Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 3 / 4 / iv / Solution

Let ${\alpha }_1,{\alpha }_2$ be distinct simple roots. Their difference is not a root, since its simple-root coefficients have opposite signs. The root string through ${\alpha }_2$ in the direction ${\alpha }_1$ therefore starts at ${\alpha }_2$ and has the formHere we use the root-string theorem: if the full string runs from $\beta -p\alpha$ through $\beta +q\alpha$, then $p-q=⟨\beta ,{\alpha }^{\vee }⟩$; in this case $p=0$. Thus ${\alpha }_1+{\alpha }_2$ is a root exactly when $A_{12}<0$, equivalently when the corresponding Dynkin diagram nodes are connected.

The root-space bracket facts are $[{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta }]\subseteq {\mathfrak{g}}_{\alpha +\beta }$, with a zero space when the sum is not a root, and $[{\mathfrak{g}}_{\alpha },{\mathfrak{g}}_{\beta }]={\mathfrak{g}}_{\alpha +\beta }\ne 0$ when the sum is a root and $\beta \ne -\alpha$. Applying these stated structural facts proves

If the two names denote the same simple root, its one-dimensional root space has zero bracket with itself, and a Dynkin diagram has no self-edge. Thus the equivalence also holds in that case.