[proofplan]
We use the Harish-Chandra isomorphism in the normalization where a central element $z\in Z(U(\mathfrak g))$ corresponds to a $W$-invariant polynomial function on $\mathfrak h^*$ whose value at $\lambda+\rho$ is the scalar by which $z$ acts on the highest weight module $L(\lambda)$. This converts equality of central characters into equality of all $W$-invariant polynomial values at the two shifted weights $\lambda+\rho$ and $\mu+\rho$. The remaining point is the invariant-theoretic fact that polynomial invariants of a finite Weyl group separate $W$-orbits. Applying that separation theorem gives the shifted orbit condition, and the converse follows immediately from $W$-invariance.
[/proofplan]
[step:Normalize the Harish-Chandra isomorphism and its action on highest weight modules]
Let
\begin{align*}
\gamma:Z(U(\mathfrak g))\to\mathbb C[\mathfrak h^*]^W
\end{align*}
denote the Harish-Chandra isomorphism in the shifted normalization characterized by the following property: for every $\lambda\in\mathfrak h^*$, every $z\in Z(U(\mathfrak g))$, and every nonzero highest weight vector $v_\lambda\in L(\lambda)$ of weight $\lambda$,
\begin{align*}
z\cdot v_\lambda=\gamma(z)(\lambda+\rho)v_\lambda.
\end{align*}
Here $\mathbb C[\mathfrak h^*]^W$ denotes the algebra of polynomial functions
\begin{align*}
f:\mathfrak h^*\to\mathbb C
\end{align*}
satisfying $f(w\nu)=f(\nu)$ for every $w\in W$ and every $\nu\in\mathfrak h^*$.
This is the standard Harish-Chandra isomorphism with the $\rho$-shift built into the evaluation convention. The irreducible highest weight module $L(\lambda)$ is generated as a $U(\mathfrak g)$-module by any nonzero highest weight vector $v_\lambda$. If $u\in U(\mathfrak g)$, then centrality of $z$ gives
\begin{align*}
z\cdot(u\cdot v_\lambda)=u\cdot(z\cdot v_\lambda)=\gamma(z)(\lambda+\rho)(u\cdot v_\lambda).
\end{align*}
Thus $z$ acts on all of $L(\lambda)$ by the scalar $\gamma(z)(\lambda+\rho)$. Hence, for every $z\in Z(U(\mathfrak g))$,
\begin{align*}
\chi_\lambda(z)=\gamma(z)(\lambda+\rho).
\end{align*}
[guided]
The first issue is the normalization. The Harish-Chandra isomorphism may be written either as an isomorphism onto ordinary $W$-invariant polynomials evaluated at $\lambda+\rho$, or as an isomorphism onto invariants for the affine dot action evaluated at $\lambda$. We choose the first convention because it matches the statement of the theorem.
Let
\begin{align*}
\gamma:Z(U(\mathfrak g))\to\mathbb C[\mathfrak h^*]^W
\end{align*}
be the Harish-Chandra isomorphism in this convention. Thus $\gamma$ is an algebra isomorphism from the center of the universal enveloping algebra to the algebra of ordinary $W$-invariant polynomial functions on $\mathfrak h^*$. Explicitly, if
\begin{align*}
f:\mathfrak h^*\to\mathbb C
\end{align*}
lies in $\mathbb C[\mathfrak h^*]^W$, then
\begin{align*}
f(w\nu)=f(\nu)
\end{align*}
for every $w\in W$ and every $\nu\in\mathfrak h^*$.
The shifted normalization says that for every central element $z\in Z(U(\mathfrak g))$, every weight $\lambda\in\mathfrak h^*$, and every nonzero highest weight vector $v_\lambda\in L(\lambda)$, one has
\begin{align*}
z\cdot v_\lambda=\gamma(z)(\lambda+\rho)v_\lambda.
\end{align*}
This is the precise place where the $\rho$-shift enters the proof.
It remains to connect this scalar on the highest weight vector with the action on the whole module. We do not use [Schur's lemma](/theorems/2414) here, because $L(\lambda)$ need not be finite-dimensional. Instead, we use the defining generation property of a highest weight module: $L(\lambda)$ is generated as a $U(\mathfrak g)$-module by $v_\lambda$. Thus every vector of $L(\lambda)$ is a finite sum of vectors of the form $u\cdot v_\lambda$ with $u\in U(\mathfrak g)$. For such a vector, centrality of $z$ gives
\begin{align*}
z\cdot(u\cdot v_\lambda)=u\cdot(z\cdot v_\lambda)=u\cdot\left(\gamma(z)(\lambda+\rho)v_\lambda\right)=\gamma(z)(\lambda+\rho)(u\cdot v_\lambda).
\end{align*}
By linearity, $z$ acts on all of $L(\lambda)$ as multiplication by $\gamma(z)(\lambda+\rho)$. The central character
\begin{align*}
\chi_\lambda:Z(U(\mathfrak g))\to\mathbb C
\end{align*}
is exactly the map recording this scalar. Therefore, for every $z\in Z(U(\mathfrak g))$,
\begin{align*}
\chi_\lambda(z)=\gamma(z)(\lambda+\rho).
\end{align*}
[/guided]
[/step]
[step:Translate equality of central characters into equality of invariant polynomial values]
Assume first that
\begin{align*}
\chi_\lambda=\chi_\mu.
\end{align*}
For every $z\in Z(U(\mathfrak g))$, the formula from the previous step gives
\begin{align*}
\gamma(z)(\lambda+\rho)=\chi_\lambda(z)=\chi_\mu(z)=\gamma(z)(\mu+\rho).
\end{align*}
Since $\gamma$ is surjective onto $\mathbb C[\mathfrak h^*]^W$, this is equivalent to saying that every $W$-invariant polynomial function $f\in\mathbb C[\mathfrak h^*]^W$ satisfies
\begin{align*}
f(\lambda+\rho)=f(\mu+\rho).
\end{align*}
[/step]
[step:Use invariant polynomials to separate Weyl group orbits]
We now use the finite reflection group invariant-theoretic separation theorem: if $\xi,\eta\in\mathfrak h^*$ satisfy
\begin{align*}
f(\xi)=f(\eta)
\end{align*}
for every $f\in\mathbb C[\mathfrak h^*]^W$, then $\eta=w\xi$ for some $w\in W$. Applying this with
\begin{align*}
\xi:=\lambda+\rho
\end{align*}
and
\begin{align*}
\eta:=\mu+\rho
\end{align*}
gives an element $w\in W$ such that
\begin{align*}
\mu+\rho=w(\lambda+\rho).
\end{align*}
This proves the forward implication.
[/step]
[step:Prove the converse from Weyl invariance]
Conversely, suppose that there exists $w\in W$ such that
\begin{align*}
\mu+\rho=w(\lambda+\rho).
\end{align*}
Let $z\in Z(U(\mathfrak g))$ be arbitrary. Since $\gamma(z)\in\mathbb C[\mathfrak h^*]^W$, the defining $W$-invariance of $\gamma(z)$ gives
\begin{align*}
\gamma(z)(\mu+\rho)=\gamma(z)(w(\lambda+\rho))=\gamma(z)(\lambda+\rho).
\end{align*}
Using again the Harish-Chandra central-character formula,
\begin{align*}
\chi_\mu(z)=\gamma(z)(\mu+\rho)=\gamma(z)(\lambda+\rho)=\chi_\lambda(z).
\end{align*}
Because this holds for every $z\in Z(U(\mathfrak g))$, we have
\begin{align*}
\chi_\lambda=\chi_\mu.
\end{align*}
Combining the forward and reverse implications proves the theorem.
[/step]
