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Let $\Phi$ be a root system of $E$. $\alpha,\beta\in \Phi$.

Let $\lbrace \beta+i\alpha | i\in \mathbb{Z}\rbrace\cap \Phi$, $\alpha$-string through $\beta$, be $\beta-r\alpha,\ldots,\beta+q\alpha$, and let $\lbrace \alpha+i\beta | i\in \mathbb{Z}\rbrace\cap \Phi$, called $\beta$-string through $\alpha$, be $\alpha-r'\beta,\ldots,\alpha+q'\beta$.

I want to prove that $\frac{q(r+1)}{(\beta,\beta)}=\frac{q'(r'+1)}{(\alpha,\alpha)}$, where $(,)$ is an inner product.

What I can show is the following:

  1. Let $\sigma_\alpha$ be a reflection w.r.t. $\alpha$. Then, $\sigma_\alpha$ reverses the string, hence, $\sigma_\alpha(\beta+q\alpha)=\beta-r\alpha$. Left hand side is = $\beta -<\beta,\alpha>\alpha-q\alpha$. Hence, $r-q=<\beta,\alpha>$, where $<\beta,\alpha>=2\frac{(\beta,\alpha)}{(\alpha,\alpha)}$. (Here, I used $\sigma_\alpha(\beta)=\beta-<\beta,\alpha>\alpha$ and $\sigma(\alpha)=-\alpha$.)

  2. By 1, $\frac{r-q}{(\beta,\beta)}=\frac{<\beta,\alpha>}{(\beta,\beta)}=\frac{2(\beta,\alpha)}{(\beta,\beta)(\alpha,\alpha)}=\frac{<\alpha,\beta>}{(\alpha,\alpha)}=\frac{r'-q'}{(\alpha,\alpha)}$.

1 Answers 1

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Here is what I just found. You have to check the equality for the 4 different kinds of root systems, which are the only possibilities. For example, for the case $\lVert \beta\rVert^2$/$\lVert \alpha\rVert^2$= -3,$<\beta,\alpha>=3$, we find that: $r-q=<\beta,\alpha>=-3$ and that $r'-q'=<\alpha,\beta>=1$.

In the diagram corresponding to this situation, i.e. G2 , we can see that $q=4$ and $q'=1 $ and the result holds for this case by a straightforward calculation.

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    Ok. Fine. Thanks2017-08-24