Timeline for $G/[G,G]$, irreps and conjugacy classes
Current License: CC BY-SA 3.0
20 events
| when toggle format | what | by | license | comment | |
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Sep 14, 2012 at 11:34 | comment | added | Alexander Chervov | @Volodya Is it "Bijection between irrep..." ? mathoverflow.net/questions/102879/… (but \tiled A_5 is not discussed there) or some other ? Can you give a link ? | |
| Sep 13, 2012 at 16:59 | comment | added | Vladimir Dotsenko | @Alexander: Qiaochu in one of his comments refers to "a previous MO discussion", what exactly is not clear? | |
| Sep 8, 2012 at 11:25 | comment | added | Alexander Chervov | @Volodya You write: "@Qiaochu Yuan: yes I am aware of that discussion, " what is "that discussion" ? | |
| Jul 25, 2012 at 2:37 | comment | added | Will Sawin | Yes, that's what I meant. Sorry for the confusion. | |
| Jul 24, 2012 at 18:57 | comment | added | Geoff Robinson | @Will: My problem was what you meant by ``up to multiplication by a $1$-dimensional representation." If you meant that there is one map for each $G/[G,G]$-orbit (strictly the dual group), acting by multiplication of irreducible characters, then it's clear to me. | |
| Jul 24, 2012 at 0:01 | comment | added | Will Sawin | I did not claim that the determination of a homomorphism to $PGL_n(C)$ was unique, nor that, if the original map was faithful, the second map would be as well. | |
| Jul 23, 2012 at 18:53 | comment | added | Geoff Robinson | @Will: up to multiplication by scalars, not up to multiplication by a $1$-dimensional representation. For example ${\rm SL}(2,5)$ has a faithful $2$-dimensional complex representation, but no non-trivial one-dimensional representation. However ${\rm SL}(2,5)$ is not isomorphic to a subgroup of ${\rm PGL}(2,\mathbb{C}).$ | |
| Jul 23, 2012 at 18:26 | comment | added | Will Sawin | On the positive side, an $n$-dimensional representation up to multiplication by $1$-dimensional representations gives a map $G \to PGL_n(\mathbb C)$. Given a conjugacy class of $G$, how would one produce a map $G \to PGL_n(\mathbb C)$? | |
| Jul 23, 2012 at 18:24 | comment | added | Will Sawin | One might be able to make Qiaochu's argument damning if the associated representations of the outer automorphism group of $A_5$ were the same. Unfortunately since that is just $\mathbb Z/2$ this won't happen, but are there other perfect groups with larger/less commutative outer automorphism groups? | |
| Jul 23, 2012 at 16:26 | comment | added | Vladimir Dotsenko | (in the previous comment $2\cos(\pm2\pi/5)$ must be $1+2\cos(\pm2\pi/5)$, of course) | |
| Jul 23, 2012 at 16:19 | comment | added | Vladimir Dotsenko | @Qiaochu Yuan: as for $A_5$, it is natural to me to assign to the 5-cycle $(12345)$ the representation for which the character of the corresponding element is $2\cos(\pm 2\pi/5)$ since in the geometric representation of the pentagon the corresponding cycle does rotate through $\pm2\pi/5$. ;-) I know more imagination is required further, I am just saying that there is no need to discard this idea right away. | |
| Jul 23, 2012 at 16:01 | comment | added | Vladimir Dotsenko | @Qiaochu Yuan: yes I am aware of that discussion, but somehow my question is different. I am not saying I have a clear reason to think that $G/[G,G]$ acts naturally on conjugacy classes, but I have a flicker of hope.. | |
| Jul 23, 2012 at 15:58 | comment | added | Qiaochu Yuan | I also disagree that $G/[G, G]$ is responsible for the ambiguities. Can you write down a natural bijection from the conjugacy classes of the binary icosahedral group $\tilde{A_5}$ (which is perfect) to its irreps? | |
| Jul 23, 2012 at 15:48 | comment | added | Qiaochu Yuan | The group to me that most obviously acts on conjugacy classes is $\text{Out}(G)$, and previous discussion on MO (mathoverflow.net/questions/21606/…) already shows that actions of $\text{Out}(G)$ on conjugacy classes and irreps are different in general. | |
| Jul 23, 2012 at 15:24 | history | edited | Vladimir Dotsenko | CC BY-SA 3.0 |
deleted 168 characters in body
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| Jul 23, 2012 at 15:22 | comment | added | Vladimir Dotsenko | @David: yes you are right. There seems to be a problem. I shall for the moment formulate a less specific version of my question, and think of how to repair it. | |
| Jul 23, 2012 at 14:49 | comment | added | David E Speyer | I know you want the answer to be "it takes it to the class of the $n$-cycle", but I don't understand how you get that from abstract group theory. | |
| Jul 23, 2012 at 14:48 | comment | added | David E Speyer | I'm confused. How does $G/[G,G]$ act on conjugacy classes? For example , if $G$ is $S_n$, how does the nontrivial element of $S_n/[S_n, S_n]$ act on the identity? | |
| Jul 23, 2012 at 14:37 | history | asked | Vladimir Dotsenko | CC BY-SA 3.0 |