How to imagine "tensoring with Serre's twisted sheaf"

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

Solution 1:

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 $Y$.

Solution 2:

I cannot really speak for algebraic geometers, but one can imagine this operation coming from graded $A$-modules (when you work with the scheme $\mathrm{Proj}(A)$), where $\mathcal{O}(1)$ is associated to $A[1]$ and tensoring with $A[1]$ just shifts the degrees (to the left). So technically speaking nothing much happens. And we can always undo this move by tensoring with $A[-1]$. On the other hand, Serre's Theorem tells us that if we make this move often enough, the coherent sheaf will be generated by global sections, which places us in the affine case. This corresponds to the fact (and is proven by) that a fraction $a/f^k$ can be "cleared" when we multiply with $f^k$.

Solution 3:

Suppose first that $L$ is a locally free sheaf of rank $1$ on $\mathbb P^n$.
Then on sufficiently small affine open subsets $U\subset \mathbb P^n$ the sections $\Gamma(U,L)$ of $L$ correspond to the regular functions $\mathcal O(U)$ on $U$.
In contrast the sections $\Gamma(U,L\otimes_\mathcal O \mathcal O(1))$ of $F$ correspond to rational functions on $U$ having poles of order at most $1$ on $U$ but not necessarily regular: twisting produces more sections.
If $F$ is an arbitrary coherent sheaf on $\mathbb P^n$, it is a quotient $\oplus L_i\to F\to 0$ of a finite direct sum $\oplus L_i$ of locally free sheaves of rank $1$ (cf. Serre's FAC, page 248, Corollaire) and the above intuition applies (albeit less vividly...).