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126 votes

What happened to Suren Arakelov?

There are memoirs by Mikhail Zelikin in Russian. He knew Arakelov personally and quite explicitly describes what happened to him. Через несколько дней произошло следующее. Был арестован Солженицын....
R W's user avatar
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24 votes
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Why are Green functions involved in intersection theory?

The Green's function is used not to measure distances in the surface but to measure distances in the line bundle. A Green's function on $X_{\mathbb C}$ that blows up at $D$ can be used to measure ...
Will Sawin's user avatar
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14 votes
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Meaning of the determinant of cohomology

It doesn't so much represent a dimension of a cohomology group as it does an Euler characteristic. More precisely, it's based on Grothendieck's generalization of the Riemann-Roch theorem to families. ...
Will Sawin's user avatar
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13 votes

Why are Green functions involved in intersection theory?

It is a good idea to restrict first to understand the case of arithmetic surfaces, which is much easier than the general case. For example, you do not have to care about Green currents (only Green ...
Robert's user avatar
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13 votes
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Analogies between classical geometry on complex surfaces and Arakelov geometry

These are indeed good questions, and while there is a very good corpus of answers to them, the analogy is not perfect. 0. The non-archimedean analogy First of all, I would like to go back to the ...
ACL's user avatar
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11 votes
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Is there any definition of $H^1$ in one dimensional Arakelov geometry

There is a definition of $h^{1}(\overline{D})$ by many people, and a definition of $H^{1}(\overline{D})$ by Alexrander Borisov using the notion of ghost spaces of the second kind. But neither is the ...
Bombyx mori's user avatar
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10 votes

Durov approach to Arakelov geometry and $\mathbb{F}_1$

Regarding, Did anyone find applications outside the theory itself? The approaches of Durov (and a number of similar methods) are guided by getting an elegant philosophy, or getting some ...
Oliver's user avatar
  • 151
8 votes

Weil height vs Moriwaki height

One can define a height with respect to a special metric so that some particular useful equation involving the height holds exactly, instead of approximately as for the Weil height. For example, one ...
Will Sawin's user avatar
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8 votes

Durov approach to Arakelov geometry and $\mathbb{F}_1$

Let me just give a quick list of references and some brief comments: for a survey of the various approaches to $\mathbb{F}_1$ see the excellent paper "$\mathbb{F}_1$ for everyone" by Lorscheid (doi:...
Riccardo Pengo's user avatar
8 votes

Why are Green functions involved in intersection theory?

Here is a rather low-brow way of tracing through Arakelov's original ideas. Recall that the intersection of two ordinary divisors $D,E$ can be written as $$ (D.E)_{v}=\sum^{r}_{i=1}-\log \lVert (f|...
Bombyx mori's user avatar
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8 votes
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Why it is difficult to define cohomology groups in Arakelov theory?

The special fiber of $X$ might be a union of one or more irreducible curves. The local ring at the generic point of each of those points is a discrete valuation ring (being a regular local ring of ...
Will Sawin's user avatar
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6 votes
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Is there a notion of hyperbolicity for number rings?

My intuition is that there will not be a precise definition of a hyperbolic number field. However, there there may be some number fields you can confidently say are hyperbolic. Consider the following ...
Will Sawin's user avatar
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6 votes
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The notions of $H^0(\widehat{ D})$ and $h^0(\widehat{D})$ are not satisfactory

The Arakelov-theoretical analogue of Riemann's inequality is Minkowski's theorem. As Felipe Voloch's answer indicates, Tate's approach makes the Riemann-Roch theorem a consequence of the Poisson ...
ACL's user avatar
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5 votes

The notions of $H^0(\widehat{ D})$ and $h^0(\widehat{D})$ are not satisfactory

This issue already comes up in Tate's proof of the functional equation for zeta functions. The functional equation should come out of some version of Riemann-Roch and, for function fields, it does. (...
Felipe Voloch's user avatar
5 votes
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Arakelov divisor on $\operatorname{Spec } O_F$: places or embeddings?

