I am trying to show that $SO_3(\mathbb Q)$ contains a free group using 5-adic numbers, and more precisely using the matrices $M_1=\left(\begin{array}{ccc}1 &0&0\\0&\frac 35&-\frac{4}5\\0&\frac 45& \frac 35\end{array}\right)$ and $M_2=\left(\begin{array}{ccc}\frac 35&-\frac{4}5&0\\\frac 45& \frac 35&0\\0&0& 1\end{array}\right)$ each having eigenvectors $e_1=\left(\begin{array}{l}1\\0\\0\end{array}\right)$, $v_1=\left(\begin{array}{r}0\\1\\i\end{array}\right)$ and $u_1=\left(\begin{array}{r}0\\i\\1\end{array}\right)$ for $M_1$ and $v_2=\left(\begin{array}{r}1\\i\\0\end{array}\right)$, $u_2=\left(\begin{array}{r}i\\1\\0\end{array}\right)$ and $e_3=\left(\begin{array}{l}0\\0\\1\end{array}\right)$ for $M_2$, the vectors being written in the canonical basis $(e_1, e_2, e_3)$ of $\mathbb{Q}_5^3$ for $\mathbb{Q}_5$ the 5-adic numbers. Here $i^2=-1$ in $\mathbb{Z}_5$, such that $|2-i|_5<1$. In this way the eigenvalues are $1$, $l=\frac {2+i}{2-i}$ and $\frac 1l$ for $M_1$ and $l$,$\frac 1l$ and $1$ for $M_2$ with $|l|_5=5$ and $|\frac 1l|_5=1/5$.
I take now the distance between lines of $\mathbb{Q}_5^3$ defined as $$d(D,D')=inf\{|v-v'|; v \in D, v' \in D', |v|=|v'|=1\},$$ for $|v|=max(|x|_5,|y|_5,|z|_5)$ the norm induced by the absolute value on $\mathbb{Q}_5$. I can find sufficiently small neighbourhoods $B_1^+$ of $Vect(v_1)$, $W_1^+$ of $Vect(u_1, e_1)$, $B_1^-$ of $Vect(u_1)$, $W_1^-$ of $Vect(v_1, e_1)$, $B_2^+$ of $Vect(v_2)$, $W_2^+$ of $Vect(u_2, e_3)$, $B_2^-$ of $Vect(u_2)$, $W_2^-$ of $Vect(v_2, e_3)$, all given by lines at distance at most $\varepsilon$ from the spaces considered, such that all the $B$'s are disjoint and $B_1^+$ is disjoint from $W_1^+$, and $B_1^-$ is disjoint from $W_1^-$ and the same with index $2$ and sign $\pm$ (and of course $B_1^-$ sits inside $W_1^+$, $B_1^+$ in $W_1^-$ and similarly for $B_2^-$ and $B_2^+$). A picture would have explained the situation better than all this text...
I would now like to apply the ping-pong lemma to $K_1=(B_1^+\cup B_1^-)$ and $K_2=(B_2^+\cup B_2^-)$ by finding a common (big) power $g_1=M_1^m$ of $M_1$ and $g_2=M_2^m$ of $M_2$ such that $g_1$ sends the complementary $C_1^+$ of $W_1^+$ to $B_1^+$, (and then $g_1^{-1}$ will send the complementary $C_1^-$ of $W_1^-$ to $B_1^-$) and $g_2$ and its inverse send the analogue sets indexed by $2$ accordingly.
I can show that being in $B_1^+$ (that is being close to $Vect(v_1)$) means for a line $Vect(v)$, with $v=a_1v_1+a_2u_1+a_3e_1$, that $max(|\frac{a_2}{a_1}|_5,|\frac{a_3}{a_1}|_5)<c_1\varepsilon$, for $c_1$ a strictly positive constant. I don't know how to show that being in $C_1^+$ means for a line $Vect(v)$ something like $min(|\frac{a_1}{a_2}|_5,|\frac{a_1}{a_3}|_5)>c_2\varepsilon$, for $c_2$ a strictly positive constant.
Applying $M_1$ means making the $v_1$-coordinate of a vector (written in the basis formed by the eigenvectors of $M_1$) bigger and bigger in front of the other two coordinates. If the condition on $C_1^+$ is the one I mentioned, or similar, that will allow me to find a convenient power of $M_1$ sending $C_1^+$ to to $B_1^+$ and I'm done. But I'm not sure about the condition on $C_1^+$ and I don't know how to deal correctly with this set.
I am sorry for the long introduction, and thank you to those who followed me up to here, and who will eventually help me !
