Let $\mathfrak{g}$ be a simple complex Lie algebra. Let $V_1,\cdots, V_n$ be the fundamental representations (the irreducible ones with fundamental weights $\omega_1,\cdots,\omega_n$). Take a $k$-tensor product of these representations: $V_{\lambda_1}\otimes\cdots\otimes V_{\lambda_k}$ (with each $\lambda_i\in\{\omega_1,\cdots,\omega_n\}$).

Decompose this product into irreducible representations. Let $\lambda=\sum n_i\omega_i$ be the highest weight of such a simple summand. Can we conclude $\sum n_i\leq k$?

I can show it holds for type $A_n$ and $C_n$.

  • $\begingroup$ Can we not conclude immediately that $\sum n_i = k$ for the "highest highest" weight? The tensor product of $V_\lambda$ and $V_\mu$ contains $V_{\mu + \lambda}$ and all other irreducible subrepresentations have lower highest weight. Extending this to $V_{\lambda_1} \otimes \cdots \otimes V_{\lambda_k}$ we see that the highest weight present overall is $\lambda_1 + \cdots + \lambda_k$ (so all other highest weights are lower than this) from which the result you want follows. $\endgroup$
    – Callum
    Commented Nov 28, 2022 at 14:32
  • $\begingroup$ That is true. But some lower weight of the form ($\sum n_i w_i -$ simple roots) may have larger $\sum n_i$. $\endgroup$
    – Jun Yang
    Commented Nov 28, 2022 at 15:53
  • 1
    $\begingroup$ Ah good point and indeed playing around with some examples produces a counter example: in $D_5$, $V_{\omega_1}\otimes V_{\omega_2}\otimes V_{\omega_3}\otimes V_{\omega_5}$ contains a subrepresentation of the form $V_{2\omega_1 + \omega_4+ 2\omega_5}$ $\endgroup$
    – Callum
    Commented Nov 28, 2022 at 18:06

1 Answer 1


Actually one can show that if $\sum n_i \lambda_i$ is a highest weight in a $k$-tensor product of fundamental representations, we have $\sum n_i\leq \beta \cdot k$ for some $\beta$ uniquely determined by the simple type of $\mathfrak{g}$.


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