Classifying finite-dimensional C*-algebras by posets of their commutative C*-subalgebras

We consider the functor C that to a unital C*-algebra A assigns the partial order set C(A) of its commutative C*-subalgebras ordered by inclusion. We investigate how some C*-algebraic properties translate under the action of C to order-theoretical properties. In particular, we show that A is finite dimensional if and only C(A) satisfies certain chain conditions. We eventually show that if A and B are C*-algebras such that A is finite dimensional and C(A) and C(B) are order isomorphic, then A and B must be *-isomorphic.


Introduction
Given a C*-algebra A with unit 1A, let C(A) be the set of all commutative unital C*subalgebras C of A such that 1A ∈ C. Then C(A) becomes a poset 1 if we order it by inclusion. Now, one could consider the following question: is it possible to recover the structure of A as a C*-algebra from the poset C(A)? More precisely, if A and B are C*-algebras such that C(A) and C(B) are isomorphic as posets, can we find an *-isomorphism between A and B?
Apart from its mathematical relevance, this problem is of considerable importance for the so called quantum toposophy program, where one tries to describe quantum mechanics in terms of topos theory (see e.g., [6], [9], [23], [41]). In this program, the central objects of research are the topoi Sets C(A) and Sets C(A) op . The motivation behind at least [23] and [41] is Niels Bohr's doctrine of classical concepts, which, roughly speaking, states that a measurement provides a "classical snapshot of quantum reality". Mathematically, this corresponds with an element of C(A), and knowledge of all classical snapshots should provide a picture of quantum reality, that is as complete as (humanly) possible. The possiblility of reconstructing A from C(A) would assure the soundness of this doctrine.
For commutative C*-algebras, Mendivil showed that the answer to the question is affirmative (see [32]). It turns out that more can be said about the commutative case. As we will see, C turns out to be a functor from unital C*-algebras to posets. Hence if f : A → B is a *isomorphism, then C(f ) : C(A) → C(B) is an order isomorphism. In [17], Hamhalter gave not only a different proof of Mendivil's statement that an order isomorphism ψ : C(A) → C(B), with A and B commutative C*-algebras, yields an *-isomorphism f : A → B, but he showed as well that this *-isomorphism can be constructed in such a way that ψ = C(f ). Moreover, he proved that as long as A is not two dimensional, there is only one *-isomorphism that induces ψ in this way.
For non-commutative C*-algebras however, the answer is negative, since Connes [7] showed the existence of a C*-algebra Ac (actually even a von Neumann algebra) that is not isomorphic to its opposite algebra A op c . Here the opposite algebra is the C*-algebra with the same Proof. By Lemma 1.2, we have A = Cq[C(Y )] with Y = X/ ∼A and q : X → Y the quotient map. So each f ∈ A is of the form g • q for some g ∈ C(Y ), meaning that f is constant on [x]A for each x ∈ X. Conversely, let f be constant on [x]A for each x ∈ X. Define g : Y → X by g([x]A) = f (x). Then g is well defined, since f is constant on [x]A, and f = g • q. Moreover, let U ⊆ C be open. Then g −1 [U ] is open in Y if and only if q −1 [g −1 [U ]] is open in X by definition of the quotient topology. Since f = g • q, we find g −1 [U ] is open in Y if and only if f −1 [U ] is open in X. Since f is continuous, we find that g −1 [U ] is open, so g ∈ C(X). Hence f lifts to a function on C(Y ), so f ∈ A. Thus 2 The functor C Definition 2.1. Let A be a C*-algebra with unit 1A. We denote the set of its commutative C*-subalgebras containing 1A by C(A).
If we denote the category of unital C*-algebras with unital *-homomorphisms as morphisms by uCStar and the category of posets with order morphisms as morphisms by Poset, Proof. Let C ∈ C(A). Then the restriction of f to C is a *-homomorphism with codomain B. It follows from the First Isomorphism Theorem for C*-algebras (see for instance [ • C(f ), and if IA : A → A is the identity morphism, then C(IA) = 1 C(A) , the identity morphism of C(A). Thus C is indeed a functor.

Lemma 2.3. Let
A be a C*-algebra. Then C(A) has all non-empty meets, where the meet is given by the intersection operator. In particular, C(A) is a meet-semilattice, and has a least element Proof. Elementary.

Lemma 2.4. [3, Proposition 14] Let
A be a C*-algebra. Then the following statements are equivalent.
(i) A is commutative; (ii) C(A) is bounded; (iii) C(A) is a complete lattice.
Proof. This follows immediately from the observation that A ∈ C(A) if and only if A is commutative, and the fact that if C(A) has a greatest element, it has all meets, so it must be a complete lattice.
