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I am currently reading the (french) book [1] by Matsumoto. In (2.1.11) the following statement (reformulated in my own words/notation) appears:

Let $(W,S)$ be a Coxeter system, let $R$ be a commutative ring, let $q=(q_s)\in R^S$ be an element with $q_s=q_t$ if $s$ and $t$ are conjugate in $W$ and denote by $R[W,q]$ the corresponding Iwahori-Hecke algebra, i.e. the unique $R$-algebra freely generated by a basis $\{T_w \mid w\in W\}$ with \begin{eqnarray} {T}_{s}{T}_{\mathbf{w}}=\begin{cases} {T}_{s\mathbf{w}} & \text{, if }\left|s\mathbf{w}\right|>\left|\mathbf{w}\right|\\ q_{s}{T}_{s\mathbf{w}}+(q_{s}-1){T}_{\mathbf{w}} & \text{, if }\left|s\mathbf{w}\right|<\left|\mathbf{w}\right| \end{cases} \end{eqnarray} for all $s\in S$, $w \in W$. Let $S^\prime \subseteq S$ be a subset such that the special subgroup $W_{S^\prime}:=\left\langle S^{\prime}\right\rangle \subseteq W$ is finite. Further write $R[W,q,W^\prime]$ for the subalgebra of $R[W,q]$ of bi-invariant functions over $W^\prime$, i.e. $R[W,q,W^\prime]$ consists of all finite sums of the form $x=\sum_{w\in W} x(w)T_w$ with $x(w)\in R$ and $x(uwu^\prime)=x(w)$ for all $u,u^\prime \in W^\prime$, $w \in W$. Assume that there exists an automorphism $\sigma$ of $W$ such that $\sigma(S)=S$, such that $q_{\sigma(s)}=q_s$ for all $s\in S$ and such that $\sigma(x^{-1}) \in W^\prime x W^\prime$ for all $x \in W$. It is then claimed that $R[W,q,W^\prime]$ is commutative.

I don't see a proper argument for this statement and unfortunately the book doesn't provide a proof. It is only mentioned that the lemma is an analogue to Selberg's lemma. Can anyone give me a hint at how to prove this?

[1] H. Matsumoto, Analyse harmonique dans les systèmes de Tits bornologiques de type affine, Lecture Notes in Mathematics, Vol. 590, Springer-Verlag, Berlin-New York, 1977. i+219 pp.

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If you want a general $R$, you can't use it directly, but you can take strong inspiration from Lemma 2.3 from Vershik–Okounkov - A New Approach to the Representation Theory of the Symmetric Groups. 2. It's used there to prove a special case of this where $q=1$, $W=S_n$ and $W'=S_{n-1}$.

The Hecke algebra has an anti-automorphism induced by $T_s\mapsto T_{\sigma(s)}$. That is, $T_{w}\mapsto T_{\sigma(w)^{-1}}$. By assumption, this preserves $R[W,q,W']$ as a subalgebra (since it sends each double coset $W'wW'$ to $W'\sigma(w^{-1})W'$). On this subalgebra it acts by the identity. If the identity of a ring is an anti-automorphism, then the ring is commutative.

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  • $\begingroup$ Sorry for the typo in the paper title when I edited! Weirdly, it's there in the arXiv version …. $\endgroup$
    – LSpice
    Commented Jan 14, 2022 at 21:23

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