Timeline for Is this modified bound quiver algebra necessarily representation-finite?
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Dec 23, 2015 at 1:35 | comment | added | Rachel | @DagOskarMadsen - Oh wow, I didn't realize that's what that meant! I see it in the definitions now. I was thinking that the length of the path was counting the vertices, not the number of arrows. Thanks for the clarification. | |
| Dec 21, 2015 at 11:23 | comment | added | Dag Oskar Madsen | This means you don't allow commutative triangles in your quiver then (path of length 2 never equal to path of length 1). | |
| Dec 21, 2015 at 10:59 | comment | added | Rachel | @DagOskarMadsen - Since the ideal $I$ is just the ideal generated by the commutative relations from the arrows in the quiver, then it is admissible, yes. | |
| Dec 19, 2015 at 22:34 | vote | accept | Rachel | ||
| Dec 18, 2015 at 9:29 | answer | added | Jeremy Rickard | timeline score: 9 | |
| Dec 18, 2015 at 9:14 | comment | added | Dag Oskar Madsen | In your definition of bound quiver algebra, is $I$ supposed to be contained in the square of the arrow ideal? | |
| Dec 17, 2015 at 19:26 | history | edited | Rachel | CC BY-SA 3.0 |
grammar
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| Dec 17, 2015 at 17:03 | history | asked | Rachel | CC BY-SA 3.0 |