The tempered Langlands classification for affine Hecke algebras

Its finite nonempty weight set admits a weight $t$ maximizing $r(\lambda (t))$. Put $\nu =p(\lambda (t))$, let $P$ be its simple-root zero set and write $\lambda (t)=\nu +\eta$ with $\eta \in K_P$ by [F1]. Then $\nu \in C_P^{++}$. Let $U=H_P{V}_t\subset V$. Every weight $t^{\mathrm{\prime }}$ of $U$ is $wt$ for some $w\in W_P$ by [F2]; because $W_P$ fixes $\nu$, its logarithmic absolute value is $\nu +w\eta$, with $w\eta \in E_P$. By [F1] its $r$-value is at least $\mathrm{\parallel }\nu \mathrm{\parallel }$, whereas maximality of $t$ in $V$ gives the reverse inequality. The equality clause in [F1] therefore forces $w\eta \in K_P$. Thus every weight of $U$ lies in $\nu +K_P$.

Choose a nonzero $H_P$-submodule of $U$ of least dimension and denote it by $\sigma$; it is irreducible, and its weights are among those of $U$. step 1.1 proves that $\sigma$ is essentially tempered of real exponent $\nu$ and in strictly positive position. The inclusion $\sigma \hookrightarrow V$ defines an $H$-linear map $I_P(\sigma )\to V$, $h\otimes v\mapsto hv$: the equality $ha\otimes v=h\otimes av$ for $a\in H_P$ proves that this formula is well defined. Its image is nonzero and $H$-stable, so irreducibility of $V$ makes it surjective. By [F3] it identifies $V$ with $L(P,\sigma )$. This proves exhaustivity.

Suppose $V\cong L(P,\sigma )\cong L(Q,\rho )$ with real exponents $\nu ,\mu$. The largest $r$-value among the weights of $V$ is intrinsic, and [F3] identifies it with both $\mathrm{\parallel }\nu \mathrm{\parallel }$ and $\mathrm{\parallel }\mu \mathrm{\parallel }$. Every weight attaining that value has projection both $\nu$ and $\mu$, so $\nu =\mu$. Their zero sets are therefore equal, giving $P=Q$. The sum of the generalized weight spaces attaining that maximum is also intrinsic; [F3] identifies its $H_P$-action with both $\sigma$ and $\rho$. Hence $\sigma \cong \rho$, proving full uniqueness rather than uniqueness modulo an unspecified Weyl conjugacy.

A vector belongs to $K_B$ exactly when its projection to $C$ is zero by [F1]. Therefore $V$ is tempered exactly when the largest $r$-value of its weights is zero. In its unique Langlands datum this value is $\mathrm{\parallel }\nu \mathrm{\parallel }$ by [F3]; vanishing implies $\nu =0$ and its zero set is $B$. Conversely $P=B$ and $\nu =0$ mean $\sigma$ is tempered and $I_B(\sigma )=\sigma =V$. The twisting assertion follows from [F2]. All selections in this proof concern a finite set or a finite-dimensional space; no arbitrary-index choice principle is used.