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

In the crystallographic convention for this course, an admissible Coxeter graph comes from linearly independent unit vectors $v_1,\dots ,v_r$ in a Euclidean space, such that for $i\ne j$,The vertices $i,j$ have no connection when $n_{ij}=0$, and a single, double or triple connection when $n_{ij}=1,2,3$. Equivalently, the symmetric Gram matrixmust be positive-definite. This equivalence follows because a positive definite Gram matrix can be realized by linearly independent vectors, while such vectors necessarily have a positive definite Gram matrix. In the edge-labelled Coxeter graph convention, these connections have labels $3,4,6$, since the corresponding inner products are $-\cos (\pi /3)$, $-\cos (\pi /4)$ and $-\cos (\pi /6)$.

Suppose the underlying graph has a cycle through $k\ge 3$ distinct vertices, and let $v_1,\dots ,v_k$ be their unit vectors from admissibility. Adjacent vertices have $(v_i,v_j)\le -1/2$, and every other pair has nonpositive inner product. ThereforeBut linear independence makes the sum nonzero, whose squared norm in a Euclidean space must be positive. This contradiction proves that the underlying graph of an admissible Coxeter graph has no cycles. Double and triple lines record edge multiplicities; they are not cycles of length two. The argument also tolerates chords, which only make the bound smaller.

In the finite-dimensional complex semisimple Lie algebra setting implicit in this root-system construction, the given Cartan subalgebra and root spaces determine a reduced crystallographic root system $R$. The ordering divides $R$ into positive roots and their negatives. The simple roots are the positive roots that cannot be written as a sum of two positive roots; a standard root-system fact is that they form a basis and that every positive root is a nonnegative integral combination of them.

Index the simple roots as ${\alpha }_1,\dots ,{\alpha }_{\ell }$ and form the Cartan matrixFor $i\ne j$, these entries are nonpositive integers and $A_{ij}{A}_{ji}\in \{0,1,2,3\}$. Construct one vertex per simple root and join distinct vertices by $A_{ij}{A}_{ji}$ lines. If root lengths differ, direct the arrow toward the shorter simple root. This labelled and oriented graph is the Dynkin diagram. Its unoriented Coxeter graph is admissible: normalizing the simple roots to unit length gives $4(v_i,v_j{)}^2=A_{ij}{A}_{ji}$. A general Lie algebra without the semisimple root-system hypotheses need not produce such a diagram.

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.

If the three simple roots are distinct and every pair of root spaces had nonzero Lie bracket, part (iv) would put all three connections in the Dynkin diagram. Their underlying Coxeter graph would contain a triangle, contradicting part (ii). Hence at least one pair satisfies $\boxed{[{\mathfrak{g}}_{{\alpha }_i},{\mathfrak{g}}_{{\alpha }_j}]=0}$. If two of the listed roots coincide, their common root space is one-dimensional, so its bracket with itself is already zero. Thus the conclusion also covers that case.