0
$\begingroup$

Are there any good description of the isometry group $O(U\oplus E_8)$? Here $U$ denotes the hyperbolic lattice and $E_8$ the root lattice of type $E_8$.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Yes. See Lattices and Codes by Wolgang Ebeling. In the second edition, this is Exercise 4.4 on page 134. I do not believe this information was in the first edition; further, there is a third edition now. Anyway, the isometry group, or automorphism group, is generated by reflections in the roots of $L$ and $\pm 1.$ Essentially you use the proof of Theorem 4.6, due to Conway. The same techniques shows that, for any integral even lattice with covering radius strictly below $\sqrt 2,$ the class number of the lattice is one. In particular, it is not necessary to have unimodularity for this latter result.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ judging from the link on your earlier question, you knew of this... $\endgroup$
    – Will Jagy
    Commented Aug 5, 2013 at 22:56

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .