Let $X$ be a quasi-projective variety, and let $L_1,...,L_n$ be ample line bundles on $X$. Is that true that if $E= \oplus L_i$, then

$$\mathbb{P}(E) \cong Proj(\bigoplus_{i_j \in \mathbb{N}^n} H^0(L_1^{\otimes i_1} \otimes...\otimes L_n^{\otimes i_n}))?$$

It is not true if $L_i$ are not ample, because then a priori $H^0(L_1^{\otimes i_1} \otimes...\otimes L_n^{\otimes i_n})$ may be 0 for any choice of $i_j$ (take for example on $\mathbb{P}^1$, $L_i=O(-i)$). I think that if $X$ is projective then it is true, maybe we could use that $H^0(O_X(1)) \otimes H^0(O_X(n)) \to H^0(O_X(n+1))$ is surjective for $n$ big enough, but I haven't been able to do it though.

  • 1
    $\begingroup$ Have you tried seeing if this works in the simplest example? :-| $\endgroup$ – Mariano Suárez-Álvarez May 1 '16 at 4:06
  • $\begingroup$ What do you mean? It seems to be working if you take $X$ projective and $L$ very ample, or $X=\mathbb{A}^n=D_{\mathbb{P}^n}(X_0) \subseteq \mathbb{P}^n$ and $L$ the restriction of $O(1)$, these are the simplest examples that are coming to my mind $\endgroup$ – User2 May 1 '16 at 4:32

I feel like I'm basically reiterating what Dtseng said, but maybe this will clear some things up.

Fix $X$ a scheme, and a set of line bundles $L_1,\ldots,L_n$. Denote $E = \bigoplus_{i=1}^n L_i$. We start by giving your graded ring a name.

Definition. The section ring or ring of sections of $L_1,\ldots,L_n$ is defined to be $$R(X;L_1,\ldots,L_n) := \bigoplus_{i_j \in \mathbf{N}^n} H^0\!\left(X,L_1^{\otimes i_1} \otimes L_2^{\otimes i_2} \otimes \cdots \otimes L_n^{\otimes i_n}\right).$$

The original reference for section rings is [EGAII, 4.5], but I also recommend [PAGI, §2.1] and [dFEM, §1.8].

We will need the following Lemma in both sections below:

Lemma. There is a decomposition $$\operatorname{S}(E) \cong \bigoplus_{i_j \in \mathbf{N}^n} L_1^{\otimes i_1} \otimes L_2^{\otimes i_2} \otimes \cdots \otimes L_n^{\otimes i_n},$$ where $\operatorname{S}(E) = \bigoplus_{d \ge 0} \operatorname{S}^d(E)$ denotes the symmetric algebra of $E$. In particular, we have $$H^0\!\left(X,\operatorname{S}(E)\right) \cong R(X;L_1,\ldots,L_n).$$

Proof. We have the following chain of isomorphisms: $$ \operatorname{S}(E) \cong \bigotimes_{i=1}^n \operatorname{S}(L_i) = \bigoplus_{i_j \in \mathbf{N}^n} \operatorname{S}^{i_1}L_1 \otimes \operatorname{S}^{i_2}L_2 \otimes \cdots \otimes \operatorname{S}^{i_n}L_n \cong \bigoplus_{i_j \in \mathbf{N}^n} L_1^{\otimes i_1} \otimes L_2^{\otimes i_2} \otimes \cdots \otimes L_n^{\otimes i_n} $$ The first isomorphism follows by using the universal property for symmetric algebras: see [EGAII, 1.7.1], or just note that this is the vector bundle version of [Lang, XVI, Prop. 8.2]. The second equality is just by definition. The last isomorphism is a consequence of the fact that for a line bundle $L$, we have $L^{\otimes n} \overset{\sim}{\to} \operatorname{S}^n L$ (one way to see this is that the canonical map from left to right is surjective, and both bundles have the same rank, so the map must be an isomorphism).

The last claim follows by taking global sections on both sides. $\blacksquare$

Now we can answer your question in the affirmative in some cases encompassing your examples. Recall that if $E = \bigoplus_{i=1}^n L_i$, by definition [Hartshorne, II, §7], we have $$\mathbf{P}(E) = \operatorname{\mathbf{Proj}}\!\left( \operatorname{S}(E) \right)$$ where $\mathbf{Proj}$ is the relative Proj construction over $X$.

$X$ an affine scheme, $L_i$ arbitrary

Suppose $X$ is an affine scheme. Then, the proof of your statement is fairly short:

Proposition. We have an isomorphism $$\mathbf{P}(E) \cong \operatorname{Proj}\!\left(R(X;L_1,\ldots,L_n)\right).$$

Proof. In the affine case, for any graded quasi-coherent $\mathcal{O}_X$-algebra $\mathscr{S}$, we have $$\operatorname{\mathbf{Proj}}(\mathscr{S}) \cong \operatorname{Proj}\!\left(H^0(X,\mathscr{S})\right)$$ by the construction for relative $\operatorname{\mathbf{Proj}}$. In our case, letting $\mathscr{S} = \operatorname{S}(E)$, we have $H^0\!\left(X,\operatorname{S}(E)\right) = R(X;L_1,\ldots,L_n)$ by the Lemma, and so the isomorphism follows by the definition of $\mathbf{P}(E)$. $\blacksquare$

$X$ a normal complete variety over a field $k$, $L_i$ ample

Suppose $X$ is a normal complete variety over a field $k$. We first use the Lemma above to reduce to the case when $n = 1$, i.e., there is only one line bundle.

