# Orbits of vectors under the action of $\mathrm{GL}_n(\mathbb Q)$

Context.

While working on a larger proof, I would love to have the following lemma, but I can't even decide if it's true or not.

The question.

We consider the action of $$\mathrm {GL}_n(\mathbb Q)$$ on $$\mathbb R^n$$ such that $$\varphi\cdot x=\varphi(x)$$ for $$\varphi\in \mathrm {GL}_n(\mathbb Q)$$ and $$x\in\mathbb R^n$$.

Let $$A$$ be a subspace of $$\mathbb R^n$$ of dimension $$2$$.

Does there exist $$u,v\in A$$ linearly independent such that $$v$$ is in the orbit of $$u$$ under the action of $$\mathrm{GL}_n(\mathbb Q)$$?

Remarks.

• We can reformulate the question this way: does there exist a rational transformation $$\varphi\in\mathrm{GL}_n(\mathbb Q)$$ such that $$\varphi(u)=v$$?

• I managed to prove this result for $$n=3$$ by constructing a rational rotation which sends $$u$$ to $$v$$.

• With a reasoning of cardinality, we can prove that this result is false if we fix $$u\in A$$, and we try to find $$\varphi\in\mathrm{GL}_n(\mathbb Q)$$ and $$v\in A$$ such that $$v=\varphi(u)$$.

• If $$A$$ contains a rational vector $$x$$, the symmetry with respect to $$x$$ will do the trick, so we can assume that $$A$$ does not contain any rational vector.

• Moreover, if $$A$$ intersect non-trivially a rational plane $$B$$, then we can consider the rotation of axis $$B^\perp$$ (of dimension $$n-2$$) and of angle $$\pi/2$$, and a little work shows this result. So we can assume now that for all rational planes $$B$$, $$A\cap B=\{0\}$$.

• I have no idea if this result is even true when $$n\geqslant 4$$.

• Any hints, references or proofs would be much appreciated.

• To be clear: you’re asking if it is possible to find any pair of two (I’m assuming distinct) vectors in $A$ so that one gets taken to the other via a rational transformation? – Santana Afton Aug 8 at 13:27
• @Dzoooks The vectors $u$ and $v$ must me in $A$ which is given – E. Joseph Aug 8 at 15:55
• In the case $n=3$ I might try and argue as follows. Consider the set of three rational matrices $\phi_1,\phi_2,\phi_3$ representing 90 degree rotations about the respective coordinate axes. For all $j=1,2,3$ we have, by dimensional considerations, $\dim(\phi_j(A)\cap A)\ge1$. So for all $j$ there exist non-zero vectors $u_j,v_j\in A$ such that $u_j=\phi_j(v_j)$. If $\{u_j,v_j\}$ is linearly dependent they need to lie on the axis of rotation. But $A$ can contain at most two of the axes. – Jyrki Lahtonen Aug 8 at 16:31
• An idea I had was the following. Let $e_j,j=1,2,3,4,$ be algebraically independent numbers (actually $e_j=\pi^{10^j}$ would work equally well). Then let $A=\langle u=(1,0,e_1,e_2),v=(0,1,e_3,e_4)\rangle$, and $P$ be a random 4x4 rational matrix. I was trying to prove that if the determinant of the matrix with columns $u,v,Pu,Pv$ vanishes, then $P$ must be singular. This would imply that $A$ and $PA$ intersect trivially for all non-singular $P$ making $A$ a counterexample. – Jyrki Lahtonen Aug 8 at 18:16
• @JyrkiLahtonen Nice plan, though a negative result would be quite a disappointment, it would be nice to know for sure that it is false! Don't hesitate to add info here if you make any progress. – E. Joseph Aug 8 at 19:26

This is false for $$n\geq 4$$. Consider the Grassmannian $$\mathrm{Gr}(2,n)$$ of all two-dimensional subspaces of $$\mathbb{R}^n$$, and recall that $$\mathrm{Gr}(2,n)$$ is a compact manifold of dimension $$2n-4$$. For each $$\varphi\in\mathrm{GL}_n(\mathbb{Q})$$, let $$S_\varphi = \{A\in \mathrm{Gr}(2,n) \mid \varphi u=v \text{ for some linearly independent }u,v\in A\}.$$ By the Baire category theorem, $$\mathrm{Gr}(2,n)$$ cannot be expressed as a union of countably many closed, nowhere dense sets. Therefore, it suffices to prove that each $$S_\varphi$$ is closed and nowhere dense in $$\mathrm{Gr}(2,n)$$.

