Timeline for Invertible matrices satisfying $[x,y,y]=x$
Current License: CC BY-SA 2.5
12 events
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| Nov 10, 2010 at 15:49 | comment | added | user6976 | Remark: In order to apply the algorithm from the paper I mentioned, you will need to represent the group as $\langle a,b,y\mid yay^{−1}=ab,yby^{−1}=ba\rangle$ (it is easy to see that these two groups are isomorphic). Thus you need to find a pair of matrices $(a,b)$ which is conjugate to the pair $(ab,ba)$. | |
| Nov 10, 2010 at 15:41 | comment | added | user6976 | @Guntram: About your other comment: David tries to do everything in the whole ring of matrices. The denominators are going to be products of determinants. We can get rid of them because we can assume that the determinants are not 0. Of course we will get extra solutions. That is why after all solutions are found, we will need to check that the matrices are not singular. This is easier that assuming that the matrices are in SL because we get fewer equations (no equations $det(x)=1, det(y)=1$). Still David's idea does not work even for $n=3$. At least Maple refuses to find a solution. | |
| Nov 10, 2010 at 15:37 | comment | added | user6976 | @Guntram: I only just noticed your comment. For some reason it was not sent to my inbox, Anyway, if you use the trace identities that I mentioned, you will find all pairs of matrices $(x,y)$ up to conjugation that satisfy the relation $[x,y,y]=x$. If I remember correctly, it is a finite set. The algorithm can be found in my paper with C. Drutu: Druţu, Cornelia, Sapir, Mark. Non-linear residually finite groups. (English summary) J. Algebra 284 (2005), no. 1, 174–178. | |
| Nov 5, 2010 at 18:41 | comment | added | Guntram | @David: You don't need to clear denominators, as you can suppose that y is in SL_n. The degree of the polynomials will be n^2+n, though, not 3n. @Mark: Playing around, I found that there is a matrix x in SL_2(C) of order 6 and y of order 8 such that [x,y,y]=x - this is of course not an answer to your question as x^6 is unipotent. Do you have an explanation for this example, though? | |
| Nov 4, 2010 at 20:11 | comment | added | user6976 | @David: My favorite CAS (Maple) refuses to deal even with the 3-dim case. What is your favorite CAS that can do it? | |
| Nov 4, 2010 at 2:59 | comment | added | David E Speyer | OK, got it. Yeah, trace identities would be the way to do this for $n=2$, and maybe for $n=3$. I think just writing out the equations should win for $n=4$, though I haven't tried it. But my point was just that you should be doing these basic low dimensional checks, and it sounds like you are. | |
| Nov 4, 2010 at 2:23 | comment | added | user6976 | @David: Just to clarify my previous comment. Every pair of $2\times 2$- invertible matrices of det 1 is determined by the traces $tr(a), tr(b), tr(ab)$ up to conjugacy. There are polynomial identities allowing to compute the trace of the word $tr(w(a,b))$ if you know $tr(a), tr(b), tr(ab)$. Then the relation $[x,y,y]x^{-1}=1$ gives that certain trace is equal to 2, etc. | |
| Nov 4, 2010 at 1:37 | comment | added | user6976 | I did it for $n=2$, of course. The conjecture is true in that case. For $n=2$ you can use the trace identities. That reduces dimension to 3 (every pair of matrices is determined by three traces, if I remember correctly). It is written in the paper with Drutu which I mentioned above. For other $n$'s I did not check. There are no trace identities and the computation is too large. | |
| Nov 4, 2010 at 1:26 | history | undeleted | David E Speyer | ||
| Nov 4, 2010 at 1:25 | history | edited | David E Speyer | CC BY-SA 2.5 |
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| Nov 4, 2010 at 1:10 | history | deleted | David E Speyer | ||
| Nov 4, 2010 at 1:09 | history | answered | David E Speyer | CC BY-SA 2.5 |