Tomita–Takesaki theory

Remarks

Example: let $M = M_n(\C)$ act by left multiplication on $H = M_n(\C)$ with inner product $\langle \xi, \eta \rangle = \operatorname{tr}(\eta^* \xi)$, and let $\Omega = \rho^{1/2}$ for an invertible density matrix $\rho$. Then $\Omega$ is cyclic and separating, $\Delta \xi = \rho \xi \rho^{-1}$ and $J \xi = \xi^*$. Check: $J \Delta^{1/2}(x \rho^{1/2}) = J(\rho^{1/2} x) = x^* \rho^{1/2} = S(x\rho^{1/2})$. Here $JMJ$ is the algebra of right multiplications, which is $M'$, and $\sigma_t(x) = \rho^{it} x \rho^{-it}$.

Non-example: let $M = B(H)$ with $\dim H \ge 2$, let $\Omega$ be a unit vector, and let $\eta$ be a unit vector orthogonal to $\Omega$. Every nonzero vector is cyclic for $B(H)$, but $\Omega$ is not separating. Indeed, the operator $x = \langle \cdot, \eta \rangle \Omega$ is nonzero and $x\Omega = 0$. Its adjoint is $x^* = \langle \cdot, \Omega \rangle \eta$, so $x^*\Omega = \eta \ne 0$. Since $x\Omega = 0\Omega$ but $x^*\Omega \ne 0^*\Omega$, the formula for $S_0$ does not define a map.

The vector state $\varphi(x) = \langle x\Omega, \Omega \rangle$ is a KMS state for the modular automorphism group, and Takesaki showed that $\sigma$ is the only one-parameter automorphism group of $M$ with this property.