9
$\begingroup$

In a recent preprint I obtained a nice characterization of root systems as a side product. I can imagine that this was known before, and that a source for this statement can shorten the proof of my main result.

Question: Can you point me to a source, or a simple proof of the facts I list below?

In the following, let $R\subset \Bbb R^d$ be finite and centrally symmetric, i.e. $R=-R$.

Some definitions:

  • $R$ is a root system if for all $r\in R$ the set $R$ is invariant w.r.t. reflection on the hyperplane $r^\bot$.
  • A half (or semi-star in the preprint) is the intersection of $R$ with a halfspace, so that the intersection contains exactly half the elements of $R$. (see figure)
  • The norm of a half is the norm of the sum of the vectors it contains.

$\qquad\qquad\qquad\qquad\qquad\qquad$

The two characterizations now read as follows:

Result.

  1. (Corollary 5.1) All halves of $R$ are congruent (i.e. they are related by orthogonal transformations) if and only if $R$ is a root system.
  2. (Theorem 5.2) If all vectors in $R$ are of the same length, and all halves of $R$ are of the same norm, then $R$ is a root system.

In point 1. the direction that proofs that $R$ is a root system is the non-trivial part.

Improve this question
$\endgroup$
2
  • 3
    $\begingroup$ That's a nice result! I had a quick look, but, without tracing the chain of implications all the way back, it's not apparent: do you use a classification result, or is this from first principles? (Also, you're missing a period at the end of Section 5.) $\endgroup$
    – LSpice
    Commented Jan 6, 2020 at 15:21
  • 4
    $\begingroup$ @LSpice Thank you! I would say it is from first principles, just with a detour over zonotopes, so probably using some well-known facts about polytopes (but no classification). In particular, the classification of root systems plays no role for this characterization. $\endgroup$
    – M. Winter
    Commented Jan 6, 2020 at 15:30

0

Reset to default

You must log in to answer this question.