# Rotational symmetry group of QxQ

What is the rotational symmetry group of $\mathbb{Q}\times \mathbb{Q}$, the subset of the real plane consisting of rational points? i.e. are there infinite rotations, is this a named group ?

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What does "symmetry group" mean? Please be clear. –  Igor Rivin Sep 11 '11 at 14:12
I mean rotations. –  user17731 Sep 11 '11 at 14:23
$SO(2,\mathbb{Q})$. –  Donu Arapura Sep 11 '11 at 14:38
When you ask whether there are "infinite rotations", I wonder what an infinite rotation is. Could you have meant "infinitely many rotations"? That, I understand. –  Michael Hardy Sep 11 '11 at 18:05
I am guessing unknown (yahoo) means an element of infinite order. –  Todd Trimble Sep 13 '11 at 12:42

Let $G$ denote the group of rotations of the plane fixing the set of rational points. This group has multiple "names":

• it is $SO(2,\mathbb{Q})$, as Donu already commented;

• it is the group $\mathbb{Q}(i)^\times / \mathbb{Q}^\times$, as quid already commented;

• it is the group $\mathbb{Z}/4\mathbb{Z} \oplus \bigoplus_{i=1}^\infty \mathbb{Z}$, as André already commented.

Here, I want to justify the isomorphisms of $G$ with $\mathbb{Q}(i)^\times / \mathbb{Q}^\times$ and $\mathbb{Z}/4\mathbb{Z} \oplus \bigoplus_{i=1}^\infty \mathbb{Z}$: the argument is a combination of quid's answer and the comments to his answer.

The group $G$ is the group of matrices $\begin{pmatrix} x & -y \cr y & x \end{pmatrix}$ with $x,y \in \mathbb{Q}$ satisfying $x^2+y^2=1$. We identify this group with the elements of norm one of the multiplicative group $\mathbb{Q}(i)^\times$ assigning to the above matrix the element $x+iy$. There is a surjective group homomorphism $$q \colon \mathbb{Q}(i)^\times \longrightarrow G$$ mapping $a$ to $q(a) = a \cdot \overline{a}^{-1}$, where $\overline{a}$ is the complex conjugate of $a$. The kernel of $q$ is $\mathbb{Q}^\times$, so that we obtain $G \simeq \mathbb{Q}(i)^\times / \mathbb{Q}^\times$. To conclude we analyze the group $\mathbb{Q}(i)^\times / \mathbb{Q}^\times$.

Every element of $\mathbb{Q}(i)^\times$ can be written uniquely as a product of powers of primes of $\mathbb{Z}[i]$ times a unit (of $\mathbb{Z}[i]$). As is well-known, the splitting of the primes of $\mathbb{Z}$ in $\mathbb{Z}[i]$ is of one of three different kinds:

• primes congruent to 3 mod 4 stay irreducible,

• primes congruent to 1 mod 4 split as a product of two distinct primes,

• 2 splits as $i \cdot (1-i)^2$.

Using unique factorisation in $\mathbb{Z}[i]$, we encode elements of $\mathbb{Q}(i)^\times$ by the exponents of their prime factors (the units are irrelevant for our purposes). We obtain that the contributions of the three kinds of primes to $\mathbb{Q}(i)^\times / \mathbb{Q}^\times$ are

• nothing for the primes dividing a prime congruent to 3 mod 4,

• a copy of $\mathbb{Z}$ for the primes dividing a prime congruent to 1 mod 4,

• a copy of $\mathbb{Z}/4\mathbb{Z}$ for the prime $(1-i)$.

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@a-fortiori: Yes, it's elementary. But I've enjoyed reading MP's answer. I don't see anything wrong with working out this exercise here. On the contrary, I think that a lot of people might benefit from this nicely written answer. –  André Henriques Sep 12 '11 at 12:53
a-fortiori, one could argue that in principle the entire question would be better on math.SE. However, why you single out this particular answer for your comments is unclear to me. To me it seems it is the least elementary of all three answers. MP, thanks for writing this up. –  quid Sep 12 '11 at 14:45
a-fortiori, I do not know how you read the question, but it asks 'what is ...' so giving various alternate descriptions of the group with different merits seems on-topic to me. –  quid Sep 12 '11 at 17:09
Oh, it's a fine answer to an elementary question. Arguably the question should be at math.se rather than here. But given that it's here, this and the other answers provide a very clear and satisfying picture of the structure of this group. –  Todd Trimble Sep 13 '11 at 12:53
One might also observe that the surjectivity of $\mathbb{Q}(i)^\times \to G$ is a miniature application of Hilbert's theorem 90. –  Todd Trimble Sep 13 '11 at 14:40

Over the reals one has a parametrization of the rotations given by $$\begin{pmatrix} \cos t & - \sin t \\ \sin t & \cos t \end{pmatrix}$$

In other words $$\begin{pmatrix} x & - y \\ y & x \end{pmatrix}$$ with $x^2 + y^2 =1$ with reals $x,y$. Now if one wishes to restrict to the rationals one asks for rational solutions of $x^2 + y^2 =1$ (every such solution will give a rotation and one must not have any irrational entries in the matrix; also cf. André Henriques answer). There is a well-know rational parametrization of these solutions given by $$x(t) = \frac{1-t^2}{1+t^2} \, , \, y(t) = \frac{2t}{1+t^2}$$ with $t$ rational and the one additional solution $(-1, 0)$. (The geometric idea is that the $t$ is the slope of a line through $(-1, 0)$ and the respective solution the other intersection point of this line with the circle.) So one would get all the rotations as $$\begin{pmatrix} \frac{1-t^2}{1+t^2} & - \frac{2t}{1+t^2} \\ \frac{2t}{1+t^2} & \frac{1-t^2}{1+t^2} \end{pmatrix}$$ and $$\begin{pmatrix} -1 & 0\\ 0 & -1 \end{pmatrix}$$ This is also closey related to parametrization of Pythagorean triples.

