5
$\begingroup$

On the blow-up of $2$-dimensional complex projective space, $\mathbb{P}^2$, I know that $$Pic(\mathbb{\tilde{P}^2})=\mathbb{Z}[\tilde{H}]+\mathbb{Z}[E]$$ where $\tilde{H}$ is the blow-up of the hyperplane bundle of $\mathbb{P}^2$, and $E$ is the exceptional divisor on $\tilde{\mathbb{P}}^2$. Note that $[\cdot]$ denotes numerical equivalence class.

I already know that the set $\{[\tilde{H}], [E]\}$ form a basis for $\tilde{\mathbb{P}}^2$. but how do the irreducible curves on the blow-up look like?

$\endgroup$

1 Answer 1

Reset to default
4
$\begingroup$

They are either strict transforms of irreducible curves from $\mathbb{P}^2$, or they are the exceptional divisor $E$.

$\endgroup$
5
  • $\begingroup$ Maybe I am getting some wires crossed: Is it true that $\{[\tilde{H}], [E]\}$ is a basis for $\tilde{\mathbb{P}}^2$? If so, does this mean that any can be written in the form of $\mathbb{Z}[\tilde{H}]+\mathbb{Z}[E]$? $\endgroup$
    – Alvey
    Mar 25, 2015 at 2:01
  • $\begingroup$ You mean a basis for the Picard group? Yes, that is correct. Every divisor is linearly equivalent to some combination of the pullback of a line, and the exceptional divisor. $\endgroup$
    – Jake Levinson
    Mar 25, 2015 at 2:16
  • $\begingroup$ This helps a lot. So... if we take an irreducible curve, say $C$, on $\mathbb{P}^2$, and its strict transform, $\tilde{C}$ in $\tilde{\mathbb{P}}^2$...then $\tilde{C}$ can be written in the form of $\mathbb{Z}[\tilde{H}]+\mathbb{Z}[E]?$. What confuses me is that if this is true, then, it appears $\tilde{C}$ won't be irreducible since it can be written in terms of the classes of $\tilde{H}$ and $E$. $\endgroup$
    – Alvey
    Mar 25, 2015 at 2:30
  • $\begingroup$ Unless, an irreducible curve $\tilde{C}$ on $\tilde{\mathbb{P}}^2$ is either of the form $\mathbb{Z}[\tilde{H}]+0[E]$ or of the form $0[\tilde{H}]+\mathbb{Z}[E]$....? $\endgroup$
    – Alvey
    Mar 25, 2015 at 3:12
  • 5
    $\begingroup$ I think you are confusing linear equivalence with irreducible decomposition. An irreducible curve can be linearly equivalent to a combination of $\tilde H$ and $E$. $\endgroup$
    – Jake Levinson
    Mar 25, 2015 at 4:41

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.