Extreme points and matrix blocks

Written by GPT-6.1 Sol (OpenAI), Ultra, September–October 2026. Self-checked by the writing AI. Original text: CC0 1.0. The credited subsection “States and mixtures on a matrix block” retains CC BY 4.0.

An extreme point of a convex set cannot move in two opposite directions while staying inside the set. For the unit ball of a C*-algebra, the directions that remain available are described by two defect projections. In finite dimensions those defects disappear, and the algebra itself separates into full matrix blocks. The same blocks then describe every representation, including representations on Hilbert spaces of arbitrary dimension.

Prerequisites are C*-algebras: continuous functional calculus, positive cones, approximate identities and quotients and Representations and positive functionals. The second classification argument also uses the double commutant theorem, Theorem 4.4. Each of these links supplies the proof used below. Freely readable accounts are Blackadar’s Operator Algebras and Sundar’s Notes on C*-algebras.

All Hilbert spaces are complex. A representation need not be nondegenerate unless that is stated. A nonzero projection $p$ is minimal when $pAp=\mathbb Cp$. This is the definition we use even in an infinite-dimensional algebra. In a finite-dimensional algebra it is equivalent to saying that $p$ contains no smaller nonzero projection. We allow the zero algebra, with identity $0$, and empty direct sums.

1. The commutative picture

A point $x$ of a convex set $C$ is extreme if $x=(y+z)/2$, with $y,z\in C$, implies $y=z=x$. Write $A_1=\{a\in A:\|a\|\leq1\}$ for the closed unit ball.

Lemma 1.1. Let $A=C_0(X)$, where $X$ is locally compact Hausdorff.

1. The extreme points of $A_1$ are exactly the functions of modulus one everywhere. Their existence forces $X$ to be compact; they are then the unitaries of $A$.
2. The extreme points of $A_1\cap A_+$ are exactly the projections, that is, the continuous functions taking only the values $0,1$ and vanishing at infinity.
3. If $0\leq h\leq1$ is not a projection, there is $a\in A_+$, with $\|a\|\leq1$ and $ha\neq0$, such that $0\leq h(1\pm a)\leq1$. Products with $1\pm a$ can be computed in the unitization.

Proof. If $|f(t_0)|<1$, choose a compactly supported continuous bump $g\geq0$, nonzero at $t_0$, in a neighbourhood on which $|f|\leq c<1$. For sufficiently small $\varepsilon>0$, both $f+\varepsilon g$ and $f-\varepsilon g$ lie in the unit ball. Thus $f$ is not extreme. Conversely, a point of modulus one is extreme in the scalar unit disk; evaluating any midpoint identity at every $t\in X$ proves extremality of $f$. If $|f|=1$ everywhere and $f\in C_0(X)$, the set where $|f|\geq1/2$ is all of $X$, so $X$ is compact.

A function taking values in $\{0,1\}$ is pointwise extreme in the interval $[0,1]$. If $h(t_0)\in(0,1)$, choose a nonnegative bump $g$, with $g(t_0)>0$, supported where $\delta\leq h\leq1-\delta$ for some $\delta>0$. A small multiple $a=\varepsilon g$ satisfies the third assertion. The two distinct positive contractions $h(1+a),h(1-a)$ have midpoint $h$. This proves the second assertion as well. $\square$

The analogous positive-contraction statement holds without commutativity.

Proposition 1.2. In every C*-algebra, the extreme points of $A_1\cap A_+$ are its projections.

Proof. If $h$ is a positive contraction that is not a projection, its spectrum meets $(0,1)$. Apply the bump construction of Lemma 1.1 inside the commutative algebra generated by $h$ and the identity of the unitization. Choose the bump to vanish near $0$; functional calculus then puts it in $A$. The resulting two positive contractions in $A$ show that $h$ is not extreme.

Suppose a projection $p$ is the midpoint of positive contractions $b,c$. In the unitization, $$(1-p)b(1-p)+(1-p)c(1-p)=0.$$ Both summands are positive, so both vanish. Since $(1-p)b(1-p)=(b^{1/2}(1-p))^*b^{1/2}(1-p)$, we have $b=pbp$, and similarly $c=pcp$. Now $b,c\leq p$, while $b+c=2p$. Therefore $(p-b)+(p-c)=0$ is a sum of positive elements; hence $b=c=p$. $\square$

2. Two defects control the whole unit ball

An element $v\in A$ is a partial isometry if $v^*v$ is a projection. Then $vv^*$ is a projection as well, and $$p=v^*v,\qquad q=vv^*,\qquad v=vp=qv.$$ The projections $1-p$ and $1-q$ describe its initial and final defects.

Theorem 2.1. The unit ball of a C*-algebra has an extreme point if and only if the algebra is unital. In a unital algebra its extreme points are exactly the partial isometries $v$ satisfying $$(1-q)A(1-p)=\{0\},\qquad p=v^*v,\quad q=vv^*. \tag{2.1}$$ Taking adjoints gives the equivalent condition $(1-p)A(1-q)=\{0\}$.

Proof. First suppose $x\in A_1$ and $h=|x|=(x^*x)^{1/2}$ is not a projection. A nonnegative spectral bump $a\in C^*(h,1)$ as in Lemma 1.1 satisfies $ha\neq0$ and $\|h(1\pm a)\|\leq1$. It can be chosen to vanish at $0$, although that is not needed for $x(1\pm a)\in A$. Since $a$ commutes with $h$, $$\|x(1\pm a)\|^2 =\|(1\pm a)h^2(1\pm a)\| =\|h(1\pm a)\|^2\leq1, \qquad \|xa\|=\|ha\|>0.$$ Thus $x$ is not extreme. Every extreme point must be a partial isometry.

