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What is the smallest simplest(non-trivial) $E_8$ -module ?

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    $\begingroup$ $\mathbb C\ \ \ $ $\endgroup$ Jan 19, 2013 at 15:20
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    $\begingroup$ I agree with Hassan; the smallest simplest $E_8$-module is actually $(0)$. $\endgroup$ Jan 19, 2013 at 16:11
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    $\begingroup$ Hassan's question combined with his comment makes the whole thing look trivial indeed. Anyway, the smallest faithful representations of all simple Lie algebras are worked out in the old literature and not very hard to relate to the fundamental weights. What is the point of the question? (And why the extra tags?) $\endgroup$ Jan 19, 2013 at 16:22
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    $\begingroup$ This is somewhat beside the point, but the difference between Allen Knutson's and Robert Bryant's answers seems to hinge on whether the word "simple" is meant in a strict mathematical sense. Since $(0)$ is a unit under direct sum, it is often more convenient to adopt the convention that it is a non-simple module. $\endgroup$
    – S. Carnahan
    Jan 20, 2013 at 0:21
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    $\begingroup$ Zuckerman calls these "the trivial representation" and "the pre-trivial representation". $\endgroup$ Jan 20, 2013 at 21:43

2 Answers 2

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Cartan showed that the lowest dimensional (nontrivial) $E_8$-module is ${\frak{e}}_8$ itself, i.e., the adjoint representation, which has dimension $248$. The next smallest nontrivial irreducible module is considerably larger dimension, $3875$, and I think that the next one after that has dimension $30380$.

At http://www-math.univ-poitiers.fr/~maavl/LiE/form.html, you can check these online using LiE.

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  • $\begingroup$ like always very niceeeeeeee $\endgroup$
    – user21574
    Jan 19, 2013 at 16:22
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Let $\mu(L)$ denote the minimal dimension of a faithful module of $L$. The complex simple Lie algebra $E_8$ satisfies $\mu(L)=\dim (L)$, as Cartan showed. Indeed, $E_8$ is the only complex simple Lie algebra with this property. There are more results in this direction (which are perhaps interesting): suppose that $L$ is a complex semisimple Lie algebra satisfying $\mu(L)=\dim (L)$. Then $L\simeq E_8\oplus \cdots \oplus E_8$. Even more general, let $L$ be a Lie algebra, where the solvable radical $rad(L)$ is abelian. Then always $\mu(L)\le \dim (L)$, and equality holds if and only if $L$ is abelian of dimension less than $5$, or $L$ is isomorphic to $E_8\oplus \cdots \oplus E_8$. (For references see arXiv:1006.2062).

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  • $\begingroup$ niceeeeeeeeeeee $\endgroup$
    – user21574
    Apr 8, 2013 at 22:03
  • $\begingroup$ @Hassan, there is very little need to use so many es. $\endgroup$ Apr 8, 2013 at 22:39
  • $\begingroup$ Dear Mariano, Yes, I believe to your comment. $\endgroup$
    – user21574
    Apr 8, 2013 at 23:21
  • $\begingroup$ In a sense, the smallest thing you could study in order to study $E_8$ is $E_8$ itself. :) $\endgroup$ Jun 19, 2014 at 15:58

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