# 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" !

• By "Algebraic Grothendieck group", do you mean the Grothendieck group of vector bundles? – David E Speyer Jun 11 '13 at 17:40
• Yes , it is the Grothendieck group of vector bundles. The one of coherent sheaves is well-known. – Al-Amrani Jun 11 '13 at 17:51
• It is mentioned in the above paper that it was the initial idea to construct such examples with huge $K_0$ among weighted projective spaces, but so far without success. – Dietrich Burde Jun 11 '13 at 18:53
• Yes, that is right. But recently, Adam Massey ( KH-Theory of Complete Simplicial Toric Varieties and Algebraic K-theory of Weighted Projective Spaces ) obtained some progress in very particular case of weights, that is (1,...,1,q). – Al-Amrani Jun 11 '13 at 19:25
• In other words, the question is an open problem that looks well known among the people that know this sort of problems. Notice there is a minisection in the FAQ mathoverflow.net/faq about this. – Mariano Suárez-Álvarez Jun 11 '13 at 19:40

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.

• Thank you for letting me know about your result on algebraic Grothendieck group of weighted projectif spaces .1) As I understood, the case of singularities not isolated is still open . Am I right ? 2) Topological complex K-theory of w.p.s.'s is well known ( Al-Amrani JPPA, 93(2) 1994). Did you study comparison between the algebraic and the topological K-groups ( in case of islated singulaties ) ?. – Al-Amrani Sep 5 '19 at 13:00
• Yes, the non-isolated case is open, and seems much more difficult. In the isolated case, I am pretty sure that the map from algebraic K-theory to topological K-theory is an isomorphism integrally but I haven't checked the details of this. With rational coefficients this does follow from our work because rationally G_0 and K_0 of wps with isolated singularities are isomorphic. – Evgeny Shinder Sep 6 '19 at 13:43

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

• Please, what is the meaning of part (d) in Theorem 1.1. (Introduction) in your paper ? – Al-Amrani Jul 15 '15 at 15:47
• That's the statement that for $n \leq 0$ that $K_{n}(P(1,...,1,a))$ is isomorphic to $K_{n}(P^{d})$ (where $P(1,...,1,a)$ is a WPS of dim d and where only one weight is bigger than 1). This comes from a few results. First, one can show that $P(1,...,1,a)$ is $K_{0}$-regular (so $(\mathcal{F}_{K})_{n} = 0$ for $n \leq 0$; follows from part (c) of Thm 1.1). Second, one can show that for WPS of that form that there is an easy resolution of singularities that implies $KH_{n}(P(1,...,1,a))$ is isomorphic to $KH_{n}(P^{d})$ for $n \leq 0$ (this is Corollary 6.4). Was there a specific question? – A. Masssey Jul 15 '15 at 23:53
• Reading your answer below, it seems you got the correct meaning. – A. Masssey Jul 15 '15 at 23:57
• Thank you for your detailed replay. If you forget the realm of WPS's, and you deal only with cones projecting Veronese varieties , do the computations change or not ? – Al-Amrani Jul 19 '15 at 13:35
• Well I never really thought of the $KH$ component in this way, but that is how I approached the calculation of the $\mathcal{F}_{K}$ components. The $(\mathcal{F}_{K})_{n}(P(1,,,1,a))$ will be $(\mathcal{F}_{K})_{n}(R)$ where $R$ is the affine coordinate ring of the cone over the degree a Veronese embedding, and this was the approach I used to show $K_{0}$-regularity As far as I am aware, the latest that's been done towards addressing K-Theory of these kinds of rings was here: arxiv.org/abs/0905.4642 I'd probably start there if interested. – A. Masssey Jul 20 '15 at 0:45

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) ?