Is the algebraic Grothendieck group of a weighted projective space finitely generated ? This is to be confronted with Joseph Gubeladze' paper : "Toric varieties with huge Grothendieck group" !
 A: In our paper with N. Pavic we have proved that over an algebraically closed field of char. 0, if the weights $a_0, \dots, a_n$ are coprime, so that singularities of  $\mathbb{P}(a_0, \dots, a_n)$ are isolated, then $K_0(\mathbb{P}(a_0, \dots, a_n)) = \mathbb{Z}^{n+1}$, see Application 3.2 in 
https://arxiv.org/pdf/1809.10919.pdf
Our proofs rely on comparison of $K_0(X) = K_0(Perf(X))$ with $G_0(X) = K_0(D^b(X))$ using $K$-groups of Orlov's singularity category and completion arguments. The same methods apply to compute $K_0(X)$ for any quasi-projective variety $X$ with isolated quotient singularities. 
A: I never did determine whether this was true or not (it was this point that unfortunately required me to use the BOT construction in my paper).
However, at the time I was working on this, it was my suspicion that the torsion part of the $KH_{0}$ of a WPS would be 0.  If true, then (at least in characteristic 0) my paper would imply that the question of whether or not $K_{0}$ of a WPS is finitely generated boils down to whether or not its $(\mathcal{F}_{K})_{0}$ is finitely generated.
However, if $KH_{0}$ has torsion, then the problem could potentially take on a whole new level of complexity (or could be equivalent, depending on whether or not it can be shown that the torsion part of $KH_{0}$ is at least controlled).
I hope someone does come along to pick this up again.  I no longer work in math professionally, but I would be very interested in seeing further progress on this.  -Adam
A: Adam Massey showed that K°(P(1,...,1,q)) = K°(P(1,...,1)).On the other hand P(1,...,1,q) is the cone with wertex (0, ...,0,1) which projects the Veronese variety Vq. Who knows any other particular nice geometrical exemples (small dimensions) of weighted projective spaces whith finitely generated algebraic GROTHENDIECK group (vector bundles) ?
