# C*-Algebras: Spectral Theory and the Gelfand Representation

> Spectra in Banach algebras are nonempty and compact; from this we derive the spectral radius formula, identify commutative C*-algebras with C0(X), and build the functional calculus.
> https://rikai.mugen-giken.com/en/mathematics/noncommutative-geometry/c-star-algebras

## 0. Key points

- The spectrum $\sigma(a)$ of an element $a$ of a complex Banach algebra is always a nonempty compact set. All the analysis of the subject is concentrated in this single point, which is proved through Liouville's theorem; everything that follows is a consequence of it.
- The axiom that defines a $C^*$-algebra is the single identity $\|a^*a\| = \|a\|^2$. From it alone the norm is determined uniquely by the $*$-algebraic structure. This is the starting point of noncommutative geometry: all the geometric information already sits inside the algebra.
- A commutative $C^*$-algebra is nothing other than $C_0(X)$ for a locally compact Hausdorff space $X$, and $X$ can be recovered from the algebra as its character space (the commutative Gelfand–Naimark theorem). "Locally compact Hausdorff space" and "commutative $C^*$-algebra" carry exactly the same information.
- Noncommutative geometry keeps the algebraic side of this dictionary, discards commutativity, and reads a general $C^*$-algebra as a *noncommutative topological space*.
- The continuous functional calculus $C(\sigma(a)) \to A$ for a normal element $a$ is the common foundation for all the later tools: positivity, projections, unitaries, and hence $K$-theory and spectral triples.

## 1. Motivation: why make an algebra of functions noncommutative?

When we study a topological space $X$, we very often study not $X$ itself but the ring $C(X)$ of continuous functions on it. In algebraic geometry this policy is carried through completely: an affine variety carries exactly the same information as its coordinate ring. Can the same be said of topological spaces? That is, can $X$ be recovered from $C(X)$?

In the late 1930s Gelfand built a framework that answers this question. His 1941 paper "Normierte Ringe" introduced the construction that extracts a *character space* from a commutative Banach algebra. Then in 1943 Gelfand and Naimark showed that imposing the single identity $\|a^*a\| = \|a\|^2$ on a Banach algebra with involution already forces it, in the commutative case, to be $C_0(X)$ and nothing else. The space $X$ is recovered from the algebra completely. A space no longer needs to be presented as a set of points: the algebra of functions is enough.

At about the same time, quantum mechanics was demanding algebras from a different direction. Observables do not commute — position and momentum satisfy $[q, p] = i\hbar$ — so the classical picture of a "point of phase space" simply fails. This is why von Neumann began the study of operator algebras. And here a natural question arises. If a commutative $C^*$-algebra *is* a space, what is a noncommutative one?

Connes' answer is that it too is a space, only a noncommutative one. This point of view acquires its force exactly where the classical construction collapses. The leaf space of a foliation and the orbit space of a group action, given the quotient topology, fail to separate points, and almost no continuous functions survive on them. The corresponding noncommutative $C^*$-algebras, by contrast, are rich; $K$-theory and index theorems work on them. For details see [Motivation for noncommutative geometry](/en/mathematics/noncommutative-geometry/motivation), and in particular <Ref to="mathematics/noncommutative-geometry/motivation#ex-irrational-rotation" text="the orbit space of the circle under an irrational rotation" />.

This article assembles the foundations. It also answers the question of why one works with $C^*$-algebras rather than Banach algebras. The answer is <Ref to="cor-norm-unique" />. The norm of a Banach algebra is extra data supplied from outside the algebra, whereas a $C^*$-norm is determined uniquely by the algebraic structure alone. It is precisely because the topology comes along automatically that one may define a noncommutative space by an algebra.

<Figure caption="The logical flow of this article. We start from Banach algebras at the top left and descend in a single line to the continuous functional calculus at the bottom right.">
<Mermaid code={`flowchart TD
  BA["Banach algebra"] --> SPEC["the spectrum is a nonempty compact set"]
  SPEC --> RAD["spectral radius formula"]
  BA --> INV["introduce an involution"]
  INV --> CSTAR["C*-identity"]
  RAD --> NORMAL["for normal elements, norm = spectral radius"]
  CSTAR --> NORMAL
  NORMAL --> UNIQ["uniqueness of the C*-norm"]
  BA --> GEL["Gelfand transform"]
  GEL --> GN["commutative Gelfand–Naimark theorem"]
  NORMAL --> GN
  GN --> FC["continuous functional calculus"]
  FC --> NCG["towards K-theory and spectral triples"]`} />
</Figure>

## 2. Banach algebras

From now on all algebras are over the field $\mathbb{C}$ of complex numbers. Over the reals <Ref to="thm-spectrum-compact" /> fails — the element $i$ of the real algebra $\mathbb{C}$ has empty real spectrum — and spectral theory never gets off the ground. Working over $\mathbb{C}$ is not a convenience but a necessity.

<Definition id="def-banach-algebra" title="Banach algebra">
Let $A$ be an associative algebra over $\mathbb{C}$ equipped with a norm $\|\cdot\|$. We call $A$ a **Banach algebra** if the following two conditions hold.

1. $(A, \|\cdot\|)$ is a Banach space (complete as a normed space).
2. $\|ab\| \le \|a\|\,\|b\|$ for all $a, b \in A$ (submultiplicativity).

If in addition there is a multiplicative identity $1 \in A$ with $\|1\| = 1$, we call $A$ **unital**. Commutativity is not assumed.
</Definition>

Submultiplicativity is not a mere technical condition. It yields the estimate $\|ab - a_0b_0\| \le \|a - a_0\|\,\|b\| + \|a_0\|\,\|b - b_0\|$, so that the multiplication $A \times A \to A$ is jointly continuous. In other words, it is the statement "$A$ is a topological algebra" translated into the language of norms.

<Example id="ex-c0x" title="The algebra of continuous functions vanishing at infinity">
Let $X$ be a locally compact Hausdorff space. A continuous function $f : X \to \mathbb{C}$ **vanishes at infinity** if for every $\varepsilon > 0$ there is a compact set $K \subset X$ with $|f(x)| < \varepsilon$ for $x \notin K$. Write $C_0(X)$ for the set of all such $f$, with pointwise operations and the supremum norm $\|f\|_\infty = \sup_{x}|f(x)|$.

Submultiplicativity follows from $\|fg\|_\infty \le \|f\|_\infty \|g\|_\infty$, and completeness from the fact that a uniform limit of continuous functions is continuous together with the fact that $C_0(X)$ is a closed subspace of $C_b(X)$: choose $N$ with $\|f_N - f\|_\infty < \varepsilon/2$ and a compact $K$ with $|f_N| < \varepsilon/2$ off $K$; then $|f| < \varepsilon$ off $K$. The product is pointwise, hence commutative.

The algebra $C_0(X)$ is unital exactly when $X$ is compact. If $X$ is compact, the constant function $1$ is a unit and $\|1\|_\infty = 1$. Conversely, if $1 \in C_0(X)$, take the compact set $K$ associated with $\varepsilon = 1/2$; for $x \notin K$ we would get $1 < 1/2$, a contradiction, so $X = K$. When $X$ is compact we write $C_0(X) = C(X)$.
</Example>

<Example id="ex-bh" title="Bounded operators on a Hilbert space, and matrix algebras">
Let $H$ be a complex Hilbert space (<Ref to="mathematics/real-analysis/lp-spaces#def-hilbert" />) and let $B(H)$ be the set of all bounded linear operators on $H$, with composition as product and the operator norm $\|T\| = \sup_{\|x\| \le 1}\|Tx\|$. From $\|STx\| \le \|S\|\|Tx\| \le \|S\|\|T\|\|x\|$ the norm is submultiplicative, and completeness of $B(H)$ in the operator norm is standard. The identity operator $I$ is a unit with $\|I\| = 1$, so $B(H)$ is a unital Banach algebra.

When $H = \mathbb{C}^n$ we have $B(H) = M_n(\mathbb{C})$. For $n \ge 2$ this is not commutative: with the matrix units $e_{ij}$ (entry $1$ in position $(i,j)$, zero elsewhere) we have $e_{12}e_{21} = e_{11} \ne e_{22} = e_{21}e_{12}$. If $\dim H \ge 2$ the same argument runs inside a two-dimensional subspace, so $B(H)$ is not commutative either.
</Example>

<Remark id="rem-unitization" title="Adjoining a unit">
A Banach algebra $A$ without a unit can always be given one formally. As a vector space put $\tilde{A} = A \oplus \mathbb{C}$, and define the product and norm by
$$
(a, \lambda)(b, \mu) = (ab + \lambda b + \mu a,\ \lambda\mu), \qquad \|(a,\lambda)\| = \|a\| + |\lambda| .
$$
Associativity is checked by direct computation; $(0,1)$ is a unit of norm $1$. Submultiplicativity is verified by
$$
\|(a,\lambda)(b,\mu)\| = \|ab + \lambda b + \mu a\| + |\lambda\mu| \le \|a\|\|b\| + |\lambda|\|b\| + |\mu|\|a\| + |\lambda||\mu| = \|(a,\lambda)\|\,\|(b,\mu)\|,
$$
and completeness follows from completeness of a direct sum. The map $a \mapsto (a,0)$ is an isometric injection whose image is a closed two-sided ideal of codimension $1$. From now on we identify $A \subset \tilde{A}$ and write $(a,\lambda)$ as $a + \lambda 1$. Note that when $A$ is a $C^*$-algebra the norm $\|a\| + |\lambda|$ does not satisfy the $C^*$-identity, so unitization in the category of $C^*$-algebras requires a different norm (constructed in the Appendix).
</Remark>

## 3. The spectrum and the spectral radius

The spectrum is what plays the role of "the set of eigenvalues" for an element $a$ of an algebra. In finite dimensions one can define it as the set of roots of the characteristic polynomial, but in infinite dimensions that route is unavailable, so we define it through invertibility.

