Determine the thinnest lattice covering of $\mathbb{R}^6$ (improve on $E_6^*$-type coverings)

A lattice $L \subset \mathbb{R}^6$ gives a covering of $\mathbb{R}^6$ by congruent balls of radius equal to its covering radius $\mu(L)$ (the largest distance from any point of space to the nearest lattice point). Its covering density is $\Theta(L) = \mu(L)^6 V_6 / \sqrt{\det \mathrm{Gram}(L)}$, where $V_6 = \pi^3/6$ is the volume of the unit 6-ball. Find the lattice $L$ minimizing $\Theta(L)$ (the thinnest lattice covering). In dimension 6 the minimum is unknown: the best-known covering lattices (related to $E_6^*$) are only proved locally optimal. Exhibit a 6-dimensional lattice with covering density strictly below the current best-known value, or prove that the best-known lattice is globally optimal.

Acceptance. FULLY RESOLVES: a proof that a specific 6-dimensional lattice attains the minimal lattice covering density (global optimality over all 6-D lattices). PARTIAL (improve record): a positive-definite $6\times6$ Gram matrix $G$ whose lattice has covering density $\Theta < \Theta_{best}$; verifier -- compute the covering radius $\mu$ from the lattice's Delaunay decomposition (finite: the covering radius is the largest circumradius among Delaunay cells) and evaluate $\Theta = \mu^6 V_6 / \sqrt{\det G}$. PARTIAL: a certified lower bound on the 6-dimensional lattice covering density approaching the best-known upper bound.