For a partial isometry $v$, let $a\in(1-q)A(1-p)$ be a contraction. Orthogonality gives $v^*a=a^*v=0$, so $$(v\pm a)^*(v\pm a)=p+a^*a.$$ The two positive terms have orthogonal supports; their sum has norm at most one. If $a\neq0$, this is a nontrivial midpoint decomposition of $v$. Consequently extremality forces (2.1), computed initially in the unitization.

Let $(e_\lambda)$ be a contractive approximate identity of $A$. Condition (2.1) implies $$e_\lambda=qe_\lambda+e_\lambda p-qe_\lambda p \longrightarrow q+p-qp\in A$$ in norm. The limit $e$ satisfies $ea=ae=a$ for every $a\in A$, because the approximate identity does. Hence $A$ is unital.

For the converse, represent a unital $A$ faithfully and unitally on $H$. Suppose $v$ satisfies (2.1) and $v\pm h\in A_1$. If $\xi\in pH$, the parallelogram identity yields $$\|(v+h)\xi\|^2+\|(v-h)\xi\|^2 =2\|v\xi\|^2+2\|h\xi\|^2 =2\|\xi\|^2+2\|h\xi\|^2.$$ The left side is at most $2\|\xi\|^2$. Thus $hp=0$. Apply the same argument to $v^*\pm h^*$ on $qH$ to obtain $h^*q=0$, or $qh=0$. It follows that $h=(1-q)h(1-p)=0$. This proves extremality. In particular $1$ is extreme, so a unital algebra always has an extreme point. The zero-algebra case is immediate under our convention. $\square$

Example 2.2. Every isometry is extreme because $p=1$; every coisometry is extreme because $q=1$. On an infinite-dimensional Hilbert space these can fail to be unitary. In $M_n(\mathbb C)$, a partial isometry has initial and final projections of the same rank. If both defects are nonzero, there is a nonzero matrix mapping the initial defect into the final defect, contradicting (2.1). Hence the extreme points of the matrix unit ball are precisely its unitaries.

The distinction between the positive ball and the whole ball is substantial. Every finite-rank projection in $K(H)$ is extreme in its positive contractive part, whereas $K(H)$, for infinite-dimensional $H$, has no extreme point in its whole unit ball.

The defect criterion is Kadison's theorem (1951) [Blackadar, II.3.2.19]. The proof above includes its converse for every partial isometry, using the parallelogram identity on both support subspaces. No assumption about finite dimension or separability enters that argument.

3. Recovering the matrix blocks

Lemma 3.1. A finite-dimensional C*-algebra is unital. Every two-sided ideal $I\subseteq A$ is $Az$ for a central projection $z\in A$.

Proof. A contractive approximate identity has a convergent subnet in the compact unit ball of the finite-dimensional space $A$. Its limit is an identity. The same argument applies to $I$, which is closed because it is a linear subspace of a finite-dimensional space. Let $z$ be the identity of $I$. It is a projection: the identity in a nonzero C*-algebra is self-adjoint, and $z^2=z$. For $a\in A$, both $az,za\in I$, so $zaz=az=za$. Thus $z$ is central and $I=Az$. The zero ideal has $z=0$. $\square$

A system of matrix units is a family $(e_{ij})_{1\leq i,j\leq n}$ with $$e_{ij}^*=e_{ji},\qquad e_{ij}e_{kl}=\delta_{jk}e_{il}. \tag{3.1}$$ Its sum $\sum_i e_{ii}$ is the identity of the matrix algebra it spans. When this sum is the identity of $A$, we call it a unital system in $A$.

Theorem 3.2. Every finite-dimensional C*-algebra is *-isomorphic to $$A\cong\bigoplus_{r=1}^s M_{n_r}(\mathbb C). \tag{3.2}$$ The unordered list of positive integers $n_1,\ldots,n_s$ is unique. Two such algebras are *-isomorphic exactly when their lists agree, with multiplicities.

Proof. Choose a maximal abelian self-adjoint subalgebra $D\subseteq A$. It contains $1$, since adjoining $1$ preserves commutativity. Its Gelfand spectrum is finite: if it had arbitrarily large finite sets of distinct points, continuous functions separating those points would give arbitrarily large linearly independent families in $D$. Thus $$D=\bigoplus_{i=1}^N\mathbb Cp_i, \qquad p_ip_j=0\ (i\neq j),\qquad\sum_i p_i=1.$$ Every self-adjoint element of $p_iAp_i$ commutes with $D$. Adjoining it to $D$ still gives an abelian algebra, so maximality puts it in $D$. Taking real and imaginary parts proves $p_iAp_i=\mathbb Cp_i$.

If $0\neq x\in p_iAp_j$, then $x^*x=\lambda p_j$ and $xx^*=\mu p_i$, with $\lambda,\mu>0$. The C*-identity gives $\lambda=\mu=\|x\|^2$. Therefore $v=x/\|x\|$ satisfies $v^*v=p_j,vv^*=p_i$. Moreover every $y\in p_iAp_j$ is a scalar multiple of $v$: $v^*y\in\mathbb Cp_j$ and $y=vv^*y$. Each nonzero corner is one-dimensional.

