# How local is the exponent in the definition of a function with analytic/algebraic singularities?

In Demailly's Analytic Methods in Algebraic Geometry (available on his web page), the definition of a (plurisubharmonic) "function with analytic singularities" is a (plurisubharmonic) function $u: X\to \mathbb{R}$ on a complex manifold $X$ that can be locally written as $$u = \frac{\alpha}{2} \log (|f_1|^2 + |f_2|^2 + \ldots + |f_N|^2) + v$$ where $\alpha \in \mathbb{R}_+$, $v$ is locally bounded and the $f_j$ are holomorphic. (Algebraic singularities require $f_j$ regular, $\alpha$ rational, and Zariski-locality.)

So far no problem.

However, next, he defines the global(!) ideal sheaf $\mathscr{I}(u/\alpha)$ of germs of holomorphic functions $h$ such that $|h| \le C e^{u/\alpha}$ for some constant $C$, i.e. $|h| \le C (|f_1| + \ldots + |f_N|)$. This would require a global constant $\alpha$ (on each connected component).

Edit: As for any positive integer $n$, we have $$\frac {\alpha} 2 \log \left(\sum |f_j|^2 \right) +v = \frac {\alpha / 2} {n} \log \left(\sum |f_j^n|^2 \right) +v',$$ even locally, $\alpha$ is at most determined up to division by integers. So the only "canonical" local $\alpha$ would be the largest possible such $\alpha$.

Therefore, in the compact case with rational local $\alpha$, we can find a global common $\alpha$.

Question: How does this work for non-rational $\alpha$ and non-compact $X$? Can one show (assuming connectedness) that one can choose a global exponent $\alpha$?

Idea: Show that all $\alpha$ are relatively rational.

Better: Starting with one local $\alpha$, show that this $\alpha$ works globally.

Edit: After a week without result, I have now asked the same question at mathoverflow.

In the algebraic setting, $X$ is quasi-projective, so by the Noetherian property, we have Zariski-compactness, so we can choose $\alpha$ globally.
In the analytic setting, (the maximal choice of) $\alpha$ can get arbitrarily small when approaching infinity, so a global choice is not in general possible.