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In Iwahori-Matsumoto's paper the Iwahori Hecke Algebra for $G=GL_n(F)$ is generated by $X_{s_0}, X_{s_i},i\in\{0,...,n-1\}$ and $ X_{\rho}$ with the relations:
$ 1) (X_{s_{i}}-q)(X_{s_{i}}+1)=0\:,\;\;\;i=0,1,...,n-1 \\ 2) X_{\rho}^n=1, \\ 3) X_{\rho}X_{s_i}=X_{s_{i+1}}X_{\rho} \\ 4) X_{s_i}X_{s_j}X_{s_i}=X_{s_j}X_{s_i}X_{s_j}, i\equiv \pm1 \mod n \\ 5) X_{s_i}X_{s_j}=X_{s_j}X_{s_i}, i\not\equiv \pm1 \mod n $

Quetsion:
If I consider the Steinberg representation $\pi$ as the unique irreducible subquotient $V$ of $Ind_B^G1$ (using normalized induction here) then this corresponds to the sign character of $H(G,J)$ acting on the one dimensional space $V^J$ i.e. $X_{s_i}(v)=-v$ for every simple reflexion $S$. My question is what is the eigenvalue of $X_{\rho}$? Of course it has to be an $n$-th root of unity but which one is it? Iwahori and Matsumoto define the sign character by sending $X_{\rho}\to 1$ but is this the only choice possible?

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  • $\begingroup$ I do not think that the relation $X_\rho^n =1$ is correct. $\endgroup$
    – Paul Broussous
    Commented Oct 2 at 20:15

1 Answer 1

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Yes, it is the only one possible. Your relations are actually wrong - in 1) you should replace -1 by +1. Then in Steinberg every $X_{s_i}$ acts by -1 and 3) immediately implies that $X_{\rho}$ acts by 1.

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  • $\begingroup$ Oops indeed, fixed it. $\endgroup$
    – idocomb
    Commented Jun 9, 2023 at 17:58
  • $\begingroup$ @alexanderbraverman I think this is not true : $X_\rho$ acts by $(-1)^{n-1}$. $\endgroup$
    – Paul Broussous
    Commented Oct 2 at 20:13

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