# Orthogonal Complements of Root Lattices in E_8

I have a rather stupid lattice theory question. Suppose $L$ is a root lattice that can be primitively embedded in the $E_8$ lattice. Is the orthogonal complement of $L$ in $E_8$ unique up to isomorphism, or for different primitive embeddings could I get non-isomorphic complements?

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You can get different orthogonal complements for different embeddings. There are two different embeddings of $A_{7}$ in $E_{8}$ so that for the first embedding the orthogonal complement is the lattice $A_{1}$, and for the second embedding the orthogonal complement is the lattice $\langle 8 \rangle$.
Sorry, I should have said that PRIMITIVE embeddings. I think then that the discriminant groups of $L$ and $L^\perp$ will be isomorphic, so at least I still have some hope. I've edited the question to reflect this. –  user4192 Jun 21 '11 at 12:21
According to Dynkin's classification of root sublattices in $E_{8}$, there are only five root subblattices in $E_{8}$ (modulo the Weyl group action) that admit more than one embedding in $E_{8}$. Moreover each of the five admits exactly two non-equivalent embeddings. The five exceptions (and their possible complements) are: –  Tony Pantev Jun 21 '11 at 13:00
$A_{1}^{\oplus 4}$, ($D_{4}$, $A_{1}^{\oplus 4}$); \linebreak $A_{3}\oplus A_{1}^{\oplus 2}$, ($A_{3}$, $A_{1}^{\oplus 2}\oplus \langle 4\rangle$); \linebreak $A_{3}^{\oplus 2}$, ($A_{1}^{\oplus 2}$, $\langle 4 \rangle^{\oplus 2}$), \linebreak $A_{5}\oplus A_{1}$, ($A_{2}$, $A_{1}\oplus \langle 6 \rangle$); \linebreak $A_{7}$, ($A_{1}$, $\langle 8\rangle$). \linebreak So for the primitive embeddings you get a uniquely determined orthogonal complement. –  Tony Pantev Jun 21 '11 at 13:09