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 ...
Alex B.'s user avatar
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5 votes
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Why does the Manin-Mumford conjecture over number fields imply the conjecture over arbitrary fields of characteristic 0?

What you want to do is, by induction on the theorem in the number field $K$ case, prove that all torsion points in $\mathcal Z_K$ lie in a finite union of torsion translates of abelian varieties (...
Will Sawin's user avatar
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4 votes
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Definition of intersection pairing on an arithmetic surface

You should have a look at Chapter 4 of Moriwaki's Book "Arakelov Geometry". In particular, your question seems to be answered in Proposition 4.19.
Robert Wilms's user avatar
3 votes
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Deligne's example of $\deg \pi_{*}\Omega_{X/Y}<0$

This example should be the elliptic curve with j-invariant 0 and can be found in Deligne's paper Preuve des conjectures de Tate et de Shafarevitch (p. 29). If you are more interested in small values ...
Robert Wilms's user avatar
3 votes
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Günter Tamme's course "Arakelov theory and Grothendieck-Riemann-Roch"

Here is a link of Gunter Tamme's course. I take a brief look and it seems centering around proving Gronthendieck-Riemann-Roch using K-theory machinery. I did not see Arakelov theory anywhere. The ...
Bombyx mori's user avatar
  • 6,249
3 votes

Meaning of the determinant of cohomology

Let us consider the arithemetic surface case, which is already very difficult (see the recent work by Gerald Montplet, for example). In this case Faltings-Riemann-Roch established that $$ \chi(O_{X}(D)...
Bombyx mori's user avatar
  • 6,249
2 votes

Why Green functions and not Neron functions?

A very short answer I learned from summer school - because they want to construct local-global correspondence. A Neron function correspond exactly to the local part and Riemann-Roch is a local-global ...
2 votes

Why it is difficult to define cohomology groups in Arakelov theory?

There is already an issue when looking at global sections of, say, an Arakelov line bundle $(\mathcal{L},||\cdot||)$ thought as a line bundle on $X = \widehat{\mathrm{Spec}~ \mathbb{Z}}$. We see that $...
Artur Jackson's user avatar
2 votes

Arakelov divisor on $\operatorname{Spec } O_F$: places or embeddings?

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 ...
Myshkin's user avatar
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2 votes

What analysis should I know for studying Arakelov Theory?

I think the amount of analysis you need to know is fairly modest. If you know distribution theory and some introductory material on pseudo-differential operators (say first half of Shubin), you should ...
Bombyx mori's user avatar
  • 6,249
2 votes

Discrete Gaussian free field for a closed manifold

Exactly what you want is here : https://arxiv.org/pdf/1809.03382.pdf
Aftermath 12345's user avatar
2 votes

Green currents in Arakelov theory

If you say that the choice of a principal divisor for moving into transversal position is "auxiliary", and you're asking for the "non-auxiliary part" of the data of a Green current,...
Robert Wilms's user avatar
1 vote

Why do Chern forms show up in Arakelov geometry?

I apologize for answering late. I think the 1D case has been discussed multiple times in the forum already. The high dimensional case you suggested was first defined by Bost. See page 63 in below: ...
Bombyx mori's user avatar
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1 vote
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Arithmetic ampleness and scalings of the metric

Let $\overline{L}$ be any arithmetically ample line bundle. In the way you have written it down, $\overline{L}_{\alpha}$ is not arithmetically ample for $\alpha$ sufficiently large, more precisely for ...
Robert Wilms's user avatar
1 vote
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Dualizing sheaf and determinant of cohomology

Recall that Serre duality says that $H^{0}(X,F)\cong H^{1}(X, K\otimes F^{*})^{*}$. So the determinant of cohomology just follows from this. However we do need slightly more machinery than Serre ...

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