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Let $H_1$, $H_2$ be two separable Hilbert spaces. Let $\mathscr{A}_1$, $\mathscr{A}_2$ be two isomorphic finite-dimensional $C^*$-algebras of operators acting on $H_1$, $H_2$ respectively. Suppose that any eigenvalue of any operator from these algebras has infinite multiplicity. (Suppose also that the identity operators belong to the algebras.)

Can we say that there is a unitary $\mathcal{U}:H_1\to H_2$ such that $\mathfrak{n}(\mathcal{A})=\mathcal{U}\mathcal{A}\mathcal{U}^{-1}$ for any $\mathcal{A}\in\mathscr{A}_1$, where $\mathfrak{n}:\mathscr{A}_1\to\mathscr{A}_2$ is a $*$-isomorphism?

The question was inspired by the comments to Two isomorphic $C^*$-algebras. What is the isomorphism between corresponding Hilbert spaces?

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If you allow the units of the algebras to not be the unit of $B(H_j)$, then the answer is no.

So assume that both subalgebras are unital in their respective $B(H_j)$. Since $\mathscr A_j$ is finite-dimensional, we have $$ \mathscr A_j=\bigoplus_{n=1}^m M_{k(n)}(\mathbb C),\ \ \ j=1,2. $$ Consider the canonical matrix units for these algebras: we have$\{ E_{st}^{n}\}$, $n=1,\ldots,m$, $s,t=1,\ldots,k(n)$ matrix units for $\mathscr A_1$, with $\mathscr A_1=\operatorname{span}\{E_{st}^n:\ n,s,t\}$ and the usual matrix unit relations $$\tag1 E_{st}^nE_{ab}^p=\delta_{np}\,\delta_{ta}\,E_{sb}^n,\ \ \ \ (E_{st}^n)^*=E_{ts}^n. $$ Also, $\sum_{j=1}^{k(n)}E_{ss}^n=I_{k(n)}$, and $\mathscr A_1=\operatorname{span}\,\{E^n_{st}:\ n,s,t\}$. For $\mathscr A_2$ Let $F^n_{st}=\mathfrak{n}(E_{st}^n)$; it is clear that we have $\{F_{st}^n\}$ with the exact same properties as $(1)$.

The hypothesis is that $E_{ss}^nH_1$ is infinite-dimensional, and similarly $F_{ss}^nH_2$ for all $n$ and all $s$. Now construct, for each $n$ and $s$, orthonormal bases $\{e^{n,s}_\alpha\}$ of $E_{ss}^nH_1$ and $\{f^{n,s}_\alpha\}$ of $F_{ss}^nH_2$, in the following way: fix $n$; let $\{e_\alpha^{n,1}\}_\alpha$ be an orthonormal basis of $E^n_{11}H_1$. Define, for $s>1$, $e_\alpha^{n,s}=E^n_{s1}e_\alpha^{n,1}$; since $E^n_{s,1}$ is an isometry from $E^n_{11}H_1$ to $E^n_{ss}H_1$, it follows that $\{e_\alpha^{n,s}\}_\alpha$ is an orthonormal basis of $E^n_{ss}H_1$. Now use the same idea to produce $\{f_\alpha^{n,s}\}_\alpha$.

As $I_{H_1}=\sum_n\sum_s E^n_{ss}$, we get that $\{e_\alpha^{n,s}\}_{\alpha,n,s}$ is an orthonormal basis of $H_1$. Similarly with $\{f_\alpha^{n,s}\}_{\alpha,n,s}$ and $H_2$.

Define a unitary $V:H_1\to H_2$ by $Ve^{n,s}_\alpha=f^{n,s}_\alpha$.

Now \begin{align} VE_{st}^n e^{m,v}_\alpha&=VE_{st}^nE_{v1}^me_\alpha^{m,1}=\delta_{nm}\,\delta_{vt}\,VE_{s1}^ne_\alpha^{n,1}=\delta_{nm}\,\delta_{vt}\,Ve_\alpha^{n,s} =\delta_{nm}\,\delta_{vt}\,f_\alpha^{n,s}\\ \ \\ &=\delta_{nm}\,\delta_{vt}\,F^n_{s1}f_\alpha^{n,1} =F^n_{st}F^m_{v1}f_\alpha^{m,1} =F^n_{st}f_\alpha^{m,v} =F^n_{st}Ve_\alpha^{m,v}. \end{align} Thus $VE_{st}^n=F_{st}^nV$, and then $F_{st}^n=VE_{st}^nV^*$. That is, $$\tag2 \mathfrak n(E_{st}^n)=VE_{st}^nV^*. $$ As the matrix units span $\mathscr A_1$, the equality $(2)$ holds for all $T\in\mathscr A_1$.

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  • $\begingroup$ Thank you. Now, it looks like a complete classification of finite dimensional $C^*$-algebras with infinite eigenspaces. I mean, It is not difficult to construct one instance of such algebras, say $\mathscr{A}$. All others are $\mathcal{U}\mathscr{A}\mathcal{U}^{-1}$, where $\mathcal{U}$ are isometries between Hilbert spaces. I believe that such type of classifications exists also for the case of finite-dimensional $C^*$-algebras with mixed infinite/finite-dimensional eigenspaces. But I am less sure about such type of classifications for AF-algebras. $\endgroup$ – AAK Mar 6 at 21:28

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