Motivating Noncommutative Geometry: Gelfand Duality and the Slogan 'Space = Algebra of Functions'

0. Key points

• The topology of a locally compact Hausdorff space $X$ is encoded, without any loss, in the commutative C*-algebra $C_0(X)$ of continuous functions on $X$ that vanish at infinity. Conversely, every commutative C*-algebra is of this form. This is the Gelfand–Naimark theorem.
• The correspondence is a contravariant equivalence of categories, so that a complete dictionary between spaces and algebras becomes available: compactness is the existence of a unit, connectedness is the absence of nontrivial idempotents, points are maximal ideals, vector bundles are finitely generated projective modules, and so on.
• If “commutative C*-algebra” and “locally compact Hausdorff space” are synonyms, then a C*-algebra without the commutativity assumption deserves to be called a noncommutative space. This is the starting point of noncommutative geometry.
• Two situations force the generalisation upon us. In quantum theory position and momentum fail to commute, so the algebra of functions on phase space is replaced by a noncommutative algebra; and for quotients, a “bad space” such as an orbit space collapses as a set of points, while surviving as a rich object in the form of a crossed product C*-algebra.
• The central example is the orbit space of the circle under an irrational rotation. Its quotient topology is the indiscrete one and the only continuous functions on it are the constants, yet the corresponding noncommutative torus $A_\theta$ is a simple infinite-dimensional C*-algebra whose K-theory recovers the angle $\theta$.
• In the world of bounded operators the relation $ab - ba = \mathbf{1}$ never holds. To treat the commutation relations inside a C*-algebra one needs the Weyl form, and the resulting relation is exactly the defining relation of the noncommutative torus.

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1. Motivation: from space as a set of points to space as an algebra of functions

1.1. The idea of a coordinate ring

Since Descartes, a space has been a set of points and geometry has been the study of the relations among them. The mathematics of the twentieth century, however, rediscovered the opposite viewpoint again and again: take as the primary object not the space itself, but the ring of functions on it.

In algebraic geometry this shift was completed earliest. Assigning to an affine algebraic variety $V$ over a field $k$ its coordinate ring $k[V]$ makes the points of $V$ correspond to the maximal ideals of $k[V]$ and the morphisms of varieties to ring homomorphisms, with all arrows reversed (Hilbert’s Nullstellensatz). Grothendieck pushed this to its limit and regarded the set $\operatorname{Spec} R$ of prime ideals of any commutative ring $R$ as a space. Rings and spaces are no longer distinguished (Ideals and quotient rings). Measure theory behaves in the same way: assigning $L^\infty(X,\mu)$ to $(X,\mu)$ translates measurable sets into idempotents and the measure into a linear functional (L^p spaces and an introduction to functional analysis).

What, then, about topology? It is natural to assign to a topological space $X$ the ring $C(X)$ of complex-valued continuous functions on it. The question is how faithful this assignment is, that is, whether $X$ can be recovered from $C(X)$. The answer is that it can, provided one carries along a suitable norm and involution, and this is the theorem established by Israel Gelfand and Mark Naimark in 1943. If $X$ is compact Hausdorff, the single piece of data $C(X)$ determines $X$ up to homeomorphism.

1.2. Two difficulties: quantum theory and bad quotients

Read the Gelfand–Naimark theorem as the equation “space $=$ commutative C*-algebra” and one question presents itself at once: what happens if commutativity is dropped?

The question does not arise out of curiosity alone. Rather, two places where existing mathematics runs aground both point towards noncommutative algebras.

The first is quantum mechanics. On the phase space of classical mechanics the observables are functions and their product is commutative. In quantum mechanics, however, position $q$ and momentum $p$ satisfy $qp - pq = i\hbar$ and do not commute. Heisenberg’s insight of 1925, that observables are matrices rather than numbers, means precisely that the algebra of functions on phase space has been replaced by a noncommutative algebra. If commutative C*-algebras are spaces, then the algebra of observables of quantum mechanics is something that is not a space: a phase space without points.

The second is quotient spaces. Dividing by an equivalence relation to form $X/\sim$ occurs everywhere in geometry: orbit spaces of group actions, leaf spaces of foliations, classes of tilings modulo translation. Yet such quotients can be destroyed by the quotient topology. As we shall see in Example 5.1, the circle divided by an irrational rotation is indistinguishable, as a topological space, from a single point.

Alain Connes’s proposal is clear. Stop forming the quotient before forming the algebra of functions, and encode the equivalence relation itself into an algebra. The resulting algebra is noncommutative, but for that very reason it loses no information.

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2. Preliminaries: Banach algebras and C*-algebras

All algebras below are over the field of complex numbers. We write $\mathbb{T} = \{z \in \mathbb{C} : |z| = 1\}$ and $\mathbb{N} = \{1, 2, \ldots\}$. The algebra of all bounded linear operators on a Hilbert space $H$ is denoted $B(H)$.

Definition 2.1（Banach *-algebra）

A complex algebra $A$ equipped with a norm $\|\cdot\|$ which is complete for that norm and satisfies

$$\|ab\| \le \|a\|\,\|b\|$$

for all $a, b \in A$ is called a Banach algebra. If $A$ has a multiplicative unit $\mathbf{1}$ with $\|\mathbf{1}\| = 1$, then $A$ is called unital.

If in addition a map $*: A \to A$ satisfies

$$(a+b)^* = a^* + b^*, \quad (\lambda a)^* = \bar{\lambda} a^*, \quad (ab)^* = b^* a^*, \quad (a^*)^* = a$$

for all $a, b \in A$ and all $\lambda \in \mathbb{C}$, then $*$ is called an involution, and a Banach algebra equipped with an involution is a *Banach -algebra.

Completeness of the norm is the condition that lets us perform limiting operations inside the algebra (Completeness of the reals and Cauchy sequences, Theorem 7.3［Completeness of the Real Numbers and Cauchy Sequences］).

Definition 2.2（C*-algebra）

A Banach *-algebra $A$ satisfying

$$\|a^* a\| = \|a\|^2$$

for every $a \in A$ is called a C*-algebra, and this identity is the C*-identity.

If $A$ is commutative ($ab = ba$ for all $a, b$), then $A$ is called a commutative C*-algebra.

The C*-identity looks like a small extra condition, but it holds the key to everything. To begin with, this single equation forces the involution to be isometric.

Remark 2.3

In a C*-algebra one has $\|a^*\| = \|a\|$. Indeed, the C*-identity together with submultiplicativity gives

$$\|a\|^2 = \|a^* a\| \le \|a^*\|\,\|a\| ,$$

so if $a \ne 0$ we may divide both sides by $\|a\|$ and obtain $\|a\| \le \|a^*\|$ (for $a = 0$ both sides vanish). Replacing $a$ by $a^*$ yields $\|a^*\| \le \|a^{**}\| = \|a\|$, and the two inequalities give equality. Moreover, as Proposition 3.5 shows, the norm is recovered from the spectral radius, so the norm of a C*-algebra is uniquely determined by the algebraic structure: the object is in effect purely algebraic.

Example 2.4（The algebra of continuous functions C_0(X)）

Let $X$ be a locally compact Hausdorff space. A continuous function $f: X \to \mathbb{C}$ vanishes at infinity if $\{x \in X : |f(x)| \ge \varepsilon\}$ is compact for every $\varepsilon > 0$; the set of all such $f$ is denoted $C_0(X)$. If $X$ is compact the condition holds automatically and $C_0(X) = C(X)$.

Equip $C_0(X)$ with pointwise sum and product, the norm $\|f\|_\infty = \sup_{x \in X} |f(x)|$ and the involution $f^*(x) = \overline{f(x)}$. Completeness follows from the fact that a uniform limit of continuous functions is continuous (Sequences of functions and uniform convergence, Theorem 4.1［関数列と一様収束］). Submultiplicativity follows by taking the supremum of the pointwise inequality $|f(x)g(x)| \le \|f\|_\infty \|g\|_\infty$. The C*-identity is a direct computation:

$$\|f^* f\|_\infty = \sup_{x} |\overline{f(x)} f(x)| = \sup_x |f(x)|^2 = \left(\sup_x |f(x)|\right)^2 = \|f\|_\infty^2 .$$

Since the product is pointwise, $C_0(X)$ is a commutative C*-algebra.

