Is it true that one always have faithful normal conditional expectation from von Neumann algebra $M$ to any abelian subalgebra $A\subset M$? Does it hold for some special class of von Neumann algebras?
Abelian subalgebras of von Neumann algebras
operator-algebrasvon-neumann-algebras
Related Solutions
The second question admit (I think) a straightforward solution. Pick $x \in M_+$ and notice that, if $E$ preserves $\varphi$, we have that $$ \tau(y \, x) = \varphi(x) = \varphi(E(x)) = \tau( y \, E(x) ) = \tau( E(y) \, x) $$ since that holds for every positive $x \in M$ we have that $y = E(y)$ and so $y \in L^1(N)$. The other direction is trivial.
Question 1 has a very clear answer when $\varphi(x) = \tau(\delta \, x)$ is a trace, ie $\delta$ is central, and satisfies that $\delta$ is invertible, ie $0 < \kappa 1 \leq \delta$. In that case, the natural choice is
$$
E_\varphi(x) = {E}_\tau(\delta)^{-\frac12} \, {E}_\tau( \delta^\frac12 \, x \, \delta^\frac12 ) \, {E}_\tau(\delta)^{-\frac12}.
$$
the map is ucp and, by the centrality of $\delta$ it satisfies that $E \circ E = E$. It also holds that $\varphi$ is preserved. Removing the invertibility is just a technicality that can be solved by taking $\delta'_\epsilon = \delta + \epsilon 1$ and normalize it to $\delta_\epsilon = \delta_\epsilon'/\| \delta_\epsilon' \|_1$. Then, you can take the limit of $E_\epsilon$ as $\epsilon \to 0$ in the pointwise weak-$\ast$ topology.
I do not have a formula for the general case of $\delta$ noncentral.
If you don't require $\nu$ to be faithful, then answer is no. For instance take $M=L^\infty[0,1]$, with $\tau$ integration against Lebesgue measure. Take $N=L^\infty[0,1]\oplus L^\infty[1,2]$, with $\nu$ given by $\nu(f\oplus g)=\tau(f)$. Let $S$ be the polynomials, and $\varphi(p)=p\oplus p$. Then $\nu\circ\varphi=\tau$, but $\varphi$ is not contractive, so it cannot extend to a $*$-homomorphism of C$^*$-algebras.
If we require that $\nu$ is faithful and that $\varphi$ has dense image, then it can be extended. To see this, consider $M$ and $N$ represented in GNS for $\tau$ and $\nu$ respectively. Let $\Omega_\tau, \Omega_\nu$ denote the corresponding cyclic vectors, then: \begin{align} \frac{\|\varphi(x)\varphi(z)\Omega_\nu\|^2}{\|\varphi(z)\Omega_\nu\|^2} &=\frac{\langle \varphi(xz)\Omega_\nu,\varphi(xz)\Omega_\nu\rangle}{\langle \varphi(z^*z)\Omega_\nu,\Omega_\nu\rangle}\\[0.3cm] &=\frac{\nu(\varphi(z^*x^*xz)}{\nu(\varphi(z^*z))}=\frac{\tau(z^*x^*xz)}{\tau(z^*z)}\\[0.3cm] &=\frac{\|xz\Omega_\tau\|^2}{\|z\Omega_\tau\|^2}. \end{align} Taking the supremum all nonzero $z$, we get $$ \|\varphi(x)\|=\|x\|. $$ Note that $\varphi(z)=0$ implies $\tau(z^*z)=\nu(\varphi(z^*z))=0$, and $z=0$, so $\varphi$ is faithful.
Best Answer
By a theorem of Takesaki (Section 6 in Conditional Expectations in von Neumann Algebras) there exists a normal conditional expectation onto every maximally abelian subalgebra of $M$ if and only if $M$ is finite. So finiteness of $M$ is necessary.
If $M$ is finite, there is a faithful normal conditional expectation onto every von Neumann subalgebra $N$. This can be constructed as the restriction of the orthogonal projection $L^2(M,\tau)\to L^2(N,\tau)$, where $\tau$ is a faithful normal tracial state on $M$.