<Definition id="def-spectrum" title="Spectrum and resolvent">
Let $A$ be a unital complex Banach algebra and $a \in A$. The set
$$
\sigma_A(a) = \{\lambda \in \mathbb{C} \ :\ \lambda 1 - a \text{ is not invertible in } A\}
$$
is the **spectrum** of $a$, the set $\rho(a) = \mathbb{C} \setminus \sigma_A(a)$ is the **resolvent set**, and for $\lambda \in \rho(a)$ the element $R_a(\lambda) = (\lambda 1 - a)^{-1}$ is the **resolvent**. If $A$ is not unital we set $\sigma_A(a) := \sigma_{\tilde{A}}(a)$ using the unitization of <Ref to="rem-unitization" />. We drop the subscript when no confusion can arise.
</Definition>

If $A$ is not unital then $0$ always belongs to $\sigma(a)$. Indeed, if $a$ were invertible in $\tilde{A}$, then since $A$ is an ideal we would get $1 = a a^{-1} \in A$, that is $A = \tilde{A}$, contradicting the assumption.

<Lemma id="lem-neumann" title="Neumann series">
Let $A$ be a unital complex Banach algebra.

1. If $a \in A$ satisfies $\|a\| < 1$, then $1 - a$ is invertible, $\displaystyle (1-a)^{-1} = \sum_{n=0}^{\infty} a^n$, and $\displaystyle \|(1-a)^{-1}\| \le \frac{1}{1 - \|a\|}$.
2. The set $G(A)$ of invertible elements is open in $A$. More precisely, if $b \in G(A)$ and $\|c - b\| < \|b^{-1}\|^{-1}$, then $c \in G(A)$.
3. The map $G(A) \to G(A)$, $b \mapsto b^{-1}$, is continuous.
</Lemma>

<Proof of="lem-neumann">
**1.** Applying submultiplicativity inductively gives $\|a^n\| \le \|a\|^n$, so $\sum_n \|a^n\| \le (1-\|a\|)^{-1} < \infty$, that is, $\sum_n a^n$ converges absolutely. Since $A$ is complete (condition 1 of <Ref to="def-banach-algebra" />), absolutely convergent series converge, and we may set $s$ to be the sum. The partial sums $s_N = \sum_{n=0}^{N} a^n$ satisfy
$$
(1-a)s_N = s_N(1-a) = 1 - a^{N+1},
$$
and $\|a^{N+1}\| \to 0$; since multiplication is continuous, letting $N \to \infty$ gives $(1-a)s = s(1-a) = 1$. The norm estimate is $\|s\| \le \sum_n \|a\|^n = (1-\|a\|)^{-1}$.

**2.** Let $b$ be invertible with $\|c - b\| < \|b^{-1}\|^{-1}$ and put $x := b^{-1}(b-c)$. Then $\|x\| \le \|b^{-1}\|\,\|b-c\| < 1$ and $c = b(1-x)$. By part 1 the element $1-x$ is invertible, and $b$ is invertible, so $c$ is invertible.

**3.** With the same notation $c^{-1} = (1-x)^{-1}b^{-1}$, so using the series of part 1,
$$
\|c^{-1} - b^{-1}\| = \Big\|\Big(\sum_{n \ge 1} x^n\Big) b^{-1}\Big\| \le \frac{\|x\|}{1 - \|x\|}\,\|b^{-1}\| .
$$
As $c \to b$ we have $\|x\| \le \|b^{-1}\|\|b-c\| \to 0$, so the right-hand side tends to $0$. Hence inversion is continuous at $b$.
</Proof>

<Theorem id="thm-spectrum-compact" title="The spectrum is nonempty and compact">
Let $A$ be a unital complex Banach algebra with $A \ne \{0\}$ and let $a \in A$. Then $\sigma(a)$ is a nonempty compact subset of $\mathbb{C}$, and
$$
\sigma(a) \subset \{\lambda \in \mathbb{C} : |\lambda| \le \|a\|\} .
$$
</Theorem>

<Proof of="thm-spectrum-compact">
**Boundedness.** If $|\lambda| > \|a\|$ then $\lambda \ne 0$ and $\|a/\lambda\| < 1$, so by part 1 of <Ref to="lem-neumann" /> the element $1 - a/\lambda$ is invertible; hence so is $\lambda 1 - a = \lambda(1 - a/\lambda)$, and $\lambda \notin \sigma(a)$. The contrapositive is the asserted inclusion.

**Closedness.** The map $\Phi(\lambda) = \lambda 1 - a$ is continuous, since $\|\Phi(\lambda) - \Phi(\mu)\| = |\lambda - \mu|$, and $G(A)$ is open by part 2 of <Ref to="lem-neumann" />; hence $\sigma(a) = \Phi^{-1}(A \setminus G(A))$ is closed. Being bounded and closed, $\sigma(a)$ is compact by <Ref to="mathematics/topology/compactness#thm-heine-borel" text="the Heine–Borel theorem" />.

**Nonemptiness.** Suppose $\sigma(a) = \emptyset$. Then $R(\lambda) = (\lambda 1 - a)^{-1}$ is defined on all of $\mathbb{C}$.

We first check that $R$ is analytic. Fix $\lambda_0 \in \rho(a)$ and let $|\lambda - \lambda_0| < \|R(\lambda_0)\|^{-1}$. Then
$$
\lambda 1 - a = (\lambda_0 1 - a) + (\lambda - \lambda_0)1 = (\lambda_0 1 - a)\bigl(1 + (\lambda-\lambda_0)R(\lambda_0)\bigr),
$$
and $\|(\lambda - \lambda_0)R(\lambda_0)\| < 1$, so part 1 of <Ref to="lem-neumann" /> gives
$$
R(\lambda) = \sum_{n=0}^{\infty} (-1)^n (\lambda - \lambda_0)^n R(\lambda_0)^{n+1}
$$
for $|\lambda - \lambda_0| < \|R(\lambda_0)\|^{-1}$. Consequently, for every continuous linear functional $\varphi \in A^{*}$ the function $\lambda \mapsto \varphi(R(\lambda))$ has a convergent power series expansion around every point, and is holomorphic on all of $\mathbb{C}$.

Next we examine the decay at infinity. For $|\lambda| > \|a\|$ we have $R(\lambda) = \lambda^{-1}(1 - a/\lambda)^{-1}$, so the norm estimate in part 1 of <Ref to="lem-neumann" /> gives
$$
\|R(\lambda)\| \le \frac{1}{|\lambda|}\cdot\frac{1}{1 - \|a\|/|\lambda|} = \frac{1}{|\lambda| - \|a\|} \xrightarrow[|\lambda| \to \infty]{} 0 .
$$

Thus $\varphi \circ R$ is an entire function tending to $0$ at infinity, and in particular bounded. By Liouville's theorem it is constant, and the constant is $0$. Since $\varphi \in A^{*}$ was arbitrary, the Hahn–Banach theorem gives $R(\lambda) = 0$ for every $\lambda$. But $R(0)(0 \cdot 1 - a) = 1$ then forces $1 = 0$, contradicting $\|1\| = 1 \ne 0$. Hence $\sigma(a) \ne \emptyset$.
</Proof>

<Aside type="note">
The use of Liouville's theorem in the proof of nonemptiness is the reason the algebra has to be *complex*. Over a real Banach algebra the rotation matrix $\begin{pmatrix} 0 & -1 \\ 1 & 0\end{pmatrix}$ has empty spectrum.
</Aside>

<Corollary id="cor-gelfand-mazur" title="Gelfand–Mazur theorem">
Let $A$ be a unital complex Banach algebra in which every nonzero element is invertible (that is, $A$ is a division algebra). Then $A = \mathbb{C}1$, and $\lambda 1 \mapsto \lambda$ is an isometric algebra isomorphism of $A$ onto $\mathbb{C}$.
</Corollary>

<Proof of="cor-gelfand-mazur">
From $\|1\| = 1$ we get $1 \ne 0$, hence $A \ne \{0\}$. Given $a \in A$, <Ref to="thm-spectrum-compact" /> gives $\sigma(a) \ne \emptyset$, so we may pick $\lambda \in \sigma(a)$. Then $\lambda 1 - a$ is not invertible, so by hypothesis $\lambda 1 - a = 0$, that is $a = \lambda 1$. If $\lambda 1 = \mu 1$ then $(\lambda-\mu)1 = 0$ and $\|1\| \ne 0$ give $\lambda = \mu$, so this $\lambda$ is unique. The map $\lambda 1 \mapsto \lambda$ is a bijection; multiplicativity is $(\lambda 1)(\mu 1) = (\lambda\mu)1$ and isometry is $\|\lambda 1\| = |\lambda|$.
</Proof>

<Example id="ex-spectrum-cx" title="Spectra in the two basic examples">
**The case of $C(X)$.** Let $X$ be a compact Hausdorff space and $f \in C(X)$. Then $\sigma(f) = f(X)$, the range of $f$. Indeed, if $\lambda \notin f(X)$ then $\lambda - f$ is nowhere zero, so $1/(\lambda-f)$ is continuous and $\lambda 1 - f$ is invertible. Conversely, if $\lambda = f(x_0)$ and some $g$ satisfied $g(\lambda 1 - f) = 1$, evaluating at $x_0$ would give $0 = 1$, a contradiction. The set $f(X)$ is compact, in agreement with <Ref to="thm-spectrum-compact" />.

