Timeline for Algebra generated by transformation matrices
Current License: CC BY-SA 4.0
27 events
| when toggle format | what | by | license | comment | |
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| Apr 21, 2023 at 20:06 | vote | accept | Mare | ||
| Apr 20, 2023 at 14:55 | history | became hot network question | |||
| Apr 20, 2023 at 12:47 | comment | added | Benjamin Steinberg | I now wrote a complete answer to your question | |
| Apr 20, 2023 at 11:37 | comment | added | Mare | @BenjaminSteinberg So it seems to me that $A_n$ is just Morita equivalent to the ring of 2x2 upper triangular matrices but $B_n$ is much more interesting. | |
| Apr 20, 2023 at 11:29 | comment | added | Benjamin Steinberg | You can compute it using it is the direct sum of the modules below. Ill get it later | |
| Apr 20, 2023 at 11:07 | comment | added | Mare | @BenjaminSteinberg The connection with the permutation matrices case would be nice. And indeed, it seems $A_n$ has 2 simple modules with a connected quiver and $B_n$ has $n$ simple modules with a connected quiver. So it should be easy to guess what $A_n$ is (but I have problems since it is not basic), but $B_n$ might be more mysterious. | |
| Apr 20, 2023 at 11:06 | comment | added | Benjamin Steinberg | The radical should consist of those matrices whose rows and columns both sum to 0 I think. I guess that is dimension n-1? | |
| Apr 20, 2023 at 10:59 | comment | added | Benjamin Steinberg | I believe that algebra from your permutation matrices is always the semisimple quotient and that explains the dimensions | |
| Apr 20, 2023 at 10:57 | comment | added | Benjamin Steinberg | I'll try to sort this out after coffee and breakfast | |
| Apr 20, 2023 at 10:52 | history | edited | Mare | CC BY-SA 4.0 |
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| Apr 20, 2023 at 10:50 | comment | added | Mare | @BenjaminSteinberg It seems the centers are one-dimensional, so that the quivers are connected. | |
| Apr 20, 2023 at 10:47 | comment | added | Benjamin Steinberg | I think the radical has codimension $1$ actually | |
| Apr 20, 2023 at 10:47 | history | edited | Mare | CC BY-SA 4.0 |
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| Apr 20, 2023 at 10:47 | answer | added | Benjamin Steinberg | timeline score: 7 | |
| Apr 20, 2023 at 10:39 | comment | added | Benjamin Steinberg | Sorry I just woke up. When my kid gets to school I'll work this out properly | |
| Apr 20, 2023 at 10:38 | comment | added | Benjamin Steinberg | The radical should be the matrices that annihilate the all ones row vector | |
| Apr 20, 2023 at 10:38 | comment | added | Benjamin Steinberg | Sorry the elements of T_n fix the all ones row vector so the algebra just fixes the suspace. | |
| Apr 20, 2023 at 10:35 | comment | added | Benjamin Steinberg | So I suspect this must have finite global dimension in characteristic 0 | |
| Apr 20, 2023 at 10:34 | comment | added | Benjamin Steinberg | $T_n$ has finite global dimension in characteristic 0 and an acyclic quiver. | |
| Apr 20, 2023 at 10:33 | comment | added | Benjamin Steinberg | Try arxiv.org/abs/1502.00959 although this might not be the final version | |
| Apr 20, 2023 at 10:28 | comment | added | Mare | @BenjaminSteinberg It seems the link you posted does not work. | |
| Apr 20, 2023 at 10:26 | comment | added | Mare | @BenjaminSteinberg Good question. In the semisimple case the image of the representation is the double centraliser which helps alot to calculate the algebra. In the modular case Im not sure how to calculate the image. But what you say should prove that the algebras both have acyclic quiver, so they both have finite global dimension. | |
| Apr 20, 2023 at 10:23 | comment | added | Benjamin Steinberg | I'd have to think more on B_n. | |
| Apr 20, 2023 at 10:16 | comment | added | Benjamin Steinberg | I believe A_n is the matrices fixing the all ones for vector assuming you have functions act on the left. | |
| Apr 20, 2023 at 10:14 | comment | added | Benjamin Steinberg | These representations are both projective indecomposable modules for Tn. See link.springer.com/article/10.1007/s10468-016-9597-. The quiver if T_n is known but but published yet so far as I know but relations are only known up to n=4. Can one determine the image of an algebra under a projective indecomposable representation from the quiver and relations up to morita equivalence | |
| Apr 20, 2023 at 7:03 | history | edited | Mare | CC BY-SA 4.0 |
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| Apr 20, 2023 at 6:54 | history | asked | Mare | CC BY-SA 4.0 |