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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?)

Examples:

  • 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.

$$W(E_8)=2.D_4(2).2$$

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 $E_8$ latticeA question on an involution of $E_8$ lattice

Hope that helps, Simon

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?)

Examples:

  • 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.

$$W(E_8)=2.D_4(2).2$$

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

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?)

Examples:

  • 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.

$$W(E_8)=2.D_4(2).2$$

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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Simon Lentner
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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?)

Examples:

  • 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.

$$W(E_8)=2.D_4(2).2$$

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