Timeline for Quiver representations
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 7 at 9:03 | comment | added | Hugh Thomas | @AndrewHubery very cool, thanks! | |
| Aug 7 at 6:30 | comment | added | Andrew Hubery | For $E_8$ you can argue as follows. Let $\theta$ be the largest root. This is fixed under all but one reflection, say $s$ corresponding to the leaf on the longest arm, and $s(\theta)=\theta-e$. If $c$ is the Coxeter transformation, then one of $c^\pm(\theta)$ equals $\theta-e$. Choosing one, say $c$, we can then compute $\dim\mathrm{Hom}(\tau X,X)$ using the Euler form, where $X$ is indecomposable of dimension vector $\theta$. This has dimension 2, so the Auslander-Reiten sequence has three middle terms. | |
| Aug 7 at 5:48 | comment | added | Hugh Thomas | @AndrewHubery How do you know the representation corresponding to the highest root will be in the AR translation orbit of the projective at the central vertex? | |
| Aug 6 at 17:23 | comment | added | Andrew Hubery | A more exact solution would be take the projective (or injective) for the central vertex, and then repeatedly compute the (inverse) Auslander-Reiten translate. This construction is detailed in many places. | |
| Aug 6 at 17:19 | comment | added | Andrew Hubery | A quick addition. As Hugh mentions, if $d$ is a positive root for a Dynkin quiver $Q$, then the general representation of dimension vector $d$ is indecomposable. So a reasonably efficient way of finding explicit matrices would be to make some suitably generic choices and then computing the endomorphism ring. If you get that the endomorphism ring is the base field, then you chose wisely. | |
| Jun 24, 2014 at 5:11 | history | edited | Hugh Thomas | CC BY-SA 3.0 |
added 363 characters in body
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| Jun 24, 2014 at 2:52 | history | answered | Hugh Thomas | CC BY-SA 3.0 |