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I have Von Neumann algebra $\mathfrak{U}$ and a weakly dense *-subalgebra $A$. I have another Von Neumann Algebra $\mathfrak{V}$ and an injective *-homorphism $$\phi: A\longrightarrow \mathfrak{V}$$ such that $\phi(A)$ generates $\mathfrak{V}$. Is it possible to say that one can extend $\phi$ to an *-isomorphism from $\mathfrak{U}$ onto $\mathfrak{V}$?

P.D. In fact the case I am interested in is when A is a Hilbert algebra and $\mathfrak{V}$ is the natural Von Neumann Algebra of A as defined by Dixmier in its books C *-Algebras or Von Neumann Algebras. I however believe the question as posted has all the required ingredients.

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No. This will almost never be true (subalgebras of the compacts are the only cases I can think of where it could work). The easiest example is probably $C[0,1].$ It has an injective homomorphism to $\ell^\infty(\mathbb{N})$--by point evaluation at the rationals--that generates $\ell^\infty(\mathbb{N})$. It also sits inside $L^\infty[0,1].$ But $\ell^\infty(\mathbb{N})$ has minimal projections and $L^\infty[0,1]$ doesn't.

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  • $\begingroup$ Do you think if I know that the Von Neumann algebras involved are Type I it will make a difference? $\endgroup$ – Carlos De la Mora May 15 '17 at 18:17
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    $\begingroup$ @Carlos De la Mora No it won't make a difference. The example I gave is Type I. $\endgroup$ – Caleb Eckhardt May 15 '17 at 19:29
  • $\begingroup$ @CalebEckhardt I want to ask a silly question: Why $\phi(C[0,1])$ generates $\ell^\infty(\mathbb{N})$? $\endgroup$ – C.Ding Jul 18 '17 at 11:04
  • $\begingroup$ @C.Ding Just to be clear; the claim is that $\phi(C[0,1])$ generates $\ell^\infty(\mathbb{N})$ as a von Neumann algebra. So it suffices to show that $\phi(C[0,1])$ is w*-dense in $\ell^\infty(\mathbb{N}).$ Let $e_n$ be a minimal projection in $\ell^\infty(\mathbb{N}).$ Let $q$ be a rational corresponding to $e_n.$ Take a sequence of uniformly bounded continuous functions $f_k$ with $f_k(q)=1$ whose support shrinks to $\{ q \}.$ Then $\phi(f_k)\rightarrow e_n$ in the w*-topology. $\endgroup$ – Caleb Eckhardt Jul 18 '17 at 14:44
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You can give a characterisation of when it holds. Without loss of generality, we may suppose that $A$ is closed, i.e. is a C$^*$-algebra. Consider the universal enveloping von Neumann algebra, which I will consider as being the bidual $A^{**}$. If $\phi:A\rightarrow\mathfrak{M}$ is a $*$-homomorphism with $\phi(A)$ weak$^*$-dense in $\mathfrak{M}$ then there is a unique extension $\tilde\phi:A^{**}\rightarrow\mathfrak{M}$ which is a surjective, weak$^*$-continuous $*$-homomorphism.

As $A$ generates $A^{**}$, your question has a positive answer if and only if $\tilde\phi$ is always an isomorphism. This is equivalent (consider the kernel of $\tilde\phi$) to $A^{**}$ having no proper weak$^*$-closed ideals. In turn, this is equivalent to $A^*$ having no proper $A$-invariant closed subspaces; and is equivalent to $A^{**}$ having no non-trivial central projections.

See Volume 1 of Takesaki, Section III, Chapter 2.

(I presume that $\mathfrak{U}'$, in your question, is not the commutant of $\mathfrak{U}$. This is slightly unfortunate notation...)

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    $\begingroup$ Can anything be said on how rare or common it is for $A^{**}$ to have no non-trivial central projections? Currently the only (infinite-dimensional) example I can think of is K(H). $\endgroup$ – Yemon Choi May 16 '17 at 20:04
  • $\begingroup$ Urgh, I don't know enough. Can the universal enveloping algebra be a factor of type II or III ?? $\endgroup$ – Matthew Daws May 17 '17 at 15:16
  • $\begingroup$ The question did ask for an injective -homomorphism so this doesn't quite give a characterization. For example, the answer is certainly yes for any finite dimensional C-algebra (the double duals of which can of course have many weak*-closed ideals). $\endgroup$ – Caleb Eckhardt Jul 18 '17 at 17:59
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    $\begingroup$ @YemonChoi In the case that $A$ is separable, then $K(H)$ is the only example. In the general case this is called Naimark's problem and wades into set theory. $\endgroup$ – Caleb Eckhardt Jul 18 '17 at 18:02
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    $\begingroup$ @CalebEckhardt non-mathematical comment: the asterisk symbol is used by Markdown, or whatever SE uses, to enclose text in italics. I think the parser is fine if there is only one asterisk, but as soon as there are two, well, italics ahoy $\endgroup$ – Yemon Choi Jul 18 '17 at 18:35

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