**The case of $M_n(\mathbb{C})$.** For $a \in M_{n}(\mathbb{C})$, invertibility of $\lambda I - a$ is equivalent to $\det(\lambda I - a) \ne 0$, so $\sigma(a)$ is the set of eigenvalues. In infinite dimensions there are plenty of elements with no eigenvalues at all — the multiplication operator $(M_t f)(t) = tf(t)$ on $L^2[0,1]$ is one — so the spectrum is a genuine generalization of the set of eigenvalues.
</Example>

<Figure caption="The spectrum and the two radii. σ(a) is contained in the closed disc of radius ‖a‖, and r(a) is the radius of the smallest circle centred at 0 containing it. For normal elements of a C*-algebra the outer dashed circle coincides with the inner solid one.">
<svg viewBox="0 0 440 280" width="100%" role="img" aria-label="The spectrum and the spectral radius in the complex plane">
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    <text x="34" y="46">|λ| = ‖a‖</text>
    <text x="296" y="86">σ(a)</text>
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</Figure>

<Theorem id="thm-spectral-radius" title="Spectral radius formula (Gelfand–Beurling)">
Let $A$ be a unital complex Banach algebra with $A \ne \{0\}$, let $a \in A$, and set
$$
r(a) := \sup\{|\lambda| : \lambda \in \sigma(a)\}
$$
(by <Ref to="thm-spectrum-compact" /> the set $\sigma(a)$ is nonempty, bounded and closed, so this supremum is finite and attained). Then the limit $\lim_{n \to \infty}\|a^n\|^{1/n}$ exists and
$$
r(a) = \lim_{n \to \infty}\|a^n\|^{1/n} = \inf_{n \ge 1}\|a^n\|^{1/n} .
$$
We call $r(a)$ the **spectral radius** of $a$.
</Theorem>

<Proof of="thm-spectral-radius">
**Step 1: $r(a) \le \inf_n \|a^n\|^{1/n}$.** Fix $\lambda \in \sigma(a)$ and $n \ge 1$, and put $q = \sum_{k=0}^{n-1}\lambda^{n-1-k}a^k$. Telescoping gives
$$
(\lambda 1 - a)q = \sum_{k=0}^{n-1}\lambda^{n-k}a^k - \sum_{j=1}^{n}\lambda^{n-j}a^{j} = \lambda^n 1 - a^n,
$$
and since $q$ is a polynomial in $a$ we also have $q(\lambda 1 - a) = \lambda^n 1 - a^n$. If $\lambda^n 1 - a^n$ were invertible with inverse $b$, multiplying on the right by $b$ would give $(\lambda 1 - a)(qb) = 1$ and on the left $(bq)(\lambda 1 - a) = 1$. In general, if $xu = 1$ and $vx = 1$ then $v = v(xu) = (vx)u = u$ and $x$ is invertible; so $\lambda 1 - a$ would be invertible, contradicting $\lambda \in \sigma(a)$. Hence $\lambda^n \in \sigma(a^n)$, and <Ref to="thm-spectrum-compact" /> gives $|\lambda|^n \le \|a^n\|$, that is $|\lambda| \le \|a^n\|^{1/n}$. Taking the supremum yields $r(a) \le \|a^n\|^{1/n}$ for every $n$, hence $r(a) \le \inf_n \|a^n\|^{1/n} \le \liminf_n \|a^n\|^{1/n}$.

**Step 2: $\limsup_n \|a^n\|^{1/n} \le r(a)$.** Fix $\varphi \in A^{*}$. As in the proof of <Ref to="thm-spectrum-compact" />, the resolvent $R(\lambda)$ is analytic on $\rho(a) \supset \{|\lambda| > r(a)\}$. Put
$$
h(z) := \varphi\bigl(R(1/z)\bigr) \quad (0 < |z| < 1/r(a)), \qquad h(0) := 0
$$
(if $r(a) = 0$ we read $1/r(a) = \infty$ and the disc is all of $\mathbb{C}$). Then $h$ is holomorphic on the punctured disc. Moreover, for $|\lambda| > \|a\|$, part 1 of <Ref to="lem-neumann" /> gives
$$
R(\lambda) = \lambda^{-1}\Bigl(1 - \frac{a}{\lambda}\Bigr)^{-1} = \sum_{n=0}^{\infty}\frac{a^n}{\lambda^{n+1}},
$$
so for $|z| < 1/\|a\|$ we have the convergent power series $h(z) = \sum_{n \ge 0}\varphi(a^n)z^{n+1}$, whose value $0$ at $z = 0$ agrees with the definition of $h(0)$. Hence $h$ is holomorphic on the whole disc $|z| < 1/r(a)$, including $z = 0$, and its Taylor series converges on that whole disc; therefore
$$
\sum_{n \ge 0}\varphi(a^n)z^{n+1} \quad \text{converges for } |z| < 1/r(a) .
$$

Now take $|\lambda| > r(a)$ and put $z = 1/\lambda$. The terms of a convergent series tend to $0$, so $\varphi(a^n)/\lambda^{n+1} \to 0$, and therefore $\sup_n |\varphi(a^n/\lambda^n)| < \infty$ for every $\varphi \in A^{*}$. Viewed through the canonical embedding $A \hookrightarrow A^{**}$, the family $\{a^n/\lambda^n\}_n$ is a pointwise bounded family of functionals on the Banach space $A^{*}$, so the uniform boundedness principle gives $M_\lambda := \sup_n \|a^n/\lambda^n\| < \infty$ (the embedding is isometric by Hahn–Banach). Hence $\|a^n\|^{1/n} \le M_\lambda^{1/n}|\lambda| \to |\lambda|$, so $\limsup_n \|a^n\|^{1/n} \le |\lambda|$, and letting $|\lambda| \downarrow r(a)$ gives the claim.

**Conclusion.** Combining the two steps,
$$
r(a) \le \inf_n \|a^n\|^{1/n} \le \liminf_n \|a^n\|^{1/n} \le \limsup_n \|a^n\|^{1/n} \le r(a),
$$
so all of these are equal. In particular the limit exists and coincides with the infimum.
</Proof>

The left-hand side $r(a)$ is determined by the purely algebraic condition of invertibility, while the right-hand side is determined by the norm alone. That these two agree is the point; in the next section we combine it with the $C^*$-identity to conclude that the norm is determined by the algebraic structure.

## 4. Involutions and the axioms of a C*-algebra

Nothing so far involves the adjoint $T \mapsto T^{*}$, a structure specific to operators on a Hilbert space. The involution is its axiomatization.

<Definition id="def-cstar" title="Involution, Banach *-algebra, C*-algebra">
1. A map $a \mapsto a^{*}$ on a complex algebra $A$ is an **involution** if for all $a, b \in A$ and $\lambda \in \mathbb{C}$
   $$
   (a+b)^{*} = a^{*} + b^{*}, \quad (\lambda a)^{*} = \bar{\lambda}\,a^{*}, \quad (ab)^{*} = b^{*}a^{*}, \quad (a^{*})^{*} = a .
   $$
   A complex algebra equipped with an involution is called a $*$-algebra.
2. A Banach algebra $A$ with an involution satisfying $\|a^{*}\| = \|a\|$ for all $a \in A$ is called a **Banach $*$-algebra**.
3. A Banach algebra $A$ with an involution satisfying
   $$
   \|a^{*}a\| = \|a\|^{2} \qquad (\forall a \in A)
   $$
   is called a **$C^*$-algebra**, and this identity is the **$C^*$-identity**.

An element with $a^{*} = a$ is **self-adjoint**, one with $a^{*}a = aa^{*}$ is **normal**, and one with $a^{*}a = aa^{*} = 1$ is **unitary**. An algebra homomorphism satisfying $\pi(a^{*}) = \pi(a)^{*}$ is a $*$-**homomorphism**, and a bijective one is a $*$-**isomorphism**.
</Definition>

The $C^*$-identity is only one equation, but a great deal follows from it. First, we see that isometry of the involution need not be assumed.

<Proposition id="prop-cstar-norm" title="Consequences of the C*-identity">
Let $A$ be a $C^*$-algebra.

1. $\|a^{*}\| = \|a\|$ for every $a \in A$. Hence a $C^*$-algebra is automatically a Banach $*$-algebra.
2. If $a \in A$ is normal then $r(a) = \|a\|$. In particular this holds for self-adjoint and for unitary elements.
3. If $A \ne \{0\}$ has a unit $1$, then $1^{*} = 1$ and $\|1\| = 1$.
</Proposition>

<Proof of="prop-cstar-norm">
**1.** From $0^{*} = (0+0)^{*} = 0^{*} + 0^{*}$ we get $0^{*} = 0$, so equality holds for $a = 0$. If $a \ne 0$, the $C^*$-identity and submultiplicativity give
$$
\|a\|^{2} = \|a^{*}a\| \le \|a^{*}\|\,\|a\| ,
$$
and dividing by $\|a\| > 0$ yields $\|a\| \le \|a^{*}\|$. Applying this to $a^{*}$ gives $\|a^{*}\| \le \|(a^{*})^{*}\| = \|a\|$, whence equality.

**2.** Consider first a self-adjoint $b$. The $C^*$-identity gives $\|b^{2}\| = \|b^{*}b\| = \|b\|^{2}$. Since $(b^{m})^{*} = (b^{*})^{m} = b^{m}$, every power $b^{m}$ is self-adjoint too, and induction on $k$ gives
$$
\|b^{2^{k+1}}\| = \|(b^{2^{k}})^{2}\| = \|b^{2^{k}}\|^{2} = \bigl(\|b\|^{2^{k}}\bigr)^{2} = \|b\|^{2^{k+1}}
$$
(the case $k = 0$ being the identity just proved). By <Ref to="thm-spectral-radius" /> the limit $\lim_n \|b^{n}\|^{1/n}$ exists, so it may be computed along the subsequence $n = 2^{k}$:
$$
r(b) = \lim_{k \to \infty}\|b^{2^{k}}\|^{1/2^{k}} = \|b\| .
$$

Now let $a$ be normal. Since $a$ and $a^{*}$ commute, $(a^{*}a)^{n} = (a^{*})^{n}a^{n} = (a^{n})^{*}a^{n}$, and the $C^*$-identity gives $\|(a^{*}a)^{n}\| = \|a^{n}\|^{2}$. The element $a^{*}a$ is self-adjoint, since $(a^{*}a)^{*} = a^{*}a^{**} = a^{*}a$, so the previous paragraph together with <Ref to="thm-spectral-radius" /> gives
$$
\|a\|^{2} = \|a^{*}a\| = r(a^{*}a) = \lim_{n}\|(a^{*}a)^{n}\|^{1/n} = \lim_{n}\bigl(\|a^{n}\|^{1/n}\bigr)^{2} = r(a)^{2} .
$$
Both sides are nonnegative, so $\|a\| = r(a)$.

