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

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?

share|improve this question
    
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? –  Jim Humphreys Apr 14 '11 at 13:48
    
Thank you! I have corrected the question. –  Anton Galaev Apr 14 '11 at 17:57
add comment

1 Answer

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.

share|improve this answer
    
There is a misprint: LHS should be $V(\mu)\otimes g$. Also the formula seems to fail for $\mu =0$. –  Victor Ostrik Apr 14 '11 at 16:35
    
I have corrected the misprint. Thanks. –  Bruce Westbury Apr 14 '11 at 16:55
    
For $\mu=0$, we have $k=0$, and the sum has only one term, namely $\alpha$ is the highest root. –  Bruce Westbury Apr 14 '11 at 16:56
    
If $g$ is not simply laced you will have two summands, is not it? –  Victor Ostrik Apr 14 '11 at 16:58
    
@Victor. Yes, you are right. This formula is only for the simply laced case when the Weyl group acts transitively on the roots. –  Bruce Westbury Apr 14 '11 at 17:12
show 1 more 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.