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Let $\mathbf u\in\mathbb Z^d$ be a primitive vector (i.e. $\gcd(u_i)=1$) and let $\Pi_{\mathbf u^\perp}$ be the orthogonal projection perpendicular to $\mathbf u$. I want to understand the geometry of the $(d-1)$-dimensional lattice, $L=\{\Pi_{\mathbf u^\perp}\mathbf n\colon \mathbf n\in\mathbb Z^d\}$. Specifically:

I want to relate $\Delta:=\max_{x\in\mathbb R^d}d(\Pi_{\mathbf u^\perp}x,L)$ to the vector $\mathbf u$.

The kind of answer I'd really like might give an upper bound for $\Delta$ in terms of $\min\{\langle \mathbf v,\mathbf v\rangle\colon \mathbf v\in\mathbb Z^d\setminus\{0\}; \langle\mathbf u,\mathbf v\rangle=0\}$). Can anyone point me in the right direction?

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    $\begingroup$ It is certainly possible. The general relevant statement is like this. If $K$ is a symmetric convex body in $\mathbb R^m$ that contains no non-zero point of a dual lattice $\Lambda^*$, then $\Lambda+C(m)K^\circ=\mathbb R^m$ where $K^\circ$ is the polar body of $K$. The constant $C(m)$ that is known to work for all bodies is $Cm\log m$; $Cm$ would be the best possible for the cube $K$. You are asking about balls only, so, probably, one can do somewhat better but for this you need to analyze the proofs. $\endgroup$
    – fedja
    Nov 11, 2017 at 16:14
  • $\begingroup$ Thank you for this. Based on your comment, I located a previous MO post, mathoverflow.net/questions/34599/…. The answers there include some references for the statements you are making (and if I understand them correctly, you can improve $Cm\log m$ to $Cm(\log m)^{1/2}$ in my case). $\endgroup$ Nov 12, 2017 at 19:53
  • $\begingroup$ You are welcome :-) If you really care about the dependence on $m$ for the particular case of the ball, I can try to think of it a bit (shaving half log in a special case of a bound that is assumed to be off by the entire log even in the general case doesn't look too impressive, though who knows...) $\endgroup$
    – fedja
    Nov 12, 2017 at 21:38
  • $\begingroup$ I guess for now I’m just pleased to know that some result exists, and to have some idea where to look for it. I’m not at the stage where I need the optimal result yet. $\endgroup$ Nov 12, 2017 at 22:13

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