Example 2.5（The operator algebra B(H) and matrix algebras）

Let $H$ be a Hilbert space and let $B(H)$ be the algebra of all bounded linear operators on it, with the operator norm and the adjoint $T \mapsto T^*$. The C*-identity is proved as follows. For every $\xi \in H$,

$$\|T\xi\|^2 = \langle T\xi, T\xi \rangle = \langle T^* T \xi, \xi\rangle \le \|T^* T\|\,\|\xi\|^2$$

by the Cauchy–Schwarz inequality, whence $\|T\|^2 \le \|T^* T\|$. The reverse inequality follows from submultiplicativity and $\|T^*\| = \|T\|$: we get $\|T^*T\| \le \|T^*\|\|T\| = \|T\|^2$.

The case $H = \mathbb{C}^n$ gives $M_n(\mathbb{C})$, a finite-dimensional C*-algebra which is noncommutative as soon as $n \ge 2$. Every norm-closed -subalgebra of $B(H)$ is a C-algebra, and the converse is also true; that is the content of Theorem 4.6.

Finally we introduce the notion at the centre of this article. For $A$ a unital algebra and $a \in A$, the set

$$\sigma(a) = \{\lambda \in \mathbb{C} : \lambda \mathbf{1} - a \text{ is not invertible in } A\}$$

is called the spectrum of $a$. If $A$ is a unital Banach algebra different from $\{0\}$, then $\sigma(a)$ is a nonempty compact set (Lemma 8.5), and

$$r(a) = \max\{|\lambda| : \lambda \in \sigma(a)\}$$

is the spectral radius. For $A = M_n(\mathbb{C})$ the spectrum $\sigma(a)$ is the set of eigenvalues; for $A = C(X)$ one has $\sigma(f) = f(X)$, the range of $f$. The latter holds because $f - \lambda$ is invertible exactly when it vanishes nowhere, in which case $1/(f-\lambda)$ is continuous.

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3. The Gelfand transform

How is a space to be manufactured out of a commutative C*-algebra $A$? Let the case $A = C(X)$ guide us. A point $x \in X$ determines the evaluation map

$$\mathrm{ev}_x : C(X) \to \mathbb{C}, \qquad f \mapsto f(x) ,$$

which is a nonzero algebra homomorphism. Conversely, if we can show that every nonzero algebra homomorphism $C(X) \to \mathbb{C}$ is an evaluation map, then $X$ can be reconstructed in purely algebraic terms. That is the plot of Gelfand theory.

Definition 3.1（Characters and the Gelfand spectrum）

Let $A$ be a commutative Banach algebra. A map $\chi: A \to \mathbb{C}$ which is linear, satisfies $\chi(ab) = \chi(a)\chi(b)$ and is not identically zero is called a character of $A$, and the set $\Omega(A)$ of all characters is the Gelfand spectrum (or character space) of $A$.

We give $\Omega(A)$ the topology induced by the weak * topology of the dual space $A^*$: thus $\chi_i \to \chi$ means $\chi_i(a) \to \chi(a)$ for every $a \in A$.

Lemma 3.2（Basic properties of characters）

Let $A$ be a unital commutative Banach algebra. Then the following hold.

1. Every $\chi \in \Omega(A)$ satisfies $\chi(\mathbf{1}) = 1$, and $\chi$ is continuous with $\|\chi\| \le 1$.
2. $\Omega(A)$ is a compact Hausdorff space in the weak * topology.

Proof（Lemma 3.2）

(1) Since $\chi \ne 0$ there is an $a$ with $\chi(a) \ne 0$, and $\chi(a) = \chi(a\mathbf{1}) = \chi(a)\chi(\mathbf{1})$ gives $\chi(\mathbf{1}) = 1$. Next, if $a$ is invertible then $\chi(a)\chi(a^{-1}) = \chi(\mathbf{1}) = 1$, so $\chi(a) \ne 0$. In other words, a character never sends an invertible element to $0$.

Suppose now that $|\chi(a)| > \|a\|$ for some $a$. Put $\lambda = \chi(a)$; then $\|a/\lambda\| < 1$, so by completeness of $A$ the Neumann series $\sum_{n \ge 0} (a/\lambda)^n$ converges absolutely and its sum is $(\mathbf{1} - a/\lambda)^{-1}$. Hence $\lambda\mathbf{1} - a$ is invertible, while $\chi(\lambda \mathbf{1} - a) = \lambda - \chi(a) = 0$, contradicting what we have just proved. Therefore $|\chi(a)| \le \|a\|$ for every $a$, that is, $\chi$ is continuous with $\|\chi\| \le 1$.

(2) By (1), $\Omega(A)$ is contained in the closed unit ball $B$ of $A^*$, and $B$ is weak * compact by the Banach–Alaoglu theorem. It therefore suffices to prove that $\Omega(A)$ is weak * closed in $B$.

Let $\varphi \in B$ lie in the weak * closure of $\Omega(A)$. By the definition of the weak * topology, given $a, b \in A$ and $\varepsilon > 0$ there is a $\chi \in \Omega(A)$ for which $|\varphi(a) - \chi(a)|$, $|\varphi(b) - \chi(b)|$, $|\varphi(ab) - \chi(ab)|$ and $|\varphi(\mathbf{1}) - \chi(\mathbf{1})|$ are all less than $\varepsilon$. Using $\chi(ab) = \chi(a)\chi(b)$ and $\chi(\mathbf 1) = 1$ and letting $\varepsilon \to 0$, we obtain $\varphi(ab) = \varphi(a)\varphi(b)$ and $\varphi(\mathbf{1}) = 1$; in particular $\varphi \ne 0$, so $\varphi \in \Omega(A)$. The Hausdorff property holds because for $\varphi \ne \psi$ there is an $a$ with $\varphi(a) \ne \psi(a)$, and these values can be separated by open sets of $\mathbb{C}$.

∎

Characters and spectra are two faces of the same thing in the commutative world.

Lemma 3.3（Description of the spectrum by characters）

Let $A$ be a unital commutative Banach algebra and $a \in A$. Then

$$\sigma(a) = \{\chi(a) : \chi \in \Omega(A)\} .$$

Proof（Lemma 3.3）

($\supseteq$) Let $\chi \in \Omega(A)$ and put $\lambda = \chi(a)$, so that $\chi(\lambda\mathbf{1} - a) = 0$. As shown in the proof of Lemma 3.2, a character never sends an invertible element to $0$. Hence $\lambda\mathbf{1} - a$ is not invertible, that is, $\lambda \in \sigma(a)$.

($\subseteq$) Let $\lambda \in \sigma(a)$ and set $b = \lambda\mathbf{1} - a$, a non-invertible element. Since $A$ is commutative, $I = bA$ is an ideal. If $\mathbf{1} \in I$, then $\mathbf{1} = bc$ for some $c$, and commutativity gives $cb = bc = \mathbf 1$, making $b$ invertible, a contradiction. Hence $I$ is a proper ideal.

By Zorn’s lemma there is a maximal ideal $M$ containing $I$. By Lemma 8.8 the ideal $M$ is closed, so the quotient $A/M$ is a unital commutative Banach algebra, and maximality of $M$ makes $A/M$ a field (Ideals and quotient rings, Theorem 6.3［イデアルと剰余環］). By the Gelfand–Mazur theorem (Theorem 8.7) we get $A/M \cong \mathbb{C}$.

Writing $\chi$ for the quotient map $A \to A/M \cong \mathbb{C}$, this is a nonzero algebra homomorphism, that is, a character, and $b \in I \subseteq M = \ker\chi$ gives $\chi(b) = 0$, that is, $\chi(a) = \lambda$.

