Omitting the rho shift breaks Kostant's formula

Given: The Axiom of Choice, $\mathfrak{g}={\mathfrak{s}\mathfrak{l}}_2$ with positive root $\alpha$, fundamental weight $\omega =\alpha \mathrm{/}2$, Weyl group $W=\{1,s\}$, Weyl vector $\rho =\omega$, the weight $\lambda =2\omega$, the weight $\mu =0$, and the Kostant partition function $P$.

The Axiom of Choice is assumed; it enters through the Kostant formula used in the counterexample (The Axiom of Choice).

For ${\mathfrak{s}\mathfrak{l}}_2$ the positive root satisfies $\alpha =2\omega$ and $s\omega =-\omega$, so $\rho =\alpha \mathrm{/}2=\omega$ and $\lambda +\rho =3\omega$ (Finite Weyl root system, lattice and chamber conventions, Fundamental weights, The Weyl vector rho for a chosen positive system).

$L(2\omega )$ is the irreducible three-dimensional ${\mathfrak{s}\mathfrak{l}}_2$-module whose weights are $2\omega ,0,-2\omega$, each with multiplicity one (Finite-dimensional representations of sl_2, Integral, dominant, and strictly dominant weights).

$P(\beta )$ counts the families $(n_{\alpha })$ with $n_{\alpha }\alpha =\beta$, so $P(\alpha )=P(2\omega )=1$, while $P(-4\omega )=P(3\omega )=0$ because $-4\omega$ and $3\omega =\frac{3}{2}\alpha$ are not nonnegative integral multiples of the simple root $\alpha$; the multiplicity formula of the Statement is $m_{\lambda }(\mu )={\sum }_{w\in W}(-1{)}^{\mathrm{\ell }(w)}P(w(\lambda +\rho )-(\mu +\rho ))$ (The Kostant partition function, Kostant's weight multiplicity formula).