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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!

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    $\begingroup$ 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. $\endgroup$ Commented Mar 22, 2022 at 8:47
  • $\begingroup$ Dear @FrancescoPolizzi, thanks for the reply! I see, thanks for the tip! $\endgroup$
    – Plano
    Commented Mar 22, 2022 at 9:16
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    $\begingroup$ 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
    Commented Mar 22, 2022 at 15:53
  • $\begingroup$ Another observation: usually the Cox ring, for free Picard group, is graded over $\mathbb{Z}^k$, while in this case the grading is over the (cartesian product of) naturals. I guess this follows precisely from the choice that the closure of the cone of effective divisor is contained in the "positive orthant" $\endgroup$
    – Plano
    Commented Mar 23, 2022 at 8:46

1 Answer 1

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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}} $.

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