∎

Definition 3.4（The Gelfand transform）

Let $A$ be a unital commutative Banach algebra. For $a \in A$ define a function $\hat{a} : \Omega(A) \to \mathbb{C}$ by

$$\hat{a}(\chi) = \chi(a) .$$

By the definition of the weak * topology, $\hat{a}$ is continuous. The map

$$\Gamma : A \to C(\Omega(A)), \qquad \Gamma(a) = \hat{a}$$

is called the Gelfand transform. Since $\widehat{ab}(\chi) = \chi(ab) = \chi(a)\chi(b) = \hat a(\chi)\hat b(\chi)$, the map $\Gamma$ is an algebra homomorphism.

By Lemma 3.3 we have $\hat{a}(\Omega(A)) = \sigma(a)$ and therefore $\|\hat{a}\|_\infty = r(a)$. Thus whether $\Gamma$ is isometric depends on whether $\|a\| = r(a)$. For a general Banach algebra this fails: a nonzero nilpotent matrix in $M_2(\mathbb{C})$ has spectral radius $0$ but nonzero norm. It is here that the C*-identity comes into play.

Proposition 3.5（The norm of a normal element equals its spectral radius）

Let $A$ be a unital C*-algebra and let $a \in A$ be normal, that is, $a^* a = a a^*$. Then

$$\|a\| = r(a) .$$

In particular, if $A$ is commutative then every element is normal, so $\|a\| = r(a)$ for every $a \in A$.

Proof（Proposition 3.5）

First, if $b \in A$ is self-adjoint ($b^* = b$), then the C*-identity gives $\|b^2\| = \|b^* b\| = \|b\|^2$.

Next we show $\|a^2\| = \|a\|^2$ for $a$ normal. Applying the C*-identity to $a^2$,

$$\|a^2\|^2 = \|(a^2)^* a^2\| = \|(a^*)^2 a^2\| ,$$

and since $a$ is normal, $a^*$ and $a$ commute, so $(a^*)^2 a^2 = (a^* a)^2$. As $a^* a$ is self-adjoint, applying what we have just proved with $b = a^*a$ gives

$$\|(a^*a)^2\| = \|a^* a\|^2 = \left(\|a\|^2\right)^2 = \|a\|^4 .$$

Hence $\|a^2\| = \|a\|^2$.

If $a$ is normal then so is $a^2$ (because $a$ and $a^*$ commute), so repeating the argument gives inductively

$$\|a^{2^n}\| = \|a\|^{2^n} \qquad (n = 0, 1, 2, \ldots) .$$

In the spectral radius formula (Lemma 8.9) $r(a) = \lim_{m \to \infty} \|a^m\|^{1/m}$ the limit exists, so it may be computed along the subsequence $m = 2^n$:

$$r(a) = \lim_{n \to \infty} \|a^{2^n}\|^{1/2^n} = \lim_{n \to \infty} \|a\| = \|a\| .$$∎

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4. The Gelfand–Naimark theorem

Theorem 4.1（Gelfand–Naimark theorem (commutative, unital case)）

Let $A$ be a unital commutative C*-algebra. Then $\Omega(A)$ is a compact Hausdorff space and the Gelfand transform

$$\Gamma : A \longrightarrow C(\Omega(A)), \qquad \Gamma(a) = \hat{a}$$

is an isometric *-isomorphism: it is a bijective algebra homomorphism with $\|\Gamma(a)\|_\infty = \|a\|$ and $\Gamma(a^*) = \overline{\Gamma(a)}$.

Proof（Theorem 4.1）

That $\Omega(A)$ is compact Hausdorff is Lemma 3.2, and that $\Gamma$ is an algebra homomorphism was checked in Definition 3.4. We prove the rest in four steps.

Step 1: characters take real values on self-adjoint elements. Let $b \in A$ be self-adjoint, let $\chi \in \Omega(A)$ and write $\chi(b) = \alpha + i\beta$ with $\alpha, \beta \in \mathbb{R}$. For every $t \in \mathbb{R}$,

$$\chi(b + it\mathbf{1}) = \alpha + i(\beta + t)$$

(here we used $\chi(\mathbf 1) = 1$ from Lemma 3.2). The bound $\|\chi\| \le 1$ from the same lemma gives

$$\alpha^2 + (\beta+t)^2 = |\chi(b+it\mathbf{1})|^2 \le \|b + it\mathbf{1}\|^2 .$$

We compute the right-hand side using the C*-identity. Since $b^* = b$ we have $(b + it\mathbf{1})^* = b - it\mathbf{1}$, so

$$\|b+it\mathbf{1}\|^2 = \|(b - it\mathbf{1})(b + it\mathbf{1})\| = \|b^2 + t^2 \mathbf{1}\| \le \|b\|^2 + t^2$$

(the last step uses the triangle inequality together with $\|b^2\| \le \|b\|^2$ and $\|\mathbf 1\| = 1$). Combining the two displays,

$$\alpha^2 + \beta^2 + 2\beta t + t^2 \le \|b\|^2 + t^2, \qquad \text{that is,} \qquad \alpha^2 + \beta^2 + 2\beta t \le \|b\|^2$$

for every $t \in \mathbb{R}$. If $\beta \ne 0$, the left-hand side tends to $+\infty$ as $t \to \pm\infty$, which is absurd. Hence $\beta = 0$, that is, $\chi(b) \in \mathbb{R}$.

*Step 2: $\Gamma$ is a -homomorphism. Decompose an arbitrary $a \in A$ as

$$a = b + ic, \qquad b = \frac{a + a^*}{2}, \quad c = \frac{a - a^*}{2i} ,$$

where both $b$ and $c$ are self-adjoint (indeed $b^* = (a^* + a)/2 = b$ and $c^* = (a^* - a)/(-2i) = c$). By Step 1 we have $\chi(b), \chi(c) \in \mathbb{R}$, so from $a^* = b - ic$,

$$\chi(a^*) = \chi(b) - i\chi(c) = \overline{\chi(b) + i \chi(c)} = \overline{\chi(a)} .$$

That is, $\widehat{a^*} = \overline{\hat{a}}$.

Step 3: $\Gamma$ is isometric. Since $A$ is commutative, all its elements are normal. By Proposition 3.5 and Lemma 3.3,

$$\|\hat{a}\|_\infty = \sup_{\chi \in \Omega(A)} |\chi(a)| = \max\{|\lambda| : \lambda \in \sigma(a)\} = r(a) = \|a\| .$$

In particular $\Gamma$ is injective, and since $A$ is complete the image $\Gamma(A)$ is a closed subset of $C(\Omega(A))$.

Step 4: $\Gamma$ is surjective. The image $\Gamma(A)$ is a subalgebra of $C(\Omega(A))$, closed under complex conjugation by Step 2, and $\Gamma(\mathbf{1})$ is the constant function $1$. Moreover $\Gamma(A)$ separates the points of $\Omega(A)$: if $\chi \ne \psi$, then by definition there is an $a$ with $\chi(a) \ne \psi(a)$, so $\hat{a}(\chi) \ne \hat{a}(\psi)$.

Since $\Omega(A)$ is compact Hausdorff, the Stone–Weierstrass theorem shows that $\Gamma(A)$ is dense. By Step 3 it is closed, so $\Gamma(A) = C(\Omega(A))$.

∎

Remark 4.2

The same conclusion holds for a commutative C*-algebra $A$ without a unit. Then $\Omega(A)$ is a locally compact Hausdorff space, not necessarily compact, and the Gelfand transform gives an isometric *-isomorphism $A \cong C_0(\Omega(A))$. The proof consists in applying the theorem above to the unitisation $\tilde{A} = A \oplus \mathbb{C}\mathbf{1}$ and observing that $\Omega(\tilde A) = \Omega(A) \cup \{\infty\}$ is the one-point compactification.

4.1. Recovering the space, and the equivalence of categories

Theorem 4.1 says that a commutative C*-algebra is necessarily an algebra of functions. Let us now check the converse statement, that the space is recovered from the algebra of functions.

