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Let $S_4 = \left(\begin{array}{cc}0&-1 \\\ 1&0 \end{array}\right) \textrm{ and } S_6 = \left(\begin{array}{cc} 1&-1 \\\ 1&0\end{array}\right)$. Serre proves in his book on trees that $SL_2(\mathbb{Z}) \cong \mathbb{Z}/4 *_{\mathbb{Z}/2} \mathbb{Z}/6$, and $S_4$ and $S_6$ are the elements corresponding to the generators of $\mathbb Z/4$ and $\mathbb Z/6$ (I'm not sure if this is related to my question). Then let $a = S_4 S_6$ and $b = S_4 S_6^2$. I believe every element of $SL_2(\mathbb Z)$ can be written as $S_6^d w S_6^e$, where $w$ is a word in $a$ and $b$ but not $a^{-1}$ or $b^{-1}$.

I wrote a program (for other purposes) that seems to show that there aren't any relations between $a$ and $b$ that have length 15 or less and don't involve $a^{-1}$ or $b^{-1}$. I'm not certain that the program is right, but if it is, one might make a naive guess that these two elements generate a free group. This makes me suspicious.

1) Does $SL_2(\mathbb Z)$ contain a free group (of rank > 1)? If it does, is there an easy way to determine whether the subgroup generated by $a$ and $b$ is free?

2) A slightly less naive guess is that $a$ and $b$ generate a free monoid in $SL_2(\mathbb Z)$. Is there a reason why $SL_2(\mathbb Z)$ can't contain a free monoid, or an example showing that it does?

EDIT: Thanks for the quick replies. As Robin and Jack pointed out, $a$ and $b$ generate SL(2,Z), so clearly don't generate a free group. Also, there are free subgroups that are easy to write down. I'm still curious about #2, though.

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You have to use an extra \ to get the matrices to display correctly. Similarly, to display { } use \\{ \\}. –  Douglas Zare Mar 18 '10 at 7:34
    
Thanks for the tip, Doug. –  Robin Chapman Mar 18 '10 at 7:43
1  
First you should work in $PSL_2(\mathbb{Z})$ just to get rid of the annoying $\pm 1$. As you stated the result is that $PSL$ is the free product of the images of $S_4$ and $S_6$. This means that every element has a unique representation as a reduced word in them: i.e a word of the form $S_4^{a_1} S_6^{b_1} \dots S_4^{a_n} S_6^{b_n}$ where $a_i = 1$ and $b_i \in \{1,2\}$ and we can optionally omit the leading and/or trailing factors. From this it's easy to solve any word problem in your $a$ and $b$ (almost as easy as in a free group) –  Victor Miller Mar 18 '10 at 22:04

3 Answers 3

up vote 11 down vote accepted

Certainly $\mathrm{SL}(2,\mathbb{Z})$ contains a free group. For instance $\Gamma(2)$, the subgroup of all matrices congruent to the identity modulo $2$, is free of rank $2$. The matrices $\left(\begin{array}{cc}1&2\\\ 0&1\end{array}\right)$ and $\left(\begin{array}{cc}1&0\\\ 2&1\end{array}\right)$ freely generate $\Gamma(2)$. This can be proved by considering the action on the upper half-plane or by careful examination of reduced words. There's a nice proof in chapter 18 of David Ullrich's book Complex Made Simple.

Your $a$ and $b$ don't generate a free group alas, since they generate all of $\mathrm{SL}(2,\mathbb{Z})$.

Re the edited question. Let's write $$T=\left(\begin{array}{cc}1&1\\\ 0&1\end{array}\right)\qquad \textrm{and}\qquad U=\left(\begin{array}{cc}1&0\\\ 1&1\end{array}\right).$$ As both Jack and I pointed out, $T^2$ and $U^2$ generate a free subgroup of rank $2$. Now it's an easy exercise to prove that $T$ and $U$ freely generate a free monoid of rank $2$ (because their entries are non-negative). On the other hand, they generate thw whole group $\mathrm{SL}(2,\mathbb{Z})$ which is certainly not free. Your matrices $a$ and $b$ are, if my calculations are right, $-U^{-1}$ and $-T^{-1}$. The matrix $S_4$ conjugates $T$ and $U$ into $U^{-1}$ and $T^{-1}$ so $U^{-1}$ and $T^{-1}$ freely generate a free monoid of rank $2$. The same must be true of $U^{-1}$ and $T^{-1}$, that is, of $a$ and $b$.

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Great, thanks for the nice answer. –  Peter Samuelson Mar 18 '10 at 21:45

For a broader perspective on proving that groups of matrices and what not are free, I highly recommend reading chapter II.B of Pierre de la Harpe's book "Topics in Geometric Group Theory", which gives a beautiful account with numerous references of known facts, especially for subgroups of SL_2.

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BabAba = 1, where Bb=1 and Aa=1. SL(2,Z) does contain free subgroups, for instance on [1,2;0,1] and [1,0;2,1], so there is no reason it could not contain free subgroups, it just so happens that {a,b} does not generate one. Of course {a,b} generates SL(2,Z) since S4=aBa and S6=Ab.

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