The next proposition is originally due to Spitters [38]. A similar statement for the functor V assigning to a von Neumann algebra M the poset V(M ) of its commutative von Neumann subalgebras can be found in [11].
Proof. Let D ⊆ C(A) be a directed subset. Let S = D. We show that S is a commutative *-algebra. Let x, y ∈ S and λ, µ ∈ C, there are D1, D2 ∈ D such that x ∈ D1 and y ∈ D2. Since D is directed, there is some D3 ∈ D such that D1, D2 ⊆ D3. Hence x, y ∈ D3, whence λx + µy, x * , xy ∈ D3, and since D3 is commutative, we obtain xy = yx. Since D3 ⊆ S, it follows that S is a commutative *-subalgebra of A. Now, S is a commutative C*-subalgebra of A, which is the smallest commutative C*-subalgebra of A containing every element of D, hence it is the join of D. If A is finite dimensional, all subspaces of A are closed, hence (ii) If f is injective, then C(f ) is an order embedding such that and for each non-empty subset {Ci}i∈I ⊆ C(A). If {Dj }j∈J ⊆ C(A) is a subset such that j∈J Dj exists, then j∈J C(f )(Dj ) exists and Moreover C(f ) has an upper adjoint C(f ) * : Proof.
(ii) Assume that f is injective. We first show that C(f ) has an upper adjoint C(f ) * . Let D ∈ C(B) and x, y By the injectivity of f it follows that xy = yx, so f −1 [D] is a commutative *-subalgebra of A, which is closed since f is continuous and D is closed. Moreover, since f (1A) = 1B, and 1B ∈ D, it follows that 1A . Then for each i ∈ I there is an ci ∈ Ci such that x = f (ci). Hence for each i, j ∈ I, we have f (ci) = f (cj ). By injectivity of f it follows that ci = cj , so x = f (c) with c ∈ i∈I Ci equal to ci for each i ∈ I. We conclude that f [ i∈I Ci] = i∈I f [Ci], which is exactly (1).
Let {Dj }j∈J be a collection of elements of C(A) such that j∈J Dj exists. Then Since C(f ) has an upper adjoint C(f ) * , we find that D k ≤ C(f ) * (E) for each k ∈ J, so j∈J Dj ≤ C(f ) * (E). Again using the adjunction, we find C(f ) j∈J Dj ≤ E. We conclude that C(f ) j∈J Dj is the join of {C(f )(Dj ) : j ∈ J}, and it follows automatically that (2) holds.
By injectivity of f , we find Hence there is some D ∈ C(A) such that C ⊆ C(f )(D). Then Finally, we show that C(f ) is an order embedding. Let C1 and C2 elements of C(A).
Since C(f ) is an order morphism, we have In other words, C(f ) is an order embedding.
(iii) This follows directly from the functoriality of C and the fact that f has an inverse.

Finite-dimensional C*-algebras
In this section we shall prove the following theorem. The proof of the theorem relies on the fact that every finite-dimensional C*-algebra is *-isomorphic to a finite sum of matrix algebras.
Theorem 3.2 (Artin-Wedderburn). Let A be a finite-dimensional C*-algebra. Then there are k, n1, . . . , n k ∈ N such that The number k is unique, whereas the numbers n1, . . . , n k are unique up to permutation.
Proof. [39,Theorem I.11.2] The strategy for proving Theorem 3.1 is the following. First, we find an order-theoretic property of C(A) that corresponds with the finite-dimensionality of A. Then we find a method of retrieving the numbers k, n1, . . . , n k from C(A). 2) All non-empty filtered subsets of C have a least element; 3) C satisfies the descending chain condition: if we have a sequence of elements C1 ≥ C2 ≥ . . . in C, i.e., a countable descending chain, then the sequence stabilizes, i.e., there is an n ∈ N such that C k = Cn for all k > n.
In order to see that these conditions are equivalent, assume that (1) holds and let F a non-empty filtered subset of C. Then F must have a minimal element C. Now, if F ∈ F, then there must be an G ∈ F such that G ≤ C, F . Since C is minimal, it follows that G = C, so C ≤ F , whence C is the least element of F.
For (2) implies (3), let C1 ≥ C2 ≥ C3 ≥ . . . be a descending chain. Then F = {Ci} i∈N is clearly a directed subset, so it has a least element, say Cn. So we must have C k = Cn for all k > n, hence C satisfies (3).
Finally, we show by contraposition that (3) follows from (1). So assume that C does not satisfy the descending chain condition. Hence we can construct a sequence C1 ≥ C2 ≥ . . . that does not terminate. The set F = {Cn : n ∈ N} is then a non-empty subset of C without a minimal element. Thus C does not satisfy (1).
