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Let $L_0, L_1$ be Euclidean lattices (say full rank) of dimension $n_i$. Let $\lambda_1(L_i)$ denote the length of the shortest vector of $L_i$, and let $\rho(L_i)$ denote the covering radius of $L_i$:

$$\rho(L_i) = \max_{x\in\mathbb{R}^{n_i}} d(x, L_i)$$

Where $d(x, L_i) = \min_{\ell\in L_i}\lVert x - \ell\rVert_2$ is the distance from $x$ to the lattice $L_i$.

It is well-known that $\lambda_1(L_0\otimes L_1) \leq \lambda_1(L_0)\lambda_1(L_1)$. Moreover, assuming $\min(n_0, n_1) \leq 43$, a result of Kitoka shows that this is an equality, and the shortest vectors of $L_0\otimes L_1$ are pure tensors $\ell_0\otimes \ell_1$, where $\lambda_i(L_i) = \lVert \ell_i\rVert_2$. Are there any analogous results known about the covering radius (and vectors which achieve the covering radius, i.e. the "deep holes")?

In particular, I have a lattice of the form $L_n\otimes B$ where $L_n$ has dimension $n\to\infty$, and $B$ is a fixed dimension "base" lattice (which I would like to set to be the Leech lattice or Gosset lattice). It is actually known how to get a good upper bound on the covering radius of this lattice (due to a result of Davenport in the 50's), but this is insufficient for my purposes --- I need to be able to find the deep holes. I can likely find the deep holes of $L_n$ and $B$ individually, but is there a way to "combine" this information to find the deep holes of $L_n\otimes B$?

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