Effect of Covering Radius on Shortest Vector - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T08:23:14Z http://mathoverflow.net/feeds/question/98322 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/98322/effect-of-covering-radius-on-shortest-vector Effect of Covering Radius on Shortest Vector Will Jagy 2012-05-30T00:40:53Z 2012-05-30T03:48:35Z <p>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?</p> <p>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.</p> <p>This is also false, in dimension 4 and above, for covering radius exactly equal to $\sqrt 2,$ such as $$F_{10}(\{0\}) \; = \;<br> \left( \begin{array}{cccc} 4 &amp; -1 &amp; -1 &amp; -1 \\ -1 &amp; 4 &amp; -1 &amp; -1 \\ -1 &amp; -1 &amp; 4 &amp; -1 \\ -1 &amp; -1 &amp; -1 &amp; 4<br> \end{array} \right)$$ from <a href="http://www.math.rwth-aachen.de/~Gabriele.Nebe/papers/CR.pdf" rel="nofollow">NEBE_PDF</a>,</p> <p>as well as the Leech lattice. </p> <p>P.S. I already know, <strong>after the fact</strong>, 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. </p> <p>EDIT: from the excellent answers at <a href="http://mathoverflow.net/questions/69428/is-the-square-of-the-covering-radius-of-an-integral-lattice-quadratic-form-always" rel="nofollow">SQUARED</a>, it seems we will be done after two items are given:</p> <p>(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? <strong>ANSWER:</strong> from Dover edition of <em>Regular Polytopes</em> 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 $\rho &lt; \sqrt 2,$ all we get is a vector with norm (squared length) no more than $$\frac{4(n+1)}{n}$$</p> <p>(B) Can we prove that a simplex in $\mathbb R^n$ with circumradius $\rho$ has an edge no longer than the answer to (A)? </p>