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For a finite-dimensional Irreducible Lie algebra representation of a complex semisimple Lie algebra with highest weight $\lambda$, this recursion computes its weight multiplicities from $m_{\lambda }(\lambda )=1$. Here $\rho$ is the half-sum of positive roots and the inner product is induced by the Killing form. Write the Casimir operator on the weight-$\mu$ space as $(\mu ,\mu +2\rho )I+2{\sum }_{\alpha >0}{F}_{\alpha }{E}_{\alpha }$. The cyclic trace identity between adjacent weight spaces gives $\mathrm{tr}(F_{\alpha }{E}_{\alpha }{|}_{V_{\mu }})={\sum }_{j\ge 1}(\mu +j\alpha ,\alpha )m_{\lambda }(\mu +j\alpha )$. Taking the trace and using the Casimir eigenvalue proves the formula. Only finitely many terms are nonzero.

Use the sl2 Lie algebra relations $[H,X]=2X$, $[H,Y]=-2Y$ and $[X,Y]=H$. A highest-weight vector satisfies $Xv=0$ and $Hv=mv$, and $H{Y}^{j}v=(m-2j)Y^{j}v$. For $k=0$, $XYv=YXv+Hv=mv$. Assuming the formula at $k-1$, one getsThus the sl2 highest-weight lowering formula isIn particular a finite-dimensional Irreducible Lie algebra representation has highest weight $m\in {\mathbb{Z}}_{\ge 0}$ and weight vectors $v,Yv,\dots ,Y^{m}v$, as in the classification of finite-dimensional sl2 representations.

Let $V_{\lambda }$ be a finite-dimensional Irreducible Lie algebra representation of a complex semisimple Lie algebra, with highest weight $\lambda$. Write $m_{\lambda }(\mu )=\dim (V_{\lambda }{)}_{\mu }$, taking it to be zero when $\mu$ is not a weight, and let $\rho$ be the half-sum of positive roots. The Killing form induces an inner product on the real span of weights. Freudenthal multiplicity formula statesThe sums are finite. For a weight $\mu \ne \lambda$, the coefficient on the left is positive: move $\mu$ into the dominant Weyl chamber, use that a weight is below $\lambda$ in dominance order, and note that $(\mu ,\rho )$ does not increase on moving back out of that chamber. The resulting recursion starts from $m_{\lambda }(\lambda )=1$.

Here is a proof using the allowed Casimir operator. Choose root vectors $E_{\alpha },F_{\alpha }$ for positive and negative roots, normalized by $B(E_{\alpha },F_{\alpha })=1$, and let $t_{\alpha }\in \mathfrak{h}$ satisfy $B(t_{\alpha },h)=\alpha (h)$. Invariance of a bilinear form on a Lie algebra gives $[E_{\alpha },F_{\alpha }]=t_{\alpha }$. If $h_i,h^i$ are dual bases of the Cartan subalgebra under $B$, the Casimir operator isOn the weight space of $\mu$, the first sum acts by $(\mu ,\mu )$. Replacing $E_{\alpha }{F}_{\alpha }$ with $F_{\alpha }{E}_{\alpha }+t_{\alpha }$ givesDefine $T_{\alpha }(\mu )=\mathrm{tr}(F_{\alpha }{E}_{\alpha }{|}_{(V_{\lambda }{)}_{\mu }})$. The cyclic trace identity between adjacent weight spaces and the commutator relation yieldIterate upwards until the weight spaces vanish to obtain $T_{\alpha }(\mu )={\sum }_{j\ge 1}(\mu +j\alpha ,\alpha )m_{\lambda }(\mu +j\alpha )$. Finally take the trace of the displayed restriction of $C$. Its Casimir eigenvalue is $(\lambda ,\lambda +2\rho )$, so subtracting $(\mu ,\mu +2\rho )m_{\lambda }(\mu )$ proves the formula. This trace argument handles weight multiplicities greater than one without choosing a separate sl2 Lie algebra string through each vector.