Let $\Phi$ an irreducible system of roots, $\Phi^{+} \subset \Phi$ a choose of positive roots. I have to prove that if $(\alpha, \beta) \ge 0$ for al $\beta \in \Phi^{+}$ then $\alpha$ is the highest among roots of the same lenght. I have a long way... is there a way that uses only some theorem? My way: Let be $\alpha '\neq \alpha$ the root of maximal lenght $||\alpha||$. Obviusly $\alpha '\in \Phi_+$, so $(\alpha,\alpha')\ge 0$. \ I want to prove that $(\alpha,\alpha')>0$. \ I show now that $$\alpha'=\sum_{\alpha_i\in\Delta}m_i\alpha_i\,\,\,\,\,\,\,\,(m_j>0)$$ We can suppose that $\alpha'$ is short. Now we suppose that $\alpha'$ is short. $\Phi$ is irreducible than it exists a simple root $\beta_i$ such that $$(\alpha',\beta_i)<0$$ so $\alpha'+\beta_i$ is a positive root longer than $\alpha'$.\ Morever $||\alpha'+\beta_i||=||\alpha'||$, because $$(\alpha'+\beta_i,\alpha'+\beta_i)=(\alpha',\alpha')+(\beta_i,\beta_i)+2(\alpha'+\beta_i)=$$ $$ =(\alpha',\alpha')+(\beta_i,\beta_i)\left[1+\frac{2(\alpha',\beta_i)}{(\beta_i,\beta_i)} \right] $$ $$\leq (\alpha',\alpha')$$ Because we have $<\alpha',\beta_i>\leq -1$, for the irreducibility of $\Phi$ we have finish.\ Now if $(\alpha,\alpha')=0$, we have $$\sum m_j(\alpha,\alpha_j)=0 \,\,\,\,\,\,\,\,\,\,\, (\alpha_j\in\Delta)$$ and so $$(\alpha,\alpha_j)=0 \,\,\,\,\,\,\,\,\,\,\,\,\, (\forall\,\alpha_j\in\Delta)$$ and it is no possible.\ We have finally that $(\alpha',\alpha)>0$, so $\alpha'-\alpha \in \Phi_+$,because $\alpha maximality$. However we had seen $(\alpha',\alpha)>0$ and $(\alpha',\alpha')/(\alpha,\alpha)=1$ we must have $<\alpha',\alpha>=1$. So we can say that $$(\alpha',\alpha)=\frac{1}{2}(\alpha,\alpha) $$ but for hipotesys we have to have $$(\alpha,\alpha'-\alpha)\ge 0 $$ and so $$(\alpha,\alpha')-(\alpha,\alpha)=-\frac{1}{2}(\alpha,\alpha)\ge 0 $$ and it si no possible. Could you give me a more easy and fast method... maybe using theorems? Is there a method that involves Weyl group?
1 Answers
Let $\alpha_0$ be the highest root of the same length as $\alpha$. Then $\alpha$ and $\alpha_0$ are in the same orbit of the Weyl group. Therefore (are you familiar with the length function on the Weyl group?) there is a sequence of simple reflections $s_{i_1}$, $s_{i_2}$, $\ldots$,$s_{i_\ell}$ such that $$ \alpha=s_{i_\ell}s_{i_{\ell-1}}\cdots s_{i_1}(\alpha_0). $$ Furthermore, by selecting the minimal number (= minimal $\ell$) of such simple reflections we have the result that the recursively defined sequence of roots $\alpha_1=s_{i_1}(\alpha_0)$, $\alpha_2=s_{i_2}(\alpha_1)$, $\ldots$, $\alpha=\alpha_\ell=s_{i_\ell}(\alpha_{\ell-1})$ is linearly ordered: $$ \alpha<\alpha_{\ell-1}<\alpha_{\ell-2}<\cdots <\alpha_0. $$ So if here $\alpha\neq\alpha_0$, or equivalently $\ell>0$, then the simple root $\beta_{i_\ell}$ corresponding to the simple reflection $s_{i_\ell}$ satisfies $(\alpha,\beta_{i_\ell})<0$. This contradicts your hypothesis.
This type of arguments are common, when describing the theory leading to the length of elements of the Weyl group. If you haven't covered that yet, then it gets messier.
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Why they are in the same orbit os the Weyl group? Coul do you insert some detail in order to expalin the sequence of simple reflection and the last statement? – ArthurStuart Feb 23 '13 at 12:20
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@Arthur: These are covered in Humphreys' book (or any treatise on Coxeter groups). I don't have my copy at home. I try to think of something :-) – Jyrki Lahtonen Feb 23 '13 at 14:35
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It would help, if I knew what has been covered in your book at this point. For example in Humphreys book (IIRC) one of the earliest facts was that all the roots are in the same orbit with a simple root. – Jyrki Lahtonen Feb 23 '13 at 14:46
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But where we used the hypotesis that $\alpha_{0}$ is the root of maximal hight. I don't undertand the step that in your text go from "then the simple root..." to "this contradicts your hypotesis". – ArthurStuart Feb 23 '13 at 17:36
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The element $w$ of the Weyl group mapping $\alpha_0$ to $\alpha$ is of length $\ell=0$ iff and $\alpha=\alpha_0$. So that assumption was used to get the existence of $\beta_{i_\ell}$. If $\ell=0$ there is no such simple root! But then we also have $\alpha=\alpha_0$. – Jyrki Lahtonen Feb 23 '13 at 18:51
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$\alpha_{\ell-1}=s_{i_\ell}(\alpha)>\alpha$ if and only if $(\alpha,\beta_{i_\ell})<0$ – Jyrki Lahtonen Feb 23 '13 at 21:44
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@JyrkiLahtonen Thanks for the neat proof. In your ordered sequence ${\alpha_k}$, is it true that $\alpha_k-\alpha_{k+1}$ is a simple root for every $k=0,1,\ldots, \ell -1$? – user Nov 01 '15 at 14:35
\\\: the correct way to do this is to leave an empty line. If you do not like the indentation (a common cause for your mistake) then ask someone who knows LaTeX how to suppress the indentation is a more civil way :-) – Mariano Suárez-Álvarez Feb 23 '13 at 02:33