Crossed Product

 Something I am working right now is with the crossed product construction. The idea that the crossed product construction has a physical meaning in AdS/CFT should be very obvious. This is primarily for two reasons: (1) there already exists a well-defined operator algebra for which outer automorphisms group can be defined, and (2) the Hamiltonian for $\mathcal{N}=4$ SYM goes like $H\sim N/g^{2}$, and has a central operator $N\times U$ defined as the difference of the Hamiltonian with the expectation value in the large $N$ limit. The first point is rather interesting and more or less provides most of the motivation needed for the crossed product construction, since time translations generated by the Hamiltonian form a group of outer automorphisms of the left and right copies of the thermofield double state, which we will discuss better in the next subsection. The second point is interesting from a more formal standpoint, where we divide the difference $H-\langle H\rangle $ by $N$ and obtain a well-defined large $N$ limit. This observation leads to the construction presented in 2112.12828, where the canonical ensemble crossed product construction was presented. We will highlight the general idea below.

Start by recalling that the group of automorphisms of an algebra $\mathcal{A}$ generated by a self-adjoint operator $\mathcal{O}$ is defined as 

\[e^{i\mathcal{O}s}\mathsf{a}e^{-i\mathcal{O}s}\in \mathcal{A}\]

for some $\mathsf{a}\in \mathcal{A}$ and $s\in \mathbb{R}$. If $e^{i\mathcal{O}s}\in \mathcal{A}$, then the group of automorphisms generated this way are said to be an ``inner" group of automorphisms, and ``outer" otherwise. We will stick with Takesaki's original convention and define our convention so that the group of automorphisms of $\mathcal{A}$ are Aut$[\mathcal{A}]$, inner automorphisms of $\mathcal{A}$ as Int$[\mathcal{A}]$ and outer automorphisms as Out$[\mathcal{A}]$. Recall that a left Haar measure is a non-zero Radon measure $\mu $ so that for Borel sets $\mathcal{B}\subset G$ and $g\in G$, 

\[ \mu \left(g\mathcal{B} \right)=\mu \left(\mathcal{B} \right)\;.\]

Let $\alpha :G\to $Aut$[\mathcal{A}]$. Then, the covariant representation of the triplet $(G, \mathcal{A}, \alpha )$ is a pair $(u, \pi )$ of a unitary representation $u:G\to \mathcal{U}(\mathcal{H})$ to the unitary group of $\mathcal{H}$ and $\pi :\mathcal{A}\to \mathcal{B}(\mathcal{H})$ (i.e. the set of bounded operators on $\mathcal{H}$) so that the covariance condition is satisfied:

\[u(g)\pi (\mathsf{a})u(g)^{*}=\pi \left(\alpha _{g}(\mathsf{a}) \right)\;.\]

We can then formally approach the crossed product as follows: taking the von Neumann algebra $\mathcal{A}$ acting on a Hilbert space $\mathcal{H}$ composing the triplet, the von Neumann algebra on $L^{2}(\mathcal{H}, G)$ generated by $u(G)$ and $\pi _{\alpha (\mathcal{A})}$ is the crossed product algebra $\mathcal{A}\rtimes G$ w.r.t $\alpha $. This may seem more complicated than necessary, but the physical picture can be found by noting that the crossed product of the type III$_{1}$ algebra $\mathcal{A}$ by Aut$[\mathcal{A}]$ is a type II$_{\infty }$ algebra $\mathcal{A}\rtimes $Aut$[\mathcal{A}]$. P.S. Go watch Dune Part 2.