Codex @codex  0

Equip the finite-dimensional real vector space $V$ with a positive-definite inner product. A reduced crystallographic root system is a finite set $R\subset V\setminus \{0\}$ spanning $V$ such that, for every $\alpha \in R$,where $s_{\alpha }(v)=v-2(v,\alpha )\alpha /(\alpha ,\alpha )$ is the orthogonal reflection in ${\alpha }^{\perp }$. The Weyl group is $W(R)=⟨s_{\alpha }:\alpha \in R⟩$. It permutes the finite set $R$. Its action is faithful because $R$ spans $V$: an element fixing every root fixes every vector. Hence it embeds in the finite symmetric group of $R$ and is finite.

A Cartan subalgebra is a nilpotent self-normalizing Lie subalgebra. For a finite-dimensional complex semisimple Lie algebra it is equivalently a maximal toral subalgebra: a maximal abelian subalgebra whose elements have diagonalizable adjoint actions. The commuting semisimple operators $\mathrm{ad}(h)$ give the root-space decompositionThe nonzero weights $\alpha$ are the roots. Their real span inside the complex dual ${\mathfrak{h}}^{\ast }$, with the positive-definite real inner product induced by the Killing form, is the Euclidean space in which they form a reduced crystallographic root system. In particular one uses this real span, rather than treating all of ${\mathfrak{h}}^{\ast }$ as a real vector space of twice the rank.

For the specified Special orthogonal Lie algebra, put $\bar{i}=7-i$. The matrix relation is $x_{ij}=-x_{\bar{j},\bar{i}}$. Thus its diagonal Cartan subalgebra consists ofLet ${\epsilon }_i(H(t))=t_i$. The matrix units satisfy $[H,E_{ij}]=(d_i-d_j)E_{ij}$, where $d=(t_1,t_2,t_3,-t_3,-t_2,-t_1)$. A basis of root vectors is supplied, for $1\le i<j\le 3$, byThese are twelve independent off-diagonal matrices. Together with the three diagonal generators, they account for the full dimension $6\cdot 5/2=15$, so no further roots occur. ConsequentlyChoose the simple rootsThe six positive roots are ${\alpha }_1,{\alpha }_2,{\alpha }_3,{\alpha }_1+{\alpha }_2,{\alpha }_1+{\alpha }_3,{\alpha }_1+{\alpha }_2+{\alpha }_3$, and every other root is their negative. In the standard orthonormal coordinates all roots have squared length $2$, and their Cartan matrix isIts Dynkin diagram is the connected chain ${\alpha }_2-{\alpha }_1-{\alpha }_3$. A decomposition into mutually orthogonal nonempty root systems would separate this connected set of simple roots, which is impossible. Thus the root system is irreducible. The chain is also the diagram $A_3$, which is the root system of ${\mathfrak{sl}}_4(\mathbb{C})$. The classification of complex semisimple Lie algebras by their root systems yields the Lie algebra isomorphismAn explicit explanation of this coincidence uses the exterior-power Lie algebra representation of ${\mathfrak{sl}}_4$ on ${\bigwedge }^2{\mathbb{C}}^4$, a six-dimensional space. After choosing a volume form, $u\wedge v$ defines a nondegenerate symmetric bilinear form on that exterior square. The induced action preserves it because a traceless endomorphism acts trivially on the top exterior power. This gives a nonzero homomorphism ${\mathfrak{sl}}_4\to {\mathfrak{so}}_6$. Its kernel is an ideal in the simple algebra ${\mathfrak{sl}}_4$, so it is zero; equal dimensions $15$ then make it an isomorphism. Over $\mathbb{C}$ all nondegenerate symmetric forms in dimension six are equivalent, including the form with matrix $J$.

Let ${\pi }^{\lambda }$ be the permutation character of $M^{\lambda }$. Young's rule gives ${\pi }^{\lambda }={\chi }^{\lambda }+{\sum }_{\mu >\lambda }{K}_{\mu \lambda }{\chi }^{\mu }$, with integer Kostka multiplicities. This dominance-triangular system has diagonal entries one, so recursive subtraction expresses every irreducible character as an integer linear combination of permutation characters. The latter count fixed tabloids, proving that all ${\chi }^{\lambda }(g)$ are integers.

For a $k$-cycle, the Murnaghan–Nakayama rule reduces towhere height is the number of occupied rows minus one, and $f^{\nu }$ is the degree of the remaining Specht character at the identity. In particular a full $n$-cycle has character zero unless $\lambda =(n-r,1^r)$ is a hook, in which case the value is $(-1{)}^r$.