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Let $L$ be a lattice associate to the Dykin matrix of type $E_{8}$. I would like to understand involutions of $L$ and their invariant $L^{+}$ and coinvariant lattice $L^-$ (I think they are isomorphic). I am sure that this has been studied in some literature. I would appreciate it if anyone could let me know the classification or some reference.

Any comments are welcome.

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The isometry group of the $E_8$ lattice is the same as the $E_8$ Weyl group (a fact that doesn't always hold for other weight lattices). So you're asking for conjugacy classes of order two elements in the $E_8$ Weyl group. Have you tried looking into the atlas of simple groups? – André Henriques Aug 11 '12 at 22:30
Not necessarily isomorphic; not even necessarily the same rank, e.g. reflection about a root has $L^+ \cong E_7$ and $L^- \cong A_1$ (and even more trivially central reflection has $L^+ = \lbrace 0 \rbrace$ and $L^- = E_8$). But it is possible for both $L^+$ and $L^-$ to be isomorphic with $D_4$. – Noam D. Elkies Aug 12 '12 at 2:34

As André already pointed out, it is sufficient to study the Weyl group $E_8$, looking in the atlas certainly works, see b).

a) BUT the Weyl group perspective makes it in my opinion enough accessible and explicit by hand:

For any involution $f$ of the root system you may proceed as for classifying Satake diagrams (just not pose the additional conditions which are in place there):

  • There exists a chamber $R^+$ (so: up to conjugation in the Weyl group) such that the number of roots transported out of $R^+$ is minimal.
  • An easy exercise shows that then $f=-id$ on these roots. The subset $S$ of simple roots $\alpha_1,....,\alpha_8$ with $f(\alpha_i)$ will be your first invariant of $f$ and determines $L_-$. Bear in mind that many parabolics $S$ are Weyl conjugate
  • Another easy exercise shows that in the quotient vector space by the previous guys $f$ acts just by permuting the remaining simple roots $\bar{S}$. This permutation $\pi$ of the sub-diagram $\bar{S}$ will be your second invariant from which you can somehow determine $L_+$ (careful, the permutation is only up to $S$). In particular if $\pi$ is trivial it will look like a restriction of the hyperplane-arrangement, those can be easily written down in terms of roots (yes, even for $E_8$ by hand, just takes a while, done it once!)

Bear in mind that I do not claim that all pairs $\{1...8\}=S\cup \bar{S},\pi$ can be realized (but I'd think so?)


  • Conjugacy class of reflections is $S=\{i\}$ for any $i$.
  • Conjugacy class of central involution $-1$ is $S=\{1...8\}$.
  • Is Noam's favourite choice (see c) $L_+=L_-=D_4$ the one with $S=D_4\subset E_8,\pi=id$ ?

b) Different approach to just look up what you want, maybe compare: By a remarkable coincidence it is a bicyclic extension of the Chevalley group of type $D_4$ over the field with two elements.


In particular the central involution $-1$ is visible. You indeed find $D_4(2)=O_8^+(2)$ in the atlas but I don't have it at hand right now.

c) As Noam pointed out there is indeed a class of involutions where both invariants and coinvariants are $D_4$. He was too humble to link to his previous post where he gives different explicit descriptions of this involutions, so I will fill in this gap: A question on an involution of $E_8$ lattice

Hope that helps, Simon

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In order to understand $E_8$, it is useful to have many different perspectives. Here are some ways of writing its eight simple roots $\alpha_1,\ldots,\alpha_8$ in various coordinate systems. I'll also list the lowest root $\alpha_0=−2\alpha_1 -3\alpha_2 -4\alpha_3 -5\alpha_4 -6\alpha_5 -4\alpha_6 -2\alpha_7 -3\alpha_8$ in those same coordinates.

Description 1: $$ \begin{matrix} \alpha_0&=& (1,-1,0,0,0,0,0,0,0)\\ \alpha_1&=& (0,1,−1,0,0,0,0,0,0)\\ \alpha_2&=& (0,0,1,−1,0,0,0,0,0)\\ \alpha_3&=& (0,0,0,1,−1,0,0,0,0)\\ \alpha_4&=& (0,0,0,0,1,−1,0,0,0)\\ \alpha_5&=& (0,0,0,0,0,1,−1,0,0)\\ \alpha_6&=& (0,0,0,0,0,0,1,−1,0)\\ \alpha_7&=& (0,0,0,0,0,0,0,1,−1)\\ \alpha_8&=& -\tfrac13( 1,1,1,1,1,1,−2,−2,−2)\,\,\,\, \end{matrix} $$

