Vrftsjtryk

Let $A subseteq cal B(H)$, $B subseteq cal B(H')$ be $C^*$-algebras, where $cal H$, $cal H'$ are Hilbert spaces and let $psi: A rightarrow B$ be a $*$-homomorphism.

My question: When does this $*$-homomorphism extend to a $*$-homomorphism $overline{psi}: A''rightarrow B''$ between the generated von Neumann algebras $A''$, $B''$ of $A$ and $B$? Is there a criterion which allows this?

There are easy examples of non-extending $ast$-homomorphisms. For instance recall that a groups homomorphism $Phi: G to H$ extends to the reduced $C^ast$-algebras iff $mathrm{ker}(Phi)$ is amenable, but it extends to the reduced von Neumann algebras iff $mathrm{ker}(Phi)$ is compact. Take the trivial character of an amenable group $G$,
$$
chi Big( sum_{gin G} a_g lambda_g Big) = sum_{gin G} a_g
$$
always extends to a $ast$-homomorphism $chi:C_{mathrm{red}}^ast G to mathbb{C}$ but never pass to the von Neumann algebra unless $G$ is finite.

In the normal case: Any $ast$-homomorphism between von Neumann algebras is spatially implemented and of the form $a mapsto p ( a otimes 1 ) p$, where $p$ is a projection commuting with $A otimes mathbb{C} 1$. Then your criterion will be something like asking the $ast$-homomorphism to be spatially implemented by a map $V: mathcal{H} to mathcal{H'}$ and take $p$ to be the range projection.

In the non-normal case: I do not have an answer. Take $A = c_0(mathbb{N}) + mathbb{C} 1 subset B(ell^2(mathbb N))$ the $C^ast$-algebra of sequences with a limit. There is a $ast$-homomorphism $psi(a_n)_n = lim_n a_n$. It does not extends to $A^{''} = ell^infty(mathbb{N})$ in a normal way but there are uncountably many non-normal extensions, one for every limit with respect to a proper ultrafilter
$$
barpsi((a_n)_n) = lim_{n, , omega} a_n
$$

The answer in general is no. Even when $Asimeq B$ and $psi$ is an isomorphism. For instance you can take $A=B=UHF(2^infty)$, and represent $B$ via GNS by the trace, and represent $A$ via a Powers' state. Then $B''=R$, the hyperfinite II$_1$-factor, and $A''$ is a type III factor. Not only $psi$ does not admit an extension: there is no nonzero homomorphism $pi:Ato B$ (proof: take a projection that halves the identity on $A$, then $operatorname{tr}pi(I)=operatorname{tr}pi(p)+operatorname{tr}pi(I-P)=2operatorname{tr}pi(I)$, so $pi(I)=0$).

Note that in the example above $psi$ is normal, so it is as good as it can be as a $*$-homomorphism; and still it is not enough to guarantee an extension.

In summary, you may find some concrete example where it works, but you cannot expect a general characterization.