There exists also a notion dual to the notion of an Artinian poset.
Definition 3.4. Let C be a poset. Then C is called Noetherian if is satisfies one of the following equivalent conditions: 1) Every non-empty subset contains a maximal element; 2) All non-empty directed subsets of C have a greatest element; 3) C satisfies the ascending chain condition: if we have a sequence of elements C1 ≤ C2 ≤ . . . in C, i.e., a countable ascending chain, then the sequence stabilizes, i.e., there is an n ∈ N such that C k = Cn for all k > n.
The following proposition can be found in [26] as Theorem 4.21.
Proposition 3.5 (Principle of Artinian induction). Let C be an Artinian poset and P a property such that: Then P(C) is true for each C ∈ C.
Proof. Assume that F = {C ∈ C : P(C) is not true} is non-empty. Since C is Artinian, this means that F has a minimal element C. Hence P(B) is true for all elements B < C, so P(C) is true by the induction step, contradicting the definition of F. Definition 3.6. Let C be a poset. Then C is called graded if one can define a function d : C → N, called a rank function such that: There is no standard definition of a graded poset. For instance, in [36] condition (i) is dropped and Z is taken as codomain of rank functions. On the other hand, [30] assumes condition (i), but not condition (ii). For our purposes, it is convenient to combine both definitions. The next three lemmas are now easy to prove. Lemma 3.7. Let C be a graded poset with rank funcion d : C → N. Then C is Artinian. If the range of d is bounded from above, C is Noetherian as well.
contradicting the minimality of n. Hence we must have F = F1 and in a similar way, we We conclude that C is Artinian. Now assume that d[C] ⊆ N has an upper bound, then the proof that C is Noetherian follows in an analogous way.
Lemma 3.8. Let C be a graded poset. Then its rank function d : C → N is unique.
Proof. By Lemma 3.7, C is Artinian. Hence every non-empty subset has a minimal element, and in particular min C = ∅. Assume that g : C → N is a rank function. By definition of a rank function we have g(M ) = d(M ) = 1 for each M ∈ min C.
Let C ∈ C such that C / ∈ min C. Assume that d(B) = g(B) for each B < C. The set {d(B) : B < C} ⊆ N is non-empty and bounded by d(C), hence it must have a maximum n. As a consequence, there is some B < C such that d(B) = n. Assume that n + 1 = d(C). Then C does not cover B, hence there is some B ′ ∈ C such that B < B ′ < C. Thus n = d(B) < d(B ′ ) contradicting the maximality of n. We conclude that d(B) + 1 = d(C), so C covers some B. Hence we obtain Lemma 3.9. Let φ : C → D be an order isomorphism between graded posets C and D with rank functions dC and dD, respectively. Then dC = dD • φ.
Thus d is a rank function on C and by Lemma 3.8, we obtain d1 = d.
Lemma 3.10. Let A be a finite-dimensional C*-algebra. Then C(A) is graded with a rank function dim : C(A) → N assigning to each element C ∈ C(A) its dimension. Moreover, the range of dim is bounded from above.
Combining Lemmas 3.7 and 3.10, we find that if A is finite dimensional, then C(A) is both Artinian and Noetherian. We shall prove that the converse holds as well.
Lemma 3.11. Let A be a C*-algebra. Then every element of C(A) is contained in a maximal element of C(A). In particular, the set max C(A) is non-empty.
Proof. Let C ∈ C(A) and let S = {D ∈ C(A) : C ⊆ D}. Then S is non-empty, so if C(A) is Noetherian, we immediately find that S contains a maximal element. If C(A) is not Noetherian, we need Zorn's Lemma. Let D = {Di}i∈I be a chain in S. Then D is certainly directed, hence it must have a join D by Proposition 2.5. Since D clearly contains C, we have D ∈ S. So for every chain C(A), there is an upper bound for the chain in S. By Zorn's Lemma it follows that S contains a maximal element M . Now, M must also be maximal in By the maximality of M with respect to S, it follows that A must be equal to M . Proposition 3.12. Let A be a C*-algebra and M a maximal commutative *-subalgebra. If M is finite dimensional, then A must be finite dimensional as well.
This statement can be found in [27] as Exercise 4.12. The solution of this exercise can be found in [28].
In particular, when A = Mn(C), there is a nice characterization of the maximal commutative C*-subalgebras of C(A).
where Dn is the commutative C*-subalgebra of A consisting of all diagonal matrices.