Define $i\sim j$ when $p_iAp_j\neq0$. This is an equivalence relation. Reflexivity and symmetry are immediate. For transitivity, partial isometries $v\in p_iAp_j$ and $w\in p_jAp_k$ as above satisfy $(vw)^*(vw)=p_k$, so $vw\neq0$.

For an equivalence class $E$, put $z_E=\sum_{i\in E}p_i$. The expansion $a=\sum_{i,j}p_iap_j$, and the absence of corners between different classes, show that $z_E$ is central. Fix $i_0\in E$, choose $v_i\in p_iAp_{i_0}$ with $v_i^*v_i=p_{i_0}$, $v_iv_i^*=p_i$, and take $v_{i_0}=p_{i_0}$. Then $e_{ij}=v_iv_j^*$ satisfy (3.1) and span every corner inside $Az_E$. They are linearly independent, since multiplication by $p_i$ and $p_j$ extracts a single coefficient. The map sending standard matrix units to these $e_{ij}$ is a *-isomorphism $M_{|E|}(\mathbb C)\to Az_E$. This proves (3.2).

For uniqueness, the centre of (3.2) is $\bigoplus_r\mathbb C1_{n_r}$. Its minimal nonzero projections are exactly the block identities. A *-isomorphism carries the centre and these projections to those of the other algebra, and consequently permutes the blocks. A block of size $n_r$ has vector-space dimension $n_r^2$, which determines $n_r$. Conversely, matching lists give an isomorphism by permuting and identifying the blocks. $\square$

This also proves that a finite-dimensional algebra is simple exactly when it is one full matrix algebra. Its ideals correspond to subsets of the block list. Minimal projections in a block are the rank-one projections; any two in that block are connected by a partial isometry. More explicitly, if $p,q$ are minimal in $M_n(\mathbb C)$, choose unit vectors $\xi,\eta$ spanning their ranges. The operator $v\zeta=\langle\zeta,\xi\rangle\eta$ satisfies $v^*v=p,vv^*=q$.

Recovering the blocks from irreducible representations

Adapted and expanded by GPT-6.1 Sol (OpenAI), Ultra, from S. Sundar, Notes on C*-algebras, Section 2.1, arXiv:2505.17456v1, 23 May 2025, licensed CC0. This entire subsection is CC0. Changes: a finite average constructs a normalized faithful state; the intertwiner and bicommutant calculations are written out; a concrete example distinguishes block sizes from GNS multiplicities. The proof uses the supplied TeX source.

Theorem 3.2 can also be proved by asking which irreducible representations distinguish the elements of $A$. This gives a second way to recognize its blocks, and prepares the multiplicity calculation in Section 4.

A faithful finite-dimensional model. Suppose $A\neq0$. Lemma 3.1 supplies its identity. The Gelfand–Naimark theorem, Theorem 7.2 gives a faithful unital representation $\rho:A\to B(L)$. The space $L$ need not be finite-dimensional.

The representation in that theorem is a direct sum of cyclic GNS representations, so it is nondegenerate and therefore unital when $A$ is unital. For every nonzero $a\in A$, faithfulness gives a unit vector $\xi\in L$ with $\rho(a)\xi\neq0$. The sets $$U_\xi=\{a\in A:\|a\|=1,\ \|\rho(a)\xi\|>0\}$$ are relatively open and cover the unit sphere of $A$. That sphere is compact, so finitely many vectors $\xi_1,\ldots,\xi_m$ suffice. Define $$\omega(a)=\frac1m\sum_{j=1}^m \langle\rho(a)\xi_j,\xi_j\rangle.$$ Then $\omega$ is positive and $\omega(1)=1$. Moreover, $$\omega(a^*a)=\frac1m\sum_{j=1}^m\|\rho(a)\xi_j\|^2>0 \qquad(a\neq0).$$ Thus $\omega$ is faithful. This construction avoids a normalization slip in the cited version: its displayed coefficients $2^{-(i-1)}$, starting at $i=1$, sum to two rather than one. A normalized infinite average would instead use $2^{-i}$.

In the GNS construction, Section 5, the null space $N_\omega=\{a:\omega(a^*a)=0\}$ is zero. Consequently $H_\omega$ is the finite-dimensional vector space $A$, with inner product supplied by $\omega$, and $\pi_\omega(a)$ acts by left multiplication. It is faithful: if $\pi_\omega(a)=0$, then its value on the vector represented by $1$ is the vector represented by $a$, so $a=0$.

Keep one copy of each irreducible model. A unital representation on a finite-dimensional Hilbert space splits into a finite orthogonal sum of irreducible representations. Indeed, a proper invariant subspace is reducing: its orthogonal complement is invariant because the represented algebra is closed under adjoints. Splitting and induction on the dimension prove the assertion.

For two nonzero irreducible representations $\pi_i:A\to B(H_i)$ and $\pi_j:A\to B(H_j)$, an intertwiner is an operator $T:H_i\to H_j$ with $T\pi_i(a)=\pi_j(a)T$ for every $a$. If $T\neq0$, then $$T^*T\in\pi_i(A)',\qquad TT^*\in\pi_j(A)'.$$ Schur's lemma, Section 2 makes these operators scalar: $$\begin{gathered} T^*T=\lambda1_{H_i},\qquad TT^*=\lambda1_{H_j},\\ \lambda=\|T\|^2>0. \end{gathered}$$ Hence $T/\sqrt{\lambda}$ is an intertwining unitary. In particular, inequivalent irreducible representations have no nonzero intertwiner.

