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I was trying to do some computations to get familiar with twisting sheaves and I ran into some questions. I would greatly appreciate any help! Thanks in advance!

  1. For line bundles, being globally generated is the same as being basepoint free. The bundle $\mathscr{O}(-1)$ on $\mathbb{P}^n$ has no global sections which means that it must have a basepoint. But I am not sure how to go about finding this basepoint. Aren't the sections of $\mathscr{O}(-1)$ just degree -1 homogeneous elements of $\mathscr{O}_{\mathbb{P}^n}$? If so, on each open $U_{x_i}$, the sections are elements of the form $\frac{f}{x_i^n}$ where $f$ is a degree $n-1$ homogeneous element of $k[x_0,\dots,x_n]$. In this case, I don't see why all sections of $\mathscr{O}(-1)$ must vanish at the same point. (And what does it mean to be a basepoint of a line bundle that has no global sections? In Vakil's notes, a base point is defined to be where all sections vanish. To find a basepoint, does it suffice to check where all the global sections vanish?)

  2. I would like to compute the global sections of $\mathscr{O}(p+q)$ where $p=[0,1],q=[1,0]\in \mathbb{P}^1$ where $\mathbb{P}^1=$Proj$k[x,y]$. We can describe $p+q$ as the Cartier divisor $(U_x,y),(U_y,x)$ (I'm using Hartshorne's definition of a Cartier divisor). Then, the sections of the line bundle $\mathscr{O}(p+q)$ are: on $U_x$, elements of the form $\frac{1}{y}\frac{f}{x^n}$ where $f\in k[x,y]$ is homogeneous of degree $n$; on $U_y$, elements of the form $\frac{1}{x}\frac{f}{y^n}$ where $f\in k[x,y]$ is homogeneous of degree $n$. Then, global sections are $g\in k[x,y]$ such that $g=\frac{1}{x}\frac{f_1}{y^n}=\frac{1}{y}\frac{f_2}{x^n}$ for some $f_1,f_2\in k[x,y]$ degree $n$. This implies that global sections must be of the form $\frac{ax+by}{xy}$ which tells us that $h^0(\mathbb{P}^1,\mathscr{O}(p+q))=2$. I am a bit confused because I thought that by degree count, $\mathscr{O}(p+q)\cong \mathscr{O}(2)$ and so must have 3-dimensional global sections.

  3. Similarly I tried to compute $\mathscr{O}(p)$ where $p=[1,0]$ by describing $p$ as the Cartier divisor $(U_x,y),(U_y,y)$. This just gave me the trivial line bundle. I think the way I am describing/corresponding divisors to Cartier divisors might be incorrect but I haven't been able to figure out how.

user153327
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  • Thanks for the reply! But if you were to take your line bundle as $\mathscr{O}(-D)$ where $D$ is some divisor, then don't all sections have to vanish on D? – user153327 May 26 '14 at 02:36
  • Well now I'm confused, or maybe always was. I want to say that there's a difference between vanishing as a rational function and as a section of a line bundle, but I'm a little worried that we use that the opposite is true when passing between Weil and Cartier. I'll try to meditate more on this. – Hoot May 26 '14 at 02:55
  • @Hoot I am very confused myself. I always thought I had a pretty good understanding of divisors and line bundles until I tried to do the computation. It's quite embarrassing to be honest... – user153327 May 26 '14 at 04:52

1 Answers1

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  1. "All sections" means all global sections here. The only global section of $\mathscr O(-1)$ is the trivial one $0$, and it vanishes on all of $\mathbb P^n$.
  2. The Cartier divisor should be $\{(U_x,y/x), (U_y,x/y)\}$. In this case, a global section would look locally like $\dfrac{x}{y}\dfrac{f_1}{x^m}$ in $U_x$ or $\dfrac{y}{x}\dfrac{f_2}{y^n}$ in $U_y$. Notice that on the intersection $U_x\cap U_y$ we get $\dfrac{f_1}{x^m} = \left(\dfrac{y}{x}\right)^2\dfrac{f_2}{y^n}$, which is exactly how we would compute a global section via the transition function $(\dfrac{y}{x})^2$ of $\mathscr O(2)\cong\mathscr O(p+q)$.
  3. Again, the Cartier divisor must be modified. You want $\{(U_x,y/x),(U_y,1)\}$. The coordinate on $U_x\cong\mathbb A^1$ is $y/x$, not $y$.
Andrew
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  • Thanks a lot for the reply. I have just one more clarifying question. For "finding" the Cartier divisor, how would you go about that? For example, what would be the divisor corresponding the Cartier divisor $(U_x,y),(U_y,x)$? The locus where this divisor vanishes is also $[1,0]$ and $[0,1]$? Is this what you were hinting at in 3? i.e. since $U_x\cong \mathbb{A}^1=$Spec$k[y/x]$ so $[1,0]$ which is the "origin" in $U_x$ is the locus where $y/x$ vanishes, not $y$? – user153327 May 26 '14 at 17:18
  • Dear @user153327, your last question in the first comment is what I meant. An effective Cartier divisor, which is what we're dealing with, can be described as a collection ${(U_i, f_i) | f_i\in\mathscr O(U_i)}$ where the $U_i$ form an open covering (+ the usual condition on $f_i/f_j$'s). E.g. the point $[0:1]\in\mathbb P^1$ is the origin in the chart $U_y = \operatorname{Spec}(k[x/y])$, i.e. is defined by the equation $x/y = 0$ there, so $[0:1] = {(U_y,x/y), (U_x,1)}$ as a Cartier divisor. – Andrew May 26 '14 at 18:09
  • that helps a lot. Thanks! One last question (for real this time!): are the sections of $K^*$ on $U_x$ elements of the form $f/g$ where $f,g\in k(x,y)$ where $f$ and $g$ are homogeneous of same degree? Thanks again for your answers!! – user153327 May 26 '14 at 18:36
  • Dear @user153327, do you mean $f,g\in k[x,y]$? I prefer to think of $\mathscr K^*(U_x)$ as the function field $k(x/y) \cong k(y/x)$ of $\mathbb P^1$. – Andrew May 26 '14 at 19:32
  • Sorry about that typo. I did mean $f,g\in k[x,y]$. I was trying to follow Hartshorne's definitions. – user153327 May 26 '14 at 19:54