Definition 3.14. Let X be a topological space with topology Ø(X). Then X is called Noetherian if the poset Ø(X) ordered by inclusion is Noetherian. Proof. First we show that every subset of X is compact. So if Y ⊆ X, let U be a cover of Y . Let V be the set of all finite unions of elements of U. Then V covers Y as well, and moreover, V is directed. Since Ø(X) is Noetherian, V has a greatest element V . Now, V contains every element of V and since V covers Y , we find that V must contain Y . It follows from the definition of V that V can be written as a finite union of elements of U, so U has a finite subcover. Thus Y is compact. Now let x ∈ X. Then X \ {x} is compact, hence closed.
Hence {x} is open, and it follows that X is discrete. Since X itself is compact, X must be finite. Proof. Assume that A is finite dimensional. By Lemma 3.10, C(A) has a rank function whose range is bounded from above. By Lemma 3.7, C(A) is both Artinian and Noetherian. Assume that A is not finite dimensional. By Lemma 3.11, C(A) has a maximal element M . By Proposition 3.12, it follows that M cannot be finite dimensional. Since M is a unital commutative C*-algebra, the Gel'fand-Naimark Theorem assures that M = C(X) for some compact Hausdorff space X, which must have an infinite number of points since A is infinite dimensional.
We construct an ascending chain in C(A) as follows. First we notice that since X is infinite and Hausdorff, Lemma 3.15 implies that X is not Noetherian. So there is an ascending chain O1 ⊆ O2 ⊆ . . . of open subsets of X that does not stabilize. For each i ∈ N, let Fi = X \ Oi. Then F1 ⊇ F2 ⊇ . . . is a descending chain of closed subsets of X, which does not stabilize. For each i ∈ I let Ci = CF i . Then Ci is a C*-subalgebra of C(X) and if i ≤ j, we have Fi ⊇ Fj, so Ci ⊆ Cj. Moreover, if i < j and Fi = Fj, then there is some x ∈ Fi such that x / ∈ Fj . By Urysohn's Lemma, there is an f ∈ C(X) such that f (x) = 0 and f (y) = 1 for each y ∈ Fj. Hence f ∈ Cj , but f / ∈ Ci. It follows that Ci = Cj, so C1 ⊆ C2 ⊆ . . . is an ascending chain that does not stabilize.
Thus C(A) contains an ascending chain as well as a descending chain, neither of which stabilizes. Hence C(A) can be neither Noetherian nor Artinian.
Recall that the center of a C*-algebra A is the set {x ∈ A : xy = yx ∀y ∈ A}, which is usually denoted by Z(A). Proof. For each y ∈ A, consider the map fy : A → A given by the assignment x → xy − yx. Clearly this is continuous and linear, so ker fy is a closed linear subspace of A. Hence Z(A) = y∈A ker fy is a closed linear subspace as well. If x, y ∈ Z(A) and z ∈ A, then xyz = xzy = zxy, so xy ∈ Z(A). Moreover, x * z = (z * x) * = (xz * ) * = zx * , so x * ∈ Z(A). Clearly xy = yx, and 1A ∈ Z(A), hence Z(A) ∈ C(A).
Let x ∈ max C(A), i.e., x ∈ M for each maximal M ∈ C(A). Let y ∈ A. Then y can be written as a linear combination of two self-adjoint elements a1, a2. If a ∈ A is self-adjoint, then C * (a, 1) is a commutative C*-subalgebra of A containing a. By Lemma 3.11, it follows that there are M1, M2 ∈ max C(A) such that ai ∈ Mi for i = 1, 2. Since x ∈ M1, M2, it follows that x commutes with both a1 and a2. Hence x commutes with y, so x ∈ Z(A). Thus max C(A) ⊆ Z(A). Now assume that x ∈ Z(A). Since x commutes with all elements of A, it commutes in particular with x * . Hence x is normal. We have x * ∈ Z(A) as well, for Z(A) is a *-subalgebra of A. Let M ∈ max C(A). Then M ∪{x, x * } is a set of mutually commuting elements, which is *-closed and contains 1A. It follows that C * (M ∪ {x, x * }), the C*-subalgebra of A generated by M ∪ {x, x * }, is commutative. Since M is maximal, C * (M ∪ {x, x * }) must be equal to M . As a consequence, x ∈ M , so we find that x is contained in every maximal commutative C*-subalgebra of A. Hence Z(A) ⊆ max C(A).
Lemma 3.18. Let A1, . . . , An be C*-algebras. Then Proof. This follows directly from the fact that multiplication on n i=1 Ai is calcultated coordinatewisely.
