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Let $R$ be a reduced, irreducible, crystallographic root system with positive roots $R^+$ and simple roots $\Delta$. Let $W$ be the Weyl group of $R$. Let $(-,-)$ be a $W$-invariant scalar product on $\mathbb{R}\Delta$ (this is unique up to non-zero scalar as far as I know).

Question 1

Let $\alpha,\beta\in R^+$ such that $(\alpha,\beta)\geq 0$ and such that $\alpha+\beta\in R$. Is then always $(\alpha+\beta)^\vee=\alpha^\vee+\beta^\vee$?

Here we denote as usual by $\alpha^\vee=2\alpha/(\alpha,\alpha)$ the dual root / coroot of a root $\alpha$.

Question 2

Let $\alpha,\beta\in R^+$ such that $(\alpha,\beta)=0$ and such that $\alpha+\beta\in R$. Is then always $(\alpha+\beta)^\vee=\alpha^\vee+\beta^\vee$?

The reason why I cannot answer this question myself is that I am not familiar with pairs of roots which are orthogonal but are not strongly orthogonal. We say $\alpha,\beta\in R$ are strongly orthogonal if $\alpha\pm\beta\notin R\cup\{0\}$.

I understand that such pairs of roots can only occur if there are two root length and if $$ (\text{long root length})^2=2\cdot(\text{short root length})^2\,, $$ i.e. only in type $\mathsf{BCF}$. But still I am not able to find them. I cannot find roots for which the hypotheses are satisfied.

NB. It also counts as an answer if you give some examples of such pairs of roots and verify Question 2 on them. In any case, please use the numbering of the simple roots as in the Bourbaki tables. It makes it much easier to follow for me.

Thank you very much for any kind of help!

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  • $\begingroup$ For my purposes it would be actually sufficient to replace the equality $(\alpha+\beta)^\vee=\alpha^\vee+\beta^\vee$ in Question 1 and Question 2 by the inequality $(\alpha+\beta)^\vee\leq\alpha^\vee+\beta^\vee$ where $\leq$ is the usual partial order on $\mathbb{Z}\Delta$ resp. $\mathbb{Z}\Delta^\vee$. $\endgroup$ – user80886 Sep 9 '16 at 17:36
  • $\begingroup$ In type $C_n$, the roots are $2e_i$, $\pm e_i \pm e_j$ ($i<j$). Thus for $i \ne j$, $\alpha=e_i+e_j$ and $\beta=e_i-e_j$ are orthogonal, but not strongly orthogonal. $\endgroup$ – Victor Protsak Sep 10 '16 at 15:20
  • $\begingroup$ Probably this question is too elementary for this site (did you try stack exchange first?). In any case, it's useful to add a tag such as 'lie-algebras' so the question doesn't get overlooked. (Even though the question itself doesn't involve simple Lie algebras directly, that's where the root system ideas originated.) $\endgroup$ – Jim Humphreys Sep 10 '16 at 15:51
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    $\begingroup$ The question is indeed elementary, and I found a satisfactory answer on my own several hours later. You can delete the whole thing, if you want. $\endgroup$ – user80886 Sep 10 '16 at 17:33
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Lemma

Let $\alpha,\beta\in R^+$ such that $(\alpha,\beta)\geq 0$ and such that $\alpha+\beta\in R$. Then we have $(\alpha+\beta)^\vee<\alpha^\vee+\beta^\vee$.

Proof

Sine $(\alpha,\beta)\geq 0$, we see that there are two root length, and that $\alpha+\beta$ is long and $\alpha,\beta$ are short. Let $$ n=\frac{(\text{long root length})^2}{(\text{short root length})^2}\,. $$ Then $n$ is a strictly positive integer, more concretely $n=2,3$. A simple computation shows that $\alpha^\vee+\beta^\vee=n(\alpha+\beta)^\vee$, in particular $(\alpha+\beta)^\vee<\alpha^\vee+\beta^\vee$.

Example

Let $R$ be of type $\mathsf{B}_2$. Then $\alpha=\alpha_1+\alpha_2$ and $\beta=\alpha_2$ are a pair of orthogonal roots which are not strongly orthogonal. Here we have $n=2$ and $\alpha^\vee+\beta^\vee=2(\alpha+\beta)^\vee=2(\alpha_1^\vee+\alpha_2^\vee)$.

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