Corollary 4.3（Points are characters）

Let $X$ be a compact Hausdorff space. The map

$$\varepsilon : X \to \Omega(C(X)), \qquad \varepsilon(x) = \mathrm{ev}_x$$

with $\mathrm{ev}_x(f) = f(x)$ is a homeomorphism.

Proof（Corollary 4.3）

Well defined. The operations of $C(X)$ are pointwise, so $\mathrm{ev}_x(fg) = f(x)g(x) = \mathrm{ev}_x(f)\mathrm{ev}_x(g)$, and likewise for sums and scalar multiples; hence $\mathrm{ev}_x$ is an algebra homomorphism. Since $\mathrm{ev}_x(\mathbf{1}) = 1 \ne 0$, it is not the zero map, so it is a character.

Injectivity. Let $x \ne y$. A compact Hausdorff space is normal, so by Urysohn's lemma（Lemma 6.1）［分離公理と距離づけ可能性］ there is an $f \in C(X)$ with $f(x) = 0$ and $f(y) = 1$ (Separation axioms and metrisability), whence $\mathrm{ev}_x \ne \mathrm{ev}_y$.

Surjectivity. Let $\chi \in \Omega(C(X))$, put $M = \ker \chi$, and let us produce a point $x$ at which every $f \in M$ vanishes simultaneously. If no such point existed, then for each $x$ we could choose $f_x \in M$ with $f_x(x) \ne 0$. The set $U_x = \{y : f_x(y) \ne 0\}$ is open and contains $x$, so $\{U_x\}$ is an open cover, and by compactness of $X$ finitely many $U_{x_1}, \ldots, U_{x_n}$ already cover it (Compactness). Put

$$g = \sum_{k=1}^{n} \overline{f_{x_k}} f_{x_k} .$$

Then $g$ is a continuous function vanishing nowhere, that is, an invertible element. But $M$ is an ideal, so $g \in M$, contradicting the fact that a character never sends an invertible element to $0$ (proof of Lemma 3.2).

Hence there is a point $x$ with $f(x) = 0$ for every $f \in M$. For any $f \in C(X)$ we have $f - \chi(f)\mathbf{1} \in M$, so at this point $f(x) - \chi(f) = 0$, that is, $\chi = \mathrm{ev}_x$.

Homeomorphism. The weak * topology is the coarsest topology making $\chi \mapsto \chi(f)$ continuous for each $f$, and $x \mapsto f(x)$ is continuous, so $\varepsilon$ is continuous. Since $X$ is compact and $\Omega(C(X))$ is Hausdorff (Lemma 3.2), a continuous bijection between them is a homeomorphism (Corollary 6.5［コンパクト性］, Continuous maps and homeomorphisms).

∎

Corollary 4.4（Homeomorphism is equivalent to *-isomorphism）

Let $X$ and $Y$ be compact Hausdorff spaces. Then $X$ and $Y$ are homeomorphic if and only if $C(X)$ and $C(Y)$ are -isomorphic as C-algebras.

Proof（Corollary 4.4）

($\Rightarrow$) For a homeomorphism $\varphi : X \to Y$, define $\varphi^*(f) = f \circ \varphi$. Then $\varphi^*$ is a *-isomorphism with inverse $(\varphi^{-1})^*$.

($\Leftarrow$) Suppose a *-isomorphism $\Phi: C(X) \to C(Y)$ is given. Its dual $\Phi^\vee(\chi) = \chi \circ \Phi$ is a map $\Omega(C(Y)) \to \Omega(C(X))$; since $\Phi$ is a bijective homomorphism, $\Phi^\vee$ is bijective as well, and by the definition of the weak * topology both $\Phi^\vee$ and its inverse are continuous, so it is a homeomorphism. By Corollary 4.3 we have $X \cong \Omega(C(X))$ and $Y \cong \Omega(C(Y))$, and composing gives $Y \cong X$.

∎

flowchart LR
X["space X"] -->|"proper continuous map φ"| Y["space Y"]
CY["commutative C*-algebra C₀(Y)"] -->|"φ* : f ↦ f∘φ"| CX["C₀(X)"]
X -.->|"C₀(−)"| CX
Y -.->|"C₀(−)"| CY

Gelfand duality: the category of geometry and the category of algebra correspond with the arrows reversed

The correspondence also holds at the level of morphisms. A proper continuous map $\varphi: X \to Y$ (one for which preimages of compact sets are compact) determines $\varphi^*: C_0(Y) \to C_0(X)$; conversely every nondegenerate *-homomorphism $C_0(Y) \to C_0(X)$ is of this form, and composition is reversed. In other words, *the category of locally compact Hausdorff spaces and proper continuous maps is contravariantly equivalent to the category of commutative C*-algebras and nondegenerate -homomorphisms.

4.2. A dictionary between geometry and algebra

Once an equivalence of categories is at hand, every geometric property translates into an algebraic one. Let us carry this out.

Proposition 4.5（Algebraic characterisations of compactness and connectedness）

1. Let $X$ be a locally compact Hausdorff space. Then $C_0(X)$ has a multiplicative unit if and only if $X$ is compact.
2. Let $X$ be a compact Hausdorff space. Then $X$ is connected if and only if the only idempotents of $C(X)$ (elements with $p^2 = p$) are $0$ and $\mathbf{1}$.

Proof（Proposition 4.5）

(1) ($\Leftarrow$) If $X$ is compact, the constant function $1$ belongs to $C_0(X)$ and is a unit.

(1) ($\Rightarrow$) Let $u \in C_0(X)$ be a unit. On a locally compact Hausdorff space, Urysohn’s lemma provides, for each $x$, a function $f \in C_0(X)$ with $f(x) \ne 0$. Evaluating $uf = f$ at $x$ and dividing by $f(x)$ gives $u(x) = 1$; as $x$ was arbitrary, $u \equiv 1$. Applying the definition of $C_0(X)$ to $u$ with $\varepsilon = 1/2$ then forces $\{x : |u(x)| \ge 1/2\} = X$ to be compact.

(2) ($\Rightarrow$) Let $p \in C(X)$ satisfy $p^2 = p$. Then $p(x)^2 = p(x)$ at every point, so $p(x) \in \{0, 1\}$. Hence $U = p^{-1}(1)$ and $V = p^{-1}(0)$ are both open and give a partition $X = U \sqcup V$. If $X$ is connected, one of them is empty and $p = \mathbf{1}$ or $p = 0$.

(2) ($\Leftarrow$) Suppose $X$ is not connected and write $X = U \sqcup V$ with $U, V$ nonempty open sets. Since $U$ is both open and closed, the indicator function $\mathbf{1}_U$ is continuous, satisfies $\mathbf{1}_U^2 = \mathbf{1}_U$, and, both $U$ and $V$ being nonempty, $\mathbf{1}_U \ne 0, \mathbf{1}$.

∎

Collecting the translations gives the following dictionary. The rightmost column lists the counterparts once commutativity is dropped.

Space $X$ (compact Hausdorff)	Commutative C*-algebra $C(X)$	General C*-algebra $A$
point	maximal ideal, character	irreducible representation, pure state, primitive ideal
compact	has a unit	has a unit
connected	only idempotents are $0, \mathbf 1$	structure of projections, $K_0$
open set $U$	closed ideal $C_0(U)$	closed two-sided ideal
closed set $F$	quotient $C(F)$	quotient C*-algebra
metrisable	separable	separable
vector bundle	finitely generated projective module (Swan’s theorem)	finitely generated projective module, $K_0$
Radon measure	positive linear functional (Riesz representation theorem)	state, trace
differential structure, metric	(invisible to the algebra of functions alone)	spectral triple $(A, H, D)$

The last two rows show the reach of the theory: measure-theoretic information translates into states and traces, differential-geometric information into spectral triples (Introduction to K-theory, Spectral triples (A, H, D)). The correspondence between vector bundles and finitely generated projective modules is furnished by the Serre–Swan theorem（Theorem 4.1）［K-理論入門］.