Ci. This means that for each i = 1, . . . , n there is a Here 1A i and 0A i denote the unit and the zero of Ai, respectively. Since 1A i , 0A i ∈ Z(Ai), Lemma 3.18 assures that e j ∈ Z(A). Since Z(A) ⊆ C, we find that e j ∈ C. It follows that and since f j ∈ C, it follows that c ∈ C. So C = n i=1 Ci. (ii) π is surjective; Proof. For each i = 1, . . . , n, let Ci ∈ C(Ai). Then C1 ⊕ . . . ⊕ Cn is clearly a commutative C*-algebra of A, which is unital since Hence the image of ι lies in C(A), so ι is well defined. Furthermore, we remark that C(pi) : C(A) → C(Ai) is an order morphism by Lemma 2.2. (iii) Let C1, . . . , Cn ∈ n i=1 C(Ai). Let C = ι( C1, . . . , Cn ). Then C = C1 ⊕ . . . ⊕ Cn, and by the calculation in (ii), we obtain π(C) = C1, . . . , Cn . Hence π Let c ∈ C. Since C ⊆ A, and A = n i=1 Ai, we have Hence c ∈ ι • π(C), so C ⊆ ι • π(C). We conclude that the inequality 1 C(A) ≤ ι • π holds.
(iv) In order to show that ι restricts to an order isomorphism Since π is an order isomorphism, it follows that π(M ) is a maximal element of k i=1 C(Mn i (C)). Clearly there are Mn i ∈ max C(Mn i (C)) for each i = 1, . . . , k such that π(M ) = Mn 1 , . . . , Mn k . It follows that ↓ π(M ) = ↓ Mn 1 × . . . × ↓ Mn k .
Since ι is the inverse of π and has codomain ↑ Z(A), we find that Hence the restriction ι : Notice that all maximal elements of C(Mn i (C)) are *-isomorphic by Lemma 3.13. More specificaly, Mn i ∈ max C(Mm i (C)) is *-isomorphic to Dn i . Since we find that Mn i is *-isomorphic to C n i . Hence there is an embedding is an order embedding with image ↓ Mn i . Thus there exists an order isomorphism between C(C n i ) and ↓ Mn i in C(Mn i (C)). The following result is also known as Hashimoto's Theorem.   Let C = φ( 1, 0 ). Then it follows that 1 = C = 0. Clearly 0, 1 is a complement of 1, 0 in C1 × C2, but it is also unique. Let D1, D2 be a complement of 0, 1 . Since meets and joins are calculated componentwise, we find hence D1 = 0, D2 = 1. Thus D1, D2 = 0, 1 , whence 1, 0 indeed has a unique complement.
Since φ is an order isomorphism, φ preserves meets and joins, hence C has a complement D = φ( 0, 1) . Now assume that C has another complement D ′ . Since φ is an order isomorphism, it follows that φ −1 (D ′ ) is a complement of 1, 0 , and by uniqueness of this complement, we obtain φ −1 (D ′ ) = 0, 1 . We find that The statement follows now by contraposition.
Proposition 3.27. Let A be a commutative finite-dimensional C*-algebra. Then C(A) is a directly indecomposable lattice.
Proof. By Lemma 2.4, C(A) is a bounded lattice. Let X be the spectrum of A. If X is a singleton set, then A is one-dimensional, hence we have C(A) = {C1A}, so C(A) = 1, the one-point lattice, and there is nothing to prove. If X is a two-point set, then C(A) = {A, C1A}. So C(A) contains no other elements than a greatest and a least one, and is therefore certainly directly indecomposable.
Assume that X has at least three points. Let B ∈ C(A), assumed not equal to C1A or A. By Lemma 1.4, we have B = x∈X C [x] B . Since X is finite, it follows that X/ ∼B is finite as well. Notice that we cannot have [x]B = {x} for all x ∈ X, otherwise B = C(X) = A. Neither can X/ ∼B be a singleton set, since otherwise B = C1A. For each element [x]B in X/ ∼B, choose a representative x. Let K be the set of representatives. Notice that K is not a singleton set, since X/ ∼B contains at least two elements. Also notice that K is not unique, since there is at least one [x]B ∈ X/ ∼B containing two or more points. Since X is discrete, it follows that K is closed.
Let f ∈ B ∩ CK and let x, y ∈ X be points such that x = y.
Combining all equalities gives f (x) = f (y). So in all cases, f (x) = f (y). So f must be constant, and we conclude that B ∩ CK = C1A.
Since C(A) is a lattice, B ∨ CK exists. Let f ∈ C(X). Define the map g : X → C by Notice that is well defined, since K is a collection of representatives. Moreover, since X is discrete, g is continuous, so g ∈ C(X). By definition, Let h = f − g. Then h ∈ C(X), and if k ∈ K, we find h(k) = f (k) − g(k) = 0, so h is constant on K. We conclude that f = g + h with g ∈ B and h ∈ CK. Hence A = C(X) = B ∨ CK.
