3
$\begingroup$

There exits an involution $\iota$ of the $E_8$ lattice such that $(E_8)^{\pm} \cong D_4$, where $(E_8)^{\pm}$ denotes the $\pm$ eigen-lattice of the involution $\iota$. Could someone kindly give me an explicit description of such involution $\iota$?

Thank you very much.

$\endgroup$

3 Answers 3

6
$\begingroup$

If you construct $E_8$ as $\lbrace{(x_1,\ldots,x_8) \in D_8^*: \sum_{i=1}^8 x_i \in 2{\bf Z} \rbrace}$ then you can use an involution such as $(x_1,x_2,x_3,x_4,x_5,x_6,x_7,x_8) \leftrightarrow (x_1,x_2,x_3,x_4, -x_5,-x_6,-x_7,-x_8)$. Recall that $D_8^* = {\bf Z}^8 \cup ({\bf Z}+\frac12)^8$; the plus and minus eigenspaces of our involution are $\lbrace(x_1,x_2,x_3,x_4,0,0,0,0)\rbrace$ and $\lbrace(0,0,0,0,x_5,x_6,x_7,x_8)\rbrace$, and those contain no $({\bf Z}+\frac12)^8$ vectors, leaving only a copy of $D_4$ in each eigenspace.

$\endgroup$
3
$\begingroup$

If you get $E_8$ as the "Construction A" lattice associated to the extended Hamming [8,4,4] code, then you can choose a weight-$4$ codeword $c$, and use the involution that multiplies by $-1$ each of the coordinates in the support of $c$. Then the plus and minus eigenlattices are each identified with the "Construction A" lattice associated to the $[4,1,4]$ repetition code, which is $D_4$. [If $C$ is a binary linear code of length then $$ \lbrace 2^{-1/2} v : v \in {\bf Z}^n, v \bmod 2 \in C \rbrace $$ is the associated lattice; see the section on Construction A in Conway and Sloane's Sphere Packings, Lattices and Groups. If you instead invert four coordinates that are not the support of a codeword then each eigenlattice is $A_1^4$.]

$\endgroup$
2
$\begingroup$

Depends on how you're constructing $E_8$. One approach is to start from $D_4 \oplus D_4$ and reconstruct $E_8$ as the sublattice of $D_4^* \oplus D_4^*$ consisting of pairs of vectors that are congruent ${\rm mod}\phantom.D_4$. Check that this is an even unimodular lattice, and thus isomorphic to $E_8$. The involution is $(v,w) \leftrightarrow (v,-w)$.

$\endgroup$

You must log in to answer this question.

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