Finally, let us record that C*-algebras are the same thing as operator algebras.

Theorem 4.6（Gelfand–Naimark theorem (general case)）

Let $A$ be an arbitrary C*-algebra. Then there exist a Hilbert space $H$ and an isometric injective *-homomorphism $\pi: A \to B(H)$ such that $\pi(A)$ is a norm-closed *-subalgebra of $B(H)$. If $A$ is separable, then $H$ may be taken separable as well.

Remark 4.7

The proof uses the GNS construction (Gelfand–Naimark–Segal). From a state $\varphi$ one forms the inner product $\langle a, b\rangle = \varphi(b^* a)$, quotients by the null space and completes to obtain a Hilbert space; performing this for sufficiently many states and taking the direct sum gives the representation (Foundations of C*-algebras).

Combining this theorem with Theorem 4.1 yields the following picture. C*-algebras are operator algebras, and the commutative ones among them are spaces. A noncommutative C*-algebra is thus an operator algebra which is not a space, and the claim of noncommutative geometry is that we should regard it as one anyway.

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5. Bad spaces: when the quotient breaks

We now display, computing everything to the end, the typical situation in which a noncommutative algebra becomes necessary.

Example 5.1（The orbit space of the circle under an irrational rotation）

Fix $\theta \in \mathbb{R} \setminus \mathbb{Q}$ and consider the rotation of the circle $\mathbb{T}$,

$$R_\theta : \mathbb{T} \to \mathbb{T}, \qquad R_\theta(z) = e^{2\pi i \theta} z .$$

This defines an action of $\mathbb{Z}$ by $n \cdot z = e^{2\pi i n\theta} z$ (Introduction to group theory: definition and examples). We consider the orbit space $\mathbb{T}/\mathbb{Z}$ with the quotient topology.

Step 1: every orbit is dense. First, the points $z_n = e^{2\pi i n \theta}$ ($n \in \mathbb{Z}$) are pairwise distinct: if $z_n = z_m$ then $(n-m)\theta \in \mathbb{Z}$, and irrationality of $\theta$ forces $n = m$.

Since $\mathbb{T}$ is compact, the infinite set $\{z_n\}_{n \in \mathbb{N}}$ has an accumulation point, so for every $\varepsilon > 0$ there are $n \ne m$ with $|z_n - z_m| < \varepsilon$. Putting $k = n - m \ne 0$ and using that rotations are isometries, $|z_k - 1| = |z_n - z_m| < \varepsilon$. Thus $z_k$ is a rotation by a nonzero angle whose chord has length less than $\varepsilon$. Consequently $\{z_{kj}\}_{j\in\mathbb{Z}}$ runs once around the circle in steps of length less than $\varepsilon$ and forms an $\varepsilon$-net in $\mathbb{T}$. As $\varepsilon$ was arbitrary, $\{z_n\}_{n \in \mathbb{Z}}$ is dense, and since rotations are isometric bijections, the orbit $\{z_n x\}$ of any $x$ is dense too.

Step 2: the quotient topology is the indiscrete topology. Let $F \subseteq \mathbb{T}$ be a nonempty $R_\theta$-invariant closed set. Take $x \in F$; by invariance the whole orbit of $x$ lies in $F$, and by Step 1 that orbit is dense. Since $F$ is closed, $F \supseteq \overline{\{z_n x\}} = \mathbb{T}$, that is, $F = \mathbb{T}$. Passing to complements, the only $R_\theta$-invariant open sets are $\emptyset$ and $\mathbb{T}$, so by the definition of the quotient topology (a set is open exactly when its preimage is an invariant open set), the topology of $\mathbb{T}/\mathbb{Z}$ is indiscrete.

Step 3: the only continuous functions are the constants. By the universal property of the quotient topology, continuous functions on $\mathbb{T}/\mathbb{Z}$ correspond bijectively to $R_\theta$-invariant continuous functions on $\mathbb{T}$. If $f \in C(\mathbb{T})$ is invariant, then $f(z_n x) = f(x)$ for all $n$, so by Step 1 the function $f$ takes the value $f(x)$ on a dense set, and continuity gives $f \equiv f(x)$. Hence

$$C(\mathbb{T}/\mathbb{Z}) = \mathbb{C} .$$

Conclusion. But $\mathbb{C}$ is the algebra of functions on a one-point space (indeed $\Omega(\mathbb{C})$ consists of the identity map alone). Seen through its algebra of functions, $\mathbb{T}/\mathbb{Z}$ is indistinguishable from a point, and the information carried by $\theta$ has been lost completely.

One orbit of the irrational rotation R_θ (θ is the fractional part of the golden ratio). Even 21 points already begin to fill the circle evenly. Since orbits are dense, the only invariant closed sets are the empty set and the whole circle

5.1. Building the algebra without taking the quotient

The reason for the failure in Example 5.1 is clear. Forming $C(\mathbb{T}/\mathbb{Z})$ amounts to extracting only the invariant elements of $C(\mathbb{T})$, and there were far too few of them.

Connes’s prescription is to add to $C(\mathbb{T})$ a new element implementing the action, instead of passing to invariants. Translated into algebra, the rotation $R_\theta$ becomes the *-automorphism

$$\alpha : C(\mathbb{T}) \to C(\mathbb{T}), \qquad \alpha(f) = f \circ R_\theta^{-1} .$$

We therefore form the largest C*-algebra containing $C(\mathbb{T})$ and a unitary element $v$ ($v^* v = v v^* = \mathbf{1}$) subject to the relation

$$v f v^* = \alpha(f) \qquad (f \in C(\mathbb{T})) .$$

This is the crossed product $C(\mathbb{T}) \rtimes_\alpha \mathbb{Z}$. It is noncommutative, since $v$ and $f$ do not commute in general.

Since $C(\mathbb{T})$ is generated by the coordinate function $u(z) = z$ (Stone–Weierstrass), the crossed product is generated by the two unitaries $u, v$, and the relation reads

$$(vuv^*)(z) = \alpha(u)(z) = u(R_\theta^{-1}z) = e^{-2\pi i\theta} z = e^{-2\pi i \theta}\, u(z) ,$$

so $vuv^* = e^{-2\pi i\theta} u$, or equivalently $uv = e^{2\pi i \theta} v u$.

Definition 5.2（The noncommutative torus (irrational rotation algebra)）

For $\theta \in \mathbb{R}$, let $A_\theta$ denote the universal C*-algebra generated by two unitary elements $u, v$ subject to the relations

$$u^* u = u u^* = \mathbf{1}, \quad v^* v = v v^* = \mathbf{1}, \quad uv = e^{2\pi i \theta}\, vu .$$

It is called the noncommutative torus, and, when $\theta$ is irrational, the irrational rotation algebra. “Universal” means that for every C*-algebra $B$ containing a pair of unitaries $(U, V)$ satisfying these relations there is exactly one *-homomorphism $A_\theta \to B$ extending $u \mapsto U$ and $v \mapsto V$. The algebra $A_\theta$ is *-isomorphic to the crossed product $C(\mathbb{T}) \rtimes_{\alpha} \mathbb{Z}$.

Example 5.3（A concrete representation of the noncommutative torus）

On $H = L^2(\mathbb{T})$, the square-integrable functions for the normalised Haar measure (L^p spaces and an introduction to functional analysis), define two operators:

$$(Uf)(z) = z f(z), \qquad (Vf)(z) = f(e^{-2\pi i \theta} z) .$$

Here $U$ is multiplication by a function of modulus $1$ and $V$ is translation by a rotation, so both are unitary. We compute

$$(UVf)(z) = z\,(Vf)(z) = z\, f(e^{-2\pi i\theta} z),$$$$(VUf)(z) = (Uf)(e^{-2\pi i\theta}z) = e^{-2\pi i\theta} z \, f(e^{-2\pi i \theta} z) .$$

Hence $VU = e^{-2\pi i\theta}\, UV$, that is, $UV = e^{2\pi i\theta}\, VU$, which is the relation of Definition 5.2. The norm-closed *-subalgebra of $B(H)$ generated by $U$ and $V$ is a concrete model of $A_\theta$ (for irrational $\theta$ this representation is automatically injective, by the simplicity established in Proposition 5.4).