We find that CK is a complement of B. However, K is not unique, and therefore neither is CK . We conclude that A and C1A are the only elements with a unique complement, so C(A) is indirectly indecomposable.
The proof of this proposition is based on the proof of the directly indecomposability of partition lattices in [37]. More can be said about C(A) when A is a commutative C*-algebra of dimension n, namely that C(A) is order isomorphic to the lattice of partitions of the set {1, . . . , n}. We refer to [22] for a complete characterization of C(A) when A is a commutative finite-dimensional C*-algebra.
We are now ready to prove the main result of this section.
Proof of Theorem 3.1. Let A be a finite-dimensional C*-algebra, and B a C*-algebra. Let φ : C(A) → C(B) an order isomorphism. By Proposition 3.16, C(A) is Noetherian, and so C(B) must be Noetherian as well. Hence Proposition 3.16 implies that B is finite dimensional. It follows from Lemma 3.10 that both C(A) and C(B) have a rank function assigning to each element its dimension. By Lemma 3.8 the rank function is unique, hence it follows from Lemma 3.9 that dim(φ(C)) = dim(C) for each C ∈ C(A). Therefore, we can reconstruct the dimensions of elements of C(A) and C(B), and the dimension is preverved by φ.
By the Artin-Wedderburn Theorem, there are unique k, k ′ ∈ N and unique Mn i (C); Without loss of generality, we may assume that the ni and n ′ i form an descending (but not necessarily strictly descending) finite sequence.
By Lemma 3.17, we have the equalities Z(A) = max C(A) and max C(B) = Z(B). Since the intersection is the meet operation in C(A) and C(B), and order isomorphism preserve both meets and maximal elements, we find that φ(Z(A)) = Z(B), so dim(Z(A)) = dim(Z(B)). Using Lemma 3.18, we find that Z(A) = n i=1 Z(Mn i (C)), and since the dimension of the center of a matrix algebra is 1, we find that dim Z(A) = k. In the same way, we find that dim Z(B) = k ′ , so we must have k = k ′ .
Let M ∈ max C(A). Then φ(M ) is a maximal element of C(B), and since φ(Z(A)) = Z(B), we find that φ restricts to an order isomorphism [Z(A), M ] → [Z(B), φ(M )]. By Proposition 3.21, we obtain an order isomorphism It is possible that for some i we have ni = 1, in which case we have C(C n i ) = 1. Since we assumed that {ni} n i=1 is a descending sequence, there is a greatest number r below k such that nr = 1. Likewise, let s be the greatest number such that n ′ s = 1. Then we obtain an order isomorphism By Proposition 3.27 and Corollary 3.24, we now find r = s, and there is a permutation π : {1, . . . , r} → {1, . . . , r} such that C(C n i ) ∼ = C(C n ′ π(i) ) for each i ∈ {1, . . . , r}. Let ψi : C(C n i ) → C(C n ′ π(i) ) be the accompanying order isomorphism. Lemma 3.10 assures that the function assigning to each element of C(C n i ) its dimension is a rank function, and similarly the dimension function is a rank function for C(C n ′ π(i) ). By Lemma 3.9, we find that dim(C) = dim(ψi(C)) for each C ∈ C(C n i ). Hence where the fact that order isomorphisms map greatest elements to greatest elements is used in the third equality. By definition of r, we must have ni = n ′ i = 1 for all i ≥ r. Hence we can extend π to a permutation {1, . . . , k} → {1, . . . , k} by setting π(i) = i for each i ≥ r. Hence k = k ′ and {n1, . . . , n k } and {n ′ 1 , . . . , n ′ k } are the same sets up to permutation. We conclude that A and B must be *-isomorphic.
We note that since the class of all finite-dimensional C*-algebras and the class of all finitedimensional von Neumann algebras are the same, a similar statement holds for the functor V assigning to a von Neumann algebra M the poset V(M ) of its commutative von Neumann subalgebras. Thus if M and N are von Neumann algebras such that M is finite-dimensional, If A is a finite-dimensional C*-algebra and B is a C*-algebra such that there is an order isomorphism φ : C(A) → C(B), then it might be the case that even though A and B are *-isomorphic, we have φ = C(f ) for more than one *-isomorphism f : It might even be the case that φ = C(f ) for each *-isomorphism f : A → B. For instance, let A = B = M2(C). Then where D2 = {diag(λ1, λ2) : λ1, λ2 ∈ C}. Furthermore, one can show that each *-isomorphism f : M2(C) → M2(C) is of the form a → uau −1 for some u ∈ U(2) [1,Theorem 4.27]. Hence C(f ) : C(M2(C)) → C(M2(C)) is given by C → uCu * for some u ∈ U(2).

