Let $\mathbb{E} \cong \mathbb{R}^n$ be a vector space with an inner product $\langle\cdot,\cdot\rangle$. A subset $R \subset \mathbb{E} \setminus \{0\}$ is called a root system, if $R$ has the following properties
- $R$ is finite and spans $\mathbb{E}$
- if $\alpha \in R$ then $-\alpha \in R$ and $\pm \alpha$ are the only multiples of $\alpha$ in $R$
- R is invariant under the reflection in the hyperplane orthogonal to any $\alpha \in R$, i.e, $\forall_{\alpha, \beta} \in R, \; s_\alpha(\beta) = \beta - 2\operatorname{proj}_\alpha(\beta) \in R$
- $\forall \alpha, \beta \in R, \; 2 \cdot \frac{\langle\beta, \alpha\rangle}{\langle\alpha,\alpha\rangle} \in \mathbb{Z}$
I am reading from this see page 2
Let $\mathbb{E} \cong \mathbb{R}^{2}$. Say $\alpha,\beta \in R$. Note the projection of $\beta$ onto $\alpha$ is $\frac{1}{2}n_{\beta \alpha}\cos(\theta)$, where $\theta$ is the angle between $\alpha$ and $\beta$, and we require this number to be an integer.
In the paper they say defining what $\theta$ is constructs the root system. Say $\theta = \frac{\pi}{3}$. So the angle between $\alpha$ and $\beta$ is $\frac{\pi}{3}$. But in $R$, if we say $\beta = -\alpha$, the angle between $-\alpha$ and $\alpha$ is $\pi$, not $\frac{\pi}{3}$. So no two roots in $R$ have the angle between them $\frac{\pi}{3}$. So how does defining what $\theta$ is define the root system?
Is it that the angle between any two adjacent vectors in $R$ is $\frac{\pi}{3}$? If so, that works I think.
I appreciate any clarification anyone can give.