TITLE: How to imagine "tensoring with Serre's twisted sheaf"
QUESTION [21 upvotes]: What has an algebraic geometer in mind when (s)he sees $\otimes \mathcal{O}(1)$? I think it has something to do with an intersection of a hypersurface...?
Thanks, Adrian

REPLY [28 votes]: Normally, when I have a coherent sheaf $\mathcal{F}$, I'm most interested in its global sections $H^0(\mathcal{F})$, since these typically correspond to "actual things": polynomials, divisors, 1-forms, etc.
As Martin Brandenburg's answer says, $\mathcal{F}$ corresponds to a graded module $M = \bigoplus_{d \geq 1} M_d$ over the homogeneous coordinate ring, and $H^0(\mathcal{F}) = M_0$, the zero-th graded piece of the module. With that in mind, I normally think of the effect that tensoring with $\mathcal{O}(1)$ has on global sections: namely, it shifts to a different graded part of the module $M$. So in particular $H^0(\mathcal{F} \otimes \mathcal{O}(1)) = M_1$ and so on. (This is really just a reformulation of the "shifting" description in Martin's answer.)
For an example plucked out of thin air, say $\mathcal{J}$ is the ideal sheaf of $Y \subseteq \mathbb{P}^n$. Then global sections of $\mathcal{J}$ are actual (homogeneous) polynomials from the homogeneous ideal $J$ of $Y$ in $k[x_0, \ldots, x_n]$:
$$H^0(\mathcal{J} \otimes \mathcal{O}(d)) = J_d = \{\text{homogeneous degree-$d$ polynomials that vanish on Y} \}.$$
So "twisting" means looking at different parts of this graded ideal. Concrete geometrical properties like, say, "$Y$ is contained in a quadric hypersurface" can be stated as $H^0(\mathcal{J} \otimes \mathcal{O}(2)) \ne 0$.
Edit: To elaborate slightly on the ideal sheaf: $\mathcal{J}$ fits in the short exact sequence
$$ 0 \to \mathcal{J} \to \mathcal{O}_{\mathbb{P}^n} \to \mathcal{O}_Y \to 0,$$
with the structure sheaf of $Y$. (The map to $\mathcal{O}_Y$ is restriction to $Y$.) 
Now tensor with $\mathcal{O}(2)$ and take global sections to get
$$ 0 \to H^0(\mathcal{J} \otimes \mathcal{O}(2)) \to H^0(\mathcal{O}_{\mathbb{P}^n}(2)) \to H^0(\mathcal{O}_Y \otimes \mathcal{O}(2)).$$
You probably know that global sections of $H^0(\mathcal{O}_{\mathbb{P}^n}(2))$ are actual homogeneous polynomials of degree 2, so this shows that $H^0(\mathcal{J} \otimes \mathcal{O}(2))$ is a subset of these: the ones whose restriction to $Y$ is zero, i.e. the ones whose vanishing loci are quadric hypersurfaces that contain $Y$.