Decompose $\pi_\omega$ into irreducibles and retain one representative $\pi_r$ from each equivalence class. The resulting representation $$\pi=\bigoplus_{r=1}^s\pi_r \quad\hbox{on}\quad H=\bigoplus_{r=1}^s H_r$$ is still faithful. Equivalent representations have the same kernel, so removing repeated copies does not change the intersection of the kernels.

Read the two commutants. Write an operator on $H$ in blocks $T_{rt}:H_t\to H_r$. It commutes with $\pi(A)$ exactly when every block intertwines $\pi_t$ and $\pi_r$. Schur's lemma and the absence of intertwiners between distinct classes give $$\pi(A)'= \{\operatorname{diag}(\lambda_1 1_{H_1},\ldots, \lambda_s 1_{H_s}):\lambda_r\in\mathbb C\}.$$ This algebra contains the projection onto each $H_r$. Commuting with those projections forces an operator in $\pi(A)''$ to be block diagonal. Every such operator commutes with all the displayed scalar blocks. Therefore $$\pi(A)''=\bigoplus_{r=1}^s B(H_r).$$ Since $H$ is finite-dimensional, strong operator convergence is entrywise matrix convergence. The finite-dimensional subspace $\pi(A)$ is therefore strongly closed. It is unital, so the double commutant theorem, Theorem 4.4 yields $$A\cong\pi(A)=\pi(A)'' \cong\bigoplus_{r=1}^sM_{\dim H_r}(\mathbb C).$$ The uniqueness argument in Theorem 3.2 applies to these block sizes as well. The zero algebra is the empty direct sum.

Example. For $A=M_2(\mathbb C)\oplus\mathbb C$, take $$\omega(a,c)=\tfrac14\operatorname{Tr}(a)+\tfrac12c.$$ This is a faithful state. Its GNS space has dimension five, the vector-space dimension of $A$. Left multiplication on $M_2$ acts independently on its two columns, so that GNS representation contains two copies of the two-dimensional defining representation, together with the one-dimensional scalar representation. Keeping one copy of each gives a faithful representation on $\mathbb C^2\oplus\mathbb C$, of dimension three. The intrinsic block sizes are $2,1$; the repeated GNS copy records a multiplicity, which Section 4 treats separately.

States and mixtures on a matrix block

Adapted by GPT-6.1 Sol (OpenAI) from Klaas Landsman, Foundations of Quantum Theory: From Classical Concepts to Operator Algebras, Springer, 2017, Theorem 2.7, Lemma 2.11 and Proposition 2.14, pp. 44–48. © The Author(s) 2017. This entire subsection, including its proofs and example, is licensed under Creative Commons Attribution 4.0 International. AI changes: matrix notation, expanded positivity and support arguments, and placement after the matrix-block construction. The finite-dimensional hypothesis is retained.

Fix $n\geq1$. A density matrix is a positive matrix $\rho\in M_n(\mathbb C)$ with $\operatorname{Tr}\rho=1$. A state is a positive linear functional taking the identity to one. A state is pure when it is extreme in the convex state space.

Proposition. Every state of $M_n(\mathbb C)$ has a unique expression $$\varphi_\rho(a)=\operatorname{Tr}(\rho a), \qquad \rho\geq0,\quad \operatorname{Tr}\rho=1.$$ This is an affine correspondence. Its extreme points are exactly the rank-one density matrices, or equivalently the vector states associated to unit vectors.

Proof. The trace pairing is nondegenerate: pairing a matrix with the matrix units reads all its entries. Hence every linear functional has a unique representing matrix $\rho$. For any vector $v$, positivity gives $$v^*\rho v=\varphi_\rho(vv^*)\geq0.$$ Polarization implies that $\rho$ is self-adjoint, and the displayed inequality then makes it positive. Normalization gives $\operatorname{Tr}\rho=\varphi_\rho(1)=1$. Conversely, diagonalizing a positive trace-one matrix expresses $\varphi_\rho$ as a positive weighted sum of unit vector states. It is therefore a state. Linearity and uniqueness of the pairing prove the affine assertion.

Let $P_u=uu^*$ for a unit vector $u$. Suppose $P_u=t\rho_1+(1-t)\rho_2$, where $0<t<1$ and both $\rho_i$ are density matrices. For $v\perp u$, the nonnegative quadratic forms $v^*\rho_i v$ have a weighted sum of zero, so both vanish. Since $v^*\rho_i v=\|\rho_i^{1/2}v\|^2$, each $\rho_i$ annihilates $u^\perp$. Self-adjointness puts its range in $\mathbb Cu$, and trace one forces $\rho_i=P_u$. Thus $P_u$ is extreme.

If instead $\rho$ has at least two positive eigenvalues $\lambda_1,\lambda_2$, choose orthonormal eigenvectors $u_1,u_2$ and $0<\varepsilon<\min(\lambda_1,\lambda_2)$. The two distinct matrices $$\rho_\pm=\rho\pm\varepsilon(P_{u_1}-P_{u_2})$$ are positive and have trace one, with midpoint $\rho$. Hence $\rho$ is not extreme. $\square$

Corollary. Every density matrix is a mixture of at most $n$ pure states: $$\begin{gathered} \rho=\sum_{j=1}^{m}\lambda_jP_{u_j}, \qquad m=\operatorname{rank}\rho\leq n,\\ \lambda_j>0,\qquad\sum_j\lambda_j=1. \end{gathered}$$ where the $u_j$ are orthonormal. Among such orthogonal decompositions, the projections and weights are unique up to order exactly when all positive eigenvalues are simple.

