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Let $\mathfrak g$ be a finite-dimensional simple Lie algebra over $\mathbb C$.

Theorem 1. The highest root is perpendicular to all but one simple root, except in the case ${\mathfrak g}={\mathfrak sl}_{n > 2}$, in which case it is perpendicular to all but two (the first and last).

Theorem 2. The space $Hom({\mathfrak g}\otimes {\mathfrak g},{\mathfrak g})$ is $1$-dimensional (spanned by the Lie bracket), except for $\mathfrak g = {\mathfrak sl}_{n > 2}$, in which case it is $2$-dimensional. (The symmetric product is $(A,B)\mapsto$ the traceless part of $AB+BA$.)

Bruce Fontaine showed me a proof of the second, using the first, via the geometric Satake correspondence. Much as I like that, I'm wondering if there is a more classical argument connecting these two facts.

(Feel free to retag if inspired.) EDIT: I forgot that ${\mathfrak sl}_2$ is the exception that proves the rule.

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@Allen: At first sight Theorem 1 doesn't seem to make sense as stated. Am I missing something? –  Jim Humphreys May 6 '13 at 16:01
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Perhaps "perpendicular to" should be replaced by "perpendicular to all but"? –  Terry Tao May 6 '13 at 17:02
    
@Jim I have edited, hopefully, to what OP intended (as Terry Tao suggested). –  Bruce Westbury May 6 '13 at 17:15
    
Thanks! $\ \ $ –  Allen Knutson May 6 '13 at 17:21
    
I'm not sure this will help, but 1. may be rephrased geometrically as: among all complex simple Lie algebras, sl_n, n>2, is the unique series such that the orbit of a highest weight line X in P(g) is such that the tangent space T_{Id}X is reducible (as a module for the stabilizing parabolic), and 2. may be phrased geometrically as: sl_n is the unique case where there is a G-invariant cubic hypersurface in P(g), as the form in 2. is completely symmetric. –  JM Landsberg May 6 '13 at 21:48
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2 Answers

up vote 4 down vote accepted

As Evan points out, "modern" technology (including Littelmann paths and canonical bases) provides an improved way to think about tensor product decompositions for simple Lie algebras. But your Theorems 1 and 2 could also be understood in purely classical terms, though I'm not sure how far anyone looked at these. The assumption is that the classification of simple Lie algebras $\mathfrak{g}$ over $\mathbb{C}$ is in hand and we ignore rank 1. Here $\mathfrak{sl}_{n+1}$ has Lie type $A_n$ with $n > 1$.

Theorem 1 then is basically a case-by-case observation, using the known description of root systems as in Bourbaki (or for exceptional types more explicitly in Springer's table here).

Then Theorem 2 is just counting the number of summands isomorphic to the adjoint module in the tensor product of this module with itself. Here the module has highest weight equal to the highest root, which I'll call $\gamma$ (Bourbaki denotes it by $\widetilde{\alpha}$). From the case-by-case study one knows (as indicated) that the multiplicity of the adjoint module here, or equivalently the dimension of the Hom space, is at least 2 for type $A_n$ and at least 1 for other types. So the remaining problem is to make these bounds exact.

It's a standard (but intricate) classical problem to work out such tensor product multiplicities for arbitrary finite dimensional highest weight representations. However, the basic approach (going back to Brauer's Comptes rendus note in 1937 and further developed by Klimyk) typically involves a huge amount of cancellation along with a summation over the entire Weyl group $W$. But the idea is quite simple: in our case, add to $\gamma$ the weights (= roots along with 0) of the second factor in the tensor product, each counted with its multiplicity: 1 for each root, $n$ for 0. This gives the full list of irreducible summands with their multiplicities, but only if you transform each non-dominant weight in the list into the (shifted) dominant Weyl chamber via the dot-action of $W$ given by $w \cdot \mu = w(\mu+\rho)-\rho$. When this unique weight is dominant in the strict sense, attach the sign of $w$ to the resulting multiplicity in the tensor product.

Brauer's method in fact implies here that we get at most $n$ occurrences of the adjoint module as summands of the tensor product. So the problem is to reduce this using Theorem 1. This one does directly using the reflections $s_i$ corresponding to the simple roots along with standard root system information such as the fact that $s_i \rho = \rho - \alpha_i$ and that $s_i \gamma = \gamma$ whenever $\alpha_i$ is orthogonal to $\gamma$. Thus $s_i \cdot (\gamma -\alpha_i) = \gamma$ (and the sign is $-1$) in the orthogonal situation. There are more details to fill in, but the point is that it's all fairly straightforward and classical even though not transparent.

UPDATE: This question led me to consult a specialist (code name SK), who recalled a more general theorem but not its source. I asked about that here. Just now my consultant has retrieved the original source in a 1996 paper by R.C. King and B.G. Wybourne here. Like most of King's other work, the paper involves classical Lie theory of interest in mathematical physics. The proof relies on techniques such as Schur functors and plethysm but not on Littelmann paths, etc. (The article itself seems to be restricted to those with library subscriptions.)

As I noted in my question, Allen's theorem 1 translates into the more general hypothesis: the highest weight of the adjoint representation (i.e., the highest root) involves in each case just one or two fundamental weights, being orthogonal to the others.

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This is not exactly classical, but…

The dimension of $\operatorname{Hom}(\mathfrak{g} \otimes \mathfrak{g}, \mathfrak{g})$ counts Littelmann paths of length zero that, when appended to the path to the highest root, remain within the dominant Weyl chamber. Such a path must necessarily be killed by the raising operator corresponding to any simple root perpendicular to the highest root. Thus, the number of such paths is at most the number of simple roots not perpendicular to the highest root. I'm not an expert on Littelmann paths, so it's not clear to me if this same sort of reasoning will show that these two numbers are actually equal, but if we take it as "obvious" that $\operatorname{Hom}(\mathfrak{g} \otimes \mathfrak{g}, \mathfrak{g})$ has dimension at least 1, the only case where the equality is not "obvious" is $\mathfrak{sl}_n$.

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Yeah, this is the sort of model that Bruce was using to index the components of the geometric Satake fiber, that gave his result. I'm still hoping for something more classical... –  Allen Knutson May 7 '13 at 3:35
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