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Let $H$ be a separable Hilbert space. Suppose that $\mathscr{A}$ and $\mathscr{B}$ are some unital $C^*$-algebras of operators acting on $H$, not necessary coinciding with $C^*$-algebra of all the possible operators acting on $H$. Suppose that they are $*$-isomorphic with the isomorphism $\mathfrak{n}:\mathscr{A}\to\mathscr{B}$.

What are the conditions for existing a unitary operator $\mathcal{U}:H\to H$, such that $\mathfrak{n}(\mathcal{A})=\mathcal{U}\mathcal{A}\mathcal{U}^{-1}$ for all $\mathcal{A}\in\mathscr{A}$?

Any conditions on the algebras $\mathscr{A}$ and $\mathscr{B}$, e.g. commutative algebras, UHF algebras, etc., when the statement can be true, are also welcome.

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Indeed, we have

The $*$-isomorphism $\mathfrak{n}:\mathscr{A}\to\mathscr{B}$ is implemented by a unitary in $B(H)$ if and only if $\mathfrak{n}$ extends to a $*$-automorphism of $B(H)$.

The forward direction is trivial, and the reverse direction follows from the fact that all automorphisms of $B(H)$ are inner.

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  • $\begingroup$ Thank you! But in the case when $\mathfrak{n}$ is defined implicitly, this condition is not easy to check. It is better to provide some conditions in terms of the algebras $\mathscr{A}$, $\mathscr{B}$, their classes, etc. $\endgroup$ – AAK Mar 3 at 11:37
  • $\begingroup$ There might be some hope for a more appealing condition, depending on how degenerate the actions of $\mathscr{A}$ and $\mathscr{B}$ on $H$ are (and probably more data), but I am not sure. I say this because if $K,K'$ are closed infinite-dimensional subspaces of $H$, then $B(K)$ and $B(K')$ sit naturally inside $B(H)$ and are $*$-isomorphic. But if the codimensions of $K$ and $K'$ in $H$ differ, then a $*$-isomorphism between $B(K)$ and $B(K')$ cannot extend to an inner $*$-automorphism of $B(H)$. $\endgroup$ – Aweygan Mar 3 at 19:30
  • $\begingroup$ Yes, you are right. Even in the commutative unital case, two algebras $$\begin{pmatrix} x & 0 & 0 & 0 \\ 0 & x & 0 & 0 \\ 0 & 0 & x & 0 \\ 0 & 0 & 0 & y \end{pmatrix}\ \ \ {\rm and}\ \ \ \begin{pmatrix} x & 0 & 0 & 0 \\ 0 & x & 0 & 0 \\ 0 & 0 & y & 0 \\ 0 & 0 & 0 & y \end{pmatrix}$$ are isomorphic, but there are no inner isomorphism. $\endgroup$ – AAK Mar 4 at 7:17

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