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 $\xi$.
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.
 A: 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, $k = 5$.
A: 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.
