Definition: Let $X$ be a normal projective variety with finitely generated Picard group. Define the Cox ring of $X$ as the multisection ring $$\text{Cox}(X)=\bigoplus_{(m_1,\ldots,m_k)\in \mathbb{N}^k} \text{H}^0(X,m_1L_1+\ldots+m_kL_k),$$ where $L_1,\ldots,L_k$ are a basis of $\text{Pic}(X)_{\mathbb{Q}}$ and whose affine hull contains $\overline{\text{Eff}(X)}$.

This is the Hu-Keel defintion of Cox ring, and I would like to understand why the second extra property is required, since by looking at the literature it looks like there are very few cases in which this condition is asked, and there is no explanation to that. In particular, these are my doubts:

  • Why do they add this condition: I know it's vague, and it is a definition so it is not correct or wrong a priori, but for istance can we always find such a basis?
  • What do they mean by affine hull: I suspect they mean the convex hull, but these are two different notions.

I apologize in advance for this low-level (and probably not research-oriented) question, I've asked the same question on MSE without receving an proper answer (I've then deleted since they were equal, and here there are some comments), thus I understand if you want to delete it. Thanks in advance!

  • 1
    $\begingroup$ Welcome to MO. I suggest that you wait a bit more (2-3 days) for answers on MSE before cross-posting your question here. Maybe the downvote (that I do not understand) is due to this. $\endgroup$ Mar 22, 2022 at 8:47
  • 1
    $\begingroup$ I think they just mean the space consisting of all positive linear combinations of the $L_i$. $Eff(X)$ is a convex cone in $Pic(X)_{\mathbb R}$, so one can always find a basis so that this cone is contained in the "first quadrant". $\endgroup$ Mar 22, 2022 at 14:52
  • 1
    $\begingroup$ Also, the Cox ring is supposed to be a ring which contains all sections of all line bundles, so it would be strange to not contain all effective line bundles in the direct sum. (Up to numerical equivalence). $\endgroup$ Mar 22, 2022 at 14:52
  • 1
    $\begingroup$ Dear @wnx, thanks for the reply! So if I understand correctly you're basically saying that, among all the possible basis for $\text{Pic}(X)_{\mathbb{Q}}$, we simply choose one such that the $\text{Eff}(X)$ is contained in $\sum_{i=1}^k \mathbb{Q}_+L_i$, right? (which looks like the convex cone generated by $L_1,\ldots,L_k$, and not the affine hull) $\endgroup$
    – Plano
    Mar 22, 2022 at 15:53
  • 1
    $\begingroup$ Yes, I think that's right! $\endgroup$ Mar 22, 2022 at 16:02

1 Answer 1


I believe that the main point is that the pseudo-effective cone may not be a rational polyhedral cone. If the pseudo-effective cone is $ \ell $-dimensional, then there may be an extremal ray $ \rho $ of the pseudo-effective cone such that $ \rho $ is not equal to $ \mathbb{R}_{+} u $ where $ u \in \mathbb{Z}^{\ell} $. It would be meaningless to speak of $ H^{0}(X, D) $ where $ D \in N^{1}(X)_{\mathbb{R}} \setminus N^{1}(X)_{\mathbb{Q}} $.

However, it is very fruitful to talk about the Cox ring in such cases. For an example of a Cox ring which does not have finitely generated support look at An Introduction to Invariants and Moduli and Counterexample to Hilbert's fourteenth Problem for the 3-dimensional additive group, both by Shigeru Mukai. In the second reference, he discusses Nagata's construction of a representation of a vector group of dimension $ n-r $ of dimension $ 2n $ such that the ring of invariants is isomorphic to the Cox ring of the blow-up of $ \mathbb{P}^{r-1}_{k} $ at $ n $-points. In the first reference he proves that for appropriately chosen $ n $ and $ r $, that the pseudo-effective cone is not a finitely generated semigroup.

Because of this difficulty, I suspect they added the latter condition. This condition is indeed what @wnx said that $ \operatorname{Eff}(X) \subseteq \mathbb{Q}_{+} \mathcal{L}_{i} $. They were forced to do this because $ \operatorname{Eff}(X) $ may not be generated by elements of $ N^{1}(X)_{\mathbb{Q}} $.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.