I have been puzzled by the following Faltings' remark in his paper *[Calculus on arithemetic surfaces](https://docs.google.com/viewer?a=v&pid=sites&srcid=ZGVmYXVsdGRvbWFpbnxhcGxhY2V0b3RoaW5rMTJ8Z3g6Njk0OTQ1ZmIxYWFjMTIwZA)* (page 394) for a few months. He says:



> 
> If $D$ is a divisor on $X$, we would like to define a hermitian scalar product on $\Gamma(X,\mathcal{O}(D))$ and $H^{1}(X,\mathcal{O}(D))$. Of course there is an obvious way to do this, namely via the square integral of the norm of a section. Unfortunately this is not good enough for us, since we are looking for the archimedean analogue of the following fact:
> 
> 
> If $V$ is a discrete valuation-ring, $K$ its field of fractions, $X$ a stable curve
> over $\textrm{Spec}(V)$, $D$ a divisor on $X\times\_{V}K$, we can extend $D$ canonically to $X$, and $\Gamma(X,\mathcal{O}(D))$ is then a lattice in $\Gamma(X\times\_{V}K, \mathcal{O}(D)).$ It consists of those meromorphic functions on $X\times\_{V}K$ which have only poles at $D$, and which are integral for certain valuations of the function field $K(X)$ of $K$, namely the valuations corresponding to the generic points of the special fibre of $X$. These valuations extend the valuation of $V$, and therefore a theorem of Gel'fand tells us that there cannot be an archimedean analogue for them.
> 
> 
> 


If I am not confused, this is precisely the difficulty that hinders a good definition of an effective cohomology theory on an arithemetic surface. And this motivated Faltings to define the volume on the determinant bundle instead without scalar products. But I do not think I understand it very well - for example, which theorem of Gelfand was Faltings talking about? Is it Gelfand–Mazur theorem, or [Gelfand-Tornheim theorem](http://planetmath.org/node/35628/pmgroup)? I also could not really understand the remark "namely the valuations corresponding to *the generic points of the special fibre of $X$*". Keep scratching my head after a few months, I decided to ask.



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 dimension $1$), and so it defines a valuation on its field of fractions, which is $K(X)$.


Let's consider a special case where the difficulty already appears. Takes $V=\mathbb Z\_p, K= \mathbb Q\_p, X = \mathbb P^1\_{\mathbb Z\_p}, K(X) = \mathbb Q\_p(x)$. There is a natural way to extend the valuation on $\mathbb Q\_p$ to a valuation of the field $K(X)$. To do this we write an element in $\mathbb Q\_p(x)$ as a ratio of two polynomials in $\mathbb Z\_p[x]$, and we define the valuation to be the highest power of $p$ dividing the numerator minus the highest power of $p$ dividing the denominator. Note that, when we restrict this valuation to $\mathbb Q\_p$, we recover the standard valuation on $\mathbb Q\_p$.


Furthermore, to explain why this is so natural, note that for any $x$ in $\overline{\mathbb F}\_p$ where neither the numerator or the denominator vanishes (mod $p$) on $x$, for any lift of $x$ to $\overline{\mathbb Z}\_p$, the valuation of the rational function calculated this way is the same as the valuation when we evaluate it at $x$, simply because the numerator and denominator remain $p$-adic units in this case. Because the nonvanishing of the numerator and the denominator is a generic condition, we say this is the valuation associated to the generic point.


The theorem that rules out an analogous construction at the infinite place is the Gelfand-Tornheim theorem, which says that there is no absolute value on $\mathbb C(X)$ for $X$ a curve over $\mathbb C$ that extends the absolute value on $\mathbb C$. 


