I'm looking for an explanation or a reference to why there is this equivelence of triangulated categories: $${D}^b(\mathrm {Coh}(\Bbb P^1))\simeq {D}^b(\mathrm {Rep}(\bullet\rightrightarrows \bullet))$$
It is my understanding that the only reason why $\Bbb P^1$ appears at all is because it is used to index the regular irreducible representations of the Kronecker quiver. I have also heard that this equivalence can be used to understand the geometry of $\Bbb P^1$.
I suppose more generally, adding arrows to the Kronecker quiver gives a similar result for $\Bbb P^n$.
Best Answer
Let $\mathcal O$ be the structure sheaf of $\mathbb P^1$. Then $\mathcal O \oplus \mathcal O(1)$ is rigid and generates the derived category of coherent sheaves on $\mathbb P^1$. Thus, it is a tilting object, and so the derived category is equivalent to the category of modules over its endomorphism ring, which is the path algebra of the Kronecker quiver.
For $\mathbb P^n$, you need $n+1$ objects to generate the derived category. You can take $\mathcal O(i)$ for $0\leq i \leq n$; this will be a tilting object again, but its endomorphism ring will not be hereditary; you will get a quiver with relations. This shouldn't be surprising, though: path algebras of quivers have global dimension one, so you shouldn't expect their derived categories to agree with derived categories of sheaves on higher-dimensional varieties.