Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am always confused about how to understand the line bundle $\mathcal{O}_{X}(k)$ on a projective scheme $X=\mathrm{Proj}(\oplus_{n=0}^\infty A_{n})$. Of course this is by definition the $\mathcal{O}$-module associated to $\oplus_{n=0}^\infty A_{n}(k)$. But I think this definition is not really geometric. My questions are following;

  1. How should I understand $\mathcal{O}_{X}(k)$ intuitively? Maybe for $k\ge0$ this is relatively easy as global sections are given by the graded piece $A_{k}$, but I still don't know any geometric picture of this.
  2. Let $X=\mathbb{P}^n$, then is it easy to tell why the universal bundle is given by $\mathcal{O}_{\mathbb{P}^n}(-1)$ (without computing the transition function)?

I would appreciate it if you could provide me with your favorite ways to see these line bundles.

share|cite|improve this question
up vote 11 down vote accepted

Here is a rather category-theoretic explanation where these Serre twists come from. When the graded ring $A$ is finitely generated by $A_1$ over $A_0$, there is a well-known universal property of $\mathrm{Proj}(A)$. Namely, morphisms $Y \to \mathrm{Proj}(A)$ over $A_0$ correspond bijectively to line bundles $\mathcal{L}$ on $Y$ together with an $A_0$-linear epimorphism $A_1 \to \Gamma(Y,\mathcal{L})$. This describes the functor of points $\hom(-,\mathrm{Proj}(A))$. Therefore, actually this may serve as a definition of the Proj construction. Now the universal element of this representable functor is a line bundle $\mathcal{O}(1)$ on $\mathrm{Proj}(A)$ together with a epimorphism $A_1 \to \Gamma(\mathrm{Proj}(A),\mathcal{O}(1))$. More generally, $\mathcal{O}(k) := \mathcal{O}(1)^{\otimes k}$ for $k \in \mathbb{Z}$.

Intuitively, Serre twists make it possible to "shift to the affine case". If $\mathcal{F}$ is a coherent sheaf on an affine scheme, there is an epimorphism $\mathcal{O}^n \twoheadrightarrow \mathcal{F}$ (global generators). This is not true in the projective case. However, for every coherent sheaf $\mathcal{F}$ on a projective scheme $X$ with a choosen ample sheaf $\mathcal{O}(1)$ there is an epimorphism $\mathcal{O}^n \twoheadrightarrow \mathcal{F}(k) := \mathcal{F} \otimes \mathcal{O}(k)$ for $k$ large enough. Intuitively, we are just clearing denominators here in order to reduce to the affine situation. It follows that there is an exact sequence

$\mathcal{O}(-k_2)^{n_2} \to \mathcal{O}(-k_1)^{n_1} \to \mathcal{F} \to 0$

In the affine case, we have $k_1=k_2=0$ and this would be a description by generators and relations. Here, this is something really similar, we just have added degrees to the generators. This also shows that the category of coherent sheaves is generated by the Serre twists.

Similarily, cohomology vanishes after shifting high enough, etc.

share|cite|improve this answer
Thank you for the answer. This give a new point of view to think about $\mathcal{O}(k)$. – M. K. Oct 18 '12 at 8:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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