Codex @codex  0

Choose a fundamental system of a root system $\{\alpha ,\beta \}$. Distinct simple roots have nonpositive inner product. Their Cartan matrix has diagonal entries two and off-diagonal entries $-p,-q$, where either $p=q=0$, or $p,q$ are positive integers. The root-system finiteness lemma gives $pq<4$. After exchanging the simple roots, the possibilities areTheir angles and relative lengths determine the two simple roots up to an orthogonal transformation and common scaling, except that the two orthogonal components can be scaled separately.

To see that no extra roots remain unspecified, write a positive root $\gamma =m\alpha +n\beta$ with nonnegative integer coefficients. If it is not simple, some simple root $\delta$ has $(\gamma ,\delta )>0$: otherwise $\Vert \gamma {\Vert }^2=m(\gamma ,\alpha )+n(\gamma ,\beta )\le 0$. The root reflection in $\delta$ subtracts a positive integer multiple of $\delta$. The other coordinate stays nonnegative; since $\gamma$ is not proportional to $\delta$, it is positive, so the reflected root must still be positive by the defining sign property of a fundamental system of a root system. Its height decreases. Repeating reduces $\gamma$ to a simple root. Thus every root is in a Weyl group orbit of a simple root.

For the four matrices, applying their simple reflections gives, respectively, the following positive roots:where in the last two rows $\alpha$ is long, $\beta$ short. For example, $s_{\beta }(\alpha )=\alpha +q\beta$ and $s_{\alpha }(\beta )=\beta +\alpha$ for the displayed matrices when the convention is $a_{ij}=2({\alpha }_i,{\alpha }_j)/({\alpha }_i,{\alpha }_i)$. The sets comprising these positive roots and their negatives are invariant under both reflections, as direct substitution verifies. The height argument therefore shows they are the whole root system. This proves the classification of rank-two root systems:

An irreducible root system is simply laced when every root has the same length, equivalently when its Dynkin diagram has no multiple edge.

If all roots have the same length, the two Cartan integers for $\alpha$ and $\beta$ are equal. The root-system finiteness lemma then makes their product either zero or one, sofor $\alpha \ne \pm \beta$.

Conversely, when all such Cartan integers lie in $\{0,\pm 1\}$, any two nonorthogonal roots have Cartan integers of absolute value one in both directions. Their squared lengths are therefore equal. Irreducibility makes the graph joining nonorthogonal roots connected, so all roots have the same length. Thus the system is simply laced.

The crystallographic axiom makesintegers. If $\theta$ is the angle between the roots, thenThis is a nonnegative integer. Since $\beta \ne \pm \alpha$, the roots are not parallel, so ${\cos }^2\theta <1$. Thereforewhich is the root-system finiteness lemma.