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changed P's to better fit literature
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Ali Caglayan
Ali Caglayan
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Why are coherent sheaves on $\mathbf$\Bbb P^1$ derived equivalent to representations of the Kronecker quiver?

I'm looking for an explanation or a reference to why there is this equivelence of triangulated categories: $${D}^b(\mathrm {Coh}(\mathbf P^1))\simeq {D}^b(\mathrm {Rep}(\bullet\rightrightarrows \bullet))$$$${D}^b(\mathrm {Coh}(\Bbb P^1))\simeq {D}^b(\mathrm {Rep}(\bullet\rightrightarrows \bullet))$$ It is my understanding that the only reason why $\mathbf P^1$$\Bbb P^1$ appears at all is because it is used to index the regular irreducible representations of the Kronecker quiver. I have also heard that this equivalence can be used to understand the geometry of $\mathbf P^1$$\Bbb P^1$.

I suppose more generally, adding arrows to the Kronecker quiver gives a similar result for $\mathbf P^n$$\Bbb P^n$.

Why are coherent sheaves on $\mathbf P^1$ derived equivalent to representations of the Kronecker quiver?

I'm looking for an explanation or a reference to why there is this equivelence of triangulated categories: $${D}^b(\mathrm {Coh}(\mathbf P^1))\simeq {D}^b(\mathrm {Rep}(\bullet\rightrightarrows \bullet))$$ It is my understanding that the only reason why $\mathbf P^1$ 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 $\mathbf P^1$.

I suppose more generally, adding arrows to the Kronecker quiver gives a similar result for $\mathbf P^n$.

Why are coherent sheaves on $\Bbb P^1$ derived equivalent to representations of the Kronecker quiver?

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 $\Bbb P^n$.

made latex adhere to 'standard'
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Ali Caglayan
Ali Caglayan
  • 1.2k
  • 10
  • 24

I'm looking for an explanation or a reference to why there is this equivelence of triangulated categories: $$\mathcal{D}^b(\mathsf{Coh}(\mathbf P^1))\simeq \mathcal{D}^b(\mathsf {Rep}(\bullet\rightrightarrows \bullet))$$$${D}^b(\mathrm {Coh}(\mathbf P^1))\simeq {D}^b(\mathrm {Rep}(\bullet\rightrightarrows \bullet))$$ It is my understanding that the only reason why $\mathbf P^1$ 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 $\mathbf P^1$.

I suppose more generally, adding arrows to the Kronecker quiver gives a similar result for $\mathbf P^n$.

I'm looking for an explanation or a reference to why there is this equivelence of triangulated categories: $$\mathcal{D}^b(\mathsf{Coh}(\mathbf P^1))\simeq \mathcal{D}^b(\mathsf {Rep}(\bullet\rightrightarrows \bullet))$$ It is my understanding that the only reason why $\mathbf P^1$ 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 $\mathbf P^1$.

I suppose more generally, adding arrows to the Kronecker quiver gives a similar result for $\mathbf P^n$.

I'm looking for an explanation or a reference to why there is this equivelence of triangulated categories: $${D}^b(\mathrm {Coh}(\mathbf P^1))\simeq {D}^b(\mathrm {Rep}(\bullet\rightrightarrows \bullet))$$ It is my understanding that the only reason why $\mathbf P^1$ 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 $\mathbf P^1$.

I suppose more generally, adding arrows to the Kronecker quiver gives a similar result for $\mathbf P^n$.

Source Link
Ali Caglayan
Ali Caglayan
  • 1.2k
  • 10
  • 24

Why are coherent sheaves on $\mathbf P^1$ derived equivalent to representations of the Kronecker quiver?

I'm looking for an explanation or a reference to why there is this equivelence of triangulated categories: $$\mathcal{D}^b(\mathsf{Coh}(\mathbf P^1))\simeq \mathcal{D}^b(\mathsf {Rep}(\bullet\rightrightarrows \bullet))$$ It is my understanding that the only reason why $\mathbf P^1$ 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 $\mathbf P^1$.

I suppose more generally, adding arrows to the Kronecker quiver gives a similar result for $\mathbf P^n$.