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 $E_8$.

$\endgroup$

1 Answer 1

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.