2
$\begingroup$

We consider $\mathbb{P}^1$ and the semiorthogonal decomposition $<\mathcal{O},\mathcal{O}(1)>=D^b(\mathbb{P}^1)$. Let $x$ be a closed point and $k(x)$ the corresponding skyscraper sheaf. Every object $\mathcal{F}$ in $D^b(\mathbb{P}^1)$ is sitting in a triangle:

$B\rightarrow \mathcal{F}\rightarrow A\rightarrow B[1]$, for $B$ in $<\mathcal{O}(1)>$ and $A$ in $<\mathcal{O}>$.

Now my question: Is it possible to describe explicitely the objects $A$ and $B$ if $\mathcal{F}=k(x)$? And is it possible to say anything about the support of the objects $A$ and $B$ in this case?

$\endgroup$
3
  • 1
    $\begingroup$ The exact sequence $0 \to \mathcal{O} \to \mathcal{O}(1) \to \mathcal{F} \to 0$ gives a triangle in $D^b(\mathbb{P}^1)$. From this you see that one can take $A = \mathcal{O}[1]$ and $B = \mathcal{O}(1)$. $\endgroup$
    – naf
    Nov 2, 2013 at 14:02
  • $\begingroup$ Oh I see. But what happens if for example we take $\mathbb{P}^2$? Then because of $<\mathcal{O},\mathcal{O}(1),\mathcal{O}(2)>$ we wuold have two triangles... $\endgroup$
    – Aleksa
    Nov 2, 2013 at 15:31
  • 1
    $\begingroup$ Parodying Woody Allen I would say that the answer is given by the exact sequence $$0\rightarrow \mathcal{O}\rightarrow \mathcal{O}(1)^2\rightarrow \mathcal{O}(2)\rightarrow k(x)\rightarrow 0$$-- but what is the question?? $\endgroup$
    – abx
    Nov 3, 2013 at 8:56

0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Browse other questions tagged or ask your own question.