The question.

Let $\xi=(\xi_1,\xi_2,\xi_3,\xi_4)\in\mathbb R^4$ be a vector with irrational coordinates.

I am interested in finding the minimal value $\mu_\xi$ of

$$\left\vert \det \begin{pmatrix} a_1 & a_2 & 0 & 1 \\ b_1 & b_2 & 1 & 0 \\ c_1 & c_2 & \xi_1 & \xi_3 \\ d_1 & d_2 & \xi_2 & \xi_4 \end{pmatrix} \right\vert,$$

for $a_1,b_1,c_1,d_1,a_2,b_2,c_2,d_2\in\mathbb Z$, in terms of the area of the parallelepiped formed by the two vectors $X_i:=(a_i,b_i,c_i,d_i)$, $i=1,2$ (which I assume linearly independent), in $\mathbb R^4$.

Let's call this area $D(X_1,X_2)$. I know we have

$$\begin{align*} D(X_1,X_2)^2 &= \Vert X_1\Vert^2\Vert X_2\Vert^2-(X_1\cdot X_2)^2 \end{align*},$$

but I have no clue on how to proceed from here.

The conjecture.

My hope (which would help the construction of another proof a lot) would be that if we chose the $\xi_i$ properly, we can show that the minimal value verifies

\begin{equation} \mu_\xi\geqslant \frac c{D(X_1,X_2)^2}\qquad\qquad (1) \end{equation}

where $c$ is a constant (it may depends on $\xi$).

Final remarks.

Despite the fact that I strongly believe that $(1)$ is true, any proof that would show that

$$\mu_\xi\geqslant \frac c{D(X_1,X_2)^\gamma}$$

for a $\gamma<4$ would be of great interest.

  • $\begingroup$ By the definition of the determinant, your minimun depends well on $\xi$ vector. If you take the two vectors the area formula $D(X_1,X_2)$ isn't right if it is the triangle area formed by $X_1,X_2$. So either you compute the expression and minimize if possible, either i guess taking particular values of $\xi$ and see a geometric proof. $\endgroup$ – Toni Mhax Sep 26 '18 at 12:56
  • $\begingroup$ @ToniMhax I edited to specify what I mean by the area formed by $X_1$ and $X_2$. Yes, the minimum depends on $\xi$, but I am interested in a particular value of $\xi$ that would solve the conjecture. $\endgroup$ – E. Joseph Sep 26 '18 at 13:04
  • $\begingroup$ I think $D(X_1,X_2)^2=||X_1||^2||X_2||^2-(X_1•X_2)^2$ $\endgroup$ – Empy2 Sep 26 '18 at 13:38
  • $\begingroup$ @Empy2 You're right, thank you, I edited. $\endgroup$ – E. Joseph Sep 26 '18 at 13:39
  • 1
    $\begingroup$ Let us continue this discussion in chat. $\endgroup$ – Toni Mhax Sep 26 '18 at 14:14

The answer is wide so we need notations, for any $(X_1,X_2)$ in ${\mathbb{Z}}^2$ distinct, there is $A$ such that $\dfrac{1}{D(X_1,X_2)}\le \dfrac{1}{A}$. $C:=area(X_1,X_2)\ge a$ (parallelogram) also $a\neq 0$ exist. The determinant is the volume of the four column vectors of the given matrix [which we suppose invertible] and equals $\mu_{\xi}=C.\alpha_{\xi}\ge \frac{c_{\xi}}{A}\ge \frac{c_{\xi}}{D(X_1,X_2)} $ for some $\alpha_{\xi}$ where $c_{\xi}= a.A.\alpha_{\xi}$.

Edit $D(X_1,X_2)$ is the volume of $(X_1,X_2, X_1\times X_2)$

Finding the minimum value $a$ explicitly is the same as finding the minimum value of $A$.

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    $\begingroup$ I am really sorry, but I do not see how it is related to any kind of answer to this question... $\endgroup$ – E. Joseph Sep 26 '18 at 16:53
  • $\begingroup$ It shows for any $\xi$ the existence of $c_{\xi}$ as in the bound. $\endgroup$ – Toni Mhax Sep 26 '18 at 17:10
  • $\begingroup$ But your $c_\xi$ depends on $X_1,X_2$, which is not allowed unfortunately. $\endgroup$ – E. Joseph Sep 26 '18 at 17:29
  • $\begingroup$ Nop. Anyway we hit a wall of supposing the matrix invertible. $a$ and $A$ can be computed, (hard now) when the vectors $X_1,X_2$ are in ${\mathbb{Z}}^4$ $\endgroup$ – Toni Mhax Sep 26 '18 at 17:44
  • $\begingroup$ I could not follow any of your computations. How can you get a sharp bound when your proof basically starts saying that $D(X_1,X_2)$ is non-zero? $\endgroup$ – C. Falcon Sep 26 '18 at 19:12

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