Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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 $dimV=6$

share|improve this question
3  
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. –  Sasha Jan 17 '13 at 17:51
2  
@Hassan Can you explain why you think these decompositions exist? and are these supposed to be a grading? –  Bruce Westbury Jan 17 '13 at 19:42
1  
@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 –  Hassan Jolany Jan 17 '13 at 20:05
add comment

1 Answer

up vote 23 down vote accepted

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 $L$.)

share|improve this answer
1  
Very niceeeeeeeeeeeeeeeeee –  Hassan Jolany Jan 17 '13 at 23:23
1  
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. –  Steven Stadnicki Oct 25 '13 at 17:19
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.