1
$\begingroup$

This is a question about abstract root system, root lattice and weight lattice. Most definitions are standard as in Wikipedia or Humphreys' Lie algebra, section 13.

If $\Phi$ is an irreducible root system in $\mathbb{R}^l$, $\Delta$ a base, $\Lambda$ the weight lattice, and $\Lambda_r$ the root lattice of $\Phi$ w.r.t $\Delta$. Prove that each coset of $\Lambda$ in $\Lambda_r$ contains unique minimal dominant weight.


Definitions:

Let $\Phi$ be a root system in $E=\mathbb{R}^l$, with a base $\Delta=\{\alpha_1,\cdots,\alpha_l\}$.

Let $(-,-)$ denote positive definite inner product on $E$, and $\langle\alpha,\beta\rangle:=\frac{2(\alpha,\beta)}{(\beta,\beta)}$.

A vector $\lambda\in E$ is called a weight if $\langle \lambda,\alpha\rangle\in\mathbb{Z}$ for all $\alpha\in\Delta$ (or equivalently all $\alpha\in\Phi$).

A weight $\lambda$ is called dominant if $\langle\lambda,\alpha\rangle\geq 0$ for all $\alpha\in\Phi$. (These are precisely weights lying in Fundamental Weyl chamber w.r.t. $\Delta$.

The weights in $E$ w.r.t. $\Phi$ form a lattice of rank$=l=\dim E$; called weight lattice.

The lattice $\Lambda:=\mathbb{Z}\alpha_1 + \cdots + \mathbb{Z}\alpha_l$ is called root lattice. It is known that $[\Lambda:\Lambda_r]$ is finite.

On $\Lambda$, define partial order: $\lambda,\mu\in \Lambda$, say $\lambda\prec\mu$ if $\mu-\lambda$ is non-negative $\mathbb{R}$-combination of $\alpha_1,\cdots,\alpha_l$.

A weight $\lambda\in\Lambda$ is called minimal dominant weight if it is a dominant weight (i.e. lies in fund. Weyl chamber) and it is minimal w.r.t. ordering $\prec$, i.e. if $\mu$ is a dominant weight s.t. $\mu\prec\lambda$ then $\mu=\lambda$.

The problem is that each coset $\lambda+\Lambda_r$ of root lattice in wight lattice contains unique minimal dominant weight.

$\endgroup$
  • $\begingroup$ Fairly sure I worked this out 30 years ago. If only I could remember how I did it. $\endgroup$ – Jyrki Lahtonen Apr 20 '18 at 6:57

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.