# Two isomorphic $C^*$-algebras. What is the isomorphism between corresponding Hilbert spaces?

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.

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.
• 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. – AAK Mar 3 at 11:37
• 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 $B(H)$. – Aweygan Mar 3 at 19:30
• 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. – AAK Mar 4 at 7:17