Let $(\mathcal{M}, \tau)$ be a tracial von Neumann algebra, that is, $\tau$ is a faithful normal trace on $\mathcal{M}$. Let $S$ be a WOT dense unital $*$-subalgebra of $\mathcal{M}$.
Given another tracial von Neumann algebra $(\mathcal{N}, \nu)$ and an algebraic $*$-homomorphism $\varphi: S \to \mathcal{N}$ that preserves the trace ($\nu \circ \varphi =\tau$), can we always extend $\varphi$ to a a trace preserving $*$-homomrphism from $\mathcal{M}$ to $\mathcal{N}$?
This situation has arised many times when dealing with separable von Neumann algebras, where $\varphi$ is defined on some countable dense subalgebra, and one would like to extend to all of $\mathcal{M}$ (for example, when $\mathcal{M} = L(\Gamma)$ for some countable group $\Gamma$, and $S$ is the algebra generated by the canonical unitaries $\{u_\gamma \vert \gamma \in \Gamma\}$). Yet for some reason, in the literature I encountered, this is often ignored or considered as trivial.
My approach for explaining this was using the 2-norms induced by the traces $\Vert x \Vert_{2,\tau} = \tau (x^*x), \; \Vert \cdot \Vert_{2, \nu}$, which turn out to be complete on the unit balls of $\mathcal{M, N}$, and capture the WOT density of $S$. However, I hit a severe roadblock since I do not assume $\varphi$ respects the operator norm in any way, and I do not see how in general trace preservation should say something about this.
Any reference or help will be greatly appreciated.
Best Answer
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.