The height $ht(\alpha)$ of a positive root $\alpha$ in a (finite, crystallographic) root system $\Phi$ is $\sum_{i=1}^n c_i$ where $\alpha = \sum_{i=1}^n c_i \alpha_i$ is its decomposition as a sum of simple roots. I am interested in the polynomial $$R_{\Phi}(x):=\sum_{\alpha \in \Phi^+} x^{ht(\alpha)-1}.$$

For example (using the usual Cartan-Killing nomenclature), we have:

  • $R_{A_n}(x)=n+(n-1)x+\cdots +2x^{n-2}+x^{n-1}$
  • $R_{B_n}(x)=R_{A_n}(x^2)+xR_{A_{n-1}}(x^2)$
  • $R_{D_n}(x)=R_{B_n}(x)-\sum_{i=n-1}^{2n-2} x^i$


  1. When $\Phi$ is an irreducible root system, is $R_{\Phi}(x)$ an irreducible polynomial over $\mathbb{Q}$? This has been checked for the exceptional types ($G_2, F_4, E_6, E_7,$ and $E_8$) and for the infinite families listed above for $n \leq 500$.
  2. If the answer to (1) is "yes", is there some uniform Lie-theoretic reason that this is true (that is, a proof not relying on the classification)?
  • 7
    $\begingroup$ Last I checked, it is still open to prove that $R_{A_n}$ is irreducible for all $n$, although it is known for "most" n. See "Classes of polynomials having only one non-cyclotomic irreducible factor" by Borisov, Filaseta, Lam, Trifonov. $\endgroup$ – Gjergji Zaimi May 31 at 16:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.