Let $M$ be a Von Neumann algebra and $N$ be a von Neumann a subalgebra of $M$. Suppose there exists a normal faithful normal State $\phi$ on $M$, then we have the modular automorphism group $(\sigma_t^{\phi})_{t\in \Bbb R}$.
Since $\phi|_N$ is a normal faithful normal state on $N$, we have the modular automorphism group $(\sigma_t^{\phi|_N})_{t\in \Bbb R}$, where $\sigma_t^{\phi|_N}:N\rightarrow N$ is the $*$-automorphism.
What is the relationship between $\sigma_t^{\phi|_N}$ and $\sigma_t^{\phi}$.Does the following conclusion hold:
$\forall x\in N$, $\sigma_t^{\phi}(x)=\sigma_t^{\phi|_N}(x), t\in \Bbb R$.
They are equal, provided that $\sigma_t^\phi$ leaves $N$ invariant.
Given a weight, its modular group can be characterized as the unique one-parameter automorphism group of the algebra that commutes with the weight and satisfies the KMS condition. This is proven for instance in Theorem VIII.1.2 in Takesaki's Theory of Operator Algebras 2.