Description 2: $$ \begin{matrix} \alpha_0&=& (1,-1,0,0,0,0,0,0)\\ \alpha_1&=& (0,1,−1,0,0,0,0,0)\\ \alpha_2&=& (0,0,1,−1,0,0,0,0)\\ \alpha_3&=& (0,0,0,1,−1,0,0,0)\\ \alpha_4&=& (0,0,0,0,1,−1,0,0)\\ \alpha_5&=& (0,0,0,0,0,1,−1,0)\\ \alpha_6&=& (0,0,0,0,0,0,1,−1)\\ \alpha_7&=&-\tfrac12(1,1,1,1,1,1,1,-1)\,\,\,\,\\ \alpha_8&=& (0,0,0,0,0,0,1,1) \end{matrix} $$

Description 3: $$ \begin{matrix} \alpha_0&=& (1,-1,0,0,0)\oplus(0,0,0,0,0)\\ \alpha_1&=& (0,1,−1,0,0)\oplus(0,0,0,0,0)\\ \alpha_2&=& (0,0,1,−1,0)\oplus(0,0,0,0,0)\\ \alpha_3&=& (0,0,0,1,−1)\oplus(0,0,0,0,0)\\ \alpha_4&=& -\tfrac15[(1,1,1,1,−4)\oplus(3,3,−2,−2,−2)]\,\,\,\,\\ \alpha_5&=& (0,0,0,0,0)\oplus(0,1,-1,0,0)\\ \alpha_6&=& (0,0,0,0,0)\oplus(0,0,1,-1,0)\\ \alpha_7&=& (0,0,0,0,0)\oplus(0,0,0,1,-1)\\ \alpha_8&=& (0,0,0,0,0)\oplus(1,-1,0,0,0) \end{matrix} $$

Description 4: $$ \begin{matrix} \alpha_0&=& (1,-1,0,0,0,0,0,0)\oplus(0,0)\\ \alpha_1&=& (0,1,-1,0,0,0,0,0)\oplus(0,0)\\ \alpha_2&=& (0,0,1,-1,0,0,0,0)\oplus(0,0)\\ \alpha_3&=& (0,0,0,1,-1,0,0,0)\oplus(0,0)\\ \alpha_4&=& (0,0,0,0,1,-1,0,0)\oplus(0,0)\\ \alpha_5&=& (0,0,0,0,0,1,-1,0)\oplus(0,0)\\ \alpha_6&=& -\tfrac14[(1,1,1,1,1,1,−3,−3)\oplus(2,−2)]\,\,\,\,\\ \alpha_7&=& (0,0,0,0,0,0,0,0)\oplus(1,-1)\\ \alpha_8&=& (0,0,0,0,0,0,1,-1)\oplus(0,0) \end{matrix} $$

Description 5: $$ \begin{matrix} \alpha_0&=& (1,-1,0,0,0,0)\oplus(0,0,0)\oplus(0,0)\\ \alpha_1&=& (0,1,-1,0,0,0)\oplus(0,0,0)\oplus(0,0)\\ \alpha_2&=& (0,0,1,-1,0,0)\oplus(0,0,0)\oplus(0,0)\\ \alpha_3&=& (0,0,0,1,-1,0)\oplus(0,0,0)\oplus(0,0)\\ \alpha_4&=& (0,0,0,0,1,-1)\oplus(0,0,0)\oplus(0,0)\\ \alpha_5&=& -\tfrac16[(1,1,1,1,1,−5)\oplus(4,−2,−2)\oplus(3,−3)]\\ \alpha_6&=& (0,0,0,0,0,0)\oplus(1,-1,0)\oplus(0,0)\\ \alpha_7&=& (0,0,0,0,0,0)\oplus(0,1,-1)\oplus(0,0)\\ \alpha_8&=& (0,0,0,0,0,0)\oplus(0,0,0)\oplus(1,-1) \end{matrix} $$

Description 6: $$ \begin{matrix} \alpha_0&=& (1,-1,0,0)\oplus(0,0,0,0,0)\\ \alpha_1&=& (0,1,-1,0)\oplus(0,0,0,0,0)\\ \alpha_2&=& (0,0,1,-1)\oplus(0,0,0,0,0)\\ \alpha_3&=& -\tfrac14[(1,1,1,−3)\oplus(2,−2,−2,−2,−2)]\\ \alpha_4&=& (0,0,0,0)\oplus(1,-1,0,0,0)\\ \alpha_5&=& (0,0,0,0)\oplus(0,1,-1,0,0)\\ \alpha_6&=& (0,0,0,0)\oplus(0,0,1,-1,0)\\ \alpha_7&=& (0,0,0,0)\oplus(0,0,0,1,-1)\\ \alpha_8&=& (0,0,0,0)\oplus(-1,-1,0,0,0) \end{matrix} $$

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