Let $N$ be a ${\rm II}\_1$ factor. A maximal abelian self-adjoint subalgebra (MASA) is a $\*$-subalgebra $A \subset N$ such that $A' \cap N = A$. It is called a *Cartan subalgebra* if moreover $\mathcal{N}(A)''=N$, with the normalizer $$\mathcal{N}(A) = \{u \in N \ | \ uAu^\* = u^\*Au = A, \ u \text{ unitary} \}.$$


Let $\alpha$ be a free ergodic action of a countable discrete group $G$ on a standard $\sigma$-finite measure space $(X,\mu)$, and let $M$ be the von Neumann algebra $L^{\infty}(X,\mu) \rtimes\_{\alpha} G$. If $M$ is a ${\rm II}\_1$ factor, then $A=L^{\infty}(X,\mu)$ is a Cartan subalgebra. But assume that $M$ is a ${\rm II}\_{\infty}$ factor, then there is a ${\rm II}\_1$ factor $N$ such that $M \simeq N \otimes B(H)$. 


**Question**: Should $N$ admit a Cartan subalgebra?  

If no, could you provide a counter-example?  

If yes, what is it? 



Cartan subalgebra and group measure space construction
First of all, there is a projection $p \in M$ of finite trace such that $N \simeq pMp$, and its equivalent class depends at most on $tr(p)$. Because $G$ is countable and $M$ a ${\rm II}\_{\infty}$ factor, the measure $\mu$ must be infinite, then we can choose a subspace $Y \subset X$ with $\mu(Y)=tr(p)$, and assume that $$N = 1\_Y (A \rtimes\_{\alpha} G) 1\_Y,$$ with $1\_Y$ the indicator function of $Y$. 


Let's assume$^1$ that $A$ is a Cartan subalgebra of $M$.


By the work of Feldman and Moore [Ergodic equivalence relations, cohomology, and von Neumann algebras](https://mathscinet.ams.org/mathscinet-getitem?mr=0578730), a von Neumann algebra has a Cartan subalgebra$^2$ iff it is of the form $M(R,s)$, i.e. generated by an equivalence relation $R$ (with all equivalent classes countable) on a standard Borel space $X$ (up to a cocycle twist $s$). The Cartan subalgebra is then the "diagonal", i.e. $L^{\infty}(X)$


Now let $Y$ be a subspace of $X$, then $1\_Y M(R,s) 1\_Y = M(R\_{|Y},s)$, so has a Cartan subalgebra equals to the diagonal $L^{\infty}(Y)$.


*Warning*: The compression $pMp$ is not necessarily a group measure space construction, because even if the equivalent relation $R\_{|Y}$ is always of the form $R\_H$, the action of $H$ on $Y$ is not necessarily free, whereas freeness is used to show that $M(R\_H,1)$ is a group measure space construction. Moreover, the restriction $R\_{|Y}$ of an ergodic$^3$ equivalent relation $R$ is not necessarily ergodic. Now, $pMp$ is a factor, so if $R\_{|Y} = R\_H$ with $H$ acting freely then $R\_{|Y}$ must be ergodic. 


*Acknowledgment*: Thanks to [Jesse Peterson](https://mathoverflow.net/users/6460/jesse-peterson) for his help. 




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$^1$it is true in the ${\rm II}\_{1}$ case, but I did not check if it is always true in the ${\rm II}\_{\infty}$ case.  

$^2$the general case requires the existence of a faithful normal conditional expectation $E: M \to A$.  

$^3$every $R$-saturated Borel set has measure $0$ or $1$. 


