Let $K$ be a bounded symmetric ($-K=K$) open convex body in $\mathbb R^n$. The critical determinant $d(K)$ of $K$ is the least possible volume $|\operatorname{det}(a_1\dots a_n)|$ of the fundamental parallelepiped of a lattice $\Lambda=\{\sum_{j=1}^n m_j a_j: m_j\in\mathbb Z\}$ such that $K\cap \Lambda=\{0\}$. Clearly, if $K'\subset K$ is another symmetric convex shape, then $d(K')\le d(K)$. The convex symmetric shape $K$ is called irreducible if the inequality is strict for every proper subset $K'$ of $K$.

It is not hard to see that the unit ball $B$ is irreducible in the dimensions 1,2,3. Moreover, if $B'\subset B$ and the radius of the largest ball contained in $B'$ is $1-\delta$, then $d(B')\le d(B)-c\delta$. However, this breaks in dimension $4$. When we cut off two opposite caps of depth $\delta$ from the unit ball in $\mathbb R^4$, we still get $d(B')<d(B)$ but the difference is now merely of order $\delta^2$. This suggests that when the dimension goes up and the number of touching points increases, we may end up in the situation when the ball is no longer irreducible. My question is whether this is really the case or whether the ball stays irreducible all the way up and just gets "less ans less" so in some sense.

The reason I'm asking is that we've just proved with Yoav Kallus that the ball in $\mathbb R^3$ is a local minimizer of the optimal lattice packing density. The corresponding statement is known to be false in $\mathbb R^2$ and I wonder if there is a trivial reason (namely, reducibility) for it to be false in some or, better, all high dimensions.

Any (relevant) ideas and/or references are welcome :).