2
$\begingroup$

Suppose that $SO(\mathbb{Q}) = \{ A \in M_n{\mathbb{(Q)}}: A^TA=I, \det A = 1 \}$ is a subgroup of $n\times n$ matrices with rational entries under matrix multiplication. Show that for $n \geq 2$ this group is not finitely generated.

The cardinality of $SO(\mathbb{Q})$ is countable, so it's possible that it is finitely generated. I think I need to find an algorithmic approach that takes a set of matrices in $SO(\mathbb{Q})$ and produces another matrix that can't be generated by that set. But this seems like a bit complicated. Any ideas?

$\endgroup$
0

2 Answers 2

5
$\begingroup$

Hint. If $SO(n;\mathbb Q)$ is finitely generated, then there exist a finite set of primes $P$ (including $1$ here by convention), such that when an element $a_{ij}$ of any $A\in SO(n;\mathbb Q)$ is written as an irreducible fraction, its denominator is a product of powers of primes from $P$. Now note that $(m^2-n^2,\ 2mn,\ m^2+n^2)$ is a Pythagorean triple and there are infinitely many primes of the form $m^2+n^2$.

$\endgroup$
2
  • $\begingroup$ Thanks. To be honest, your answer was a bit confusing the first time I read it, until I read Hagen von Eitzen's answer which expanded your ideas. Now I fully appreciate the beauty of the ideas behind the solution. $\endgroup$
    – math.n00b
    May 8, 2014 at 11:22
  • $\begingroup$ @math.n00b I think von Eitzen and I just wrote our answers independently. If one thinks about the 2x2 rotation matrices first, it is natural to relate those "rational" sines and cosines to Pythagorean triples. Hence it's also quite natural to notice that $SO(2;\mathbb Q)$ can't possibly be finitely generated because there are infinitely many primes of the form $m^2+n^2$. $\endgroup$
    – user1551
    May 8, 2014 at 12:56
3
$\begingroup$

Assume you have a finite set of generators $A_1,\ldots, A_k$. For suitable $a_i\in \mathbb N$ (the common denominator), we have $a_iA_i\in M_n(\mathbb Z)$ and also $a_iA_i^{-1}\in M_n(\mathbb Z)$. Let $p$ be a prime with $p\not\mid \prod a_i$ and $p\equiv 1\pmod 4$. Then there exists a solution of $p=u^2+v^2$ with $u,v\in\mathbb Z$. Let

$$A=\begin{pmatrix}\frac{u^2-v^2}{u^2+v^2}&\frac{2uv}{u^2+v^2}&0&\cdots&0\\ \frac{-2uv}{u^2+v^2}&\frac{u^2-v^2}{u^2+v^2}&0&\cdots &0\\ 0&0&1&\cdots&0\\ \vdots&\vdots&\vdots&\ddots&\vdots\\ 0&0&0&\cdots &1\end{pmatrix} $$

If $A$ wree a product of $A_i$s and $A_i^{-1}$s, multiplying it with a power of $\prod a_i$ would make it an integer matrix, but such a factor cannot get rid of the $p$ in the denominator!

$\endgroup$
2
  • $\begingroup$ Why $a_iA_i^{-1}$ is in $M_n(\mathbb{Z})$? I don't get that part, even though I spent sometime thinking about it. And, more importantly, why did you choose $A$ that way? I mean the first two by two block, what does it do exactly? Did you create it that way only because $A$ must lie in $SO(\mathbb{Q})$? So, you used rational parametrization of the unit circle that somehow gives $AA^T=I$ and $\det A=1$. Right? $\endgroup$
    – math.n00b
    May 8, 2014 at 11:08
  • $\begingroup$ OK, now I fully understand this solution! Wow. I learned two clever tricks in this solution, I liked it. Thanks. $\endgroup$
    – math.n00b
    May 8, 2014 at 11:20

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .