarrow \mathcal{O}^*\longrightarrow \Gamma\longrightarrow 0$$

and from the equality (up to a sign) of the composition

$$K^*\longrightarrow\mathcal{Div}(\overline{\mathcal{O}}) \longrightarrow \prod_{\mathfrak{p}|\infty}\mathbb{R}$$

and the logarithmic Minkowski map

$$\lambda: f \mapsto (...,\log |f|_\mathfrak{p},...)$$

This shows quite explicitly why the definition uses necessarily all the embeddings.

For reference, this is also consistent with the use in Moriwaki's and Lang's textbooks on Arakelov theory.

  • $\begingroup$ Moriwaki book and De Jong thesis use the embeddings, on the other other hand Schoof-De Geer paper "effectivity of Arakelov..." deals with places. Also Neukirch book uses places. To be honest I thought that the "best definition" was the one with the embeddings, because when you move to Arakelov theory on surfaces over $B$ I think you need all the Riemann surfaces $X_\sigma$ for any embedding $\sigma$. Conjugate embedding modify consistently the structure of $X_\sigma$. $\endgroup$
    – Dubious
    Oct 7 '16 at 10:01
  • $\begingroup$ I am a bit confused now. Both approaches (Moriwaki vs Neukirch, for example) are the same: places are uniquely determined by embeddings, and vice versa. Or am I missing somethig? $\endgroup$
    – Myshkin
    Oct 7 '16 at 10:22
  • $\begingroup$ Two conjugate embeddings give the same place, right? So by considering all the embeddings you are considering the same place twice $\endgroup$
    – Dubious
    Oct 7 '16 at 10:28
  • $\begingroup$ @Dubious The usual convention is to consider the conjugate complex embeddings as a single pair (that is, one element). For example Neukirch uses places, but he makes this convention quite explicit, see for example the paragraph just below the definition of place [Chapter III.1, page 184]. $\endgroup$
    – Myshkin
    Oct 7 '16 at 10:38
  • $\begingroup$ @Dubious That way there is no ambiguity. Considering all embeddings (single or pairs) you only consider every place once. $\endgroup$
    – Myshkin
    Oct 7 '16 at 10:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.