Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I'm wondering if there's always a (not too complicated?) way to characterize a matrix group by conditions on the coefficients.

I know if I'm dealing with matrix groups over a field, then it's sort of a question of seeing if the matrix group is Zariski-closed, which would correspond to polynomial conditions on the coefficients, though I'm not necessarily only interested in polynomial conditions, and I want to work with integer coefficients.

More specifically, I've got four matrices $A,B,C,D$ in $SL(2,\mathbb{Z})$, and I'm looking at the subgroup of the group generated by these four matrices (which have the sole relation $DCBA = 1$), such that the number of times $A$ or $C$ appears in any word is even.

Actually, (Formatting looks weird, these are meant to be $2\times2$ matrices and the coefficients are ordered as top left, top right, bottom left, bottom right) $$A = \left[\begin{array}{ll} 1 & 5 \\ 0 & 1\end{array}\right]$$ $$B = \left[\begin{array}{ll} 1 & 0 \\ -1 & 1\end{array}\right]$$ $$C = \left[\begin{array}{ll} 11 & 20 \\ -5 & -9\end{array}\right]$$ $$D = \left[\begin{array}{ll} 11 & 25 \\ -4 & -9\end{array}\right]$$

($A,B,C,D$ actually generate the congruence group $\Gamma^1(5)$)

In fact, if we let $$M = \left[\begin{array}{ll} -2 & -5 \\ 1 & 2\end{array}\right]$$ then $C = MAM^{-1}$, and $D = MBM^{-1}$.

share|improve this question
    
I took the liberty of fixing the formatting (see the sidebar item "Basic solution") –  Yemon Choi Oct 16 '12 at 4:46
1  
I took the liberty of changing $B$ to a matrix with determinant $1$ rather than $0.$ –  Will Jagy Oct 16 '12 at 4:56
    
Note that your matrices do not seem to fit the group, see en.wikipedia.org/wiki/… –  Will Jagy Oct 16 '12 at 5:19
1  
@Will: $\Gamma^1(5)$, not $\Gamma_1(5)$; $\Gamma^1$ has the top right entry divisible by 5. I corrected $B$ again, so it is now in $\Gamma^1(5)$ and the claimed relation $D = MBM^{-1}$ holds. But it's not totally clear to me what the question here is, and as far as I can see it has nothing whatsoever to do with modular forms. –  David Loeffler Oct 16 '12 at 9:04

1 Answer 1

The question is rather vague, but I hope that the following is helpful.

You seem to want a solution to the membership problem for your subgroup of $SL(2,\mathbb{Z})$. Note that such a solution is known not to exist in $SL(4,\mathbb{Z})$, which contains a copy of $F_2\times F_2$; I suspect its existence is open for $SL(3,\mathbb{Z})$.

However, $SL(2,\mathbb{Z})$ is group-theoretically much nicer, being as it is virtually free. The membership problem is certainly solvable here quite efficiently, though I don't know an implementation off the top of my head. The methods involved go back to Stallings's notion of folding. A google search turned up the paper 'Membership problem for the modular group' by Gurevich and Schupp.

share|improve this answer
1  
By the way, if you are correct and the subgroup generated by $A,B,C,D$ is free on $A,B,C$ (I haven't checked), then 'the subgroup ... such that the number of times A or C appears in any word is even' is generated by $A^2, C^2, AC, B, ABA^{-1}$. –  HJRW Oct 16 '12 at 10:22
    
Note also that the fact that the membership problem is not solvable in general shows that there are subgroups which are not cut out by a family of polynomials. –  HJRW Oct 16 '12 at 13:09

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.