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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.

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    $\begingroup$ 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
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    $\begingroup$ @Dzoooks The vectors $u$ and $v$ must me in $A$ which is given $\endgroup$ – E. Joseph Aug 8 at 15:55
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ @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
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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 $\mathrm{Gr}(2,n)$.

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  • $\begingroup$ 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
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    $\begingroup$ 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
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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.

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  • $\begingroup$ 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

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