Outlook
We have shown that C(A) is a complete invariant for finite-dimensional C*-algebras, whereas Mendivil and Hamhalter showed that C(A) completely determine commutative C*-algebras. The question is whether there are more classes of C*-algebras which can be classified by C(A). An interesting class might be that of AF-algebras, i.e., C*-algebras A that can be approximated by finite-dimensional C*-algebras. Usually one considers only separable AF-algebras, which are C*-algebras A such that . is an ascending chain of finite-dimensional C*-subalgebras of A. It is well known that this class of AF-algebras can be classified by Bratteli diagrams [5] and by K-theory [13].
One could also look at C*-algebras A such that A = D for some directed set D consisting of finite-dimensional C*-subalgebras of A. In this case A need not be separable, and therefore C*-algebras of these form are called non-separable AF-algebras. It turns out that neither Bratteli diagrams nor K-theory can completely classify this class of C*-algebras [14], [29]. However, as one might have noticed, the framework of C(A) might be suitable in order to classify non-separable AF-algebras if one compares the definition of non-separable AF-algebras with the content of Proposition 2.5. If this is indeed the case, then C(A) might be an interesting alternative for K-theory.
Since C(A) is a dcpo, and domain theory (see for instance [15]) deals with various properties of dcpos, a first step might be the study of the domain-theoretical properties of C(A). For details, we refer to [24]. It might be interesting to compare the domain-theoretical properties of C(A) with those of V(M ), the poset V(M ) of commutative von Neumann subalgebras of a von Neumann algebra M . For the von Neumann case, we refer to [11].
It might be interesting to look at non-unital C*-algebras as well. The reason why we did not consider non-unital C*-algebras lies within quantum toposophy, from which this research evolved. In quantum toposophy one is forced to work constructively; and whereas constructive Gel'fand duality for unital commutative C*-algebras holds (see for instance [2] and [8]), it was not known yet whether the non-unital version holds as well. However, Henry recently proved a non-unital version of constructive Gel'fand duality [20], which suggests that non-unital C*-algebras can be incorporated within quantum toposophy as well.
In the non-unital case one could proceed as follows. If CStar denotes the category of C*algebras with *-homomorphisms as morphisms, we can define the functor C0 : CStar → Poset as follows. Given a C*-algebra A, we denote the poset of commutative C*-algebras by C0(A), and if f : The functor C0 shares some properties with C, for instance Proposition 2.6 holds as well if we replace C by C0. It is even the case that we can describe injectivity of a *-homomorphism f : A → B completely in order theoretic properties of C0(f ). This is possible, since 0, the C*-algebra consisting of only one element 0, is always an element of C0(A). We expect that Theorem 3.1 holds as well if we replace C by C0. Some minor details in the proofs must be adjusted, but we expect that most lemmas still hold, since each finitedimensional C*-algebra A is automatically unital, hence C(A) can be regarded as subposet of C0(A).
However, it might be difficult to prove a non-unital version of Mendivil and Hamhalter's theorem to the effect that C0(A) determines a commutative C*-algebra A up to *-isomorphism, since it is desirable that we can identify C*-ideals of A as elements of C0(A) in order to reconstruct A, and it is not clear how to make this identification. This is already visible if we consider C0(C 2 ) = {0, C1, C2, C3, C 2 }, where The least element and the greatest element of C0(C 2 ) are 0 and C 2 , respectively, and C1, C2, C3 are mutually incomparable. Here C1 and C2 are the only elements that correspond to ideals of C 2 , but it is not possible to distinguish them from C3 in an order theoretical way.
Thus C0(A) has some advantages as well as disadvantages with respect to C(A). If A is unital, it could be useful to consider both posets at the same time. In this case, C(A) can be considered a subposet of C0(A). It might be interesting to remark that in quantum toposophy, a pair (C, D) of a poset C and a subposet D of C exactly corresponds to a site (C, J), i.e., a poset C equipped with a Grothendieck topology, such that the category Sh(C, J) of J-sheaves is equivalent to Sets D op . Hence if A is unital, then the pair (C0(A), C(A)) corresponds to a site (C0(A), J) such that Sh(C0(A), J) ∼ = Sets C(A) op . Since one usually studies the topos Sets C(A) op , it follows that one can integrate C0(A) in an elegant way in the usual framework of quantum toposophy. For details on Grothendieck topologies and sheaves on posets, we refer to [31].

A Order-theoretical notions
We recall some definitions in order theory and refer to [12] for a detailed exposition.
A poset (C, ≤) is a set C equipped with a (partial) order ≤. That is, ≤ is a binary relation, which is reflexive, antisymmetric and transitive. We often write C instead of (C, ≤) if it is clear which order is used. A poset C becomes a category if we consider its elements as objects, and taking a unique morphism C → D if and only if C ≤ D for each C, D ∈ D.