**3.** For any $a$ we have $a\,1^{*} = \bigl((1^{*})^{*}a^{*}\bigr)^{*} = (1\,a^{*})^{*} = a$, and similarly $1^{*}a = a$, so $1^{*}$ is also a unit; by uniqueness of the unit, $1^{*} = 1$. Then $\|1\| = \|1^{*}1\| = \|1\|^{2}$ gives $\|1\| \in \{0,1\}$, and $A \ne \{0\}$ forces $\|1\| = 1$.
</Proof>

<Aside type="caution">
Normality cannot be dropped in part 2; <Ref to="ex-nilpotent" /> is a counterexample. The $C^*$-identity is information about the pair consisting of $a$ and its adjoint, not a device for computing the norm of $a$ alone from its spectrum.
</Aside>

<Corollary id="cor-norm-unique" title="Uniqueness of the C*-norm">
Let $A$ be a $*$-algebra. There is at most one norm on $A$ making it a $C^*$-algebra: if $\|\cdot\|_{1}$ and $\|\cdot\|_{2}$ both make $A$ a $C^*$-algebra, then $\|\cdot\|_{1} = \|\cdot\|_{2}$.
</Corollary>

<Proof of="cor-norm-unique">
Whether an element is invertible depends only on the algebraic structure of $A$, not on the norm. The unitization $\tilde{A} = A \oplus \mathbb{C}$ is also defined as an algebra without reference to a norm, so $\sigma(a)$, and hence $r(a)$, do not depend on the norm either. Given $a \in A$, the element $a^{*}a$ is self-adjoint, so part 2 of <Ref to="prop-cstar-norm" /> applies to each $\|\cdot\|_{i}$ ($i = 1, 2$), and together with the $C^*$-identity
$$
\|a\|_{i}^{2} = \|a^{*}a\|_{i} = r(a^{*}a) \qquad (i = 1, 2) .
$$
The right-hand side does not depend on $i$, so $\|a\|_{1} = \|a\|_{2}$.
</Proof>

<Remark id="rem-algebraic" title="What this uniqueness means">
<Ref to="cor-norm-unique" /> is decisive for noncommutative geometry. The norm is not extra data but a quantity determined uniquely by the $*$-algebraic structure, so the policy of *defining* a noncommutative space to be a $C^*$-algebra introduces no analytic choices. For general Banach $*$-algebras the situation is different. The algebra $\ell^{1}(\mathbb{Z})$ of <Ref to="ex-wiener" /> is a $*$-subalgebra densely embedded in $C(\mathbb{T})$, but the $\ell^{1}$ norm and the supremum norm do not agree. Freedom in choosing the norm means that geometric information lies outside the algebra. The $C^*$-identity removes that freedom.
</Remark>

Let us check the basic examples, beginning with $B(H)$ from <Ref to="ex-bh" />. For $T \in B(H)$ the Riesz representation theorem produces a unique adjoint $T^{*}$, characterized by $\langle Tx, y\rangle = \langle x, T^{*}y\rangle$, and the axioms of an involution follow from the properties of the inner product. Isometry follows from
$$
\|T^{*}y\| = \sup_{\|x\| \le 1}|\langle T^{*}y, x\rangle| = \sup_{\|x\| \le 1}|\langle y, Tx\rangle| \le \|y\|\,\|T\| ,
$$
which gives $\|T^{*}\| \le \|T\|$, and the reverse inequality by applying this to $T^{*}$. The $C^*$-identity follows from the Cauchy–Schwarz inequality: from
$$
\|Tx\|^{2} = \langle Tx, Tx\rangle = \langle T^{*}Tx, x\rangle \le \|T^{*}Tx\|\,\|x\| \le \|T^{*}T\|\,\|x\|^{2}
$$
we get $\|T\|^{2} \le \|T^{*}T\|$, while the reverse is $\|T^{*}T\| \le \|T^{*}\|\|T\| = \|T\|^{2}$. Hence $B(H)$ is a $C^*$-algebra; in particular $M_{n}(\mathbb{C})$ is one with the conjugate transpose $(A^{*})_{ij} = \overline{A_{ji}}$ as adjoint, and it is noncommutative for $n \ge 2$. Every norm-closed $*$-subalgebra inherits all the axioms, so the algebra $K(H)$ of compact operators is a $C^*$-algebra as well (without a unit when $\dim H = \infty$).

Next take $C_{0}(X)$ from <Ref to="ex-c0x" />. With complex conjugation $f^{*}(x) = \overline{f(x)}$ as involution we get $\|f^{*}f\|_{\infty} = \sup_{x}|f(x)|^{2} = \|f\|_{\infty}^{2}$, so the $C^*$-identity holds. As we shall see in the next section, there are no other commutative $C^*$-algebras.

<Example id="ex-nilpotent" title="For non-normal elements the norm is invisible to the spectrum">
Let $A = M_{2}(\mathbb{C})$ and $a = e_{12} = \begin{pmatrix} 0 & 1 \\ 0 & 0\end{pmatrix}$.

Since $\det(\lambda I - a) = \lambda^{2}$, <Ref to="ex-spectrum-cx" /> gives $\sigma(a) = \{0\}$ and hence $r(a) = 0$ (the same conclusion follows from $a^{2} = 0$ and <Ref to="thm-spectral-radius" />). On the other hand $a(x_{1}, x_{2})^{\mathsf{T}} = (x_{2}, 0)^{\mathsf{T}}$, so $\|a\boldsymbol{x}\| = |x_{2}| \le \|\boldsymbol{x}\|$ with equality at $\boldsymbol{x} = (0,1)^{\mathsf{T}}$, whence $\|a\| = 1$. Therefore
$$
r(a) = 0 < 1 = \|a\| .
$$
The $C^*$-identity itself is not violated: $a^{*} = e_{21}$, $a^{*}a = e_{22}$, and $\|a^{*}a\| = 1 = \|a\|^{2}$. What fails is normality, since $aa^{*} = e_{11} \ne e_{22} = a^{*}a$. This is exactly why part 2 of <Ref to="prop-cstar-norm" /> requires normality.
</Example>

<Remark id="rem-gns" title="The noncommutative Gelfand–Naimark theorem">
The examples above — norm-closed $*$-subalgebras of $B(H)$ — are in fact all of them. For every $C^*$-algebra $A$ there exist a Hilbert space $H$ and an isometric $*$-isomorphism $A \cong \pi(A) \subset B(H)$. The proof goes through the GNS construction, which builds representations out of positive functionals (Murphy, *C\*-Algebras and Operator Theory*, Chapter 3). We do not use this result here, but it is the justification for the intuition that "a $C^*$-algebra is an operator algebra".
</Remark>

<Remark id="rem-star-hom-contractive" title="*-homomorphisms are automatically continuous">
Let $A$ and $B$ be unital $C^*$-algebras and $\pi : A \to B$ a unit-preserving $*$-homomorphism. Then $\pi$ is contractive. Indeed, if $a$ is invertible then $\pi(a)\pi(a^{-1}) = \pi(a^{-1})\pi(a) = 1$ shows $\pi(a)$ is invertible, so $\sigma(\pi(a)) \subset \sigma(a)$ and hence $r(\pi(a)) \le r(a)$. Both $a^{*}a$ and $\pi(a)^{*}\pi(a) = \pi(a^{*}a)$ are self-adjoint, so part 2 of <Ref to="prop-cstar-norm" /> gives
$$
\|\pi(a)\|^{2} = \|\pi(a)^{*}\pi(a)\| = r(\pi(a^{*}a)) \le r(a^{*}a) = \|a^{*}a\| = \|a\|^{2} .
$$
This is the version for maps of the statement that the norm is determined by the algebraic structure (<Ref to="cor-norm-unique" />).
</Remark>

## 5. The Gelfand representation and the commutative Gelfand–Naimark theorem

We now assemble, in the commutative case, the procedure that recovers the space. The key idea is to regard the set of homomorphisms from the algebra to $\mathbb{C}$ as the set of points. In $C(X)$, evaluation $f \mapsto f(x)$ at a point $x$ is such a homomorphism, so this amounts to replacing a point by the algebraic operation "evaluation at that point".

<Definition id="def-character" title="Characters and the Gelfand spectrum">
Let $A$ be a commutative Banach algebra. A nonzero map $\chi : A \to \mathbb{C}$ that is linear and satisfies $\chi(ab) = \chi(a)\chi(b)$ is called a **character** of $A$. The set of all characters is written $\Omega(A)$; with the topology inherited from the weak-$*$ topology of the dual space $A^{*}$ it is called the **Gelfand spectrum** (character space, maximal ideal space) of $A$.
</Definition>

<Lemma id="lem-character" title="Basic properties of characters">
Let $A$ be a commutative Banach algebra and $\chi \in \Omega(A)$.

