# Eigenvalue of Iwahori Hecke Algebra element for the Steinberg

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?

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.