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Background. My questions are motivated by the following:

A. Conway and Sloane in "On the covering multiplicity of lattices" (Discrete and Computational Geometry, 8 (1992) 109-130) considered the following quantity associated with a lattice $L$ in ${\mathbb R}^n$: Take the smallest $R$ such that the union of closed $R$-balls centered at the points of $L$ is the entire ${\mathbb R}^n$. Then define $CM(L)$ to be the multiplicity of the resulting covering, i.e., the maximum number of balls with nonempty intersection minus 1.

Note: Their definition is slightly different but equivalent to mine. One can also define $CM(L)$ as $$ \min \mu({\mathcal B}_R) $$ where the minimum is taken over multiplicities of coverings of ${\mathbb R}^n$ (by open balls) of the form $$ {\mathcal B}_R=\{B(x, R): x\in L\}. $$

Conway and Sloane proved that for $n\le 8$, the multiplicity of the lattice of type $A_n^\ast$ (the dual lattice of $A_n$) is equal to $n$ and conjectured that for every lattice $L$ of rank $n\ge 9$, $CM(L)> n$. (The inequality $CM(L)\ge n$ holds for all rank $n$ lattices $L$; this is a simple corollary of the fact that the covering dimension of ${\mathbb R}^n$ equals $n$.) They made a number of computations supporting their conjecture. For instance, $CM(A_9^\ast)\ge 11$. For the standard cubic lattice $L_n$ they proved that $$ CM(L_n)\sim 2.089097\ldots^{n + O(\sqrt{n})} $$

Question 1. Was there any progress on this conjecture?

Looking at mathscinet did not reveal anything useful.

B. The notion of covering dimension $\dim(X)$ of a metric space $(X,d)$ is defined as $$ \liminf_{\epsilon\to 0} \mu({\mathcal U}) $$ where the infimum is taken over multiplicities $\mu({\mathcal U})$ of open $\epsilon$-covers ${\mathcal U}$ of $X$ in the sense that the diameter of each member of ${\mathcal U}$ is at most $\epsilon$. (The covering dimension depends only on the topology of $X$.)

Loosely speaking, my second question is: What happens if in this definition we take the infimum over all coverings by metric balls of variable radii? Let us call the resulting quantity $\dim_d(X)$: $$ \dim_d(X)= \liminf_{\epsilon\to 0} \mu({\mathcal B}) $$ where the infimum is taken over all coverings ${\mathcal B}$ of $X$ by open balls of diameter $\le \epsilon$.

Here are more specific questions:

Question 2. Is $\dim_d(X)=\dim(X)$ for every metric space?

This sounds too good to be true, thus, here is a modified version of this question:

Question 2'. Is it true that every metrizable topological space $X$ admits a metric $d$ for which $\dim_d(X)=\dim(X)$?

(This question is motivated by the relation of covering and Hausdorff dimension.)

To avoid pathological examples, let us assume that spaces in questions are locally compact. Then Question 2' has positive answer in the case of spaces of covering dimension $0$.

Lastly, to connect A and B and to illustrate my ignorance in these matters:

Question 3. Is it true that for the Euclidean metric $d$ on ${\mathbb R}^n$, $$ \dim_d({\mathbb R}^n)=n \ ? $$ Note that the answer is yes for $n\le 8$ since one can use rescalings of a multiplicity $n$ ball covering ${\mathcal B}_R$ associated with $A_n^\ast$.

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    $\begingroup$ You might mention somewhere that by multiplicity you understand the number of intersecting balls (open subsets) minus $1$. Otherwise, your formulae in item B concerning dimensions look wrong. $\endgroup$ Commented Jan 7, 2014 at 3:03
  • $\begingroup$ Some stupid idea about Question 3. Let us call a nondegenerate $n$-simplex $\sigma\subset{\mathbb R}^n$ $good$ if it is covered by the open unit balls centred at its vertices and each such ball does not intersect the face of $\sigma$ opposite to the centre of the ball. Note that a triangulation of ${\mathbb R}^n$ by good simplices provides an affirmative answer to Question 3. In order to construct such a triangulation, it suffices to triangulate by good simplices an appropriate nondegenerate parallelepiped. $\endgroup$ Commented Jan 7, 2014 at 3:06
  • $\begingroup$ In your definition of $\dim_d$, do you want all balls in a covering to be of the same diameter? If it is to be analogous to the usual covering dimension, then you should allow balls of various diameters - as long they are small. This detail is crucial for your questions 2, 2' and 3. $\endgroup$ Commented Jan 7, 2014 at 4:20
  • $\begingroup$ In the def of CM(L), you should take only the interior of the balls, otherwise already for n=1 you would have CM(L)=2. $\endgroup$
    – domotorp
    Commented Jan 7, 2014 at 4:57
  • $\begingroup$ Dear Sasha, Wlodek and domotorp. The answer to all your comments is yes, I edited the question accordingly. $\endgroup$
    – Misha
    Commented Jan 7, 2014 at 5:16

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