# Tomita-Takesaki versus Frobenius: where is the similarity?

I've often heard Alain Connes say that the modular flow of Tomita-Takesaki theory should be thought of as a characteristic zero analog of the Frobenius endomorphism.
... can anyone justify this claim?

Given a von Neumann algebra $M$, its modular flow is a canonically defined homomorphism $$\mathbf{\Phi}: i\mathbb R\quad\longrightarrow\quad \text{BIM}^\times(M)$$ that, in the presence of a state (or weight), lifts to a homomorphism $i\mathbb R\to Aut(M)$. Here, $\text{BIM}^\times(M)$ denotes the 2-group of invertible $M$-$M$-bimodules. The bimodule $\mathbf{\Phi}(it)$ is the non-commutative $L^p$-space for the value $\frac 1 p=it$.

Given a ring $R$ of characteristic $p$, its Frobenius is a canonically defined homomorphism $$\mathbf{F}:\mathbb N\quad\longrightarrow\quad End(R)$$ such that $\mathbf{F}(1)$ sends $x$ to $x^p$. More generally, $\mathbf{F}(n)$ sends $x$ to $x^{p^n}$.

So far, the only analogy I can see is that both $\mathbf{F}$ and $\mathbf{\Phi}$ are canonically defined actions...

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minor nitpick: typo in title –  Yemon Choi Jun 20 '11 at 9:48
Thanks: Typo fixed –  André Henriques Jun 20 '11 at 9:58
That's an interesting question. Personally, I don't understand/know what "the $\mathbb F_1$ Frobenius" should give (just archimedean factors or the whole completed zeta function?) But in the framework of Endomotives Connes (together with Consani and Marcolli) developed a (co)homological approach (based on cyclic homology) which leads to a spectral realization of zetas in form of a canonical $R$-action on a certain homology group which might be considered as an analogue of Frobenius action on l-adic cohomology. This is explained in alainconnes.org/docs/bookwebfinal.pdf pp. 556. –  user5831 Jun 20 '11 at 10:06
This may not be of much help (and perhaps you already know the article better than I do) but in his article for the "IMU: Visions and Perspectives" book, Connes starts with some big-picture overview of how one might start with number-theoretic considerations (class field theory) and be led to thinking about modular flow. See pages 1-6 of alainconnes.org/docs/imufinal.pdf –  Yemon Choi Jun 20 '11 at 21:15
In what sense does the Frobenius provide a spectral realization of a zeta-function? The eigenvalues of the Frobenius operator (acting on cohomology groups if you like) are the zeros of the $\zeta$-function, is this sufficient? –  Junkie Jun 21 '11 at 3:02

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 $$S z\xi=Fr(z)S\xi\quad\hbox{for all}\quad z\in F_{p^n},~\xi\in V.$$ Given such a map $T$ we can extract an $F_{p^n}$-linear map $T^n$ (the analogue of $T\rightarrow |T|^{2i}$ 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.