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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 $R$?

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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.

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  • $\begingroup$ Thank you for the simple example. $\endgroup$
    – TVS_integration
    Commented Jul 15, 2019 at 21:35
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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.

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    $\begingroup$ Simpler is $C \times C $ embedded in $2 \times 2$ matrices over $C$ ($C$ = complexes). $\endgroup$
    – David Handelman
    Commented Jul 15, 2019 at 16:58
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    $\begingroup$ @DavidHandelman: would you please post your comment as an answer? The answer given above is ridiculous overkill for the given question. $\endgroup$
    – Nik Weaver
    Commented Jul 15, 2019 at 17:30

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