# Extension of trace on von Neumann subalgebra

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 $$R$$?

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