1. If $A$ is unital then $\chi(1) = 1$.
2. For every $a \in A$ we have $\chi(a) \in \sigma(a)$, and in particular $|\chi(a)| \le r(a) \le \|a\|$. Hence $\chi$ is continuous with $\|\chi\| \le 1$, and $\|\chi\| = 1$ if $A$ is unital.
3. If $A$ is a unital commutative $C^*$-algebra then $\chi(a^{*}) = \overline{\chi(a)}$ for every $a$; that is, $\chi$ is a $*$-homomorphism.
</Lemma>

<Proof of="lem-character">
**1.** Since $\chi \ne 0$ there is a $b$ with $\chi(b) \ne 0$; dividing $\chi(b) = \chi(b \cdot 1) = \chi(b)\chi(1)$ by $\chi(b)$ gives $\chi(1) = 1$.

**2.** Suppose first that $A$ is unital. If $a$ is invertible then $1 = \chi(aa^{-1}) = \chi(a)\chi(a^{-1})$, so $\chi(a) \ne 0$. Put $\lambda := \chi(a)$; then $\chi(\lambda 1 - a) = 0$, so by the contrapositive of what we just proved $\lambda 1 - a$ is not invertible, that is $\lambda \in \sigma(a)$. Hence $|\chi(a)| \le r(a) \le \|a\|$ by <Ref to="thm-spectrum-compact" />. Being linear and satisfying $|\chi(a)| \le \|a\|$, the map $\chi$ is continuous with $\|\chi\| \le 1$; in the unital case $\chi(1) = 1$ and $\|1\| = 1$ give $\|\chi\| = 1$.

If $A$ is not unital, it suffices to observe that $\tilde{\chi}(a + \lambda 1) := \chi(a) + \lambda$ is a character of $\tilde{A}$. Multiplicativity follows from
$$
\tilde{\chi}\bigl((a+\lambda 1)(b + \mu 1)\bigr) = \chi(ab + \lambda b + \mu a) + \lambda\mu = \chi(a)\chi(b) + \lambda\chi(b) + \mu\chi(a) + \lambda\mu = \tilde{\chi}(a + \lambda 1)\tilde{\chi}(b + \mu 1) .
$$
Applying the unital case to $\tilde{\chi}$ gives $\chi(a) = \tilde{\chi}(a) \in \sigma_{\tilde{A}}(a) = \sigma(a)$.

**3.** Let $a$ be self-adjoint first, and write $\chi(a) = \alpha + i\beta$ with $\alpha, \beta \in \mathbb{R}$. For every $t \in \mathbb{R}$, part 2 and <Ref to="thm-spectrum-compact" /> give
$$
|\chi(a + it1)|^{2} = |\alpha + i(\beta + t)|^{2} = \alpha^{2} + (\beta+t)^{2} \le \|a + it1\|^{2} .
$$
We estimate the right-hand side by the $C^*$-identity. Since $(a + it1)^{*} = a^{*} - it1^{*} = a - it1$ (using $1^{*} = 1$ from part 3 of <Ref to="prop-cstar-norm" />),
$$
\|a + it1\|^{2} = \|(a - it1)(a + it1)\| = \|a^{2} + t^{2}1\| \le \|a\|^{2} + t^{2}
$$
by the triangle inequality together with $\|1\| = 1$ and $\|a^{2}\| \le \|a\|^{2}$. Combining the two, for every $t \in \mathbb{R}$,
$$
\alpha^{2} + \beta^{2} + 2\beta t + t^{2} \le \|a\|^{2} + t^{2}, \qquad \text{that is} \qquad 2\beta t \le \|a\|^{2} - \alpha^{2} - \beta^{2} .
$$
The right-hand side is a constant independent of $t$, so if $\beta \ne 0$ we reach a contradiction by letting $t \to \pm\infty$. Hence $\chi(a) = \alpha \in \mathbb{R}$. For general $a$, the elements $h = (a + a^{*})/2$ and $k = (a - a^{*})/(2i)$ are self-adjoint with $a = h + ik$ and $a^{*} = h - ik$, and $\chi(h), \chi(k) \in \mathbb{R}$, so
$$
\chi(a^{*}) = \chi(h) - i\chi(k) = \overline{\chi(h) + i\chi(k)} = \overline{\chi(a)} .
$$
</Proof>

<Definition id="def-gelfand-transform" title="Gelfand transform">
Let $A$ be a commutative Banach algebra. For $a \in A$ the map
$$
\hat{a} : \Omega(A) \to \mathbb{C}, \qquad \hat{a}(\chi) = \chi(a)
$$
is the **Gelfand transform** of $a$, and $\Gamma : a \mapsto \hat{a}$ is the **Gelfand representation** of $A$.
</Definition>

<Theorem id="thm-gelfand-transform" title="The Gelfand representation">
Let $A$ be a unital commutative Banach algebra with $A \ne \{0\}$.

1. $\Omega(A)$ is a nonempty compact Hausdorff space in the weak-$*$ topology.
2. Each $\hat{a}$ is continuous, and $\Gamma : A \to C(\Omega(A))$ is a unit-preserving algebra homomorphism.
3. $\chi \mapsto \ker\chi$ is a bijection from $\Omega(A)$ onto the set of maximal ideals of $A$.
4. $\sigma(a) = \hat{a}(\Omega(A))$ for every $a$; hence $\|\hat{a}\|_{\infty} = r(a) \le \|a\|$ and $\Gamma$ is contractive.
</Theorem>

<Proof of="thm-gelfand-transform">
**2.** The function $\hat{a}$ is the restriction to $\Omega(A)$ of the weak-$*$ continuous function $\varphi \mapsto \varphi(a)$, hence continuous. Linearity and multiplicativity of $\Gamma$ follow from $\widehat{a+b}(\chi) = \chi(a)+\chi(b)$ and $\widehat{ab}(\chi) = \chi(a)\chi(b)$, and unitality from $\hat{1}(\chi) = \chi(1) = 1$ (part 1 of <Ref to="lem-character" />).

**1.** By part 2 of <Ref to="lem-character" />, $\Omega(A)$ is contained in the closed unit ball $B$ of $A^{*}$, which is weak-$*$ compact by the Banach–Alaoglu theorem. Moreover
$$
\Omega(A) = \{\varphi \in B : \varphi(1) = 1\} \cap \bigcap_{a, b \in A}\{\varphi \in B : \varphi(ab) - \varphi(a)\varphi(b) = 0\}
$$
(the condition $\varphi(1) = 1$ guarantees $\varphi \ne 0$). For each $c \in A$ the map $\varphi \mapsto \varphi(c)$ is weak-$*$ continuous, and products and differences of continuous functions are continuous, so the right-hand side is an intersection of closed sets and hence closed. A closed subset of a compact set is compact (<Ref to="mathematics/topology/compactness#thm-closed-in-compact" />), so $\Omega(A)$ is compact, and it is Hausdorff because the weak-$*$ topology is. Nonemptiness follows from part 3 together with the existence of maximal ideals in a unital commutative algebra (Zorn's lemma).

**3.** Since $\chi(1) = 1$, the map $\chi$ is an algebra homomorphism onto $\mathbb{C}$, so $A/\ker\chi \cong \mathbb{C}$ is a field, and as $A$ is commutative and unital, $\ker\chi$ is a maximal ideal.

*Injectivity.* Suppose $\ker\chi = \ker\psi =: M$. Since $\chi(a - \chi(a)1) = 0$ we have $A = M + \mathbb{C}1$. Both $\chi$ and $\psi$ vanish on $M$ and send $1$ to $1$, so they agree on all of $A$.

*Surjectivity.* Let $M$ be a maximal ideal. By part 1 of <Ref to="lem-neumann" /> the open ball $\{b : \|b-1\| < 1\}$ consists of invertible elements, and a proper ideal $M$ contains no invertible element (if $u \in M$ were invertible then $1 = u^{-1}u \in M$ and $M = A$), so $\|m - 1\| \ge 1$ for every $m \in M$. This inequality is preserved under taking closures, so $1 \notin \overline{M}$; and $\overline{M}$ is an ideal by continuity of the operations, so maximality of $M$ gives $\overline{M} = M$, that is, $M$ is closed.

Hence $A/M$ is a Banach space with the quotient norm $\|a + M\| = \inf_{m \in M}\|a-m\|$. For $m_{1}, m_{2} \in M$ we have $ab - (a-m_{1})(b-m_{2}) \in M$, so
$$
\|(a+M)(b+M)\| = \inf_{m \in M}\|ab - m\| \le \|a - m_{1}\|\,\|b - m_{2}\| ,
$$
and taking infima over $m_{1}, m_{2}$ gives submultiplicativity. From $\|m-1\| \ge 1$ we get $\|1+M\| \ge 1$, and $\|1+M\| \le \|1\| = 1$, so $\|1+M\| = 1$. Since $M$ is maximal and $A$ is commutative and unital, $A/M$ is a field, so by <Ref to="cor-gelfand-mazur" /> there is an isometric isomorphism $\iota : A/M \to \mathbb{C}$. Composing with the quotient map $\pi$ gives a character $\chi := \iota\circ\pi$ with $\ker\chi = M$.

