---- > [!theorem] Theorem. ([[angles and lengths in a root system]]) > Let $(\Phi, E)$ be a [[root system]]. If $\alpha, \beta \in \Phi$ with $\beta \neq \pm \alpha$, then $\langle \beta, \alpha^{\vee} \rangle \ \langle \alpha, \beta^{\vee} \rangle \in \{ 0,1,2,3 \}.$ > This severely constrains the possibilities for what $\langle \beta, \alpha^{\vee} \rangle$ and $\langle \alpha, \beta^{\vee} \rangle$ can be, yielding the following table of consequences (we have assumed WLOG $(\beta,\beta)\geq(\alpha,\alpha)$ ): > > > > | **$\langle \beta, \alpha^{\vee} \rangle$** | **$\langle \alpha, \beta^{\vee} \rangle$** | **$\frac{(\beta, \beta)}{(\alpha,\alpha)}$** | **$\theta$** | > | ------------------------------------------ | ------------------------------------------ | -------------------------------------------- | ----------------------------- | > | 0 | 0 | ? | $\frac{\pi}{2}=90{\degree}$ | > | 1 | 1 | 1 | $\frac{\pi}{3}=60{\degree}$ | > | -1 | -1 | 1 | $\frac{2\pi}{3}=120{\degree}$ | > | 2 | 1 | 2 | $\frac{\pi}{4}=45{\degree}$ | > | -2 | -1 | 2 | $\frac{3\pi}{4}=135{\degree}$ | > | 3 | 1 | 3 | $\frac{\pi}{6}=30{\degree}$ | > | -3 | -1 | 3 | $\frac{5\pi}{6}=150{\degree}$ | > > (We call $\frac{(\beta,\beta)}{(\alpha,\alpha)}=\frac{\|\beta\|^{2}}{\|\alpha\|^{2}}=\frac{\|\beta\|}{\|\alpha\|}$ the **length ratio**.) > [!proposition] Corollaries. > 1. If $\alpha,\beta \in \Phi$ with $\beta \neq \pm \alpha$ and $(\alpha,\beta)<0$ (i.e., the angle between them is obtuse), then $\alpha+\beta \in \Phi$. > 2. If $\Phi$ is [[reducible root system|irreducible]], there are at most two root lengths: $|\{ (\alpha,\alpha): \alpha \in \Phi \}|=2.$ > 3. [[root string|Root strings]] have size at most four. > 4. If $\Delta$ a [[root basis]] and $\alpha,\beta \in \Delta_{}$, $\alpha \neq \beta$, then $(\alpha, \beta) \leq 0$ (angle is at least 90$\degree$). ^proposition > [!proof]- Proof. ([[angles and lengths in a root system]]) > This is fun. Write $\|v\|=(v,v)^{1/2}$ for $v \in E$. Let $0 \leq \theta < \pi$ be the angle between $\alpha$ and $\beta.$ Then $(\alpha, \beta)=\|\alpha\| \|\beta\| \cos \theta$, hence $\langle \alpha, \check \beta \rangle \langle \beta, \check \alpha \rangle = \frac{2(\alpha, \beta)}{(\beta, \beta)} \cdot \frac{2(\beta, \alpha)}{(\alpha, \alpha)}= 4 \cos ^{2} \theta.$ > By the [[root system]] axioms, this must be an integer (as a product thereof). On the other hand, $\cos ^{2} \theta \in [0,1]$ in general, and here $\cos ^{2} \theta \neq 1$ since we assumed $\beta \neq \pm \alpha$. The result follows. ---- #### ----- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```