Lemma. Consider the projective space bundle $\mathbf{P}(E)$. Then, we have an isomorphism of $k$-algebras $$R(X;L_1,\ldots,L_n) \cong R\!\left(\mathbf{P}(E);\mathcal{O}_{\mathbf{P}(E)}(1)\right).$$

Proof. Let $\pi\colon \mathbf{P}(E) \to X$ denote the projection morphism. By [EGAIII, Prop. 2.1.15] or [AVB, Lem. 3.1], there is a graded isomorphism of algebras $$\operatorname{S}(E) \cong \bigoplus_{d \ge 0} \pi_*(\mathcal{O}_{\mathbf{P}(E)}(1))$$ and taking global sections yields $R(\mathbf{P}(E);\mathcal{O}_{\mathbf{P}(E)}(1))$ on the left-hand side. The previous Lemma gives the right-hand side. $\blacksquare$

We can then conclude:

Proposition. If $L_1,\ldots,L_n$ are ample, then $$\mathbf{P}(E) \cong \operatorname{Proj}\!\left(R(X;L_1,\ldots,L_n)\right).$$

Proof. By the Lemma above, it suffices to show $$\mathbf{P}(E) \cong \operatorname{Proj}\!\left(R\!\left(\mathbf{P}(E);\mathcal{O}_{\mathbf{P}(E)}(1)\right)\right).$$ By [AVB, Lem. 2.2] or [PAGII, Prop. 6.1.13(i)], we have that $E$ is an ample vector bundle, and so $\mathcal{O}_{\mathbf{P}(E)}(1)$ is ample. Thus, the Proposition reduces to showing the following

Claim. If $Y$ is a complete normal variety over a field $k$, and $L$ is an ample line bundle on $Y$, then $$Y \cong \operatorname{Proj}\!\left(R(Y;L)\right).$$

and applying it to $Y = \mathbf{P}(E)$, $L = \mathcal{O}_{\mathbf{P}(E)}(1)$, since $\mathbf{P}(E)$ is normal if $X$ is. I can flesh out the details of the Claim if you'd like, but there is already a proof on Math.SE using Serre vanishing, and you can also prove this by using [Hartshorne, II, Exc. 5.13–14]. You can also look at this other Math.SE answer. $\blacksquare$

I'll end by remarking that even if $n = 1$, if we weaken the condition to $L$ semi-ample, that is, $L^{\otimes m}$ is globally generated for all $m \gg 0$, the isomorphism does not hold. The situation instead is that there is a canonical morphism $f\colon X \to \operatorname{Proj}\!\left(R(X;L)\right) =: Y$ such that $f_*\mathcal{O}_X = \mathcal{O}_Y$, but the morphism is only birational: a curve $C \subset X$ is contracted if and only if $(L \cdot C) = 0$. See [dFEM, Rem. 1.8.5] for this claim. We illustrate this with a concrete

Example. Let $X = \operatorname{Bl}_0\mathbf{P}^2$ be the blow-up of $\mathbf{P}^2$ at the origin, and consider $L = \mathcal{O}_X(\ell)$ the line bundle corresponding to the strict transform of a line $\ell$ in $\mathbf{P}^2$, not passing through $0$. Then, $L$ is semi-ample (since $\lvert \ell \rvert$ is already base-point free) but not ample since $(L \cdot E) = 0$, where $E$ is the exceptional divisor. Moreover, $$R(X;L) = \bigoplus_{d \ge 0} H^0(X,L^{\otimes d}) = \bigoplus_{d \ge 0} H^0\!\left(X,\mathcal{O}_X(d\ell)\right) = \bigoplus_{d \ge 0} H^0\!\left(\mathbf{P}^2,\mathcal{O}_{\mathbf{P}^2}(d\ell)\right) \cong k[x_0,x_1,x_2],$$ but certainly $\operatorname{Bl}_0\mathbf{P}^2 \not\cong \operatorname{Proj}(k[x_0,x_1,x_2]) = \mathbf{P}^2$ (since, e.g., $\operatorname{Bl}_0\mathbf{P}^2$ has a $(-1)$-curve while $\mathbf{P}^2$ does not).


I was so sure that this result was going to be false, and wrote something below saying to look at using relative proj instead. If that's what you were looking for, the original answer is at the bottom.

