1
$\begingroup$

If we consider crystallographic root systems, then for each $k$ such that $n \leq k \leq d-1$ where $d$ is the Coxeter number, it seems to be the case that there is exactly one root of height $k$ with nonzero coefficients with respect to each simple root.

It seems to work if you check it case by case, but I would be interested to hear of a general way of arguing the point.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

This does not hold in general. In type E6, for example, there are roots for which the "middle" simple root has multiplicity 2 and so does one of the two other simple roots corresponding to a vertex adjacent to it, while all the other simple roots have multiplicity one.

Improve this answer
$\endgroup$
4
  • $\begingroup$ Is there any place on the internet or in the literature where all the roots in the $E_6$, $E_7$ or $E_8$ root systems are written out explicitly as linear combinations of simple roots? $\endgroup$
    – Rupert
    Aug 3, 2015 at 9:34
  • 2
    $\begingroup$ Several places. There are appendices to Bourbaki's Lie theory chapters 4-6, which are a standard reference. I didn't have that ready to hand, and found them via google. Here is E6, for instance: maths.nuigalway.ie/~pbrowne/personal/research/roots/e6_rt.pdf Let me also record that I didn't feel like I understood the positive roots until I knew the "knitting" procedure from quiver representations, by which you can calculate the positive roots in a canonical order about as quickly as you can write the results down. $\endgroup$
    – Hugh Thomas
    Aug 3, 2015 at 14:58
  • $\begingroup$ Do you have a reference for the "knitting" procedure? $\endgroup$
    – LSpice
    Sep 5, 2022 at 10:24
  • 1
    $\begingroup$ @LSpice Section 10 of Representations of Finite-Dimensional Algebras by Gabriel and Roiter; there are also versions on the web (e.g., eprints.lancs.ac.uk/id/eprint/155415/1/coelho_simoes.pdf). What is not so easy to find is a version explained for people who just want to know the algorithm, without the representation theory. I think this might be worth explaining and I could try to do so if you post a separate question. $\endgroup$
    – Hugh Thomas
    Sep 6, 2022 at 2:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.