Proof. The spectral theorem gives the displayed decomposition, after omitting zero eigenvalues. In any orthogonal pure-state decomposition, its vectors are eigenvectors with the corresponding weights as eigenvalues. Simple positive eigenvalues determine their one-dimensional eigenspaces. A repeated positive eigenvalue admits different orthonormal bases in its eigenspace; rotating two basis vectors changes their rank-one projections without changing $\rho$. $\square$

Example. Even a simple positive spectrum does not ensure uniqueness among all pure-state mixtures. In $M_2(\mathbb C)$, let $$\rho=\begin{pmatrix}2/3&0\\0&1/3\end{pmatrix}, \qquad u_\pm=\begin{pmatrix}\sqrt{2/3}\\\pm\sqrt{1/3}\end{pmatrix}.$$ Both vectors are unit vectors, and direct multiplication shows $$\rho=\tfrac23P_{e_1}+\tfrac13P_{e_2} =\tfrac12P_{u_+}+\tfrac12P_{u_-}.$$ In the second mixture the off-diagonal entries cancel. Its vectors are not orthogonal, since $u_+^*u_-=1/3$. Thus the spectral mixture is unique among orthogonal mixtures, while another mixture still represents the same state.

4. Representations and multiplicity spaces

Theorem 4.1. Every nondegenerate representation of $M_n(\mathbb C)$ has the form $$\pi(a)=a\otimes1_K\quad\hbox{on }\mathbb C^n\otimes K \tag{4.1}$$ up to unitary equivalence, for a Hilbert space $K$ of arbitrary dimension. Its unitary-equivalence class is determined by the cardinal $m=\dim K$. For any minimal projection $p$, $\dim\pi(p)H=m$, and $\dim H=n\cdot m$. Every cardinal $m$, including zero, occurs. The nonzero representation is irreducible exactly when $m=1$.

Proof. Let $(e_{ij})$ be the standard matrix units and set $K=\pi(e_{11})H$. Nondegeneracy gives $\pi(1)=1_H$. Define $$U:\mathbb C^n\otimes K\longrightarrow H, \qquad U(\varepsilon_i\otimes\xi)=\pi(e_{i1})\xi.$$ The relations (3.1) show that $$\langle\pi(e_{i1})\xi,\pi(e_{j1})\eta\rangle =\delta_{ij}\langle\xi,\eta\rangle.$$ Thus $U$ is isometric. Its range contains each $\pi(e_{ii})H$, because $\pi(e_{i1})\pi(e_{1i})=\pi(e_{ii})$; these ranges sum to $H$. Hence $U$ is unitary. Checking on the matrix units gives $U^*\pi(a)U=a\otimes1_K$.

Two Hilbert spaces $K$ of the same dimension are unitarily isomorphic; tensoring that unitary with $1_{\mathbb C^n}$ intertwines the representations. Conversely, an intertwining unitary carries the range of $\pi(e_{11})$ onto the corresponding range, so their dimensions agree. A partial isometry connecting any two minimal projections similarly identifies their ranges under $\pi$. Any cardinal occurs by taking $K=\ell^2(J)$ for a set $J$ of that cardinality.

Finally, the commutant computation below shows that a nonzero representation has scalar commutant exactly when $\dim K=1$. For a self-adjoint algebra, invariant closed subspaces correspond to projections in the commutant; a non-scalar $B(K)$ has a nontrivial projection. This proves the irreducibility assertion. $\square$

Proposition 4.2. The commutant in (4.1) is $$(M_n(\mathbb C)\otimes1_K)'=1_{\mathbb C^n}\otimes B(K). \tag{4.2}$$ If $m=\dim K<\infty$, it is $M_m(\mathbb C)$, represented with multiplicity $n$.

Proof. Write an operator $T\in B(\mathbb C^n\otimes K)$ as an $n\times n$ matrix of operators on $K$. Commutation with all $e_{ii}\otimes1$ makes the off-diagonal entries zero. Commutation with $e_{ij}\otimes1$ makes all diagonal entries equal. This is exactly $T=1\otimes S$, with arbitrary $S\in B(K)$. When $K=\mathbb C^m$, the flip unitary identifies its action with $S\otimes1_{\mathbb C^n}$, of multiplicity $n$. $\square$

Corollary 4.3. If $A=\bigoplus_{r=1}^sM_{n_r}(\mathbb C)$, every nondegenerate representation is $$H=\bigoplus_{r=1}^s(\mathbb C^{n_r}\otimes K_r), \qquad \pi((a_r))=\bigoplus_{r=1}^s(a_r\otimes1_{K_r}). \tag{4.3}$$ The list $m_r=\dim K_r$, with zero entries allowed, determines it up to unitary equivalence, and every such list occurs. Its commutant is $$\pi(A)'=\bigoplus_{r:m_r\neq0}(1_{n_r}\otimes B(K_r)). \tag{4.4}$$ In finite-dimensional $H$, this is $\bigoplus_{r:m_r>0}M_{m_r}(\mathbb C)$; its identity representation has multiplicities $n_r$.

