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Let $\alpha $,$ \beta$ $\in$ $\Phi$. Let the $\alpha$-string through $\beta$ be $\beta-r\alpha$,$\ldots$,$\beta+q\alpha$,and let the $\beta$-string through $\alpha$ be $\alpha-r'\beta$, $\ldots$,$\alpha+q'\beta$.Prove that $$\frac{q(r+1)}{(\beta,\beta)}=\frac{q'(r'+1)}{(\alpha,\alpha)}.$$

$r-q=\langle \beta,\alpha \rangle$

Valentin
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user63788
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1 Answers1

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Note that the reflaction along $\alpha$, which maps $\alpha\mapsto-\alpha$, $\beta\mapsto \beta-2\frac{\langle\alpha,\beta\rangle}{\langle\alpha,\alpha\rangle}\alpha$ leaves the $\alpha$ string invariant, hence exchanges the ends of the string, i.e. $\beta-r\alpha\mapsto\beta+q\alpha$. On the other hand, by linearity, $\beta-r\alpha\mapsto \beta-2\frac{\langle\alpha,\beta\rangle}{\langle\alpha,\alpha\rangle}\alpha+r\alpha$ so that we conclude $$r-q = 2\frac{\langle\beta,\alpha\rangle}{\langle\alpha,\alpha\rangle}. $$ This shows that the second result $$ r-q=\langle\beta,\alpha\rangle$$ holds if and only if $\langle\alpha,\beta\rangle=0$ or $\langle\alpha,\alpha\rangle=2$.

Hagen von Eitzen
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  • Reflection about $\alpha$ maps $\beta+q\alpha$ to $\beta-r\alpha$, at the same time ,it equals $\beta-\langle\beta,\alpha \rangle\alpha-q\alpha $. so $r-q=\langle\beta,\alpha \rangle$ it is irrelevant to 2 or 0 It only needs to prove the fist equality. – user63788 Mar 04 '13 at 04:16
  • There is a confusion of notations here. In the OP, $( \cdot, \cdot)$ means a scalar product, whereas $\langle \cdot, \cdot \rangle$ is a pairing (non-symmetric if we have roots of different length) which is actually defined by $\langle \beta, \alpha \rangle := 2\frac{(\beta, \alpha)}{(\alpha, \alpha)}$. So your first conclusion is actually $r-q = 2\frac{(\beta, \alpha)}{(\alpha, \alpha)} = \langle \beta, \alpha \rangle$. – Torsten Schoeneberg Nov 17 '17 at 17:29