Timeline for Is this generalized character always a character?
Current License: CC BY-SA 4.0
25 events
| when toggle format | what | by | license | comment | |
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| Sep 23 at 19:27 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
typo corrected
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| Sep 23 at 12:11 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
updated with further cases of a positive answer
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| S Sep 19 at 15:04 | history | bounty ended | CommunityBot | ||
| S Sep 19 at 15:04 | history | notice removed | CommunityBot | ||
| Sep 19 at 4:18 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Modified unduly pessimistic statement
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| Sep 17 at 11:16 | comment | added | Geoff Robinson | @TomWIlde : I have put some comments related to this in the body of the question now. | |
| Sep 17 at 8:15 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Corrected omission
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| Sep 17 at 4:49 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
latex fix
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| Sep 16 at 12:48 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Included further general observations
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| Sep 16 at 9:59 | comment | added | Geoff Robinson | @TomWIlde : Yes, that is true, and I was aware of it, thanks . I had played with Adams operations a bit before, and also I have played with FS-indicators quite a lot in my time. In the case $p =2, $ in general, the FS indicator shows that certain weakenings of the definition of \Psi give generalized characters which are not characters. | |
| Sep 16 at 8:39 | comment | added | Tom WIlde | One last remark - hopefully correct if obvious - when $p=2$ and the Sylow $2$-subgroups of $G$ are elementary abelian, the assertion is (as per Will Sawin's comment) equivalent to $\nu(\chi)\ge 0$ for all $\chi\in\mathrm{Irr}(G).$ This is I think well known (it follows from e.g. Theorem 3 in doi.org/10.1016/0021-8693(76)90096-X). | |
| Sep 13 at 20:08 | comment | added | Geoff Robinson | @TomWIlde : To bad- it's good to pursue ideas, but they don't always work out- that's an occupational hazard. | |
| Sep 13 at 19:52 | comment | added | Tom WIlde | @GeoffRobinson : I'm afraid my ``conjecture'' fails for a group of order 864. My wishful thinking - mea culpa. | |
| Sep 13 at 13:30 | comment | added | Geoff Robinson | @TomWIlde : I would have to think about your first question., it doesn't seem obvious at first sight, and it would be a non-standard approach to the p-solvable case, but that might be good. I was aware of the second aside. Also, how $\chi_{0}$ decomposes into irreducibles is (sort of) well understood, but I think decomposing $\chi_{0}^{*}$ is a different matter. | |
| Sep 13 at 13:00 | comment | added | Tom WIlde | As a (probably irrelevant) aside, for any group $G,$ $\sum_{\chi\in\mathrm{Irr}(G)}[\chi^*_0,\chi]$ is the number of $p$-regular classes of $G.$ This follows immediately from second orthogonality, I have sometimes wondered if it is significant. There's a general question as to what irreducibles appear in $\chi^*_0,$ which again might be relevant. | |
| Sep 13 at 12:55 | comment | added | Tom WIlde | I wonder if the following might conceivably be true: $[\chi^*_0,1]_G\ge[\chi^*_0,1]_U$ whenever $G$ is $p$-solvable for odd $p,$ $\chi\in\mathrm{Irr}(G)$ and $U$ is a maximal subgroup of $G$ of index a power of $p.$ Here $\chi^*_0$ is defined as in the question. Using GAP I find so far that this holds for all groups of order dividing $432,$ when $p=3.$ This would give the desired result by induction for $p$-solvable $G$ ($p$ odd). | |
| Sep 13 at 0:19 | comment | added | Geoff Robinson | Thanks David. It has been checked by GAP, etc, for many groups (not by me, but by some people who were present when I asked it in a problem session at a meeting (where you were present). I believe it is true for "theoretical" reasons, but since I know no proof, I am not really entitled to that belief. | |
| Sep 12 at 23:21 | comment | added | David A. Craven | I don't know what you have tried, but the result holds for $PSL_2(q)$, $SL_2(q)$, $q\leq 31$, $PSL_3(q)$, $SL_3(q)$, $PSU_3(q)$, $PSp_4(q)$, $q\leq 9$, $A_n$ and $S_n$, $n\leq 15$, and all groups of order at most 200. | |
| S Sep 11 at 13:47 | history | bounty started | Geoff Robinson | ||
| S Sep 11 at 13:47 | history | notice added | Geoff Robinson | Draw attention | |
| Sep 11 at 2:16 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
Pointed out that stronger version holds in special case that $G$.has a normal $p$-complement.
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| Sep 10 at 11:15 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
added extra sentence to addendum
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| Sep 9 at 16:21 | history | edited | Geoff Robinson | CC BY-SA 4.0 |
pointed out extreme cases
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| Sep 9 at 14:26 | comment | added | Will Sawin | Your last criterion is equivalent the claim $\langle \psi^{p^k}, 1 \rangle \geq 0$ for all $\chi$ irreducible complex characters of $G$, $k$ sufficiently large, and $\psi^{p^k}$ the Adams operation: This operation produces $\chi_0^*$ up to a Galois conjugation that permutes the irreducible complex characters. But I don't think this makes proving it any easier. | |
| Sep 9 at 13:17 | history | asked | Geoff Robinson | CC BY-SA 4.0 |