To that end, we decompose $$S_\varphi$$ as a disjoint union $$T_\varphi \uplus U_\varphi$$, where

• $$T_\varphi$$ is the set of all $$A\in S_\varphi$$ for which $$\varphi(A) \ne A$$, and

• $$U_\varphi$$ is the set of all $$A\in S_\varphi$$ for which $$\varphi(A) = A$$.

It suffices to prove that $$T_\varphi$$ and $$U_\varphi$$ are closed and nowhere dense in $$\mathrm{Gr}(2,n)$$.

Claim. $$T_\varphi$$ is either empty or is a submanifold of $$\mathrm{Gr}(2,n)$$ of dimension $$n-1$$.

Proof: Suppose $$T_\varphi$$ is nonempty. If $$A\in T_\varphi$$, then $$A\cap\varphi^{-1}(A)$$ is a one-dimensional subspace of $$A$$, and this contains exactly one pair $$\{u,-u\}$$ of unit vectors. Such a $$u$$ has the property that $$u,\varphi u\in A$$ and $$\{u,\varphi u,\varphi^2 u\}$$ are linearly independent. Let $$\widetilde{T_\varphi} = \{u\in \mathbb{R}^n : \|u\|=1\text{ and }u,\varphi u,\varphi^2 u\text{ are linearly independent}\}.$$ Then $$\widetilde{T_\varphi}$$ is an open subset of the unit $$(n-1)$$-sphere in $$\mathbb{R}^n$$ and the map $$p\colon \widetilde{T_\varphi}\to T_\varphi$$ defined by $$p(u) = \mathrm{Span}\{u,\varphi u\}$$ is a degree two covering map, which proves the claim. $$\square$$

Since $$n-1 < 2n-4$$ for $$n\geq 4$$, this gives us the following.

Corollary. $$T_\varphi$$ is closed and nowhere dense in $$\mathrm{Gr}(2,n)$$ as long as $$n\geq 4$$.

Claim. $$U_\varphi$$ is a union of finitely many submanifolds of $$\mathrm{Gr}(2,n)$$, all of dimension at most $$n-2$$.

Proof: We separate the possible $$A\in U_\varphi$$ into three types, based on the eigenvalues of the restriction of $$\varphi$$ to $$A$$:

1. The restriction of $$\varphi$$ to $$A$$ has two distinct real eigenvalues $$\lambda,\mu$$.
2. The restriction of $$\varphi$$ to $$A$$ has one real eigenvalue $$\lambda$$ and is not diagonalizable.
3. The restriction of $$\varphi$$ to $$A$$ has two complex eigenvalues $$\lambda,\overline{\lambda}$$.

In each case the eigenvalues of the restriction must also be eigenvalues of $$\varphi$$, of which there are only finitely many. Our strategy is to analyze the set of all $$A$$ of a given type corresponding to a given eigenvalue or pair of eigenvalues.

For type (1), let $$\lambda$$ and $$\mu$$ be distinct real eigenvalues of $$\varphi$$, and let $$E_\lambda$$ and $$E_\mu$$ be the corresponding eigenspaces. Then any $$A$$ corresponding to $$\lambda$$ and $$\mu$$ can be written uniquely as the sum of a one-dimensional subspace of $$E_\lambda$$ and a one-dimensional subspace of $$E_\mu$$. If $$\dim(E_\lambda) = d_\lambda$$ and $$\dim(E_\mu) = d_\mu$$, then the set of all such $$A$$ is homeomorphic to $$\mathrm{Gr}(1,d_\lambda) \times \mathrm{Gr}(1,d_\mu)$$, which is a manifold of dimension $$d_\lambda+d_\mu - 2$$. In particular, since $$d_\lambda+d_\mu \leq n$$, the set of all such $$A$$ for a given pair $$\lambda,\mu$$ is a submanifold of $$\mathrm{Gr}(2,n)$$ of dimension at most $$n-2$$.

