Exotic 2-adic lifts of mod $2$ Steinberg idempotent

Denote $B_n$ the Borel subgroup of $Gl_n(Z/2)$, i.e., the subgroup of
upper triangular matrices, $\Sigma _n$ the subgroup of permutation matrices.
The (conjugate) Steinberg idempotent is defined to be
$$e_n’=\frac{1}{q_n}\Sigma _{g\in \Sigma _n}sgn(g)g \Sigma _{g\in B _n}g
\in Z_{(2)}[Gl_n(Z/2)]$$
where $q_n$ is an appropriate odd number, $sgn$ is the signature of the permutation.

Now, let’s forget about the signature and consider
$$x_n=\Sigma _{g\in \Sigma _n}g \Sigma _{g\in B _n}g
\in Z_{(2)}[Gl_n(Z/2)].$$
This agrees with $e_n’$ modulo $2$, so one can apply usual procedure
(applying the map $a\rightarrow -2a^3+3a^2$ repeatedly and taking the limit) to produce an idempotent $$\widetilde{e_n’}\in Z_{2}^{\wedge}[Gl_n(Z/2)]$$ which agrees with $x_n$, thus with $e_n’$ modulo $2$.

When $n=2$ one can write down explicitly $\widetilde{e_2′}$, and one has
$$\widetilde{e_2′}=\frac{1}{3}\{(e+\sigma)(e+b)-2b\sigma(e+b)\}$$ $$\mbox{ where } e=\left( \begin{array}{rr} 1 & 0 \\ 0 &1 \end{array} \right), b=\left( \begin{array}{rr} 1 & 1 \\ 0 &1 \end{array} \right), \mbox{ and }\sigma=\left( \begin{array}{rr} 0 & 1 \\ 1 &0 \end{array} \right).$$
The general theory says that $e_n’$ and $\widetilde{e_n’}$ are conjugate, but when $n=2$ one can say more. We have
$$e_2’\widetilde{e_2′}=e_2’$$ which implies that $e_2’$ and $\widetilde{e_2′}$
provide an identical splitting of $Z_{2}^{\wedge}[Gl_n(Z/2)]$.

My question is: is there something known about
$e_n’$? In particular, do we always have $e_n’\widetilde{e_n’}=e_n’$?

Answer

Attribution
Source : Link , Question Author : user43326 , Answer Author : Community

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