However, I think the claim is true if $X$ is projective, so, if you're interested, please check my work on this. If $X$ is quasi-projective, you have way too many sections, so there's no chance it'd work. For example, we could let $X=\mathbb{A}^1$ and $\mathscr{E}=\mathscr{O}$, in which case you're taking Proj of a ring that is not even finitely generated in degree 1.

First, we use the concept of an ample vector bundle, which was given in the answer here: How do we define ample vector bundles. One of the equivalent definitions of an ample vector bundle $\mathscr{E}$ over $X$ was that the tautological line bundle $\mathscr{O}_{\mathbb{P}\mathscr{E}}(1)$ is ample on $\mathbb{P}\mathscr{E}$.

So now there are two things to see:

1) The direct sum of ample vector bundles is ample.

2) $H^{0}(Sym^r\mathscr{E})=H^0(\mathscr{O}_{\mathbb{P}\mathscr{E}}(r))$.

It's probably possible to see this directly, but it's late, so I'm just going to dig through the article (http://archive.numdam.org/ARCHIVE/PMIHES/PMIHES_1966__29_/PMIHES_1966__29__63_0/PMIHES_1966__29__63_0.pdf ) given in the linked answer above to see that it's true. Claim 1) is given in Proposition 2.2, and this I think can be done directly from the definition.

Claim 2) is a special case of Lemma 3.1, where we let $F=\mathscr{O}_X$ to get $Sym^r\mathscr{E}=\pi_{*}\mathscr{O}_{\mathbb{P}\mathscr{E}}(r)$ where $\pi: \mathbb{P}\mathscr{E}\rightarrow X$ is the projection.

Finally, putting the two together, we get that $\mathscr{O}_{\mathbb{P}\mathscr{E}}(1)$ is ample, so $\mathbb{P}\mathscr{E}=Proj(\oplus_{r\geq 0}{H^0(\mathscr{O}_{\mathbb{P}\mathscr{E}}}(r)))=Proj(\oplus_{r\geq 0}{H^{0}(Sym^r\mathscr{E})})$, which is equal to the formula you gave above in the case where $\mathscr{E}$ is a direct sum of ample line bundles.

Original answer:

It seems to me that what you're looking for is the relative proj construction. In the relative proj construction, instead of having a graded ring over some base ring $A$, we have a sheaf of graded algebras over a scheme $X$. Over each open affine ${\rm Spec}(A)$ in $X$, the relative proj agrees with the usual version, and it glues together to a projective scheme over $X$.

Given that, if we have a locally free sheaf $\mathscr{F}$, we have $\mathbb{P}\mathscr{F}=\mathcal{Proj}({\rm Sym}^{\cdot}\mathscr{F})$ (whether we take the dual $\mathscr{F}^{\vee}$ inside of the symmetric power depends on your convention for what $\mathbb{P}\mathscr{F}$ means). If you assume $\mathscr{F}$ splits as a direct sum of line bundles and expand, you get something similar to your formula, but as a sheaf of graded rings over $X$ instead of just one graded ring. For a reference, this is definition 17.2.3 in Vakil's notes (http://math.stanford.edu/~vakil/216blog/FOAGdec2915public.pdf).

  • $\begingroup$ I am sorry to bother you again but I don't understand a few things.. I don't understand why is it not true for $\mathbb{A}^1$. I was thinking that with $\mathbb{A}^1$ you should get $Proj(k[x,t])$ where $x$ has degree 1 and $t$ has degree 0, so it is $\mathbb{P}^0_{\mathbb{A}^1}=\mathbb{A}^1$, where am I wrong? Moreover in your answer, how do you get $\mathbb{P}(E)=Proj(\bigoplus H^0(Sym^r(E)))$? (claim 2 is Hartshorne III.8.4 a) $\endgroup$ – User2 May 1 '16 at 11:39
  • $\begingroup$ Okay, you have to make a consistent choice of what you're taking Proj over. In the first answer, I'm assuming we're working with schemes over $k$ (I think we can generalize, but it's better to start with this), and all our Proj constructions are over $k$. With $\mathbb{A}^1$, if we take Proj over $k$, the degree 1 part is already $k[x]$, which is an infinite $k$-vector space, and we're already doomed. $\endgroup$ – DCT May 1 '16 at 13:35
  • $\begingroup$ Now, you might say that taking proj over $k$ was stupid, we should really take proj over $k[x]$, so we get the right answer. Then in the case where $X$ is not affine, say $\mathbb{P}^1$, we need to use relative proj, which is what I suggested in the original answer. $\endgroup$ – DCT May 1 '16 at 13:36
  • $\begingroup$ As for your second question, in general, if you have $X$ a projective variety over $k$ and $L$ a very ample line bundle, then $X=Proj(\oplus_{r\geq 0}{H^r(L^{\otimes r})})$. $\endgroup$ – DCT May 1 '16 at 13:38
  • $\begingroup$ Okay but in this case you don't know that $O_{\mathbb{P}(E)}(1)$ is very ample over $k$, it is just ample over $X$ $\endgroup$ – User2 May 1 '16 at 13:55

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