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This is a follow-up to a recent question by Allen Knutson here, involving a special type of tensor product multiplicity for a simple Lie algebra $\mathfrak{g}$ over $\mathbb{C}$ (or other algebraically closed field of characteristic 0). Here the dominant integral weioghts $\lambda$ index up to isomorphism the finite dimensional simple $\mathfrak{g}$-modules $V(\lambda)$. A tensor product of two such modules decomposes as the direct sum of others, with multiplicities which have been studied classically and in more recent times. The multiplicity is equally well given by $\dim \mathrm{Hom}_\mathfrak{g}\big(V(\nu), V(\lambda) \otimes V(\mu)\big)$ (or a version with the two terms switched).

The adjoint module, which we denote by $\mathfrak{g}$, is one of these simple modules. Allen formulates in his Theorem 2 a rule for the multiplicity of $\mathfrak{g}$ in its tensor product with itself. Since this wasn't familiar to me, I asked a more experienced specialist whether it looked familiar. The reply was that a more general result is known, but with an unremembered source: $$ \dim \mathrm{Hom}_\mathfrak{g} \big(\mathfrak{g}, \mathrm{End}\,V(\lambda)\big) = d(\lambda),$$ where $d(\lambda)$ is the number of fundamental weights occurring with positive coefficient in the standard expression of $\lambda$ as a $\mathbb{Z}^+$-linear combination of those weights. Note $\mathrm{End}\,V(\lambda) \cong V(\lambda) \otimes V(\lambda)$ when this module is self-dual; this occurs for the adjoint module (nondegeneracy of Killing form), or in general when $\lambda = -w_\circ \lambda$ ($w_\circ$ the longest element of the Weyl group).

To recover Allen's Theorem 2, recall from Bourbaki's tables that the highest root (= highest weight of the adjoint module) is twice the first fundamental weight for type $C$, the sum of the first and last fundamental weights for type $A$, and a fundamental weight for all other irreducible root systems.

If the general result is true as I stated it above, what is the original source? (And is the proof "classical" or "modern", relying or not on knowledge of how many simple roots are orthogonal to the highest root?).

ADDED: My question has just been answered, as I've outlined in an update to Allen's original question. It's a reference I wouldn't easily have tracked down. Since the title and math physics style of the King-Wybourne paper may be hard to penetrate at first, maybe I should emphasize the difference between working with $V \otimes V$ and working with $V^* \otimes V$ (the End space). In their paper self-dual is usually called selfcontragredient. Often, but not always in types $A_n (n>1), D_n (n \text{ odd}), E_6$, simple modules for a simple Lie algebra are automatically self-dual. The distinction is easy to see if you tensor the standard 3-dimensional module for $\mathfrak{sl}_3$ with itself (which gives summands of dimensions 6, 3) or with its dual (which gives a trivial summand along with the adjoint module).

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Obviously I didn't know this more general result, but Evan Jenkins' modern proof generalizes easily enough. Restate as ${\rm Hom}(V(\lambda),{\mathfrak g}\otimes V(\lambda))$. Then we're looking at appending a $\Theta$-path of weight $0$ to a dominant $\lambda$-path, keeping the whole inside the dominant chamber (here $\Theta$) is the highest root. There are $rank(\lie{g})$ many $\Theta$-paths of weight $0$, of course, and the Lakshmibai-Seshadri ones go $0 \to -\alpha/2 \to 0$. –  Allen Knutson May 10 '13 at 12:43
    
@Allen: Yes, it seems reasonable to get this from the recent methods, but I'm curious about who first wrote this down (and where, why, how). There's a fairly extensive literature, hard to search: Dynkin, Kostant, Dixmier, Joseph, .... Kempf too had papers on tensor product decompositions, from the geometric viewpoint. –  Jim Humphreys May 11 '13 at 13:27
    
P.S. The paper by King-Wybourne requires the added hypothesis that $V(\lambda)$ (in the above formula I was told about) is self-dual. This is true for the adjoint module in your special case, but I'm uncertain whether the added hypothesis is really needed or not. I wonder whether the newer methods make that clear? –  Jim Humphreys May 20 '13 at 12:20
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