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Do anybody know , why can we write the following decompositions for exceptional Lie algebras $E_6$, $E_7$ and $E_8$

1) $E_8=(V^{\star}\otimes \wedge^{8}V^{\star})\bigoplus \wedge^{6}V^{\star}\bigoplus \wedge^{3} V^{\star} \bigoplus \mathfrak{g}\mathfrak{l}(V)\bigoplus \wedge^{3} V\bigoplus \wedge^{6} V \bigoplus(V\otimes \wedge^{8}V)$, here $dimV=8$

2) $E_7= \wedge^{6}V^{\star}\bigoplus \wedge^{3} V^{\star} \bigoplus \mathfrak{g}\mathfrak{l}(V)\bigoplus \wedge^{3} V\bigoplus \wedge^{6} V $,here $dimV=7$

3) $E_6= \wedge^{6}V^{\star}\bigoplus \wedge^{3} V^{\star} \bigoplus \mathfrak{g}\mathfrak{l}(V)\bigoplus \wedge^{3} V\bigoplus \wedge^{6} V $,here $dimV=6$

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    $\begingroup$ I don't know but I would guess that you should consider $A_{n-1} \subset E_n$ and consider the Lie algebra as its representation. This should give your formulas. $\endgroup$
    – Sasha
    Jan 17, 2013 at 17:51
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    $\begingroup$ @Hassan Can you explain why you think these decompositions exist? and are these supposed to be a grading? $\endgroup$ Jan 17, 2013 at 19:42
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    $\begingroup$ @Dear Bruce Yes these are supposed to be grading($E_6, E_7$ in $Z_2$ grading). I have seen this fact without proof in a Russian paper $\endgroup$
    – user21574
    Jan 17, 2013 at 20:05

1 Answer 1

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The $E_6$ and $E_7$ decompositions you list are explained in Cartan's 1894 thesis (see pages 89–92 for these formulae). For $E_8$, Cartan instead gives a decomposition (a $\mathbb{Z}_3$-grading) of the form $$ {\frak{e}}_8 = \Lambda^3(W^\ast)\oplus {\frak{sl}}(W)\oplus \Lambda^3(W) $$ for a $9$-dimensional space $W$. From this, you can get the decomposition you list for $E_8$ by writing $W = L\oplus V$ where $L$ has dimension $1$ and $V$ has dimension $8$. Then, using the fact that $L\otimes \Lambda^8(V)\simeq \Lambda^9(W)$, which is trivial under $\mathrm{SL}(W)$, one finds that, under the action of $\mathrm{GL}(V)\subset \mathrm{SL}(W)$, these three spaces break up into the $7$ spaces (actually, $8$, since ${\frak{gl}}(V)$ has a center) you have listed for $E_8$.

There are other places in Cartan's papers where he explains this further. For example, the $E_6$ case is explained at greater length in his paper Les groupes réels, simples, finis, et continus (Ann. Éc. Norm. 31 (1914), 263–355). See the formulae on pp. 298 & 299.

It's probably worth adding that, in this same paper, beginning on page 313, Cartan gives a similarly nice $\mathbb{Z}_2$-graded decomposition $$ {\frak{e}}_7 = {\frak{sl}}(W)\oplus \Lambda^4(W) $$ where $W$ is a vector space of dimension $8$. (The even part is the ${\frak{sl}}(W)$ subalgebra, and the odd part is the $\Lambda^4(W)$, which is isomorphic to $\Lambda^4(W^\ast)$ as an $\mathrm{SL}(W)$-module.) To get the decomposition you list, you use the same symmetry-breaking idea as worked for $E_8$: Write $W = L \oplus U$ for $L$ a $1$-dimensional subspace and $U$ a $7$-dimensional subspace, then, under $\mathrm{GL}(U)$, we have $L\simeq \Lambda^7(U^\ast)$, and the above decomposition breaks into $$ {\frak{e}}_7 = \bigl(\mathbb{R}\oplus {\frak{sl}}(U)\oplus L{\otimes}U^\ast\oplus L^\ast{\otimes}U\bigr)\oplus \bigl(\Lambda^4(U)\oplus L{\otimes}\Lambda^3(U)\bigr), $$ so now, if you set $V = K\otimes U$, where $K^{\otimes 3}= L$, the pieces line up with what you have listed after you permute them around a bit. (The only tricky part is seeing that you can take a cube root of the line $L$.)

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    $\begingroup$ Very niceeeeeeeeeeeeeeeeee $\endgroup$
    – user21574
    Jan 17, 2013 at 23:23
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    $\begingroup$ IIRC, that core dissection of $E_8$ goes by the name 'triality' and John Baez has written some pretty fantastic (if fairly high-level) exposition of it over on This Week's Finds. $\endgroup$ Oct 25, 2013 at 17:19
  • $\begingroup$ Sorry, which week? $\endgroup$ Jun 6, 2017 at 23:13

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