Generation of $\mathrm{SO}(n,\mathbb{Q})$ by coordinate subgroups

Question

$\DeclareMathOperator\SO{SO}\SO(n,\mathbb{Q})$ is the group of $n\times n$ matrices $A$ with rational entries such that $AA^t=I$ and $\hbox{det}(A)=1$.

The $n$ coordinate subgroups of $\SO(n,\mathbb{Q})$ are the subgroups $H_i$ consisting of those matrices that fix the vector $e_i$, where $e_1,\ldots,e_n$ is the standard basis for $\mathbb{Q}^n$. Each such subgroup is isomorphic to $\SO(n-1,\mathbb{Q})$.

For which values of $n$ is $\SO(n,\mathbb{Q})$ generated by its coordinate subgroups?

Stefan Witzel has shown me a proof that this doesn't happen for $n=3$ and some other results, but I will let him describe these.

Answer 1

$\newcommand{\Z}{\mathbf{Z}}\newcommand{\Q}{\mathbf{Q}}$Yes for $n\ge 6$. Indeed, let $A$ be a matrix in $\mathrm{SO}(n,\mathbf{Q})$. First, if $Ae_i=e_i$ for some $i$, there is nothing to do, and if $Ae_i=-e_i$, say for $i=1$, multiplying $A$ with the diagonal matrix $(-1,-1,1,\dots,1)$ allows to conclude.

Write $Ae_1=(x_1,\dots,x_n)$. If $x_1=\pm 1$, we are thus done. Otherwise, $\sum_{i\ge 2}x_i^2$ is a positive rational, hence sum of $4$ squares $y_2^2+\dots+y_5^2$, and Witt's theorem ensures that there is some element of $\mathrm{O}(n-1)(\mathbf{Q})$ mapping $(x_2,\dots,x_n)$ to $(y_2,\dots,y_5,0,\dots,0)$. Hence composing $A$ with an element of $\mathrm{SO}_n(\mathbf{Q})$ mapping $e_1$ to $\pm e_1$ yields $A'$ mapping $(x_1,\dots,x_n)=Ae_1$ to $(x_1,y_2,\dots,y_5,0,\dots,0)$. Using $n\ge 6$ and again Witt's theorem, we find an element of $\mathrm{SO}_n(\mathbf{Q})$, mapping $e_n$ to $\pm e_n$ mapping $A'Ae_1$ to $(1,0,\dots,0)$.

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Edit: It is also true for $n=5$. As in the previous case, the point is showing that every orbit of the given subgroup (the subgroup $H$ generated by basis element stabilizers) in the rational 1-sphere meets an element with a zero coordinate.

Start from a tuple $(n_1,\dots,n_5)$ with $\sum n_i=1$. First suppose that all $n_i$ belong to $\Z_2$. Modulo 4, each $n_i^2$ is $1$ or $0$, hence either all are odd, or a single one, say $n_1$, is odd. If all are odd, all are $1$ modulo $8$, and we have a contradiction. So a single one is $1$. Then $n_1^2+n_2^2+n_3^2+n_4^2$ is equal to $1$ modulo $4$, hence, by Legendre's theorem, is a sum of 3 squares, and hence we can conclude in the same way as we did in the case $n\ge 6$ using Witt's theorem: the $H$-orbit of $(n_1,\dots,n_5)$ meets an element with a zero coordinate.

Otherwise, some $n_i$ is not in $\Z_2$. Let $2^k$ be a common denominator in $\Q_2$ for the $n_i$. Write $m_i=2^kn_i$. Then $\sum m_i^2=4^k$, the $m_i$ are in $\Z_2$ and not all even. Modulo $4$, since the $m_i^2$ are all $0$ or $1$ and sum to $0$, we deduce that exactly four of the $m_i$ are odd, say $m_5$ is even. Then $(m_1^2+m_2^2+m_3^2+m_4^2)/4$ is a positive element of $\Q\cap\Z_2$, hence is a sum of 4 squares in $\Q\cap\Z_2$, namely $q_1^2+q_2^2+q_3^2+q_4^2$. Write $q_5=m_5/2$ Hence using Witt's theorem, $(m_1,m_2,m_3,m_4,m_5)/2^k$ is in the $H$-orbit of $(q_1,q_2,q_3,q_4,q_5)/2^{k-1}$. Arguing by induction on $k$, each orbit in the rational 1-sphere contains an orbit contained in $\Z_2^5$, and we are done.

This completes the picture, since the answer is a trivial no for $n=2$ and is also no for $n=3,4$ (see this answer and this answer).

To summarize, for $n\ge 6$ we used the fact that any positive rational is a sum of 4 rational squares, and for $n=5$ we used the fact (Legendre) that every positive element of $\Q\cap\Z_2$ that is $1$ modulo 4 is a sum of 3 squares in $\Q\cap\Z_2$. (The latter follows from the same statement in $\Z$, just multiplying by an odd square.)

