Twisted sheaf $\mathcal{F}(n)$.

Question

Let $\mathcal{F}$ be a sheaf on a scheme $X$ and $O_X(k)$ as usual. We define $\mathcal{F}(n) = \mathcal{F} \otimes_{O_X} O_X(n)$, I don't undertand this definition. What is this tensor product? Then, if we can try to find examples, if $n=1$ and $\mathcal{F}=\mathcal{O}_{\mathbb{A}^1_k}$, who is $\mathcal{O}_{\mathbb{A}^1_k}(1)$?, and $\mathcal{O}_{\mathbb{A}^1_k}(n)$?

Answer 1

As already noted, the twisted sheaf $\mathcal O(n)$ is defined at first on projective space $\mathbb P^r$. The basic way to think about is that the global sections of $\mathcal O(n)$ on $\mathbb P^r$ is precisely precisely the vector space of homogeneous polynomials of degree $n$. So $\mathcal O(n)$ is a way of talking about homogeneous polynomials in a more geometric way.

You ask how to think about $\mathcal F(n)$ for a coherent sheaf $\mathcal F$. One basic example of a coherent sheaf is an ideal sheaf $\mathcal I_X$ cutting out some projective variety $X$ in $\mathbb P^r$. Then $\mathcal I_X(n)$ will be the sheaf whose global sections are the homogeneous polynomials of degree $n$ that vanish on $X$.

For an example of how to use these ideas:

Suppose that $X$ is a rational curve of degree $3$ in $\mathbb P^3$ (rational curve means that $X$ is isomorphic to $\mathbb P^1$). Degree $3$ means that a generic hyperplane in $\mathbb P^3$ cuts $X$ in $3$ points, which can be rephrased by saying that $\mathcal O_X(1)$ is a degree $3$ invertible sheaf on $X$. Thus $\mathcal O_X(n)$ is a degree $3n$ invertible sheaf.

We have the short exact sequence $$0 \to \mathcal I_X \to \mathcal O_{\mathbb P^3} \to \mathcal O_X \to 0,$$ and if we twist this becomes $$0 \to \mathcal I_X(n) \to \mathcal O(n) \to \mathcal O_X(n) \to 0.$$ Taking global sections gives a left exact sequence $$0 \to \text{degree $n$ hom. polys. vanishing on } X \to \text{ all degree $n$ hom. polys.} \to \Gamma( X, \text{degree $3n$ invertible sheaf}) $$ Now the dimension of the space of all degree $n$ hom. polys. on in $4$ variables is $n+3 \choose 3$, while the number of degree $3n$ hom. polys. in $2$ variables is $3n+1$. Thus this left exact sequence can be rewritten as $$0 \to \text{degree $n$ hom. polys. vanishing on } X \to {n+3 \choose 3} \text{-dimensional vector space } \to (3n+1)~\text{-dimensional vector space}.$$

Thus we see that if $n \geq 2,$ then the space of degree $n$ hom. polys. vanishing on $X$ is non-zero, and conclude for example that any rational cubic curve in $\mathbb P^3$ is contained in a (possibly singular) quadric.

There are lots more examples of this kind worked out in Chapter IV of Hartshorne, and that is a good place to look to get an idea of how to use these tools geometrically.

Answer 2

It only makes sense to talk about $\mathcal F(k)$ and $\mathcal O_X(k)$ if you have an embedding $i:X \hookrightarrow \mathbb P^n$ for some $n$. Then $\mathcal O_X(k)$ is the pullback of the $k$'th twist of the structure sheaf on $\mathbb P ^n$. In symbols, $\mathcal O_X(k) := i^* \mathcal O_{\mathbb P ^n}(k)$. For details, see Hartshorne's book, Chapter II, section 5 (page 120).

Thus, it doesn't make sense to talk about $\mathcal O_{\mathbb A^1}(k)$ without specifying an embedding $i:\mathbb A^1 \hookrightarrow \mathbb P ^n$ such that $i ^* \mathcal O_{\mathbb P ^n}(k)$ is invertible (I´m not sure if this is possible).

For an example, consider the projective line $\mathbb P ^1$. It can be realized as the homogeneous zero set of $\mathsf {Proj} \, k[x,y]$. Then to give a sheaf on $\mathbb P ^1$ is equivalent to giving a graded $k[x,y]$-module $M$ and sheafify to get a sheaf $\mathcal F = \tilde M$. Then $\mathcal F(n)$ is defined as $\mathcal F \otimes \mathcal O_{\mathbb P^1}(n)$, but this is just $\tilde{ M(n)}$, the sheafification of the twisted module. (Proposition 5.12 in Hartshorne, Chapter II)

To sum up: finitely-generated graded modules correspond to coherent sheaves on projective spaces and their twist corresponds to twisting the grading of the modules. (recall that the twisted module $M(k)$ is the same module, but with grading $M(k)_l = M_{l+k}$, so the zeroth piece of $M(k)$ is $M_k$)

Answer 3

I am sorry, but I think $O_X(n)$ can be defined when the underlying scheme is projective. So I think the proper example can be $\mathcal{O}_{\mathbb{P}^1_k}(1) $,and this is just tautological line bundle which is made from the dgree one parts of the oringinal graded ring.