**4.** Let $\lambda \in \sigma(a)$. Then $(\lambda 1 - a)A$ is a proper ideal: if $(\lambda 1 - a)b = 1$ then, $A$ being commutative, $b$ would be a two-sided inverse and $\lambda 1 - a$ would be invertible. The union of a chain of proper ideals is a proper ideal, so by Zorn's lemma this ideal is contained in a maximal ideal $M$. By part 3 there is a $\chi$ with $M = \ker\chi$, and then $0 = \chi(\lambda 1 - a) = \lambda - \chi(a)$, that is $\hat{a}(\chi) = \lambda$. Conversely, if $\lambda = \chi(a)$ then $\lambda \in \sigma(a)$ by part 2 of <Ref to="lem-character" />. Hence $\sigma(a) = \hat{a}(\Omega(A))$ and
$$
\|\hat{a}\|_{\infty} = \sup\{|\lambda| : \lambda \in \sigma(a)\} = r(a) \le \|a\|
$$
(the last inequality by <Ref to="thm-spectrum-compact" />), so $\Gamma$ is contractive.
</Proof>

<Example id="ex-wiener" title="The Wiener algebra ℓ¹(Z): the Gelfand transform is the Fourier series">
Let $A = \ell^{1}(\mathbb{Z})$ with the convolution product $(a * b)_{n} = \sum_{k}a_{k}b_{n-k}$ and the norm $\|a\|_{1} = \sum_{n}|a_{n}|$. The inequality $\|a*b\|_{1} \le \|a\|_{1}\|b\|_{1}$ follows from the triangle inequality and Tonelli's theorem. The algebra is commutative with unit $\delta_{0}$, and the involution $(a^{*})_{n} = \overline{a_{-n}}$ is isometric, so $A$ is a Banach $*$-algebra.

**Determining all characters.** Let $\chi \in \Omega(A)$ and put $z := \chi(\delta_{1})$. Since $\delta_{1} * \delta_{-1} = \delta_{0}$, the element $\delta_{1}$ is invertible, so part 2 of <Ref to="lem-character" /> gives $|z| \le \|\delta_{1}\|_{1} = 1$ and $|z^{-1}| = |\chi(\delta_{-1})| \le 1$, whence $|z| = 1$. Each $\delta_{n}$ is an $|n|$-fold convolution of $\delta_{\pm 1}$, so $\chi(\delta_{n}) = z^{n}$ for all $n \in \mathbb{Z}$; the finitely supported sequences are dense and $\chi$ is continuous, so
$$
\chi(a) = \sum_{n \in \mathbb{Z}} a_{n}z^{n} .
$$
Conversely, for $|z| = 1$ the right-hand side converges absolutely and defines a character (multiplicativity from the definition of convolution). Hence $\Omega(\ell^{1}(\mathbb{Z}))$ is identified with the unit circle $\mathbb{T}$, and the Gelfand transform $\hat{a}(z) = \sum_{n}a_{n}z^{n}$ is precisely the absolutely convergent Fourier series.

**Wiener's theorem.** If $f(z) = \sum_{n}a_{n}z^{n}$ with $\sum_{n}|a_{n}| < \infty$ has no zero on $|z| = 1$, then $1/f$ also has an absolutely convergent Fourier series. Indeed, by part 4 of <Ref to="thm-gelfand-transform" /> we have $\sigma(a) = f(\mathbb{T}) \not\ni 0$, so $a$ is invertible in $\ell^{1}(\mathbb{Z})$, and $b = a^{-1}$ satisfies $\hat{b}\hat{a} = 1$, that is $\sum_{n}b_{n}z^{n} = 1/f(z)$ with $\sum_{n}|b_{n}| < \infty$. A purely algebraic argument about invertibility yields an analytic conclusion.

**But this is not a $C^*$-algebra.** Take $a = \delta_{0} + \delta_{1} - \delta_{2}$, so that $\|a\|_{1} = 3$. Computing the autocorrelation $(a^{*}*a)_{m} = \sum_{j}\overline{a_{j}}a_{j+m}$ gives
$$
(a^{*}*a)_{0} = 3, \quad (a^{*}*a)_{\pm 1} = 0, \quad (a^{*}*a)_{\pm 2} = -1,
$$
so $\|a^{*}*a\|_{1} = 5 \ne 9 = \|a\|_{1}^{2}$ and the $C^*$-identity fails. The same thing is visible on the Gelfand side: $|\hat{a}(e^{i\theta})|^{2} = \widehat{a^{*}*a}(e^{i\theta}) = 3 - 2\cos 2\theta$, so $r(a) = \|\hat{a}\|_{\infty} = \sqrt{5} < 3 = \|a\|_{1}$, whereas in a commutative $C^*$-algebra these two would have to agree by part 2 of <Ref to="prop-cstar-norm" />. The $C^*$-completion of $\ell^{1}(\mathbb{Z})$ is exactly $C(\mathbb{T})$.
</Example>

<Theorem id="thm-gelfand-naimark" title="Gelfand–Naimark theorem (commutative case)">
1. If $A$ is a unital commutative $C^*$-algebra with $A \ne \{0\}$, then $\Gamma : A \to C(\Omega(A))$ is a surjective isometric $*$-isomorphism.
2. If $X$ is a compact Hausdorff space, then $x \mapsto \mathrm{ev}_{x}$, where $\mathrm{ev}_{x}(f) = f(x)$, is a homeomorphism $X \to \Omega(C(X))$.
3. If $A$ is a commutative $C^*$-algebra with $A \ne \{0\}$, not necessarily unital, then $\Omega(A)$ is a locally compact Hausdorff space and $\Gamma : A \to C_{0}(\Omega(A))$ is a surjective isometric $*$-isomorphism. Moreover $A$ is unital if and only if $\Omega(A)$ is compact.
</Theorem>

<Proof of="thm-gelfand-naimark">
**1.** *$*$-homomorphism.* By part 3 of <Ref to="lem-character" /> we have $\widehat{a^{*}}(\chi) = \overline{\hat{a}(\chi)}$ for every $\chi$, so $\Gamma(a^{*}) = \Gamma(a)^{*}$. That $\Gamma$ is an algebra homomorphism was proved in part 2 of <Ref to="thm-gelfand-transform" />.

*Isometry.* Since $A$ is commutative, every element is normal. Part 2 of <Ref to="prop-cstar-norm" /> gives $\|a\| = r(a)$ and part 4 of <Ref to="thm-gelfand-transform" /> gives $r(a) = \|\hat{a}\|_{\infty}$, so $\|\Gamma(a)\|_{\infty} = \|a\|$; in particular $\Gamma$ is injective.

*Surjectivity.* The image $B := \Gamma(A)$ is a subalgebra containing the constants, since $\Gamma(1) = 1$, and closed under complex conjugation as just seen. Moreover, if $\chi \ne \psi$ then by definition there is an $a$ with $\chi(a) \ne \psi(a)$, so $B$ separates the points of $\Omega(A)$. By part 1 of <Ref to="thm-gelfand-transform" /> the space $\Omega(A)$ is compact Hausdorff, so the Stone–Weierstrass theorem makes $B$ dense in $C(\Omega(A))$. On the other hand $\Gamma$ is isometric and $A$ is complete, so $B$ is a closed subspace. Being dense and closed, $B = C(\Omega(A))$.

**2.** Linearity and multiplicativity are the very definition of the pointwise operations, and $\mathrm{ev}_{x}(1) = 1 \ne 0$, so $\mathrm{ev}_{x}$ is a character.

*Injectivity.* If $x \ne y$, then since a compact Hausdorff space is normal, <Ref to="mathematics/topology/separation-axioms#lem-urysohn" text="Urysohn's lemma" /> provides an $f$ with $f(x) = 1$ and $f(y) = 0$, so $\mathrm{ev}_{x} \ne \mathrm{ev}_{y}$.

*Continuity.* If $x_{i} \to x$ then $f(x_{i}) \to f(x)$ for every $f$, which is precisely $\mathrm{ev}_{x_{i}} \to \mathrm{ev}_{x}$ in the weak-$*$ topology.

*Surjectivity.* Let $\chi \in \Omega(C(X))$ and put $M = \ker\chi$. Suppose that for every $x \in X$ there were an $f_{x} \in M$ with $f_{x}(x) \ne 0$. Since $M$ is an ideal, $g_{x} := |f_{x}|^{2} \in M$ with $g_{x} \ge 0$ and $g_{x}(x) > 0$. The open sets $U_{x} = \{y : g_{x}(y) > 0\}$ cover $X$, so by compactness $X = \bigcup_{i \le n}U_{x_{i}}$, and $g := \sum_{i}g_{x_{i}} \in M$ is everywhere positive. Then $g$ is invertible, contradicting the properness of $M$. Hence there is some $x_{0}$ with $f(x_{0}) = 0$ for all $f \in M$, that is $M \subset \ker\mathrm{ev}_{x_{0}}$. By part 3 of <Ref to="thm-gelfand-transform" /> both are maximal ideals, so $M = \ker\mathrm{ev}_{x_{0}}$, and by the injectivity in the same part 3, $\chi = \mathrm{ev}_{x_{0}}$.

Thus $x \mapsto \mathrm{ev}_{x}$ is a continuous bijection, $X$ is compact and $\Omega(C(X))$ is Hausdorff (part 1 of <Ref to="thm-gelfand-transform" />), so it is a homeomorphism (<Ref to="mathematics/topology/compactness#cor-compact-hausdorff-homeo" />).

**3.** This goes through the unitization; we prove it in the Appendix.
</Proof>

Putting 1 and 2 together: applying $C(-)$ to $\Omega(A)$ returns $A$, and applying $\Omega(-)$ to $C(X)$ returns $X$. The category of compact Hausdorff spaces and the category of unital commutative $C^*$-algebras are contravariantly equivalent (the correspondence of morphisms is <Ref to="exr-functoriality" />). Under this correspondence every topological notion translates into an algebraic one.

| On the side of the space $X$ | On the side of the commutative $C^*$-algebra $A = C_{0}(X)$ |
|---|---|
| $X$ is compact | $A$ has a unit |
| a point $x \in X$ | the character $\mathrm{ev}_{x} \in \Omega(A)$ (a maximal ideal of codimension $1$) |
| a proper continuous map $\varphi : X \to Y$ | a $*$-homomorphism $C_{0}(Y) \to C_{0}(X)$ |
| an open set $U \subset X$ | a closed two-sided ideal $C_{0}(U) \subset A$ |
| a closed set $F \subset X$ | the quotient algebra $A/C_{0}(X \setminus F) \cong C_{0}(F)$ |
| $X$ is compact and connected | the only projections of $A$ (elements with $p = p^{*} = p^{2}$) are $0$ and $1$ |
| $X$ is compact and metrizable | $A$ is separable |
| a complex vector bundle over $X$ | a finitely generated projective module over $A$ (<Ref to="mathematics/noncommutative-geometry/k-theory#thm-serre-swan" text="the Serre–Swan theorem" />) |

Every entry in the right-hand column can be stated without using commutativity. This is the starting point of noncommutative geometry: one regards a general $C^*$-algebra $A$ as "$C_{0}$ of a noncommutative space" and adopts the notions in the right-hand column as definitions. The last row is the gateway to $K$-theory; see [Introduction to K-theory](/mathematics/noncommutative-geometry/k-theory) for details. For a concrete noncommutative example, [the noncommutative torus](/mathematics/noncommutative-geometry/noncommutative-torus) is I think the most transparent one.

