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Notations.

Let $\xi=(\xi_1,\xi_2,\xi_3,\xi_4)\in\mathbb( R\setminus\mathbb Q)^4$ such that $\xi_1\xi_4-\xi_2\xi_3\ne 0$ and the $\xi_i$ are linearly independent over $\mathbb Q$.

I have the following linear form:

$$\begin{matrix}L\colon & \mathbb R^6 & \to & \mathbb R \\ &(\eta_1,\ldots,\eta_6) & \mapsto & \eta_1(\xi_1\xi_4-\xi_2\xi_3)-\eta_2\xi_4+\eta_3\xi_3-\eta_4\xi_2+\eta_5\xi_1-\eta_6.\end{matrix}$$

We consider the norm $\Vert\cdot\Vert$ to be the euclidean norm on $\mathbb R^6$.

The problem.

I am interesting in finding a constant $\gamma>0$, such that if we choose $\xi$ properly, then the resulting linear form $L_\xi$ will verify:

$$\forall \eta\in\mathbb Z^6\setminus\{0\},\quad L_\xi(\eta)\geqslant \frac c{\Vert\eta\Vert^\gamma}\gcd(\eta_1,\ldots,\eta_6),$$

where $c=c_\xi$ is a constant which depends only on $\xi$.

The conjecture.

There are hopes for this to be true, since if we choose $\xi$ properly (for instance badly approximated by rationals), then for $\eta\in\mathbb Z^6\setminus\{0\}$, $L_\xi(\eta)$ will have troubles being too small.

I believe that the constant $\gamma=2$ would work for a fine choice of $\xi$.

Additional remarks.

This is a part of a longer proof, and if this results happens to be true, it would help me a great deal in that other proof. Unfortunately, I don't have any clue on how to start to attack this problem, so any leads would be much appreciated.

I do believe that $\gamma=2$ would work (and it would be the best), but any proof that would work for a $\gamma<4$ would be great.

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  • $\begingroup$ Wouldn’t your assumptions imply that the property holds therefore for all $\eta \in \mathbb{R}^6\backslash \{0\}$? In this case, I guess the statement is false — taking a $\mathbb{R}^6$ vector orthogonal to the one defined by $(\xi_1\xi_4-\xi_2 \xi_3, -\xi_4,\xi_3,-\xi_2,\xi_1,-1)$ would do the job $\endgroup$ Sep 27, 2018 at 8:18
  • $\begingroup$ @JoãoRamos I thought of that, but I wasn't quite sure this implies the property holds for all $\eta\in\mathbb R^6\setminus\{0\}$. Is this the case juste because $L_\xi$ and $\Vert\cdot\Vert$ are continuous? $\endgroup$
    – E. Joseph
    Sep 27, 2018 at 8:25
  • $\begingroup$ I would say so... as both the norm and the functional are continuous - and the constants bounding $L_{\xi}$ from below are only dependent on $\xi$ -, one can take limits to a general $\eta$. $\endgroup$ Sep 27, 2018 at 8:28
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    $\begingroup$ I think that, instead of the norm (in $c/||\eta||^{\gamma}$), there must be something involving denominators of $\eta$. Otherwise, replacing $\eta$ with $\eta/N$ (with large natural $N$) leads to absurd. $\endgroup$
    – metamorphy
    Sep 27, 2018 at 8:29
  • $\begingroup$ Like metamorphy states, there’s got to be something that measures rationality of $\eta$, otherwise a that much general statement has to be false $\endgroup$ Sep 27, 2018 at 8:30

2 Answers 2

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I think this is a direct application of a Theorem of Kleinbock and Margulis https://arxiv.org/pdf/math/9810036.pdf

Let me give a little more detail. Let $f$ be the following map from $\mathbb{R}^4 \to \mathbb{R}^5$ $$f(\xi_1,\xi_2,\xi_3,\xi_4)=(\xi_1\xi_4-\xi_2\xi_3,-\xi_4,\xi_3,-\xi_2,\xi_1).$$ It is not hard to check that partial derivatives $(\partial f/\partial \xi_i)_{1\leq i\leq 4}$ together with $\partial^2 f/\partial \xi_1\partial\xi_4$ span $\mathbb{R}^5$, so the image of $f$ is is a nondegenerate manifold in the sense of this article. By Theorem A of the aforementionned paper, for almost every $\xi=(\xi_1,..,\xi_4)$, $f(\xi)$ is not very well approximable, meaning that for all $\epsilon>0$, there exist only finitely many integer vectors $q \in \mathbb{Z}^5$ such that there exist a $p\in \mathbb{Z}$ such that $$|\langle q, f(\xi) \rangle + p|. \|q\|^{5(1+\epsilon)} \leq 1.$$ Taking the infimum over this finite set of $q$ tells us that there exist a constant $c_\epsilon>0$ such that for all $q \in \mathbb{Z}^5$ and $p\in \mathbb{Z}$, $$|\langle q, f(\xi) \rangle + p|. \|q\|^{5(1+\epsilon)} \geq c_\epsilon.$$ Since $\langle q, f(\xi) \rangle + p=L_\xi(q_1,...,q_5,p)$, this gives you the kind of estimate needed.

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  • $\begingroup$ Thanks for the very interesting new way of attacking this problem. I'll study what you said with great attention (because the exponent $5$ you proved is a little too great, so I will have to reconsider some things). $\endgroup$
    – E. Joseph
    Sep 27, 2018 at 9:46
  • $\begingroup$ As metamorphy pointed out, 5 is not enough, but $5+\epsilon$ will do $\endgroup$
    – user120527
    Sep 27, 2018 at 9:48
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I'm afraid the converse is true. For any $\zeta_1, \ldots, \zeta_k \in \mathbb{R}$ and any integer $n > 0$, there exist integers $n_0, \ldots, n_k$ with absolute values at most $n$, not all zero, such that $|n_0 + n_1\zeta_1 + \ldots + n_k\zeta_k| \leq n^{-k}$ (this is plainly the pigeonhole principle applied to the set of fractional parts $\{m_1\zeta_1 + \ldots + m_k\zeta_k\}$ for all positive integers $m_1, \ldots, m_k$ with values at most $n$). In your case, $k = 5$.

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