Proposition 5.4（Simplicity and uniqueness of the trace for the irrational rotation algebra）

Let $\theta$ be irrational. Then $A_\theta$ is simple: its only closed two-sided ideals are $\{0\}$ and $A_\theta$. Moreover $A_\theta$ carries exactly one tracial state, given by

$$\tau\left(\sum_{m,n} a_{mn} u^m v^n\right) = a_{00} .$$

Remark 5.5

The proof uses the conditional expectation obtained by averaging the natural action of the torus $\mathbb{T}^2$ on $A_\theta$ given by $\gamma_{(s,t)}(u) = su$, $\gamma_{(s,t)}(v) = tv$, namely $E(a) = \int_{\mathbb{T}^2} \gamma_{(s,t)}(a)\, ds\, dt$, together with the fact that irrationality of $\theta$ makes the range of $E$ equal to $\mathbb{C}\mathbf{1}$. For the details see simplicity and uniqueness of the trace for the irrational rotation algebra（Theorem 4.5）［非可換トーラス A_θ］ (The example of the noncommutative torus) and Rieffel’s original paper (reference 5).

Read against the dictionary in which closed ideals correspond to open sets, simplicity says “there is only one point”. Yet $A_\theta$ is infinite-dimensional and its structure is nothing like that of a simple algebra such as $M_n(\mathbb{C})$. It is exactly this situation — one point only, but a point of infinite richness — that characterises a noncommutative space.

Example 5.6（What happens when θ is rational）

For $\theta = 0$ the relation becomes $uv = vu$, so $A_0$ is the universal commutative C*-algebra generated by two commuting unitaries, that is, $A_0 \cong C(\mathbb{T}^2)$. Through Gelfand duality this is nothing but the two-dimensional torus, which is the reason for calling $A_\theta$ a noncommutative torus.

For $\theta = p/q$ in lowest terms with $q \ge 1$ the situation is intermediate. Repeated use of the relation gives

$$u^q v = e^{2\pi i \theta q}\, v u^q = e^{2\pi i p}\, v u^q = v u^q$$

(since $p$ is an integer, $e^{2\pi i p} = 1$), and likewise $v^q$ commutes with $u$. Hence $u^q$ and $v^q$ are commuting unitaries and the subalgebra they generate is isomorphic to $C(\mathbb{T}^2)$. In fact the centre of $A_{p/q}$ is exactly this subalgebra, and $A_{p/q}$ is known to be isomorphic to the algebra of continuous sections of a bundle of $M_q(\mathbb{C})$‘s over the two-dimensional torus.

Thus for rational $\theta$ the noncommutativity is confined inside $q \times q$ matrices and “points” reappear as the spectrum of the centre. For irrational $\theta$ this escape route is closed (Proposition 5.4) and the noncommutativity becomes essential. This matches the difference between rational rotations, whose orbits are finite and whose quotient is an ordinary circle, and irrational rotations, which lead to the situation of Example 5.1.

Tip

That $A_\theta$ retains $\theta$ can be measured precisely by K-theory. One has $K_0(A_\theta) \cong \mathbb{Z}^2$, and the image of the map $\tau_*: K_0(A_\theta) \to \mathbb{R}$ induced by the unique trace $\tau$ is $\mathbb{Z} + \theta\mathbb{Z}$. In other words, the real number $\theta$ can be read off from the algebra, up to the action of $\mathrm{GL}_2(\mathbb{Z})$. The information destroyed in the collapsed quotient space survives in the noncommutative algebra. The computation is carried out in Introduction to K-theory and in the K-groups of the noncommutative torus and the image of the trace（Theorem 6.1）［非可換トーラス A_θ］.

The same prescription works over a wide range. The action of a discrete group $\Gamma$ on a space $X$ gives the crossed product $C_0(X) \rtimes \Gamma$; a foliation gives the foliation C*-algebra; the Penrose tilings give a groupoid C*-algebra. In every case the “bad quotient” comes back to life as a noncommutative C*-algebra. The common framework is that of groupoid C*-algebras, which may be understood as making an algebra out of the graph of the equivalence relation itself.

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6. The noncommutativity demanded by quantum theory

The other road towards noncommutative algebras comes from physics. Can Heisenberg’s canonical commutation relation $qp - pq = i\hbar \mathbf{1}$ be treated as it stands inside the framework of C*-algebras? The answer is no.

Theorem 6.1（Wielandt's theorem）

Let $A$ be a unital Banach algebra different from $\{0\}$. Then there are no $a, b \in A$ with

$$ab - ba = \mathbf{1} .$$

In particular there are no bounded operators $Q, P \in B(H)$ satisfying $QP - PQ = i\hbar I$ with $\hbar \ne 0$.

Proof（Theorem 6.1）

Suppose $a, b$ satisfy $ab - ba = \mathbf{1}$.

Step 1: induction for $ab^n - b^n a = n b^{n-1}$ ($n \ge 1$). The case $n = 1$ is the hypothesis itself (with $b^0 = \mathbf{1}$). Assuming the identity for $n$,

$$\begin{aligned} a b^{n+1} - b^{n+1} a &= (a b^n) b - b^{n+1} a \\ &= (b^n a + n b^{n-1}) b - b^{n+1} a \\ &= b^n (ab) + n b^n - b^{n+1}a \\ &= b^n (ba + \mathbf{1}) + n b^n - b^{n+1} a \\ &= b^{n+1} a + b^n + n b^n - b^{n+1}a = (n+1) b^n . \end{aligned}$$

The second line uses the induction hypothesis and the fourth uses $ab = ba + \mathbf{1}$.

Step 2: estimating norms. From Step 1 together with the triangle inequality and submultiplicativity,

$$n \|b^{n-1}\| = \|ab^n - b^n a\| \le 2\|a\|\,\|b^n\| \le 2\|a\|\,\|b\|\,\|b^{n-1}\| ,$$

so if $b^{n-1} \ne 0$ we may divide both sides by $\|b^{n-1}\| > 0$ and obtain $n \le 2\|a\|\,\|b\|$. The right-hand side is a constant independent of $n$, so the inequality fails once $n$ is large enough. Hence $b^{n-1} = 0$ for some $n$; that is, there is an integer $m \ge 0$ with $b^m = 0$.

Step 3: the contradiction. Let $m_0$ be the smallest $m \ge 0$ with $b^m = 0$. If $m_0 = 0$ then $\mathbf{1} = b^0 = 0$, so $A = \{0\}$, contrary to hypothesis. Hence $m_0 \ge 1$, and minimality gives $b^{m_0 - 1} \ne 0$. But Step 1 with $n = m_0$ yields

$$0 = a b^{m_0} - b^{m_0} a = m_0\, b^{m_0 - 1} ,$$

and since $m_0 \ge 1$ this forces $b^{m_0-1} = 0$, contradicting $b^{m_0-1} \ne 0$.

∎

The result looks negative, but it is also a guide: position and momentum cannot be realised as bounded operators, so one must either use unbounded operators or change the form of the relation. The latter road, which stays inside the C*-algebraic framework, is the Weyl form.

Example 6.2（The Weyl form and its coincidence with the noncommutative torus）

Suppose $q, p$ satisfy $qp - pq = i\hbar\mathbf 1$ and consider the exponentials

$$U_s = e^{isq}, \qquad V_t = e^{itp} \qquad (s, t \in \mathbb{R}) .$$

Being exponentials of self-adjoint operators, these are unitary. The commutator $[isq, itp]$ is central, so the Baker–Campbell–Hausdorff formula degenerates to the form

$$e^{A}e^{B} = e^{A+B}e^{[A,B]/2} .$$

Taking $A = isq$ and $B = itp$ gives $[A,B] = (is)(it)[q,p] = -st\cdot i\hbar\,\mathbf{1}$, so

$$U_s V_t = e^{A+B} e^{-i\hbar st/2}, \qquad V_t U_s = e^{A+B} e^{+i\hbar st /2} ,$$

and dividing one by the other,

$$U_s V_t = e^{-i\hbar s t}\, V_t U_s .$$

This is the Weyl form of the canonical commutation relations. No unbounded operator appears; only a relation between unitary elements remains.