## 6. The continuous functional calculus

The commutative Gelfand–Naimark theorem is a theorem about commutative $C^*$-algebras, but even in a noncommutative $C^*$-algebra, a single normal element $a$ commutes with $a^{*}$, so the subalgebra generated by $a$ is commutative. Applying the commutative theorem there produces a tool usable in the noncommutative world: we may "substitute" $a$ into a continuous function $f$.

<Remark id="rem-spectral-permanence" title="Permanence of the spectrum">
Let $A$ be a unital $C^*$-algebra and $B \subset A$ a $C^*$-subalgebra (a norm-closed $*$-subalgebra) containing the same unit. Then $\sigma_{B}(b) = \sigma_{A}(b)$ for every $b \in B$. The inclusion $\sigma_{A}(b) \subset \sigma_{B}(b)$ follows because invertibility in $B$ implies invertibility in $A$, but the reverse inclusion is nontrivial, and it is false for general Banach algebras. The disc algebra of <Ref to="exr-disc-algebra" /> may, by the maximum principle, be regarded as a closed subalgebra of $C(\mathbb{T})$; the spectrum of the coordinate function $z$ is $\mathbb{T}$ in $C(\mathbb{T})$ but the closed disc $\overline{\mathbb{D}}$ in the disc algebra, so the hole gets filled in on the side of the subalgebra. This cannot happen in a $C^*$-algebra because the spectrum of a self-adjoint element is contained in the real line and so has no hole to fill. A proof is in Murphy, *C\*-Algebras and Operator Theory*, Chapter 2. From now on we use this fact and simply write $\sigma(b)$.
</Remark>

<Corollary id="cor-functional-calculus" title="Continuous functional calculus">
Let $A$ be a unital $C^*$-algebra, let $a \in A$ be normal, and let $C^{*}(a,1)$ be the smallest norm-closed $*$-subalgebra containing $a$, $a^{*}$ and $1$.

1. $C^{*}(a,1)$ is a unital commutative $C^*$-algebra, and $\hat{a} : \Omega(C^{*}(a,1)) \to \sigma(a)$ is a homeomorphism.
2. There is exactly one unit-preserving $*$-isomorphism $\Phi : C(\sigma(a)) \to C^{*}(a,1)$ sending the identity function $\mathrm{id}$ to $a$, and it is isometric. For $f \in C(\sigma(a))$ we write $f(a) := \Phi(f)$.
3. (Spectral mapping theorem) For every $f \in C(\sigma(a))$ we have $\sigma(f(a)) = f(\sigma(a))$, and for every $g \in C(\sigma(f(a)))$ we have $(g \circ f)(a) = g(f(a))$.
</Corollary>

<Proof of="cor-functional-calculus">
**1.** The $*$-subalgebra $P$ generated by $a$, $a^{*}$ and $1$ consists of all polynomials in $a$ and $a^{*}$. Since $a$ is normal, $a$ and $a^{*}$ commute, so any two elements of $P$ commute, and by continuity of multiplication so do any two elements of its closure. Hence $B := C^{*}(a,1) = \overline{P}$ is a commutative norm-closed $*$-subalgebra, that is, a commutative $C^*$-algebra.

Applying part 4 of <Ref to="thm-gelfand-transform" /> to $B$ gives $\hat{a}(\Omega(B)) = \sigma_{B}(a)$, which equals $\sigma(a)$ by <Ref to="rem-spectral-permanence" />, so $\hat{a} : \Omega(B) \to \sigma(a)$ is surjective (continuity is part 2 of the same theorem). Injectivity goes as follows. If $\chi(a) = \psi(a)$, then part 3 of <Ref to="lem-character" /> gives $\chi(a^{*}) = \overline{\chi(a)} = \psi(a^{*})$, and part 1 gives $\chi(1) = \psi(1) = 1$, so $\chi$ and $\psi$ agree on $P$ as algebra homomorphisms; both are continuous (part 2) and $P$ is dense, so they agree on all of $B$. Since $\Omega(B)$ is compact and $\sigma(a)$ is Hausdorff, the continuous bijection $\hat{a}$ is a homeomorphism.

**2.** The pullback $\hat{a}^{\sharp} : C(\sigma(a)) \to C(\Omega(B))$, $f \mapsto f \circ \hat{a}$, is a unital isometric $*$-isomorphism (bijectivity because $\hat{a}$ is a homeomorphism, isometry from $\|f \circ \hat{a}\|_{\infty} = \|f\|_{\infty}$). By part 1 of <Ref to="thm-gelfand-naimark" /> the map $\Gamma : B \to C(\Omega(B))$ is an isometric $*$-isomorphism too, so setting
$$
\Phi := \Gamma^{-1} \circ \hat{a}^{\sharp} : C(\sigma(a)) \longrightarrow B
$$
gives a unital isometric $*$-isomorphism with $\Phi(\mathrm{id}) = \Gamma^{-1}(\hat{a}) = a$.

*Uniqueness.* Let $\Psi : C(\sigma(a)) \to A$ be a unital $*$-homomorphism with $\Psi(\mathrm{id}) = a$. Then $\Psi(\overline{\mathrm{id}}) = a^{*}$, so $\Phi$ and $\Psi$ agree on all polynomials in $z$ and $\bar{z}$. These polynomials separate the points of $\sigma(a)$, contain the constants and are closed under complex conjugation, so by the Stone–Weierstrass theorem they are dense in $C(\sigma(a))$. Both $\Phi$ and $\Psi$ are continuous (<Ref to="rem-star-hom-contractive" />), so they coincide.

**3.** Being a $*$-isomorphism, $\Phi$ preserves invertibility and hence spectra. Therefore
$$
\sigma_{B}(f(a)) = \sigma_{C(\sigma(a))}(f) = f(\sigma(a))
$$
(the last equality by <Ref to="ex-spectrum-cx" />, since $\sigma(a)$ is a compact Hausdorff space). By <Ref to="rem-spectral-permanence" />, $\sigma_{B}(f(a)) = \sigma_{A}(f(a))$.

As for composition: $f(a) \in B$ is normal, so part 2 provides a map $\Phi_{f(a)} : C(\sigma(f(a))) \to A$ for $f(a)$. On the other hand $g \mapsto \Phi_{a}(g \circ f)$ is a unital $*$-homomorphism with the same domain and codomain, being the composite of $g \mapsto g\circ f$ with $\Phi_{a}$. Both send $\mathrm{id}$ to $f(a)$, so by the uniqueness in part 2 they coincide; that is, $(g\circ f)(a) = g(f(a))$.
</Proof>

<Example id="ex-sqrt" title="Computing a square root explicitly">
**The finite-dimensional case.** Let $A = M_{2}(\mathbb{C})$ and $a = \begin{pmatrix} 2 & 1 \\ 1 & 2\end{pmatrix}$. This $a$ is self-adjoint, hence normal, and from the characteristic polynomial $(2-\lambda)^{2} - 1 = \lambda^{2}-4\lambda+3$ we get $\sigma(a) = \{1, 3\}$ by <Ref to="ex-spectrum-cx" />. The function $f(t) = \sqrt{t}$ is continuous on $\sigma(a)$, so $f(a)$ is defined. Since $\sigma(a)$ has two points, $C(\sigma(a)) \cong \mathbb{C}^{2}$ and $f$ is reproduced exactly by a polynomial. Lagrange interpolation gives $p(t) = \frac{\sqrt{3}-1}{2}t + \frac{3-\sqrt{3}}{2}$, and indeed
$$
p(1) = \frac{\sqrt{3}-1}{2} + \frac{3-\sqrt{3}}{2} = \frac{2}{2} = 1 = \sqrt{1}, \qquad p(3) = \frac{3\sqrt{3}-3}{2} + \frac{3-\sqrt{3}}{2} = \frac{2\sqrt{3}}{2} = \sqrt{3},
$$
so $p = f$ on $\sigma(a)$ and therefore $f(a) = p(a)$. Computing,
$$
f(a) = \frac{\sqrt{3}-1}{2}\begin{pmatrix} 2 & 1 \\ 1 & 2\end{pmatrix} + \frac{3-\sqrt{3}}{2}\begin{pmatrix} 1 & 0 \\ 0 & 1\end{pmatrix} = \frac{1}{2}\begin{pmatrix} \sqrt{3}+1 & \sqrt{3}-1 \\ \sqrt{3}-1 & \sqrt{3}+1\end{pmatrix} .
$$
Let us check this. With $u = (\sqrt{3}+1)/2$ and $v = (\sqrt{3}-1)/2$ we get $u^{2}+v^{2} = \frac{(4+2\sqrt{3}) + (4-2\sqrt{3})}{4} = 2$ and $2uv = 2\cdot\frac{3-1}{4} = 1$, so indeed $f(a)^{2} = \begin{pmatrix} 2 & 1 \\ 1 & 2\end{pmatrix} = a$.

