Rotational symmetry group of QxQ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T03:50:43Z http://mathoverflow.net/feeds/question/75142 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/75142/rotational-symmetry-group-of-qxq Rotational symmetry group of QxQ unknown (yahoo) 2011-09-11T14:09:41Z 2011-09-13T05:04:34Z <p>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 ?</p> http://mathoverflow.net/questions/75142/rotational-symmetry-group-of-qxq/75143#75143 Answer by AndrĂ© Henriques for Rotational symmetry group of QxQ AndrĂ© Henriques 2011-09-11T14:46:15Z 2011-09-11T15:06:13Z <p>For every solution of the equation $x^2+y^2=1$, $x,y\in \mathbb Q$, the matrix $$\begin{pmatrix} x &amp; y \\ -y &amp; x \end{pmatrix}$$ is a rotation that maps $\mathbb Q^2$ to itself.</p> <p>An example is given by the rotation $$\begin{pmatrix} \textstyle\frac45 &amp; \textstyle\frac35 \\ \textstyle-\frac35 &amp; \textstyle\frac45 \end{pmatrix}$$ which has infinite order, and thus provides an answer to your second question.</p> http://mathoverflow.net/questions/75142/rotational-symmetry-group-of-qxq/75147#75147 Answer by quid for Rotational symmetry group of QxQ quid 2011-09-11T15:57:35Z 2011-09-12T14:49:21Z <p>Over the reals one has a parametrization of the rotations given by <code>$$\begin{pmatrix} \cos t &amp; - \sin t \\ \sin t &amp; \cos t \end{pmatrix}$$</code></p> <p>In other words <code>$$\begin{pmatrix} x &amp; - y \\ y &amp; x \end{pmatrix}$$</code> with $x^2 + y^2 =1$ with reals $x,y$. Now if one wishes to restrict to the rationals one asks for <em>rational</em> 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 <code>$$x(t) = \frac{1-t^2}{1+t^2} \, , \, y(t) = \frac{2t}{1+t^2}$$</code> 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 <code>$$\begin{pmatrix} \frac{1-t^2}{1+t^2} &amp; - \frac{2t}{1+t^2} \\ \frac{2t}{1+t^2} &amp; \frac{1-t^2}{1+t^2} \end{pmatrix}$$</code> and <code>$$\begin{pmatrix} -1 &amp; 0\\ 0 &amp; -1 \end{pmatrix}$$</code> This is also closey related to <a href="http://en.wikipedia.org/wiki/Pythagorean_triples" rel="nofollow">parametrization of Pythagorean triples</a>. </p> http://mathoverflow.net/questions/75142/rotational-symmetry-group-of-qxq/75208#75208 Answer by MP for Rotational symmetry group of QxQ MP 2011-09-12T10:11:47Z 2011-09-12T10:11:47Z <p>Let $G$ denote the group of rotations of the plane fixing the set of rational points. This group has multiple "names": </p> <ul> <li><p>it is $SO(2,\mathbb{Q})$, as Donu already commented; </p></li> <li><p>it is the group $\mathbb{Q}(i)^\times / \mathbb{Q}^\times$, as quid already commented; </p></li> <li><p>it is the group $\mathbb{Z}/4\mathbb{Z} \oplus \bigoplus_{i=1}^\infty \mathbb{Z}$, as André already commented.</p></li> </ul> <p>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.</p> <p>The group $G$ is the group of matrices $\begin{pmatrix} x &amp; -y \cr y &amp; x \end{pmatrix}$ 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 $\mathbb{Q}(i)^\times / \mathbb{Q}^\times$.</p> <p>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: </p> <ul> <li><p>primes congruent to 3 mod 4 stay irreducible, </p></li> <li><p>primes congruent to 1 mod 4 split as a product of two distinct primes, </p></li> <li><p>2 splits as $i \cdot (1-i)^2$.</p></li> </ul> <p>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 </p> <ul> <li><p>nothing for the primes dividing a prime congruent to 3 mod 4, </p></li> <li><p>a copy of $\mathbb{Z}$ for the primes dividing a prime congruent to 1 mod 4, </p></li> <li><p>a copy of $\mathbb{Z}/4\mathbb{Z}$ for the prime $(1-i)$.</p></li> </ul> http://mathoverflow.net/questions/75142/rotational-symmetry-group-of-qxq/75252#75252 Answer by S. Carnahan for Rotational symmetry group of QxQ S. Carnahan 2011-09-12T19:33:49Z 2011-09-13T05:04:34Z <p>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.</p> <p>Recall that <a href="http://en.wikipedia.org/wiki/Weil_restriction" rel="nofollow">restriction of scalars</a> (also known as Weil restriction) of the multiplicative group <code>$\mathbb{G}_m$</code> 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 <em>equal</em> 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.</p> <p>Since restriction of scalars is right adjoint to base change, you get a counit homomorphism <code>$\mathbb{G}_{m, \mathbb{Q}} \to \operatorname{Res}_{\mathbb{Q}}^{\mathbb{Q}[i]} \mathbb{G}_{m, \mathbb{Q}[i]}$</code> of group schemes. The resulting quotient is an anisotropic rank one torus whose rational points are in natural bijection with elements of $\mathbb{Q}[i]^\times/\mathbb{Q}^\times$.</p> <p>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 $\mathbb{Q}[i]$.</p> <p>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 <a href="http://mathoverflow.net/questions/10879/intuition-for-group-cohomology/10903#10903" rel="nofollow">this MathOverflow answer</a>.</p>