For type (2), let $$\lambda$$ be a real eigenvalue of $$\varphi$$ with higher algebraic multiplicity than geometric multiplicity. Let $$E_\lambda$$ be the eigenspace for $$\lambda$$ and let $$E_\lambda'$$ be the nullspace of $$(\varphi-\lambda I)^2$$. Then any $$A$$ of type (2) corresponding to $$\lambda$$ has one-dimensional image in $$E_\lambda'/E_\lambda$$ and is entirely determined by this image. If $$\dim(E_\lambda) = d_\lambda$$ and $$\dim(E_\lambda') = d_\lambda'$$, then the set of all such $$A$$ is homeomorphic to $$\mathrm{Gr}(1,d_\lambda'-d_\lambda)$$, which is a manifold of dimension $$d_\lambda'-d_\lambda - 1$$. In particular, since $$d_\lambda'-d_\lambda \leq n-1$$, the set of all such $$A$$ for a given $$\lambda$$ is a submanifold of $$\mathrm{Gr}(2,n)$$ of dimension at most $$n-2$$.

For type (3), let $$\lambda$$ be a complex eigenvalue of $$\varphi$$, and let $$E_\lambda$$ be the eigenspace for $$\lambda$$ in $$\mathbb{C}^n$$. Then any $$A$$ of type (3) corresponding to $$\lambda$$ is obtained by taking a subspace of $$E_\lambda$$ of complex dimension one and taking the real part of each vector. If $$\dim_{\mathbb{C}}(E_\lambda) = d_\lambda$$, then the set of all such $$A$$ is homeomorphic to the complex Grassmannian $$\mathrm{Gr}_{\mathbb{C}}(1,d_\lambda)$$, which is a manifold of real dimension $$2d_\lambda-2$$. In particular, since $$2d_\lambda \leq n$$, the set of all such $$A$$ for a given $$\lambda$$ is a submanifold of $$\mathrm{Gr}(2,n)$$ of dimension at most $$n-2$$. $$\square$$

Corollary. $$U_\varphi$$ is closed and nowhere dense in $$\mathrm{Gr}(2,n)$$ for all $$n\geq 3$$.

Incidentally, what's going on here from an algebraic perspective should be roughly that each $$S_\varphi$$ is an algebraic subvariety of $$\mathrm{Gr}(2,n)$$ of dimension $$n-1$$, with $$T_\varphi$$ being the set of regular points of $$S_\varphi$$ and $$U_\varphi$$ being its set of singular points, but we don't need to know any of that to provide a topological proof that it's nowhere dense in $$\mathrm{Gr}(2,n)$$.

• Amazing proof ! It is non-constructive though and does not produce an explicit counterexample $A$. I wonder if it can be made constructive by digging deeper into each part of the proof. – Ewan Delanoy Sep 8 at 14:59
• @EwanDelanoy Well, since all of the equations defining each $S_\varphi$ are algebraic in nature, it must be the case that the subspace generated by $(1,0,a_3,a_4)$ and $(0,1,b_3,b_4)$ is a counterexample whenever $a_3,a_4,b_3,b_4$ are algebraically independent, and indeed there ought to be an algebraic proof of this. The identity $AD-BC=0$ that you find in your answer ought to be the start of this. Since all the coefficients need to be zero, this gives a system of polynomial equations involving the $g_{i}$, and the goal is to show that there are no nonzero solutions. – Jim Belk Sep 8 at 15:11
• Indeed. Not sure what you mean by "nonzero" solutions ; the set of solutions to $AD-BC=0$ (or the corresponding equation for dimensions larger than $4$) is obviously the matrices in $GL_{n}(\mathbb Q)$ with at least one rational eigenvalue, yielding in each case a pair $(u,v)$ with $v=\phi u$ but $(u,v)$ are not linearly independent . A putative first progress would be in finding a sort of "formula" or "explicit construction" for the value of a rational eigenvalue of a solution $G$. – Ewan Delanoy Sep 8 at 15:27
• A very nice idea. I was hoping to use the fact that the matrix group is countable somehow, but completely missed this. – Jyrki Lahtonen Sep 8 at 17:07

This is not a full answer, but is too long for a comment.

Following the idea in Jyrki Lahtonen's comment, let $$a_3,b_3,a_4,b_4$$ be four algebraically independent (over $$\mathbb Q$$) real numbers, and let $$A$$ be the plane spanned by $$a$$ and $$b$$ where

$$a=\begin{pmatrix}1 \\ 0\\ a_3 \\ a_4 \end{pmatrix}, b=\begin{pmatrix}0 \\ 1\\ b_3 \\ b_4 \end{pmatrix}$$

Suppose that we have a matrix $$G=(g_{ij})\in GL_4({\mathbb Q})$$, two nonzero vectors $$u=u_1a+u_2b, v=v_1a+v_2b$$ in $$A$$ such that $$v=Gu$$.

