Timeline for A polynomial homomorphism from Gl to the group of units is a power of the determinant
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 13, 2012 at 17:06 | comment | added | Bugs Bunny | QY, just apply the original map to the diagonal matrix $Diag(x,1,1,1...)$ | |
| Jul 13, 2012 at 14:42 | comment | added | Vladimir Dotsenko | Bugs Bunny: yes, you're right, I was typing it preoccupied with something else, and ended up saying something silly. All necessary information was there though :) | |
| Jul 13, 2012 at 13:54 | comment | added | Qiaochu Yuan | Maybe I am just being silly, but it is not clear to me that the resulting map $k^{\times} \to k^{\times}$ will be algebraic. | |
| Jul 13, 2012 at 12:53 | comment | added | Bugs Bunny | Jesko, Vladimir virtually answered your question: you have 4 2x2-matrices with $\pm 1$ on the diagonal. But there are 3 different embeddings of $K_4$ with this image... | |
| Jul 13, 2012 at 12:50 | comment | added | Bugs Bunny | Vladimir, your "embeddings" have kernels :-)) | |
| Jul 13, 2012 at 12:48 | comment | added | Bugs Bunny | David, except $n=2$ and $k=F_2$ only, which is a very special case as $k^\times$ is trivial:-)) So you dont need any extensions. If $k=F_3$, $SL_2$ is not perfect but still the commutant of $GL_2$ | |
| Jul 13, 2012 at 12:42 | comment | added | Vladimir Dotsenko | Elaborating on Bugs Bunny's Klein-4 suggestion: the group $\{\pm1\}\times\{\pm1\}$ has three nontrivial 1-dim'l representations: $\{(-1,1)\mapsto 1, (1,-1)\mapsto -1\}$, $\{(-1,1)\mapsto -1, (1,-1)\mapsto 1\}$, $\{(-1,1)\mapsto -1, (1,-1)\mapsto -1\}$. In general, finite abelian groups have a lot of embeddings into $GL_1$... | |
| Jul 13, 2012 at 12:29 | comment | added | darij grinberg | (I am aware that the less naive and closer-to-the-substance avatar of the question is about homomorphisms of algebraic groups rather than homomorphisms of abstract groups which happen to be polynomial maps; thus this question is somewhat tongue-in-cheek.) | |
| Jul 13, 2012 at 12:27 | comment | added | darij grinberg | Why could I always extend the base field without the map losing its homomorphism property? | |
| Jul 13, 2012 at 11:51 | comment | added | David E Speyer | Technical note: $SL_n(k) = [GL_n(k), GL_n(k)]$ in all cases EXCEPT $n=2$ and $k=\mathbb{F}_2$ or $\mathbb{F}_3$. Doesn't effect the end result, because you could always extend the base field before running your argument. See Lang's <i>Algebra</i>, Chapter XIII. | |
| Jul 13, 2012 at 11:02 | vote | accept | Jesko Hüttenhain | ||
| Jul 13, 2012 at 10:54 | comment | added | Jesko Hüttenhain | I don't fully understand: What exactly are these diagonal immersions? What do you mean by your last statement? | |
| Jul 13, 2012 at 9:50 | comment | added | Bugs Bunny | Off course, it is not independent. Take $G$ to be the Klein-4-group. It will have 3 diagonal immersions into $GL(2)$, each immersion different elements will have determinant 1. | |
| Jul 13, 2012 at 9:32 | vote | accept | Jesko Hüttenhain | ||
| Jul 13, 2012 at 10:54 | |||||
| Jul 13, 2012 at 9:31 | comment | added | Jesko Hüttenhain | That is, indeed, very nice and brief. Do you have any ideas about my edit? | |
| Jul 13, 2012 at 9:19 | history | answered | Bugs Bunny | CC BY-SA 3.0 |