In a finite von Neumann algebra, the unique tracial state serves as one, then for a general von Neumann algebra, does it exist?

Using direct integral decomposition, also known as reduction theory, one can reduce the problem to the case of a factor. A conditional expectation in this case is a state. Every factor admits a state, but only σfinite factors admit faithful states. Thus if you require the conditional expectation to be faithful, all factors in the direct integral decomposition must be σfinite, otherwise no additional conditions are needed to ensure the existence of a conditional expectation. 


The answer is yes, provided that $M$ has a faithful normal semifinite weight (this always exists) that is also semifinite when restricted to the centre (this I'm not so sure how easily can happen). When $M$ has a faithful normal semifinite weight $\varphi$, with $\varphi_{Z(M)}$ semifinite, consider the modular group $\sigma_t^\varphi$ associated with $\varphi$. For each $t\in\mathbb{R}$, $\sigma_t^\varphi$ is an automorphism of $M$, and in particular it preserves its centre. This means that $$ \sigma_t^\varphi(Z(M))=Z(M), \ \ t\in\mathbb{R} $$ These conditions, by Takesaki's Theorem (IX.4.2 in Takesaki 2, or JFA1972) are equivalent to the existence of a conditional expectation $E:M\to Z(M)$, with $\varphi\circ E=\varphi$. This last condition forces $E$ to be faithful and normal. 

