Timeline for Invariants of matrices (by simultaneous $\mathrm{GL}_n$ conjugation) over arbitrary rings
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Aug 26 at 23:24 | vote | accept | stupid_question_bot | ||
| Sep 17, 2021 at 1:58 | comment | added | stupid_question_bot | I think I've finally pieced it all together. Thank you so much! | |
| Sep 16, 2021 at 7:14 | comment | added | Wilberd van der Kallen | See Appendix B.8. | |
| Sep 15, 2021 at 6:47 | comment | added | Wilberd van der Kallen | It is true that the assumption is that $k$ is a field. But this is not used here in an essential way. Anyway, the answer to what filtration you should use to reduce to the $\mathbb Z$ finite modules is: Use the Donkin truncation functors for the action of $SL_2$ from the left. So let the saturated set $\pi$ of dominant weights grow in Donkin 1986. | |
| Sep 14, 2021 at 15:57 | comment | added | stupid_question_bot | Chapter II 4.20 in Jantzen only applies to fields (he makes the assumption "$k$ is a field" in the beginning of II 4). Similarly Donkin seems to only discuss this over fields, so my question was how to promote this to a result over $\mathbb{Z}$, given that the Lemma in Appendix B.9 only applies to $\mathbb{Z}$-finite $G$-modules. (For $\mathbb{Z}[n]$ you can just apply B.9 to the pieces of the degree filtration) | |
| Sep 14, 2021 at 12:22 | comment | added | Wilberd van der Kallen | See II 4.20 in the book. | |
| Sep 14, 2021 at 8:09 | comment | added | Wilberd van der Kallen | It looks like $\mathbb Z[SL_2]$ satisfies the cohomological criterion for good filtration as a $SL_2\times SL_2$ module. (Left and right action). | |
| Sep 14, 2021 at 7:33 | comment | added | Wilberd van der Kallen | First use conjugation by $SL_2$ instead. The advantage is that then it becomes the restriction of the action of a big product of $2n$ copies of $SL_2$. I believe one has a filtration of the coordinate ring whose layers are tensor products $H^0(\mu)\boxtimes H^0(\nu)$. Compare 2.2a in Donkin, On Schur Algebras and Related Algebras, 1 , JOURNAL OF ALGEBRA 104, (1986). And the diagonal inclusion into the big product is a Donkin pair. Do not have books here. | |
| Sep 14, 2021 at 4:13 | comment | added | stupid_question_bot | If we replace $\mathbb{Z}[n]$ with the coordinate ring of the product of $n$ copies of $SL_2$, do you know if this coordinate ring (with the same conjugation action by $GL_2$) should have a good filtration? I think I can see it over fields, but it's less clear to me how to deduce it over $\mathbb{Z}$. The lemma in Appendix B.9 only applies to $k$-finite $GL_2$-modules; while it's obvious how one can filter $\mathbb{Z}[n]$ w/ $k$-finite $G$-stable pieces, I don't see an obvious analog for the coordinate ring of $SL_2$. | |
| Sep 13, 2021 at 19:54 | vote | accept | stupid_question_bot | ||
| Aug 26 at 18:59 | |||||
| Sep 12, 2021 at 6:06 | comment | added | Wilberd van der Kallen | @stupid_question_bot The trivial representation is $V(0)$ and $0$ is dominant. | |
| Sep 10, 2021 at 18:10 | comment | added | stupid_question_bot | From chapter B, we know that $M$ having a good filtration implies that the higher Ext groups $Ext^i_G(V(\lambda),M)$ vanish for $i > 0$ and $\lambda$ dominant, whereas we want to see that $H^1(G,M)$ vanishes. Is the trivial representation a direct summand of $V(\lambda)$? | |
| Sep 9, 2021 at 7:36 | comment | added | Wilberd van der Kallen | Chapter B explains that it indeed is implied by the result over fields. You may also need that field extensions do not matter for good filtrations. So talking about algebraically closed fields is just meant to put the reader at ease. | |
| Sep 9, 2021 at 5:45 | comment | added | stupid_question_bot | Thank you! That is very helpful! When you say that Donkin proves that $M = \mathbb{Z}[n]$ has a good filtration, are you referring to his arguments in $\S3$ (on p399)? He doesn't seem to explicitly claim the existence of a good filtration over $\mathbb{Z}$ - he only seems to discuss it over alg. closed fields; is the result over $\mathbb{Z}$ implied by his arguments? (I am not especially familiar with representation theory so this is not clear to me) | |
| Sep 8, 2021 at 8:59 | history | answered | Wilberd van der Kallen | CC BY-SA 4.0 |