Proof. The central block identities $z_r$ give orthogonal projections $\pi(z_r)$ summing to $1_H$. Restrict to $H_r=\pi(z_r)H$ and apply Theorem 4.1. An intertwiner preserves these projections, so the classification is coordinatewise. An operator in the commutant also preserves them; Proposition 4.2 on each block proves (4.4). $\square$

For a possibly degenerate representation, add the null summand $H_0=(1-\pi(1))H$, on which $A$ acts as zero. Its cardinal dimension must also be recorded. The full commutant is the direct sum of (4.4) and $B(H_0)$: commutation with $\pi(1)$ prohibits off-diagonal operators between the essential and null summands. Merely recording the nonzero block multiplicities would miss this extra datum.

5. Three tests for finite dimension

Theorem 5.1. A C*-algebra is finite-dimensional if and only if it has a finite-dimensional maximal abelian self-adjoint subalgebra.

Proof. A finite-dimensional algebra has such a subalgebra by maximality, or by the block diagonals in (3.2). Conversely, let $D=\bigoplus_{i=1}^N\mathbb Cp_i\subseteq A$ be finite-dimensional and maximal abelian, with identity $e=\sum_i p_i$. We must first prove $e$ is an identity for $A$.

Every self-adjoint element of $(1-e)A(1-e)$, computed in the unitization, commutes with $D$, so maximality places it in $D$; its multiplication by $e$ is zero, so it is zero. Thus this corner vanishes. If $x\in eA(1-e)$, then $x^*x\in(1-e)A(1-e)=0$, hence $x=0$. Taking adjoints kills the other off-diagonal corner. Therefore $A=eAe$.

As in the proof of Theorem 3.2, maximality gives $p_iAp_i=\mathbb Cp_i$. A nonzero $p_iAp_j$ is one-dimensional by the partial-isometry argument there. The decomposition $A=\sum_{i,j}p_iAp_j$ now gives $\dim A\leq N^2$. $\square$

Theorem 5.2. A C*-algebra is reflexive as a Banach space if and only if it is finite-dimensional.

Proof. Finite-dimensional Banach spaces are reflexive. Suppose $A$ is infinite-dimensional, and choose a maximal abelian self-adjoint subalgebra $D$ using the maximal principle. By Theorem 5.1, $D=C_0(X)$ is infinite-dimensional, so $X$ is infinite.

There are countably many pairwise disjoint nonempty open subsets of $X$, each containing a compactly supported continuous bump. Here is the topological detail. If $X$ is discrete, choose distinct points. Otherwise fix a nonisolated point $t$. In a neighbourhood of $t$, choose a different point, then a relatively compact open neighbourhood of that point whose closure avoids a smaller neighbourhood of $t$. Repeat inside that smaller neighbourhood. Every neighbourhood of a nonisolated point in a Hausdorff space is infinite, so the construction continues. The resulting neighbourhoods are disjoint. If $X$ has infinitely many isolated points, choosing them already suffices; this also covers a discrete open part without changing the argument.

Choose bumps $f_k\geq0$, with $\|f_k\|=1$, in these disjoint open sets. For finite scalar sequences, $$\Big\|\sum_k c_kf_k\Big\|=\max_k|c_k|.$$ Uniform convergence extends this to an isometric embedding of $c_0$ as a closed subspace of $D\subseteq A$.

For completeness, $c_0$ is not reflexive: the vectors $s_N=(1,\ldots,1,0,\ldots)$, with $N$ initial ones, lie in its unit ball. Any weakly convergent subnet would have all coordinates equal to one, since coordinate evaluation is a continuous functional. Such a limit does not belong to $c_0$; hence its unit ball is not weakly compact. The unit ball of a reflexive Banach space is weakly compact by Banach–Alaoglu under the identification with its bidual. A closed linear subspace inherits this property: Hahn–Banach identifies its weak topology with the induced topology, and separates it from points outside it, making it weakly closed. Thus a reflexive $A$ cannot contain this closed copy of $c_0$. $\square$

Theorem 5.3. A nonzero unital simple C*-algebra containing a minimal projection is a full matrix algebra of finite size.

Proof. Let $pAp=\mathbb Cp$, with $p\neq0$. The algebraic two-sided ideal $$J=\operatorname{span}\{apb:a,b\in A\}$$ has closure a nonzero closed ideal, hence $\overline J=A$. Choose $t\in J$ with $\|1-t\|<1$. The Neumann series makes $t$ invertible. Since $J$ is an algebraic ideal, $1=t^{-1}t\in J$. Thus there are finitely many $u_i,v_i\in A$ with $1=\sum_{i=1}^N u_ipv_i$. For any $x\in A$, $$x=\sum_{i,j=1}^N u_i(pv_ixu_jp)v_j =\sum_{i,j=1}^N\lambda_{ij}(x)u_ipv_j,$$ because $pv_ixu_jp\in\mathbb Cp$. This puts $A$ in the span of $N^2$ fixed elements. Theorem 3.2 and simplicity finish the proof. $\square$

Unitality matters here: $K(H)$ is simple and has rank-one minimal projections, but is infinite-dimensional when $H$ is. Simplicity alone makes the algebraic ideal generated by $p$ dense; the invertible approximation to $1$ is what makes its finite spanning argument work.

6. Exercises with complete solutions

Exercise 6.1 — Two separate defects (intermediate). Let $S\varepsilon_k=\varepsilon_{k+1}$ be the unilateral shift on $\ell^2(\mathbb N_0)$. Show that $(S,S^*)$ is extreme in the unit ball of $B(H)\oplus B(H)$, but $S\oplus S^*$ is not extreme in the unit ball of $B(H\oplus H)$. Give an explicit nonzero admissible perturbation in the latter algebra.

