# Two isomorphic finite-dimensional $C^*$-algebras with infinite eigenspaces

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?

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 $$T\in\mathscr A_1$$.

• 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. – AAK Mar 6 at 21:28