Timeline for Can we infer an isomorphism of quivers from an isomorphism of their corresponding path algebras?
Current License: CC BY-SA 3.0
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| Mar 21, 2017 at 20:05 | comment | added | Jeremy Rickard | @BenjaminSteinberg I think it's an easy calculation with explicit projective resolutions. Note that once you've chosen the modules $M_v$ you can redefine the loops in $KQ$ at each vertex $v$ by adding a suitable multiple of the idempotent $e_v$ (without changing the path algebra) so that all arrows act as zero on $M_v$, to simplify the calculation. | |
| Mar 21, 2017 at 14:53 | comment | added | Benjamin Steinberg | is it easy to see Ext1 computes the edges in the non acyclic case? | |
| Mar 21, 2017 at 12:14 | history | answered | Jeremy Rickard | CC BY-SA 3.0 |