Why it is difficult to define cohomology groups in Arakelov theory? I have been puzzled by the following Faltings' remark in his paper Calculus on arithemetic surfaces (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? 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.
 A: 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 $\mathbb C$. 
A: 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 $H^0(X,\mathcal{L})$ isn't even a group.
This doens't mean, of course, that it will be impossible to define cohomology in a meaningful way; but that one must think a bit harder.
You can check the work of Durov, New Approach to Arakelov Geometry, where he does implement some sort of cohomology theory.