Fixing $s$ and $t$, putting $u = U_s$ and $v = V_t$, and letting $\theta$ be the fractional part of $-\hbar st / 2\pi$, the relation becomes

$$uv = e^{2\pi i \theta} vu ,$$

which is precisely the relation of Definition 5.2. In other words, the noncommutative torus is nothing but the phase space of quantum mechanics restricted to translations along a lattice. A problem in dynamics, the irrational rotation, and a problem in quantum theory, the canonical commutation relations, arrive at one and the same C*-algebra.

Remark 6.3

In the classical limit $\hbar \to 0$, that is $\theta \to 0$, the algebra $A_\theta$ approaches the commutative algebra $C(\mathbb{T}^2)$ (Example 5.6). A family of commutative algebras of functions is thus “deformed” into a family of noncommutative algebras with $\hbar$ as the parameter; this is the picture of deformation quantisation, and noncommutative geometry supplies the language for treating the deformed objects geometrically. The algebra $A_\theta$ also arises as a model for two-dimensional quantum Hall systems, where the integer quantisation of the Hall conductance is derived from the K-theory of $A_\theta$ and an index theorem (the noncommutative torus and the integer quantum Hall effect（Example 4.10）［指数定理への応用］, Applications to index theorems).

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7. The road ahead

The starting point is now in place. A space is a commutative C*-algebra; drop commutativity and the phase spaces of quantum theory and the bad quotients come into view. What remains to be done is to transplant the apparatus of geometry into the noncommutative setting, and the chapters that follow carry out that work.

• Foundations of C*-algebras: positive elements, states, representations, ideals, tensor products.
• Introduction to K-theory: starting from the dictionary entry “vector bundle = finitely generated projective module”, the groups $K_0, K_1$ and Bott periodicity.
• Spectral triples (A, H, D): reading the Dirac operator $D$ as an instrument for measuring distance, and transplanting differential structure and metric.
• The example of the noncommutative torus: computations of projections, traces and K-theory for the algebra $A_\theta$ constructed here.
• Applications to index theorems: the noncommutative version of the Atiyah–Singer index theorem and its consequences.

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8. Exercises

Exercise 8.1Easy

Let $A$ be a unital commutative C*-algebra. Prove the following.

1. If $b \in A$ is self-adjoint ($b^* = b$), then $\sigma(b) \subseteq \mathbb{R}$.
2. If $w \in A$ is unitary ($w^* w = w w^* = \mathbf{1}$), then $\sigma(w) \subseteq \mathbb{T}$.

Solution

By Theorem 4.1 the map $\Gamma$ is a *-isomorphism, and by Lemma 3.3 we have $\sigma(a) = \hat{a}(\Omega(A))$.

(1) Step 1 of the proof of Theorem 4.1 showed that $\chi(b) \in \mathbb{R}$ for a self-adjoint $b$ and every character $\chi$. Hence

$$\sigma(b) = \{\chi(b) : \chi \in \Omega(A)\} \subseteq \mathbb{R} .$$

(2) Since $\Gamma$ is a *-homomorphism, $\hat{w}\,\overline{\hat{w}} = \widehat{w w^*} = 1$, that is, $|\chi(w)| = 1$ for every $\chi$. Hence

$$\sigma(w) = \{\chi(w): \chi \in \Omega(A)\} \subseteq \mathbb{T} .$$

These statements remain true without commutativity of $A$: one applies the present result to the commutative C*-subalgebra generated by $b$ (respectively $w$) and $\mathbf{1}$, and uses the fact that the spectrum computed in a subalgebra agrees with the spectrum computed in the whole algebra (spectral permanence, Remark 6.1［C*-Algebras］).

Exercise 8.2Standard

Let $X$ be a compact Hausdorff space. Show that $C(X)$ is finite-dimensional if and only if $X$ is a finite set, and that in this case $|X| = n$ implies $C(X) \cong \mathbb{C}^n$ with the componentwise product.

Solution

($X$ finite $\Rightarrow$) If $X$ has $n$ points, then, a finite Hausdorff space being discrete, every function on $X$ is continuous, so $C(X) = \mathbb{C}^X \cong \mathbb{C}^n$, of dimension $n$. The product is pointwise, so it agrees with the componentwise product of $\mathbb{C}^n$.

($\Rightarrow$ $X$ finite) Suppose $X$ contains $n+1$ distinct points $x_1, \ldots, x_{n+1}$. A compact Hausdorff space is normal, so by Urysohn’s lemma there are $f_k \in C(X)$ with

$$f_k(x_k) = 1, \qquad f_k(x_j) = 0 \ (j \ne k)$$

for each $k$ (separate $\{x_k\}$ from the finite closed set $\{x_j : j \ne k\}$). Evaluating $\sum_k c_k f_k = 0$ at $x_j$ gives $c_j = 0$, so these functions are linearly independent and $\dim C(X) \ge n+1$.

Hence if $\dim C(X) = n$ is finite, then $X$ has at most $n$ points, and combined with the estimate in the other direction we get $|X| = n$ and $C(X) \cong \mathbb{C}^n$.

Exercise 8.3Standard

Let $X, Y$ be compact Hausdorff spaces, let $\varphi: X \to Y$ be continuous, and define $\varphi^*: C(Y) \to C(X)$ by $\varphi^*(f) = f \circ \varphi$. Show that $\varphi$ is surjective if and only if $\varphi^*$ is injective.

Solution

($\Rightarrow$) Suppose $\varphi$ is surjective and $\varphi^*(f) = 0$. Then $f(\varphi(x)) = 0$ for every $x \in X$, and surjectivity gives $f(y) = 0$ for every $y \in Y$, that is, $f = 0$.

($\Leftarrow$) We argue by contraposition. If $\varphi$ is not surjective, then $\varphi(X)$ is a compact subset of $Y$ by compactness of $X$, hence closed since $Y$ is Hausdorff, so $Y \setminus \varphi(X)$ is a nonempty open set. Choose $y_0 \in Y \setminus \varphi(X)$ and use Urysohn’s lemma to produce $f \in C(Y)$ with $f(y_0) = 1$ and $f = 0$ on $\varphi(X)$. Then $f \ne 0$ while $\varphi^*(f) = 0$, so $\varphi^*$ is not injective.

One shows in the same way that $\varphi$ is injective if and only if $\varphi^*$ is surjective (using the Tietze extension theorem). This is a manifestation, at the level of morphisms, of the equivalence of categories underlying Corollary 4.4.

Exercise 8.4Hard

Let $\theta$ be irrational and let $A_\theta$ be the noncommutative torus of Definition 5.2. It is known that every element $a$ of $A_\theta$ has a formal Fourier series

$$a \sim \sum_{m, n \in \mathbb{Z}} a_{mn}\, u^m v^n, \qquad a_{mn} = \tau(a\, v^{-n} u^{-m})$$

and that the coefficients $(a_{mn})$ determine $a$ uniquely, where $\tau$ is the unique trace. Use this to show that the centre of $A_\theta$ is $\mathbb{C}\mathbf{1}$.

Solution

Let $a \in A_\theta$ belong to the centre; in particular it commutes with both $u$ and $v$.

The basic relation $uv = e^{2\pi i\theta} vu$ gives $u v u^{-1} = e^{2\pi i \theta} v$, and iterating,

$$u v^n u^{-1} = e^{2\pi i n\theta} v^n \qquad (n \in \mathbb{Z})$$

(for $n \ge 0$ by induction, for $n < 0$ by taking inverses of both sides). Since $u$ commutes with $u^m$, we get $u (u^m v^n) u^{-1} = e^{2\pi i n \theta} u^m v^n$.

