TITLE: Twisted sheaf $\mathcal{F}(n)$.
QUESTION [14 upvotes]: 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  $\mathcal{O}_{\mathbb{A}^1_k}(n)$?

REPLY [26 votes]: 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.