A linear form can not be too small on integer points

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 $$\xi$$.

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.

• 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 Sep 27, 2018 at 8:18
• @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? Sep 27, 2018 at 8:25
• 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 $\eta$. Sep 27, 2018 at 8:28
• 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. Sep 27, 2018 at 8:29
• Like metamorphy states, there’s got to be something that measures rationality of $\eta$, otherwise a that much general statement has to be false Sep 27, 2018 at 8:30

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.
• 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). Sep 27, 2018 at 9:46
• As metamorphy pointed out, 5 is not enough, but $5+\epsilon$ will do Sep 27, 2018 at 9:48
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, $$k = 5$$.