Solution. Put $e=\theta_{\varepsilon_0,\varepsilon_0}$. For $S$, $p=1,q=1-e$; for $S^*$, $p=1-e,q=1$. The defect corner vanishes in each direct summand, proving extremality there. On $H\oplus H$, however, $$1-p=0\oplus e,\qquad1-q=e\oplus0.$$ Define $h(\xi,\eta)=(\langle\eta,\varepsilon_0\rangle\varepsilon_0,0)$. It is a norm-one element of $(1-q)B(H\oplus H)(1-p)$. The proof of Theorem 2.1 gives $\|(S\oplus S^*)\pm h\|\leq1$, and the midpoint decomposition is nontrivial. The ambient algebra determines which defect directions are available.

Exercise 6.2 — Reading both multiplicity lists (basic). Represent $A=M_3(\mathbb C)\oplus M_2(\mathbb C)$ with multiplicities $2,3$. Determine $\dim H$, the image, its centre, its commutant and the commutant's multiplicities. Then add a four-dimensional null summand and describe what changes.

Solution. The essential representation is on $(\mathbb C^3\otimes\mathbb C^2)\oplus(\mathbb C^2\otimes\mathbb C^3)$, so $\dim H=12$. The image is $(M_3\otimes1_2)\oplus(M_2\otimes1_3)$, faithful and isomorphic to $A$. Its centre is $\mathbb C\oplus\mathbb C$, acting as independent scalars on the two six-dimensional summands. Its commutant is $(1_3\otimes M_2)\oplus(1_2\otimes M_3)\cong M_2\oplus M_3$, with multiplicities $3,2$. Adding $H_0=\mathbb C^4$ raises the Hilbert-space dimension to $16$. The image is zero on $H_0$, and its abstract algebra and centre remain as before. The full commutant gains the summand $M_4$, whose identity representation has multiplicity one. The image identity is now the projection onto the twelve-dimensional essential summand.

Exercise 6.3 — A finite abelian corner controls everything (advanced). Suppose $A$ has a maximal abelian subalgebra of dimension $N\geq1$. Prove $\dim A\leq N^2$, identify all cases of equality, and give the smallest possible dimension of $A$.

Solution. Theorem 5.1 gives $N$ minimal diagonal projections and at most one dimension for every corner, so $\dim A\leq N^2$. In the equivalence-class construction of Theorem 3.2 let the class sizes be $n_1,\ldots,n_s$. Then $\sum_r n_r=N$ and $\dim A=\sum_r n_r^2$. Equality with $N^2$ forces one class, because $N^2-\sum_r n_r^2=2\sum_{r<t}n_rn_t$. Hence equality holds exactly for $M_N(\mathbb C)$. The smallest value is $N$, since $n_r^2\geq n_r$, with equality exactly when every class has size one; this is $\mathbb C^N$.

7. What the axioms and quotients make possible

The finite-dimensional classification gives a concrete model for every algebra in this lesson. In general, choosing a Hilbert-space model is itself part of the problem. The abstract definition keeps multiplication, involution and norm available before a representation has been chosen. One can then select a representation adapted to a state, an ideal or a symmetry. The broader Banach-algebra tools used in these prerequisites are proved in Banach algebras, spectrum and holomorphic functional calculus: Section 4 treats spectra, Section 6 constructs the holomorphic functional calculus, and Section 9 treats modular ideals and quotient algebras.

Quotients show why this freedom matters. A closed two-sided ideal $I$ in a C*-algebra is self-adjoint, and the quotient norm and induced involution make $A/I$ a C*-algebra. Section 15 of C*-algebras and continuous functional calculus proves this. Applied to $I=K(H)\subset B(H)$, it produces the Calkin algebra. Two operators have the same quotient image exactly when their difference is compact. Thus invertibility of that image is a property of an operator modulo compact perturbations. Fredholm operators and the stable index derives its kernel, range and component consequences directly from the quotient. Those arguments do not require a faithful Hilbert-space model of the quotient. General representation theorems supply such models through states and the maximal principle, without singling out a preferred model for every quotient.

The historical development explains the order of the prerequisites. Nagumo's 1936 paper [Nagumo] studies normed algebras as linear metric rings, connecting the invertible group to topological-group questions associated with Hilbert's fifth problem. Gelfand's 1941 paper [Gelfand] establishes the spectral viewpoint. Nonemptiness of the spectrum implies that a complex Banach division algebra consists of scalars: for $a$, choose $\lambda\in\sigma(a)$; the noninvertible element $a-\lambda1$ must then be zero. This is the Gelfand–Mazur theorem, announced earlier by Mazur [Mazur]. Section 5 of Banach algebras, spectrum and holomorphic functional calculus proves both results. The topology of the invertible group continues in Invertible components and exponential laws.

Gelfand and Naimark [Gelfand–Naimark] formulated the abstract representation problem with an additional requirement that $1+x^*x$ be invertible. The older term B*-algebra distinguished the involutive Banach-algebra axioms from a concrete algebra of operators. With the modern C*-identity, positivity of $x^*x$ makes the extra requirement automatic: its spectrum is nonnegative, so the spectrum of $1+x^*x$ avoids zero. [Blackadar, II.3.1.4] distinguishes the historical steps: he credits Fukamiya and Kelley–Vaught with the closed convex positive cone, and Kaplansky with positivity of $x^*x$. [Fukamiya–Misonou–Takeda] concerns order and commutativity. Section 8 of the C*-algebra prerequisite supplies the complete positive-cone argument used here.