If either B ≤ C or C ≤ B for each B, C ∈ C, we say that ≤ is a linear order , and we call C a linearly ordered set. A linearly ordered subposet of a poset is called a chain. Given a poset C with order ≤, we define the opposite poset C op as the poset with the same underlying set C, but where B ≤ C if and only if C ≤ B in the original order.
Let D ⊆ C be a subset. Then D is called an upper set or an up-set if C ∈ D and D ≥ C implies D ∈ D for each C, D ∈ C; a lower set or an down-set if C ∈ D and D ≤ C implies D ∈ D for each C, D ∈ C; directed if for each D1, D2 ∈ D there is a D3 ∈ D such that D1, D2 ≤ D3; and filtered if for each D1, D2 ∈ D there is a D3 ∈ D such that D1, D2 ≥ D3.
If C ∈ C, we define the up-set and down-set generated by C by ↑ C = {B ∈ C : B ≥ C} and ↓ C = {B ∈ C : B ≤ C}, respectively. We can define the up-set generated by a subset D of C by ↑ D = D∈D ↑ D. Similarly, we define the down-set generated by D by ↓ D = D∈D ↓ D. Greatest and least elements are always unique. If C itself contains a least and a greatest element, usually denoted by 0 and 1, respectively, we say that C is a bounded . The set of all maximal elements of C is denoted by max C, whereas min C denotes the set of all minimal elements of C.
If D ⊆ C, then an element C ∈ C such that D ≤ C for each D ∈ D is called an upper bound of D. Similarly, C is called a lower bound of D if C ≤ D for each D ∈ D. If D has a least upper bound C, usually called the join of D, then we write C = D. Dually, if C is a greatest lower bound of D, usually called the meet of D, then we write C = D. If D is a two-point set, say D = {D1, D2}, we write D1 ∨ D2 instead of D, and D1 ∧ D2 instead of D. We say that D1 ∨ D2 and D1 ∧ D2 are the binary join and binary meet, respectively, of D1 and D2. If we consider C as a category, then the join of D is exactly the same as the coproduct of D, whereas the meet of D is exactly the product of D.
If C is a poset such that all binary meets exists, then we call C a meet-semilattice. If the join of all directed subsets of C exist, we call C a directed-complete partial order, abbreviated by dcpo. If all binary meets and joins exists, we call C a lattice. If all arbitrary meets and joins exist, then we call C a complete lattice. Notice that a complete lattice C is automatically bounded, since C is its greatest element, and C is its least element. Moreover, if C has all arbitrary meets, it is automatically a complete lattice, since the join of a subset D of C is given by D = {C ∈ C : D ≤ C ∀D ∈ D}.
Let C1, C2 be posets and φ : C1 → C2 a map. Then φ is called an order morphism if C ≤ D implies φ(C) ≤ φ(D) for each C, D ∈ C1; an embedding of posets if φ(C) ≤ φ(D) if and only if C ≤ D for each C, D ∈ C1; and an order isomorphism if it is an order morphism such that φ • ψ = 1C 2 and ψ • φ = 1C 1 for some order morphism φ : C2 → C1. Here 1C i : CCi → Ci is the identity order morphism. If φ is an order morphism and there is an order morphism ψ : C2 → C1 such that for each C1 ∈ C1 and C2 ∈ C2 we have φ(C1) ≤ C2 if and only if C1 ≤ ψ(C2), we say that ψ is the upper adjoint of φ, and φ the lower adjoint of ψ. Clearly an embedding of posets φ is injective, but the converse does not always hold. Moreover, a map φ : C1 → C2 is an order isomorphism if and only if it is a surjective order embedding. If we consider C1 and C2 as categories, then the upper adjoint is exactly the same as a right adjoint. If C1 and C2 are both lattices, then φ is called a lattice morphism if φ(C ∧ D) = φ(C) ∧ φ(D) and φ(C ∨ D) = φ(C) ∨ φ(D) for each C, D ∈ C1. If φ is bijective, then φ is called a lattice isomorphism. A lattice morphism φ : C1 → C2 is automatically an order morphism. An order isomorphism between lattices is automatically a lattice isomorphism.
Let C1, . . . , Cn be posets. Then the cartesian product of the Ci is defined as the set n i=1 Ci, sometimes also denoted as C1 × . . . × Cn, equipped by the order defined by C1, . . . , Cn ≤ D1, . . . , Dn if and only if Ci ≤ Di for each i = 1, . . . , n. If Ci is a lattice for each i = 1, . . . , n, then n i=1 Ci is a lattice as well.