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I came in to suggest "The Pythagorean group". –  Allen Knutson Sep 11 '11 at 16:05
I'm guessing that this group is abstractly isomorphic to $\mathbb Z/4\mathbb Z\oplus\bigoplus_{i=1}^\infty \mathbb Z$. How does one prove that? –  André Henriques Sep 11 '11 at 16:45
@Allen: Yes, Pythagorean and not Phytagorean. –  Gerald Edgar Sep 11 '11 at 16:48
Unless I am mistaken, this group is isomorphic to $(\mathbb{Q}(i))^\times$. If this is so, then what André says should follow from the description of the units and unique factorisation of the ring of Gaussian integers. –  M P Sep 11 '11 at 17:54
I think I see my mistake: as you said the group $\mathbb{Q}(i)^\times = \mathbb{Q}(i) \setminus \{0\}$ maps to the group $G$ you describe via $a \mapsto a/a'$ (followed by the map I wrote in a previous comment). Since the kernel of this map is $\mathbb{Q}^\times$, it follows that the group $G$ is isomorphic to $\mathbb{Q}(i)^\times/\mathbb{Q}^\times$. And now we can conclude using the splitting of primes of $\mathbb{Z}$ in $\mathbb{Z}[i]$: primes congruent to 1 mod 4 contribute a copy of $\mathbb{Z}$ to $G$, while the prime 2 contributes the cyclic group of order 4. –  M P Sep 11 '11 at 20:22

For every solution of the equation $x^2+y^2=1$, $x,y\in \mathbb Q$, the matrix $$\begin{pmatrix} x & y \\\ -y & x \end{pmatrix}$$ is a rotation that maps $\mathbb Q^2$ to itself.

An example is given by the rotation $$\begin{pmatrix} \textstyle\frac45 & \textstyle\frac35 \\\ \textstyle-\frac35 & \textstyle\frac45 \end{pmatrix}$$ which has infinite order, and thus provides an answer to your second question.

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Isn't this just the unit disc of $\mathbb Q[i]$? –  Asaf Karagila Sep 11 '11 at 15:02
@Asaf: Yes, it's the intersection of the unit circle $\{z\in\mathbb C:|z|=1\}$ with $\mathbb Q[i]\subset \mathbb C$. –  André Henriques Sep 11 '11 at 15:04

I might as well add another two interpretations of this group from the theory of algebraic tori - this should make the isomorphism between $SO_2(\mathbb{Q})$ and $\mathbb{Q}[i]^\times/\mathbb{Q}^\times$ mentioned in another answer look a bit more natural. The group you seek is the group of rational points of the norm torus of the field extension $\mathbb{Q}[i]/\mathbb{Q}$, and it is also the group of rational points of the quotient of the Weil restriction (along the same extension) of the multiplicative group by the counit subgroup.

Recall that restriction of scalars (also known as Weil restriction) of the multiplicative group $\mathbb{G}_m$ along $\mathbb{Q}[i]/\mathbb{Q}$ produces a two dimensional algebraic torus, that in particular is a subgroup of $GL_{2,\mathbb{Q}}$. The restriction of determinant to this subgroup is called the norm map, and its kernel (i.e., the intersection with $SL_2$) is the norm torus. The norm torus is equal to $SO_2$ as an algebraic subgroup of $SL_2$ - this is special to $\mathbb{Q}[i]$ among quadratic extensions, as other quadratic fields have norms that are not isomorphic to the standard diagonal quadratic form.

Since restriction of scalars is right adjoint to base change, you get a counit homomorphism $\mathbb{G}_{m, \mathbb{Q}} \to \operatorname{Res}_{\mathbb{Q}}^{\mathbb{Q}[i]} \mathbb{G}_{m, \mathbb{Q}[i]}$ of group schemes. The resulting quotient is an anisotropic rank one torus whose rational points are in natural bijection with elements of $\mathbb{Q}[i]^\times/\mathbb{Q}^\times$.

An isomorphism between these two descriptions can be seen by passing to character lattices as Galois modules (this passage is an antiequivalence - see SGA3 Exp. 9). The character lattice of $\mathbb{G}_m$ is a copy of $\mathbb{Z}$ with trivial action, and restriction of scalars corresponds to taking a tensor product with the group ring of the Galois group (i.e., it is the induced representation - there is also a way to think of this as a pushforward of étale sheaves on the corresponding spectra of fields). The norm torus construction corresponds to taking the quotient by the invariant sublattice, and the counit quotient corresponds to taking the Galois anti-invariant sublattice. As abstract Galois modules, both character lattices are free of rank one with isotropy group equal to the absolute Galois group of $\mathbb{Q}[i]$.

You can show that there are elements of infinite order by noting that $m$th roots of unity have degree $\phi(m)$ for all $m \geq 1$, so you only have to find a point that is neither a 4th root of unity nor a 6th root of unity. Any nondegenerate Pythagorean triple will suffice to produce a suitable rotation, and there is a neat characterization of such triples using cohomology given in this MathOverflow answer.

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