Let $R$ be a finite von Neumann algebra and $S$ be a von Neumann subalgebra of $R$. Does every normal tracial state on $S$ extend to a normal tracial state on $R$?
2 Answers
Let $C$ denote the complexes, and embed $C \times C$ in $M_2 C$ ($2 \times 2$ matrices) as diagonal matrices. Then $M_2 C$ has unique trace, but $C \times C$ has two extremal ones.
No, if $\Gamma$ is a group without torsion and with the Infinite conjugacy class property then the algebra $L\Gamma$ is a factor of type $II$ and has only one normal tracial state.
Now, let $x $ in $\Gamma$ different from the identity. The von Neumann algebra $S$ generated by $x$ is isomorphic to $L^\infty(S^1)$, the algebra of measurable functions on the circle. It has many normal tracial states.

4$\begingroup$ Simpler is $C \times C $ embedded in $2 \times 2$ matrices over $C$ ($C$ = complexes). $\endgroup$ Jul 15, 2019 at 16:58

1$\begingroup$ @DavidHandelman: would you please post your comment as an answer? The answer given above is ridiculous overkill for the given question. $\endgroup$ Jul 15, 2019 at 17:30