By looking at the first two coordinates in this equation $$v=Gu$$, we already obtain

$$\begin{array}{lcl} v_1 & = & g_{11}u_1+g_{12}u_2+g_{13}(u_1a_3+u_2b_3)+g_{14}(u_1a_4+u_2b_4) \\ v_2 & = & g_{21}u_1+g_{22}u_2+g_{23}(u_1a_3+u_2b_3)+g_{24}(u_1a_4+u_2b_4) \end{array}\tag{1}$$

or rearranging terms,

$$\begin{array}{lcl} v_1 & = & (g_{11}+g_{13}a_3+g_{14}a_4)u_1+(g_{12}+g_{13}b_3+g_{14}b_4)u_2 \\ v_2 & = & (g_{21}+g_{23}a_3+g_{24}a_4)u_1+(g_{22}+g_{23}b_3+g_{24}b_4)u_2 \end{array}\tag{2}$$

Since $$v=v_1a+v_2b$$, we have $$v_3=v_1a_3+v_2b_3$$ and $$v_4=v_1a_4+v_2b_4$$ whence

$$\begin{array}{lccl} v_3 & = & & (g_{11}a_3+g_{13}a_3^2+g_{14}a_3a_4+g_{21}b_3+g_{23}a_3b_3+g_{24}a_4b_3)u_1 \\ & & + & (g_{12}a_3+g_{13}a_3b_3+g_{14}a_3b_4+g_{22}b_3+g_{23}b_3^2+g_{24}b_3b_4) u_2 \\ v_4 & = & & (g_{11}a_4+g_{13}a_3a_4+g_{14}a_4^2+g_{21}b_4+g_{23}a_3b_4+g_{24}a_4b_4)u_1 \\ & & + & (g_{12}a_4+g_{13}a_4b_3+g_{14}a_4b_4+g_{22}b_4+g_{23}b_3b_4+g_{24}b_4^2) u_2 \\ \end{array}\tag{3}$$

Now since $$v=Gu$$, we also have (compare with (2))

$$\begin{array}{lcl} v_3 & = & (g_{31}+g_{33}a_3+g_{34}a_4)u_1+(g_{32}+g_{33}b_3+g_{34}b_4)u_2 \\ v_4 & = & (g_{41}+g_{43}a_3+g_{44}a_4)u_1+(g_{42}+g_{43}b_3+g_{44}b_4)u_2 \end{array}\tag{2'}$$

Combing (3) with (2'), we obtain the system

$$\left\lbrace\begin{array}{lcl} Au_1+Bu_2 & = & 0 \\ Cu_1+Du_2 & = & 0 \\ \end{array}\right.\tag{4}$$

where

$$\begin{array}{ll} A & = & -g_{31}+(g_{11}-g_{33})a_3-g_{34}a_4+g_{21}b_3+g_{13}(a_3^2)+g_{14}a_3a_4+g_{23}a_3b_3+g_{24}a_4b_3 \\ B & = & -g_{32}+(g_{22}-g_{33})a_3+g_{12}a_3-g_{34}b_4+g_{23}(b_3^2)+g_{13}a_3b_3+g_{14}a_3b_4+g_{24}b_3b_4 \\ C & = & -g_{41}+(g_{11}-g_{44})a_4+g_{43}a_3+g_{21}b_4+g_{14}(a_4^2)+g_{13}a_3a_4+g_{23}a_3b_4+g_{24}a_4b_4 \\ D & = & -g_{42}+(g_{22}-g_{44})b_4+g_{12}a_4-g_{43}b_3+g_{24}(b_4^2)+g_{13}a_4b_3+g_{14}a_4b_4+g_{23}b_3b_4 \\ \end{array}\tag{5}$$

So we have $$AD-BC=0$$, and using the algebraic independence hypothesis we obtain a rather complicated polynomial system in the $$g_{ij}$$'s.

• I really hope you can make more progress. This is a bit different than what I got. IIRC I could show that a few 2x2-minors of $G$ must vanish, but nothing conclusive. – Jyrki Lahtonen Sep 8 at 17:06