Reduced crystallographic Euclidean root system

Let $E$ be a finite-dimensional real inner product space with inner product $(\cdot ,\cdot )$ (Real and complex inner product spaces, with the inner product linear in the first argument), and let $\mathrm{\Phi }\subseteq E\setminus \{0\}$ be a finite subset which spans $E$ over $\mathbb{R}$. For $\alpha \in \mathrm{\Phi }$ define $s_{\alpha }:E⟶E,s_{\alpha }(x)=x-\frac{2(x,\alpha )}{(\alpha ,\alpha )}\alpha \mathrm{.}$ Since $(\alpha ,\alpha )>0$ for $\alpha \mathrm{\ne }0$, the scalar $2(x,\alpha )\mathrm{/}(\alpha ,\alpha )$ is well defined, and substitution gives $s_{\alpha }(\alpha )=-\alpha ,s_{\alpha }(x)=x\text{whenever }(x,\alpha )=0.$ A reduced crystallographic root system in $E$ (equivalently, a reduced abstract root system) is such a pair $(E,\mathrm{\Phi })$ satisfying:

1. $s_{\alpha }(\mathrm{\Phi })=\mathrm{\Phi }$ for every $\alpha \in \mathrm{\Phi }$;
2. ${\displaystyle \frac{2(\beta ,\alpha )}{(\alpha ,\alpha )}}\in \mathbb{Z}$ for all $\alpha ,\beta \in \mathrm{\Phi }$ (the crystallographic, or integrality, condition);
3. $\mathbb{R}\alpha \cap \mathrm{\Phi }=\{\alpha ,-\alpha \}$ for every $\alpha \in \mathrm{\Phi }$ (the reducedness condition).

The elements of $\mathrm{\Phi }$ are its roots, and $s_{\alpha }$ is the reflection in the hyperplane orthogonal to $\alpha$: it fixes ${\alpha }^{\perp }$ pointwise and negates $\alpha$.

Every root $\alpha$ satisfies $-\alpha \in \mathrm{\Phi }$: taking $\beta =\alpha$ in condition 2 gives $2\in \mathbb{Z}$, and condition 1 gives $s_{\alpha }(\alpha )=-\alpha \in \mathrm{\Phi }$. The integer $2(\beta ,\alpha )\mathrm{/}(\alpha ,\alpha )$ is the Cartan integer attached to the ordered pair $(\beta ,\alpha )$; condition 2 is the crystallographic axiom recorded above.