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A low tech (naive?) piece of intuition comes straight from the definition of the modular operator and what happens if one tries to carry it over to finite fields.

The nontrivial automorphism $z\mapsto\overline{z}$ in $Gal(\mathbb{C}/\mathbb{R})$ is encoded in a von Neumann algebra via the existence of a *-operation $(zX)^\ast=\overline{z}X^\ast$. When $M$ is faithfully represented in a Hilbert space $H$ with cyclic and separating vector $\Omega$ we construct $SX\Omega:=X^\ast\Omega$, then $\Delta:=|S|^2$, then $\sigma_{t}(X)=\Delta^{it}X\Delta^{-it}$, so the modular group encodes $z\mapsto\overline{z}$ in some sense.

Let $n\in\mathbb{N}$ and note that the Frobenius automorphism of $F_{p^n}$ generates $Gal(F_{p^n}/F_p)$. Take an associative, unital algebra $R$ over $F_{p^n}$ equipped with a bijection $Q:R\to R$, satisfying $Q(zx)=Fr(z)Q(x)$ for all $z\in F_{p^n}$ and $x\in R$. If $R$ is faithfully represented on an $F_{p^n}$ vector space $V$ with cyclic and separating vector $\Omega$ then we obtain a map $S:V\to V$, $S x\Omega=Q(x)\Omega$, which has the property $$S z\xi=Fr(z)S\xi\quad\hbox{for all}\quad z\in F_{p^n},~\xi\in V.$$ This belongs to the ring of $F_{p}$-linear maps - given any Given such a map $T$ we can extract an $F_{p^n}$-linear map $T^n$ (the analogue of $T\rightarrow |T|^{2i}$ in characteristic zero)for an antilinear operator). Set $\Delta:=S^n$. As $\Omega$ is cyclic and separating and $Q$ is a bijection, $\Delta$ is invertible and we can form the maps $\sigma_{m}(x)=\Delta^{m} x\Delta^{-m}$ for each $m\in\mathbb{Z}$. Then we find that $\sigma_m$ is an $F_{p^n}$ algebra homomorphism and $\sigma_{m_1}\circ\sigma_{m_2}=\sigma_{m_1+m_2}$. If $R$ is a field over $F_{p^n}$ then we have the canoncial map $Q(x)=x^p$ and $\sigma_m(x)=Fr^{nm}(x)$. This extends easily to the case $R=M_n(K)$ for a field $K$. I'm not sure if a nice' $Q$ exists for a general division algebra.

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A low tech (naive?) piece of intuition comes straight from the definition of the modular operator and what happens if one tries to carry it over to finite fields.

The nontrivial automorphism $z\mapsto\overline{z}$ in $Gal(\mathbb{C}/\mathbb{R})$ is encoded in a von Neumann algebra via the existence of a *-operation $(zX)^\ast=\overline{z}X^\ast$. When $M$ is faithfully represented in a Hilbert space $H$ with cyclic and separating vector $\Omega$ we construct $SX\Omega:=X^\ast\Omega$, then $\Delta:=|S|^2$, then $\sigma_{t}(X)=\Delta^{it}X\Delta^{-it}$, so the modular group encodes $z\mapsto\overline{z}$ in some sense.

Let $n\in\mathbb{N}$ and note that the Frobenius automorphism of $F_{p^n}$ generates $Gal(F_{p^n}/F_p)$. Take an associative, unital algebra $R$ over $F_{p^n}$ equipped with a bijection $Q:R\to R$, satisfying $Q(zx)=Fr(z)Q(x)$ for all $z\in F_{p^n}$ and $x\in R$. If $R$ is faithfully represented on an $F_{p^n}$ vector space $V$ with cyclic and separating vector $\Omega$ then we obtain a map $S:V\to V$, $S x\Omega=Q(x)\Omega$, which has the property $$S z\xi=Fr(z)S\xi\quad\hbox{for all}\quad z\in F_{p^n},~\xi\in V.$$ This belongs to the ring of $F_{p}$-linear maps - given any such map $T$ we can extract an $F_{p^n}$-linear map $T^n$ (the analogue of $T\rightarrow |T|^{2i}$ in characteristic zero). Set $\Delta:=S^n$. As $\Omega$ is cyclic and separating and $Q$ is a bijection, $\Delta$ is invertible and we can form the maps $\sigma_{m}(x)=\Delta^{m} x\Delta^{-m}$ for each $m\in\mathbb{Z}$. Then we find that $\sigma_m$ is an $F_{p^n}$ algebra homomorphism and $\sigma_{m_1}\circ\sigma_{m_2}=\sigma_{m_1+m_2}$. If $R$ is a field over $F_{p^n}$ then we have the canoncial map $Q(x)=x^p$ and $\sigma_m(x)=Fr^{nm}(x)$. This extends easily to the case $R=M_n(K)$ for a field $K$. I'm not sure if a nice' $Q$ exists for a general division algebra.