Timeline for Does $SL_3(R)$ embed in $SL_2(R)$?
Current License: CC BY-SA 3.0
23 events
| when toggle format | what | by | license | comment | |
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| Apr 6, 2016 at 9:10 | answer | added | Uri Bader | timeline score: 23 | |
| S Nov 8, 2014 at 19:16 | history | suggested | Incnis Mrsi | CC BY-SA 3.0 |
refining tags: if Ī understand correctly, nobody considered non-commutative R (where a definition of SL is ambiguous-to-non-existent)
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| Nov 8, 2014 at 17:54 | review | Suggested edits | |||
| S Nov 8, 2014 at 19:16 | |||||
| Dec 4, 2010 at 15:06 | answer | added | Frogger | timeline score: 1 | |
| Nov 27, 2010 at 0:08 | answer | added | Peter Samuelson | timeline score: 1 | |
| Nov 26, 2010 at 21:30 | answer | added | Tim Dokchitser | timeline score: 6 | |
| Nov 26, 2010 at 20:32 | answer | added | Qiaochu Yuan | timeline score: 14 | |
| Nov 26, 2010 at 16:40 | comment | added | Tim Dokchitser | @Alex You're right of course, I was thinking $GL$! | |
| Nov 26, 2010 at 14:49 | comment | added | Alex B. | @Tim This is an embedding of $S_3$ into $GL_2(R)$, right? The determinant of the 2-cycles will be -1. | |
| Nov 26, 2010 at 12:46 | answer | added | Bugs Bunny | timeline score: 1 | |
| Nov 26, 2010 at 12:04 | comment | added | Kevin Buzzard | @BS: One might also need something like: "the map $SL(3,R)\to SL(2,R)$ extends to a map of rings $M(3,R)\to M(2,R)$" for this strategy to work, right? That doesn't sound so good (e.g. I don't think it's true for maps $SL(3,R)\to SL(3,R)$---things like "inverse-transpose" give trouble...)... | |
| Nov 26, 2010 at 11:59 | comment | added | Tim Dokchitser | For any (commutative) $R$ one can embed $S_3$ into $G=SL_2(R)$ using $[[0,-1],[1,-1]]$ of order 3 and $[[0,1],[1,0]]$ of order 2, and under this embedding the centralizer of $S_3$ is just the centre of $SL_2(R)$. I have a feeling it is impossible to embed $S_3$ into $SL_3(R)$ like that. Can one make this into a proof? | |
| Nov 26, 2010 at 11:35 | comment | added | BS. | If one could in some way recover the algebra $M_n(R)$ from the group $SL_n(R)$, one could try to recover $n$ via the [Amitsur-Levitzki theorem][1] [1]: eom.springer.de/a/a110570.htm | |
| Nov 26, 2010 at 11:17 | answer | added | Boris Bukh | timeline score: 2 | |
| Nov 26, 2010 at 10:32 | comment | added | Kevin Buzzard | Bah: "if $R$ is an integral domain then both these conditions" -> "if $R$ is an integral domain with $2\not=0$ then both these conditions" | |
| Nov 26, 2010 at 10:27 | comment | added | Kevin Buzzard | Actually, these comments make a good answer to another question currently kicking around MO: "what is the point of a universal object?". An $n\times n$ matrix over a general ring $S$ is a map from that ring $R$ above to $S$, reducing questions like CH for any ring, to CH for that one universal ring. One notices this when doing the first exercise I suggested: checking CH in the 2x2 case by bashing out the algebra. One has a bunch of equations in $A,B,C,D$ all of which magically work out, but instead of thinking of $A,B,C,D$ as varying through the ring one can think of them as poly variables. | |
| Nov 26, 2010 at 10:17 | comment | added | Alex B. | That works, thanks! Sorry about the off-topic comments. | |
| Nov 26, 2010 at 10:08 | comment | added | Kevin Buzzard | Alex: for the general case, consider the fact that you believe it for the polynomial ring $R=Z[A_{11},A_{12},...,A_{nn}]$ (which is an ID) and now apply it to the matrix whose $(i,j)$th entry is $A_{ij}$ and see what this tells you. | |
| Nov 26, 2010 at 10:05 | comment | added | Alex B. | I was just about to apologise for my silliness. I am still interested in the general case of $n\times n$ matrices though. | |
| Nov 26, 2010 at 10:01 | comment | added | Kevin Buzzard | Alex: If you have any doubts about Cayley-Hamilton for a 2x2 matrix over an arbitrary commutative ring, why not just bash out the algebra! | |
| Nov 26, 2010 at 9:54 | comment | added | Alex B. | @Kevin Do you have a reference for a proof of Cayley-Hamilton over arbitrary commutative rings? Or does your hint refer to the case of integral domains? Or does a special case of Cayley-Hamilton work if the matrix is of order 2 and 2 is not a zero-divisor? | |
| Nov 26, 2010 at 9:44 | comment | added | Kevin Buzzard | Your "order 2" remark I guess generalises to something like: if $R$ is a commutative ring with $2$ not a zero divisor, and if $x^2=1$ only has a finite number of solutions in $R$ (e.g. if $R$ is an integral domain then both these conditions are satisfied), then $SL(3,R)$ has more elements of order 2 then $SL(2,R)$ so there's no injection. [Hint: use Cayley-Hamilton on a matrix of order 2 to convince yourself it must be scalar, if I got it right; conversely use the fact that $-1$ isn't 1 to build more elements in $SL(3,R)$ than this] | |
| Nov 26, 2010 at 8:18 | history | asked | Jonathan Kiehlmann | CC BY-SA 2.5 |