Past exam of the Lie algebras and representation theory of symmetric groups course of the University of Oxford / 2008 / A / 4 / Solution

Use the convention $A_{ij}=⟨{\alpha }_j,{\alpha }_i^{\vee }⟩=2({\alpha }_i,{\alpha }_j)/({\alpha }_i,{\alpha }_i)$. For a finite reduced crystallographic root system, diagonal entries are two and off-diagonal entries belong to $\{0,-1,-2,-3\}$, with $A_{ij}{A}_{ji}\in \{0,1,2,3\}$. Transposing this convention merely transposes the Cartan matrix.

The $\alpha$-string through $\beta$ has the consecutive roots $\beta -p\alpha ,\dots ,\beta +q\alpha$, where $p,q\ge 0$ and $p-q=⟨\beta ,{\alpha }^{\vee }⟩$. For distinct simple roots ${\alpha }_i,{\alpha }_j$, the negative step is absent, so $p=0$ and $q=-A_{ij}$.

The matrix determines each simple reflection on the simple-root basis:It therefore determines their generated Weyl group as linear transformations. Every root is in the Weyl orbit of a simple root. For a positive nonsimple root $\beta =\sum c_i{\alpha }_i$, some $(\beta ,{\alpha }_i)$ is positive because $\sum c_i(\beta ,{\alpha }_i)=(\beta ,\beta )>0$. Its reflection subtracts a positive integer multiple of ${\alpha }_i$ and lowers height. It stays positive unless the root is simple, since the other simple-root coefficients are unchanged and at least one is positive. Induction on height reaches a simple root; negative roots follow by sign. Hence $\Phi =W\Delta$, which is completely reconstructed by those reflection matrices. This determines the system up to root-system isomorphism, rather than fixing an absolute length scale; irreducible components may be rescaled.

For positive roots for a chain Cartan matrix, symmetry and adjacency imply all simple roots have equal length; normalize their squared lengths to two. Adjacent inner products are minus one and all others zero. Since every root is Weyl-conjugate to a simple root, every root also has squared length two. Write a positive root as $\beta ={\sum }_{i=1}^l{c}_i{\omega }_i$ with nonnegative integers $c_i$, and set $c_0=c_{l+1}=0$. ThenThere must be exactly two nonzero integer differences, one $+1$ and one $-1$. Nonnegativity and the two zero endpoint values imply that the coefficients are one on one consecutive interval and zero elsewhere.

Conversely a single simple root is a root. If ${\omega }_i+\cdots +{\omega }_j$ is a root and $j<l$, its pairing with ${\omega }_{j+1}^{\vee }$ is minus one. Reflecting in ${\omega }_{j+1}$ therefore adds that simple root, making the next longer interval a root. Induction constructs every interval. Thus