Let $g$ be a simple complex Lie algebra with an irreducible representation $g\subset so(V)$ with the highest weight $\Lambda$.

In the book by Onishchik and Vinberg "Lie groups and algebraic groups" the following formula is given. If $V_\Lambda$ and $V_M$ are irreducible $g$-modules with the highest weights $\Lambda$ and $M$, then the multiplicity of $V_N$ in $V_\Lambda\otimes V_M$ equals

$\dim$ { $v\in(V_M)_{N-\Lambda}|(e_i)^{\Lambda_i+1}v=0, i=1,...,l$}

$= \dim$ {$v\in(V_N)_{\Lambda-M'}|(e_i)^{M'_i+1}v=0, i=1,...,l$},

where $l$ is the rank of $g$, $\{e_i,f_i,H_i\}$ are canonical generators of $g$, $\Lambda_i$ are labels on the Dynkin diagram, and $M'$ is the highest weight of $(V_M)^*$, $(V_M)_{N-\Lambda}$ denotes the weight space of weight $N-\Lambda$.

Taking $V_M=g$, we obtain $$V\otimes g=kV\oplus\oplus_\lambda V_\lambda,$$ where $k$ is the number of non-zero labels on the Dynkin diagram of $g$ defining the representation $g\subset so(V)$, and $V_\lambda$ are irreducible modules different from $V$ with the the highest weights $\lambda$ that are pairwise different. Of course, one of the modules $V_\lambda$ is $V_{\Lambda+\delta}$, $\delta$ is the highest root.

Are there some other formulas that may give more information about $V_\lambda$? I would like to have an expression for an element (not necessary of the highest vector) of each $V_\lambda$. Is it possible that $V_\lambda$ contains an element of the form $v\otimes A$, where $v\in V$, $A\in g$?

I am classifying irreducible subalgebras $g\subset so(V)$ that admit linear maps from $V$ to $g$ satisfying some equation. I guess that only $V_{\Lambda+\delta}$ may consist of such maps. To show that other $V_\lambda$ do not consist of such maps, I need to take some element from these modules and to check the equation.

The above formulas show that to each $V_\lambda$ there is a preferred root space in $g$ of weight $\lambda-\Lambda$ and a line in $V_\Lambda$ of weight $\lambda-\delta$. How the elements from these lines can be used?

  • $\begingroup$ The set-up here looks confusing. Is $V$ tensored with the adjoint representation? (The latter is irreducible only if the Lie algebra is simple.) Other parts of the question are also unclear to me, with a serious misprint at the end, an undefined symbol $\delta$, etc. And what is the decomposition method used here for the tensor product? $\endgroup$ – Jim Humphreys Apr 14 '11 at 13:48

I think the formula you are looking for is $V(\mu)\otimes g\cong kV(\mu) \oplus \oplus_\alpha V(\mu+\alpha)$ where the sum is over roots $\alpha$ such that $\mu+\alpha$ is a dominant weight. Here $V(\lambda)$ is irreducible with highest weight $\lambda$ for $\lambda$ an integral dominant weight.

  • $\begingroup$ There is a misprint: LHS should be $V(\mu)\otimes g$. Also the formula seems to fail for $\mu =0$. $\endgroup$ – Victor Ostrik Apr 14 '11 at 16:35
  • $\begingroup$ I have corrected the misprint. Thanks. $\endgroup$ – Bruce Westbury Apr 14 '11 at 16:55
  • $\begingroup$ For $\mu=0$, we have $k=0$, and the sum has only one term, namely $\alpha$ is the highest root. $\endgroup$ – Bruce Westbury Apr 14 '11 at 16:56
  • $\begingroup$ If $g$ is not simply laced you will have two summands, is not it? $\endgroup$ – Victor Ostrik Apr 14 '11 at 16:58
  • $\begingroup$ @Victor. Yes, you are right. This formula is only for the simply laced case when the Weyl group acts transitively on the roots. $\endgroup$ – Bruce Westbury Apr 14 '11 at 17:12

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