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For "even" integral lattices in dimension at least 4, does a covering radius strictly less than $\sqrt 2$ imply that there is a vector of norm 2, also called a root?

Note that this is simply false in dimensions 2 and 3, the three relevant even lattices corresponding to the quadratic forms $$ 2 x^2 + x y + 2 y^2, \; \; 2 x^2 + 2 y^2 + 2 z^2 + y z + 2 z x + 2 x y, \; 2 x^2 + 2 y^2 + 2 z^2 - y z + z x + x y. $$ When you double these to get an integral inner product matrix, the diagonal elements are 4.

This is also false, in dimension 4 and above, for covering radius exactly equal to $\sqrt 2,$ such as $$ F_{10}(\{0\}) \; = \; \left( \begin{array}{cccc} 4 & -1 & -1 & -1 \\\ -1 & 4 & -1 & -1 \\\ -1 & -1 & 4 & -1 \\\ -1 & -1 & -1 & 4 \end{array} \right) $$ from NEBE_PDF,

as well as the Leech lattice.

P.S. I already know, after the fact, that the answer is yes. I am hoping for some lemma that says so ahead of time. Indeed, from the complete (finite) list, we can state that every such lattice has an improper automorph, including dimensions 1,2,3,4,5,6,7,8,9,10. Improper automorphs are always available in odd dimension but uncertain in even dimension. So all that is left over after this question is the binary form $2 x^2 + x y + 2 y^2,$ which is "ambiguous" or equivalent to its own "opposite" class. Anyway, it has an evident improper automorph. Oh, in any dimension, if there is a root, reflection in the root is an improper automorph.

EDIT: from the excellent answers at SQUARED, it seems we will be done after two items are given:

(A) What is the edge length of the regular simplex in $\mathbb R^n$ with circumradius $\rho,$ or at least a fairly good upper bound? ANSWER: from Dover edition of Regular Polytopes by Coxeter, pages 157-158, with circumradius $\rho$ the edge length is $$ \rho \; \sqrt{ \frac{2(n+1)}{n} } $$ Well, nuts. If all we can say is that $\rho < \sqrt 2,$ all we get is a vector with norm (squared length) no more than $$ \frac{4(n+1)}{n} $$

(B) Can we prove that a simplex in $\mathbb R^n$ with circumradius $\rho$ has an edge no longer than the answer to (A)?

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Note that the packing radius is smaller than the covering radius, so there is a vector strictly shorter than $2 \sqrt 2,$ so squared length below 8, so squared length 6 or 4 or 2. – Will Jagy May 30 '12 at 2:24
(A) It's easiest to do this by putting the vertices at $r := \rho / \sqrt 2$ times the unit vectors $e_0,\ldots,e_n$ in ${\bf R}^{n+1}$. The circumcenter is $r (e_0 + e_1 + \cdots + e_n) / (n+1)$, which makes the circumradius $r \sqrt{n/(n+1)} = \rho \sqrt{n/(2n+2)}$. Sanity check: this gives $\rho/2$, $\rho/\sqrt{3}$, and $\rho\sqrt{3/8}$ for $n=1,2,3$; the first two are familiar, and the last is clear by inscribing the tetrahedron in a cube of side $r$. – Noam D. Elkies May 30 '12 at 3:40
@Noam, thanks, do I have the edit correct from Coxeter's book? It does not give me what I want. – Will Jagy May 30 '12 at 3:50
(B) Yes. By scaling it's enough to do the case $\rho = \sqrt{(2n+2)/n}$ which makes the circumradius $1$. Suppose $v_0,\ldots,v_n$ are unit vectors. Then $$ 0 \leq \| v_0 + \cdots + v_n\|^2 = n+1 + \sum_{0\leq i<j \leq n} \langle v_i,v_j \rangle. $$ Hence some $\langle v_i,v_j \rangle \geq -1/n$, with equality iff the $v_i$ are vertices of a regular simplex. But then $\|v_i-v_j\|^2 \geq 2 - 2(-1/n) = (2n+2)/n$, QED. Yes, this also gives an alternative proof of (A). ~ – Noam D. Elkies May 30 '12 at 3:52
What's the "edit ... from Coxeter's book"? I don't see Coxeter mentioned either here or in the "SQUARED" thread. Anyway it seems I'm bumping up against a comments-per-hour limit (because I've been editing comments by deleting and posting amended versions), so I'll have to come back here tomorrow. – Noam D. Elkies May 30 '12 at 3:59

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