Answer 2

An obstruction to this happening is this: if $\DeclareMathOperator\SO{SO}\SO(n-1,\mathbb{Q})$ is contained in $\SO(n-1,\mathbb{Z}_p)$ (meaning that no entry of a matrix has $p$ in the denominator) but $\SO(n,\mathbb{Q})$ is not contained in $\SO(n,\mathbb{Z}_p)$ then the coordinate subgroups can at best generate $\SO(n,\mathbb{Q}) \cap \SO(n,\mathbb{Z}_p)$ which is a proper subgroup of $\SO(n,\mathbb{Q})$.

The entries of a column of an element of $\SO(n,\mathbb{Q})$ give a rational solution of $x_1^2 + \ldots + x_n^2 = 1$ and if the denominator is divisible by $p$ they give an integral solution to $y_1^2 + \ldots + y_n^2 \equiv 0$ modulo $p$. If no such solution exists then $\SO(n,\mathbb{Q}) < \SO(n,\mathbb{Z}_p)$. For $n = 2$ such a solution is equivalent to $-1$ being a square modulo $p$ which is not the case if $p \equiv 3$ modulo $4$, showing that $\SO(2,\mathbb{Q}) < \SO(2,\mathbb{Z}_p)$ for $p \equiv 3$ modulo $4$. On the other hand, it is not hard to produce a ``Pythagorean quadruple'' $1^2 + 2^2 + 2^2 = 3^2$ (or $2^2 + 3^2 + 6^2 = 7^2$ this was Ian's starting point) and from it a matrix in $\SO(3,\mathbb{Q}) \setminus \SO(3,\mathbb{Z}_3)$ (respectively in $\SO(3,\mathbb{Q}) \setminus \SO(3,\mathbb{Z}_7)$). This shows that $\SO(3,\mathbb{Q})$ is not generated by its coordinate subgroups.

At this point I thought it is about whether or not $\SO(n)$ is $\mathbb{Q}_p$-isotropic or not, which has a classical answer (see for instance Cohen, Number Theory Volume 1, Theorem 5.2.16) but not quite: $\SO(n)$ is $\mathbb{Q}_2$-anisotropic for $n \in \{3,4\}$ and yet it properly contains $\SO(n,\mathbb{Z}_2)$.

Meanwhile: is there an easy condition for a $\mathbf{Z}_p$-group scheme $\mathbf{G}$ to have $\mathbf{G}(\mathbb{Z}_p)$ maximal compact in $\mathbf{G}(\mathbb{Q}_p)$? (or should this be a new question)? (Here $p = 2$ appears to be special, but that may be specific to quadratic forms?)

Answer 3

No for $n=3,4$. Indeed, consider the following property $L(n)$, for a prime $p$: there are integers $k, m_1,\dots,m_n$ with $k,\gcd(m_1,\dots,m_n)$ coprime to $p$ such that $\sum_{i=1}^n m_i^2=k^2p^2$.

For $n\ge 2$ I claim that $\mathrm{SO}(n,\mathbf{Q})$ is contained in $\mathrm{SO}(n,\mathbf{Z}_p)$ if and only $L(n)$ fails.

Indeed, suppose that the inclusion fails. Then there is a column in some such matrix $\mathrm{SO}(n,\mathbf{Q})$ that is not in $\mathbf{Z}_p$. That is, it writes as $(m_1/k,\dots,m_n/pk)$ with $p$ not dividing some $m_i$. Then $\sum m_i^2=k^2p^2$ and $L(n)$ holds. Conversely, suppose that $L(n)$ holds. So there are $k,m_i$ as given. Then by Witt's theorem, there is a matrix in $\mathrm{O}(n,\mathbf{Q})$ whose first column is $(m_1/kp,\dots,m_n/kp)$. Changing the sign of the second column, we can suppose it has determinant one. So the inclusion fails.

Hence, if $L(n)$ holds and $L(n-1)$ fails, then $\mathrm{SO}(n,\mathbf{Q})$ is not generated by stabilizers of basis elements.

It remains to see that

• for $p=3$, $L(3)$ holds and $L(2)$ fails
• for $p=2$, $L(4)$ holds and $L(3)$ fails.

Indeed for $p=3$, $L(2)$ fails because $0$ is not a sum of two nonzero squares modulo 3 (this works for any odd prime $p$ in $4\mathbf{Z}+3$). Also $L(3)$ holds: $1^2+2^2+2^2=3^2$. Concretely, any matrix in $\mathrm{O}(3,\mathbf{Q})$ whose first column is $(1/3,2/3,2/3)$ is not a product of matrices fixing basis elements.

For $p=2$, $L(3)$ fails because $0$ is not a sum of three not-all-even squares modulo 4. And $L(4)$ holds because $1+1+1+1=2^2$.