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 $\mathrm{GL}_n(\mathbb Q)$?


  • 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.

  • 1
    $\begingroup$ 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? $\endgroup$ – Santana Afton Aug 8 at 13:27
  • 1
    $\begingroup$ @Dzoooks The vectors $u$ and $v$ must me in $A$ which is given $\endgroup$ – E. Joseph Aug 8 at 15:55
  • 4
    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Aug 8 at 16:31
  • 2
    $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Aug 8 at 18:16
  • 1
    $\begingroup$ @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. $\endgroup$ – 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 $\mathrm{Gr}(2,n)$.

  • $\begingroup$ 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. $\endgroup$ – Ewan Delanoy Sep 8 at 14:59
  • $\begingroup$ @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. $\endgroup$ – Jim Belk Sep 8 at 15:11
  • 1
    $\begingroup$ 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 $G$. $\endgroup$ – Ewan Delanoy Sep 8 at 15:27
  • $\begingroup$ A very nice idea. I was hoping to use the fact that the matrix group is countable somehow, but completely missed this. $\endgroup$ – 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

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


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

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

  • $\begingroup$ 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. $\endgroup$ – Jyrki Lahtonen Sep 8 at 17:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.