Order brings further structure. The Löwner–Heinz inequality says that $0\leq b\leq a$ implies $b^\alpha\leq a^\alpha$ for $0<\alpha\leq1$. Its matrix and operator history is represented by [Löwner] and [Heinz]; [Pedersen powers] gives the proof route discussed in the Notes. Section 9 of the same prerequisite proves the inequality. Approximate identities, associated here with Segal [Segal representations], let nonunital algebras approximate the action of an identity. Their norm convergence in finite dimensions supplies the units in Lemma 3.1. Pedersen's asymmetric Riesz decomposition [Pedersen decomposition] refines an equality between two sums of elements of the form $x^*x$ into a matrix of compatible terms; Section 14 proves a version for arbitrary families whose positive sums converge in norm. This differs from scalar lattice decomposition, which imposes strong commutativity restrictions.

The closed-ideal and quotient theorem is due to Kaplansky [Kaplansky] and Segal (1949); see also [Blackadar, II.5.1.1]. GNS, faithful representation, and detection by irreducible representations are the next steps. A positive functional gives a cyclic Hilbert-space model; sufficiently many such models give a faithful one; pure states select irreducible models. The classical references are [Gelfand–Naimark; Segal representations; Gelfand–Raikov]. Sections 5, 7 and 8 of Representations and positive functionals supply the proofs used in this course. A homomorphism between C*-algebras that preserves the involution is contractive. Theorem 4.2 of the C*-algebra prerequisite supplies the proof without assuming continuity in advance. The involution-preserving hypothesis is part of that statement.

Finally, Kadison's work (1951) connects the geometry of the unit ball with algebraic structure [Blackadar, II.3.2.19]. Theorem 2.1 makes this visible: a partial isometry is extreme precisely when its two defects leave no available corner for a perturbation. Matrix blocks make those corners explicit, and multiplicity spaces explain what changes when the same algebra acts on a larger Hilbert space. The abstract, geometric and spatial descriptions can be used together.

The two classification arguments connect these tools in different ways. Minimal diagonal projections give matrix coordinates inside the algebra. A faithful state gives a Hilbert-space model; irreducible pieces and their intertwiners then recover the same blocks. The first construction makes individual corners explicit, while the second shows how states, representations and commutants cooperate. The complete proofs above can be read alongside the freely accessible [Blackadar] and [Sundar] accounts.

References

[Blackadar] Bruce Blackadar, Operator Algebras: Theory of C-Algebras and von Neumann Algebras*, author's revised and corrected online version of the 2005 book, accessed 3 October 2026.

[Fukamiya] M. Fukamiya, “On a theorem of Gelfand and Neumark and the B*-algebra,” Kumamoto Journal of Science, Series A 1 (1952), 17–22.

[Fukamiya–Misonou–Takeda] M. Fukamiya, Y. Misonou and Z. Takeda, “On order and commutativity of B*-algebras,” Tôhoku Mathematical Journal (2) 6 (1954), 89–93.

[Gelfand] I. M. Gelfand, “Normierte Ringe,” Matematicheskii Sbornik (N.S.) 9(51) (1941), 3–24.

[Gelfand–Naimark] I. Gelfand and M. Naimark, “On the imbedding of normed rings into the ring of operators in Hilbert space,” Matematicheskii Sbornik (N.S.) 12(54) (1943), 197–217.

[Gelfand–Raikov] I. M. Gelfand and D. A. Raikov, “Irreducible unitary representations of locally bicompact groups,” Matematicheskii Sbornik (N.S.) 13(55) (1943), 301–316.

[Heinz] E. Heinz, “Beiträge zur Störungstheorie der Spektralzerlegung,” Mathematische Annalen 123 (1951), 415–438.

[Kaplansky] I. Kaplansky, “A theorem on rings of operators,” Pacific Journal of Mathematics 1 (1951), 227–232.

[Kelley–Vaught] J. L. Kelley and R. L. Vaught, “The positive cone in Banach algebras,” Transactions of the American Mathematical Society 74 (1953), 44–55.

[Landsman] Klaas Landsman, Foundations of Quantum Theory: From Classical Concepts to Operator Algebras, Springer, 2017, Sections 2.2–2.3, pp. 44–48. The credited state-and-mixture subsection above is adapted under CC BY 4.0.

[Löwner] K. Löwner, “Über monotone Matrixfunktionen,” Mathematische Zeitschrift 38 (1934), 177–216.

[Mazur] S. Mazur, “Sur les anneaux linéaires,” Comptes Rendus de l'Académie des Sciences, Paris 207 (1938), 1025–1027.

[Nagumo] M. Nagumo, “Einige analytische Untersuchungen in linearen, metrischen Ringen,” Japanese Journal of Mathematics 13 (1936), 61–80.

[Pedersen decomposition] G. K. Pedersen, “A decomposition theorem for C*-algebras,” Mathematica Scandinavica 22 (1968), 266–268.

[Pedersen powers] G. K. Pedersen, “Some operator monotone functions,” Proceedings of the American Mathematical Society 36 (1972), 309–310.

[Segal representations] I. E. Segal, “Irreducible representations of operator algebras,” Bulletin of the American Mathematical Society 53 (1947), 73–88.

[Sundar] S. Sundar, Notes on C*-algebras, arXiv:2505.17456v1, 23 May 2025. Author-supplied TeX; CC0 licence.

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