Timeline for Splitting field for $\mathrm{GL}(2,p)$ - reference request
Current License: CC BY-SA 4.0
23 events
| when toggle format | what | by | license | comment | |
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| Jun 7, 2023 at 16:12 | comment | added | Benjamin Steinberg | @DavidA.Craven, thanks for the reference. | |
| Jun 7, 2023 at 16:10 | comment | added | David A. Craven | The Schur indices of the characters in question are 1 by projecteuclid.org/journals/… and so it just comes down to character values. The character table is well known, and so the result should hold for Q1. | |
| Jun 7, 2023 at 13:05 | vote | accept | Benjamin Steinberg | ||
| Jun 6, 2023 at 11:51 | answer | added | Geoff Robinson | timeline score: 4 | |
| Jun 6, 2023 at 10:13 | comment | added | Benjamin Steinberg | Thanks @DerekHolt. I meant every. | |
| Jun 6, 2023 at 8:10 | answer | added | Mare | timeline score: 2 | |
| Jun 6, 2023 at 6:08 | comment | added | Derek Holt | By the way, I find your use of "any" in 1 and 2 linguistically ambiguous. It could mean "does there exist ..." or "is this true for all ...". | |
| Jun 5, 2023 at 22:00 | comment | added | Dave Benson | @BenjaminSteinberg This sounds very interesting. I look forward to seeing the outcome of this work. | |
| Jun 5, 2023 at 18:44 | comment | added | Benjamin Steinberg | @NicholasKuhn, of course a splitting field for all the maximal subgroups of a monoid splits the monoid and so it all boils down to GL. infinite representation type is interesting for M_2(F_p) because GL(2,p) has finite representation type. An in characteristic different than p things look like pretty much like GL(2,p) and GL(1,p) put together, so this is the interesting case. For M_2(F_2) the basic algebra is a string algebra studied by Gelfand-Pomanarev but I don't have quiver presentation in general since it would connect to GL(2,p)'s quiver is a messy way. | |
| Jun 5, 2023 at 18:38 | comment | added | Benjamin Steinberg | @NicholasKuhn, yes that is what I was getting at. I think I have a nice new proof of Glover's theorem that $M_2(\mathbb F_p)$ has infinite representation type using CDE triangle for $M_2(F_p)$. To do this as cleanly as possible I wanted to choose a complete dvr with residue field $\mathbb F_p$ and field of fractions a splitting field for $M_2(\mathbb F_p)$ and I hoped that $\mathbb Q_p$ and $\mathbb Z_p$ worked, but it looks to me like $\mathbb Q_p$ doesn't have enough roots of unity so I have to cite the Cartan matrix doesn't change once you extend scalars over a splitting field. | |
| Jun 5, 2023 at 18:17 | comment | added | Nicholas Kuhn | Dave is right about your second question. And you may be interested in knowing that $\mathbb F_q$ is even a splitting field for the semigroup $M_n(\mathbb F_q)$. | |
| Jun 5, 2023 at 18:00 | comment | added | Benjamin Steinberg | @YCor, yes. Or equivalently, every simple $KG$-module remains simple when extending the scalars. | |
| Jun 5, 2023 at 17:56 | comment | added | YCor | What's a splitting field for a finite group $G$? in char. zero, does it mean a field $K$ for which $KG$ is a product of matrix algebras over $K$? | |
| Jun 5, 2023 at 17:47 | comment | added | Benjamin Steinberg | @DaveBenson, thanks. I don't really know the representation theory of $GL(n,q)$ so well, unfortunately. I found a nice paper doing the $p=2$ case and it seemed they constructed $p^2-p$ irreducibles over $\mathbb F_p$ using symmetric powers and twisting by the powers of the determinant and that seems to be the number of $p$-regular conjugacy classes if my back of the envelope calculation is correct. But the characteristic $0$ looked messier so I wasn't sure how the characters of the anisotropic torus come in. So it wasn't clear if I needed corresponding roots of units and $p^{th}$-roots | |
| Jun 5, 2023 at 17:38 | comment | added | Dave Benson | I think in general for $q$ a power of $p$, any field containing $\mathbb{F}_q$ is a splitting field for $GL(n,q)$. Doesn't this come from the construction of the irreducibles via Weyl modules? But then for example for $GL(2,p)$ the two eigenvalues of an element of order $p^2-1$ are Frobenius images of each other. So when you Brauer lift, you need a $(p^2-1)$th root of unity plus its $p$th power. | |
| Jun 5, 2023 at 17:32 | comment | added | Benjamin Steinberg | I guess I need $p^2-1$-roots of units to deal with matrices with eigenvalues in a quadratic extension. The exponent of $GL(2,p)$ is $p(p^2-1)$ but I am guessing the $p$ is not needed. | |
| Jun 5, 2023 at 17:31 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Jun 5, 2023 at 17:24 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Jun 5, 2023 at 17:24 | comment | added | Dave Benson | Looking at the Atlas, it seems to be false for $GL(2,7)$ too. | |
| Jun 5, 2023 at 17:17 | comment | added | David Loeffler | Isn't this false for $p = 3$? | |
| Jun 5, 2023 at 17:14 | comment | added | Dave Benson | Have you tried Cédric Bonnafé's book? | |
| Jun 5, 2023 at 17:10 | history | edited | Benjamin Steinberg | CC BY-SA 4.0 |
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| Jun 5, 2023 at 17:04 | history | asked | Benjamin Steinberg | CC BY-SA 4.0 |