The map $\gamma(x) = uxu^{-1}$ is a *-automorphism of $A_\theta$, and since $\tau$ is the only tracial state we have $\tau \circ \gamma = \tau$. Hence the Fourier coefficients satisfy $(\gamma(a))_{mn} = e^{2\pi i n\theta} a_{mn}$. If $a$ commutes with $u$ then $\gamma(a) = a$, so uniqueness of the coefficients gives

$$a_{mn}\left(e^{2\pi i n \theta} - 1\right) = 0 \qquad (\forall m, n) .$$

As $\theta$ is irrational, $e^{2\pi i n\theta} = 1$ only for $n = 0$, so $a_{mn} = 0$ whenever $n \ne 0$.

Similarly, consider $v x v^{-1}$. Sandwiching $uv = e^{2\pi i\theta}vu$ between $v^{-1}$‘s gives $v u v^{-1} = e^{-2\pi i \theta} u$, and iterating, $v (u^m v^n) v^{-1} = e^{-2\pi i m\theta} u^m v^n$; so if $a$ commutes with $v$, the same argument gives $a_{mn} = 0$ whenever $m \ne 0$.

Together, all Fourier coefficients of $a$ vanish except $a_{00}$. The element $\mathbf{1}$ has coefficient $a_{00} = 1$ and all others $0$, so by uniqueness $a = a_{00}\mathbf{1}$; that is, the centre is $\mathbb{C}\mathbf{1}$.

Supplement. For $\theta = p/q$ one has $e^{2\pi i n\theta} = 1$ for all $n \in q\mathbb{Z}$, so the argument cannot eliminate the terms in which $m$ and $n$ are both multiples of $q$, and the copy of $C(\mathbb{T}^2)$ generated by $u^q, v^q$ remains in the centre (Example 5.6). This makes it clear exactly where irrationality is used.

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References

1. A. Connes, Noncommutative Geometry, Academic Press, 1994 — Chapter I (examples of noncommutative spaces: Penrose tilings, foliations, quotients by group actions) and Chapter II. This book is the blueprint for the present article; the full text is available on the author’s site (alainconnes.org).
2. G. J. Murphy, C*-Algebras and Operator Theory, Academic Press, 1990 — Chapter 1 (Banach algebras and Gelfand theory) and Chapter 2 (C*-algebras, the commutative Gelfand–Naimark theorem, the GNS construction). The standard textbook for §2–§4 of this article.
3. K. R. Davidson, C*-Algebras by Example, Fields Institute Monographs 6, American Mathematical Society, 1996 — the chapter on irrational rotation algebras, which contains detailed proofs of Proposition 5.4 and Example 5.6.
4. I. Gelfand and M. Naimark, “On the imbedding of normed rings into the ring of operators in Hilbert space”, Matematicheskii Sbornik 12 (1943), 197–213 — the original paper for Theorem 4.1 and Theorem 4.6.
5. M. A. Rieffel, “C*-algebras associated with irrational rotations”, Pacific Journal of Mathematics 93 (1981), 415–429. doi:10.2140/pjm.1981.93.415 — the construction of projections in irrational rotation algebras and their K-theory.
6. H. Wielandt, “Über die Unbeschränktheit der Operatoren der Quantenmechanik”, Mathematische Annalen 121 (1949), 21 — the original paper for Theorem 6.1.

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Appendix: lemmas from spectral theory

We collect here the basic facts about Gelfand theory for commutative Banach algebras that were used in the text. All of them are standard, and proofs may be found in Chapter 1 of reference 2.

Nonemptiness of the spectrum. The spectrum of an element of a unital Banach algebra is a nonempty compact set.

Lemma 8.5（Nonemptiness of the spectrum）

Let $A$ be a unital complex Banach algebra different from $\{0\}$ and let $a \in A$. Then $\sigma(a)$ is a nonempty compact set and $\sigma(a) \subseteq \{\lambda : |\lambda| \le \|a\|\}$.

Remark 8.6

Boundedness follows because for $|\lambda| > \|a\|$ the element $\lambda\mathbf{1} - a = \lambda(\mathbf{1} - a/\lambda)$ is invertible by the Neumann series, and closedness follows because the set of invertible elements is open.

Nonemptiness rests on complex analysis. If $\sigma(a) = \emptyset$, then the resolvent $\lambda \mapsto (\lambda\mathbf{1} - a)^{-1}$ is an $A$-valued function holomorphic on all of $\mathbb{C}$ whose norm tends to $0$ as $|\lambda| \to \infty$, hence bounded. Liouville’s theorem, applied after composing with an arbitrary $\varphi \in A^*$, shows that it is the constant function $0$; but the resolvent takes invertible values and so never vanishes.

The Gelfand–Mazur theorem. This was the key step in the proof of Lemma 3.3.

Theorem 8.7（Gelfand–Mazur theorem）

Let $A$ be a unital complex Banach algebra in which every nonzero element is invertible. Then $\lambda \mapsto \lambda \mathbf{1}$ is an isometric algebra isomorphism from $\mathbb{C}$ onto $A$; that is, $A \cong \mathbb{C}$.

Proof（Theorem 8.7）

Take any $a \in A$. By Lemma 8.5 there is a $\lambda \in \sigma(a)$, and by definition $\lambda\mathbf{1} - a$ is not invertible. By hypothesis the only non-invertible element is $0$, so $a = \lambda\mathbf{1}$.

Hence $A = \mathbb{C}\mathbf{1}$ and $\lambda \mapsto \lambda\mathbf{1}$ is surjective; injectivity follows from $\mathbf 1 \ne 0$, and isometry from $\|\lambda\mathbf{1}\| = |\lambda|\,\|\mathbf{1}\| = |\lambda|$.

∎

Closedness of maximal ideals. This was needed to make the quotient a Banach algebra.

Lemma 8.8（Maximal ideals are closed）

Let $A$ be a unital Banach algebra and let $M \subseteq A$ be a maximal proper ideal (two-sided; simply an ideal when $A$ is commutative). Then $M$ is closed.

Proof（Lemma 8.8）

First, a proper ideal $I$ contains no invertible element: if $x \in I$ were invertible, then $\mathbf{1} = x^{-1}x \in I$ and $I = A$. If $\|\mathbf{1} - y\| < 1$ then $y$ is invertible by the Neumann series, so the open ball $B$ of radius $1$ centred at $\mathbf{1}$ consists of invertible elements. Hence $I \cap B = \emptyset$, and since $B$ is open the closure $\bar{I}$ also misses $B$, so $\mathbf{1} \notin \bar{I}$.

On the other hand, continuity of addition and multiplication makes the closure of an ideal an ideal. Therefore $\bar{M}$ is a proper ideal containing $M$, and maximality of $M$ gives $\bar{M} = M$, that is, $M$ is closed.

∎

The spectral radius formula. This was used in Proposition 3.5.

Lemma 8.9（Beurling–Gelfand spectral radius formula）

Let $A$ be a unital Banach algebra and $a \in A$. Then the limit $\lim_{n\to\infty}\|a^n\|^{1/n}$ exists and

$$r(a) = \lim_{n \to \infty} \|a^n\|^{1/n} = \inf_{n \ge 1} \|a^n\|^{1/n} .$$

Remark 8.10

Existence of the limit follows from the subadditivity of $\log\|a^n\|$ and Fekete’s lemma. The inequality $r(a) \le \lim \|a^n\|^{1/n}$ is obtained from the factorisation $\lambda^n \mathbf{1} - a^n = (\lambda\mathbf{1} - a)(\lambda^{n-1}\mathbf{1} + \cdots + a^{n-1})$, which yields $\lambda \in \sigma(a) \Rightarrow \lambda^n \in \sigma(a^n)$, together with the boundedness in Lemma 8.5. The reverse inequality is obtained by estimating the coefficients of the Laurent expansion of the resolvent, which converges for $|\lambda| > r(a)$. The details are in reference 2.