**The infinite-dimensional case.** Let $H = L^{2}[0,1]$ and $a = M_{t}$, where $(M_{t}\xi)(t) = t\,\xi(t)$. The operator $M_{t}$ is self-adjoint; for $\lambda \notin [0,1]$ the operator $M_{1/(\lambda-t)}$ is an inverse, while for $\lambda \in [0,1]$ the range of $M_{\lambda-t}$ is dense but not all of $H$, so it is not invertible. Hence $\sigma(a) = [0,1]$. The map $f \mapsto M_{f}$ is a unital $*$-homomorphism $C[0,1] \to B(H)$ sending $\mathrm{id}$ to $M_{t}$, so the uniqueness in part 2 of <Ref to="cor-functional-calculus" /> gives $f(a) = M_{f}$, and in particular $\sqrt{a} = M_{\sqrt{t}}$. Since $\sqrt{t}$ agrees with no polynomial on $[0,1]$, the element $\sqrt{a}$ is obtained only as a limit of uniform approximations. One sees here that the continuous functional calculus is "the completion of polynomial calculus".
</Example>

The continuous functional calculus is the workhorse of the theory. If $a$ is self-adjoint one can show $\sigma(a) \subset \mathbb{R}$ (by part 3 of <Ref to="lem-character" /> and part 4 of <Ref to="thm-gelfand-transform" />, since characters take real values on the commutative subalgebra); if moreover $\sigma(a) \subset [0,\infty)$ we call $a$ **positive**, and then $a = b^{*}b$ with $b = \sqrt{a}$. The functional calculus is a machine for producing projections ($\sigma(a) \subset \{0,1\}$) and unitaries ($\sigma(a) \subset \mathbb{T}$), out of which the $K$-theory groups $K_{0}$ and $K_{1}$ are assembled. The operators $|D|$ and $D(1+D^{2})^{-1/2}$ used in spectral triples are defined by the same calculus (in its unbounded extension). See [Spectral triples (A, H, D)](/mathematics/noncommutative-geometry/spectral-triples) (<Ref to="mathematics/noncommutative-geometry/spectral-triples#def-spectral-triple" />) and [Applications to index theorems](/mathematics/noncommutative-geometry/index-theorem).

## 7. Exercises

<Exercise id="exr-spectrum-basics" difficulty="Easy">
Let $A$ be a unital complex Banach algebra, $a \in A$ and $\mu \in \mathbb{C}$.

1. Show that $\sigma(\mu 1 + a) = \mu + \sigma(a) := \{\mu + \lambda : \lambda \in \sigma(a)\}$.
2. Show that if $a$ is invertible then $0 \notin \sigma(a)$ and $\sigma(a^{-1}) = \{\lambda^{-1} : \lambda \in \sigma(a)\}$.

<Solution>
**1.** Since $\lambda 1 - (\mu 1 + a) = (\lambda - \mu)1 - a$, the conditions $\lambda \in \sigma(\mu 1 + a)$ and $\lambda - \mu \in \sigma(a)$, that is $\lambda \in \mu + \sigma(a)$, are equivalent.

**2.** If $a$ is invertible then so is $0 \cdot 1 - a = -a$, so $0 \notin \sigma(a)$; for the same reason $0 \notin \sigma(a^{-1})$. For $\lambda \ne 0$ we have the identity
$$
-\lambda\, a\,(\lambda^{-1}1 - a^{-1}) = -\lambda\bigl(\lambda^{-1}a - aa^{-1}\bigr) = -a + \lambda 1 = \lambda 1 - a .
$$
Here $-\lambda \ne 0$ is a scalar and $a$ is invertible, and multiplying by invertible elements does not affect invertibility. Hence $\lambda 1 - a$ and $\lambda^{-1}1 - a^{-1}$ are invertible or not together; taking contrapositives, $\lambda \in \sigma(a)$ is equivalent to $\lambda^{-1} \in \sigma(a^{-1})$. Neither spectrum contains $0$, so $\lambda \mapsto \lambda^{-1}$ is a bijection from $\sigma(a)$ onto $\sigma(a^{-1})$.
</Solution>
</Exercise>

<Exercise id="exr-selfadjoint-nilpotent" difficulty="Standard">
Let $A$ be a $C^*$-algebra.

1. Show that if $a \in A$ satisfies $a^{*}a = 0$ then $a = 0$.
2. Show that if $b \in A$ is self-adjoint and $b^{n} = 0$ for some $n \ge 1$, then $b = 0$.
3. Show by a counterexample that the conclusion of 2 fails without the hypothesis of self-adjointness.

<Solution>
**1.** The $C^*$-identity (part 3 of <Ref to="def-cstar" />) gives $\|a\|^{2} = \|a^{*}a\| = 0$, hence $a = 0$.

**2.** In the proof of part 2 of <Ref to="prop-cstar-norm" /> it was shown that a self-adjoint $b$ satisfies $\|b^{2^{k}}\| = \|b\|^{2^{k}}$ for $k = 0,1,2,\ldots$. Choosing $k$ with $2^{k} \ge n$ gives $b^{2^{k}} = b^{n}\,b^{2^{k}-n} = 0$, so $\|b\|^{2^{k}} = 0$, that is $b = 0$. The same follows from part 2 of <Ref to="prop-cstar-norm" /> together with <Ref to="thm-spectral-radius" />: $\|b\| = r(b) = \lim_{m}\|b^{m}\|^{1/m} = 0$, since $b^{m} = 0$ for $m \ge n$.

**3.** The element $a = e_{12} \in M_{2}(\mathbb{C})$ of <Ref to="ex-nilpotent" /> is a counterexample: $a^{2} = 0$ and $a \ne 0$, and $a$ is not self-adjoint ($a^{*} = e_{21} \ne e_{12}$). Indeed $\|a\| = 1$ while $r(a) = 0$, so $\|b\| = r(b)$ fails. In a commutative $C^*$-algebra $C_{0}(X)$ both 1 and 2 say the obvious thing, namely "if $|f|^{2} = 0$ then $f = 0$"; the strength of the $C^*$-identity is that this follows from the axioms alone.
</Solution>
</Exercise>

<Exercise id="exr-disc-algebra" difficulty="Standard">
Let $\mathbb{D} = \{z \in \mathbb{C} : |z| < 1\}$ with closure $\overline{\mathbb{D}}$. The algebra
$$
A(\mathbb{D}) = \{f \in C(\overline{\mathbb{D}}) : f|_{\mathbb{D}} \text{ is holomorphic}\}
$$
is called the **disc algebra**.

1. Show that $A(\mathbb{D})$ is a unital commutative Banach algebra under the supremum norm.
2. Show that for each $w \in \overline{\mathbb{D}}$ the map $\mathrm{ev}_{w}$ is a character of $A(\mathbb{D})$. You may take for granted the converse, that every character of $A(\mathbb{D})$ is of this form (so that $\Omega(A(\mathbb{D})) \cong \overline{\mathbb{D}}$).
3. Show that $A(\mathbb{D})$ can never be a $C^*$-algebra under the supremum norm.

<Solution>
**1.** The disc $\overline{\mathbb{D}}$ is compact, so $C(\overline{\mathbb{D}})$ is a unital commutative Banach algebra by <Ref to="ex-c0x" />; sums, products and scalar multiples of holomorphic functions are holomorphic and the constant $1$ is holomorphic, so $A(\mathbb{D})$ is a subalgebra. If $f_{n} \in A(\mathbb{D})$ converges uniformly to $f$, then it converges uniformly on $\mathbb{D}$ as well, so by Weierstrass' theorem that a locally uniform limit of holomorphic functions is holomorphic we get $f \in A(\mathbb{D})$; that is, the subalgebra is closed. A closed subalgebra is a Banach algebra, which gives the claim.

**2.** The map $\mathrm{ev}_{w}$ is linear with $\mathrm{ev}_{w}(fg) = \mathrm{ev}_{w}(f)\mathrm{ev}_{w}(g)$, and $\mathrm{ev}_{w}(1) = 1 \ne 0$, so it is a character.

**3.** Suppose $A(\mathbb{D})$ were a $C^*$-algebra under the supremum norm and some involution. It is commutative, so by part 1 of <Ref to="thm-gelfand-naimark" /> the Gelfand representation $\Gamma$ is an isometric $*$-isomorphism onto $C(\Omega(A(\mathbb{D})))$. Under the identification granted in 2 we have $\hat{f}(\mathrm{ev}_{w}) = f(w)$, so $\Gamma$ sends $f$ to itself, and surjectivity would mean $A(\mathbb{D}) = C(\overline{\mathbb{D}})$.

But this is false. The function $g(z) = \bar{z}$ is continuous on $\overline{\mathbb{D}}$, yet writing $z = x+iy$ we have $u(x,y) = x$ and $v(x,y) = -y$, so the Cauchy–Riemann equation $u_{x} = v_{y}$ reads $1 = -1$ and $g$ is not holomorphic on $\mathbb{D}$. Hence $g \in C(\overline{\mathbb{D}}) \setminus A(\mathbb{D})$, a contradiction.

This example also shows that "the Gelfand transform is isometric" and "the algebra is a $C^*$-algebra" are different statements. Under the identification above, the Gelfand transform of $A(\mathbb{D})$ is the identity map and so is isometric, yet the image is not closed under complex conjugation, so the algebra is not a $C^*$-algebra. It is exactly here that the $C^*$-identity does its work (part 3 of <Ref to="lem-character" />).
</Solution>
</Exercise>

<Exercise id="exr-functoriality" difficulty="Hard">
Let $X$ and $Y$ be compact Hausdorff spaces.

1. Show that for a continuous map $\varphi : X \to Y$ the map $\varphi^{\sharp} : C(Y) \to C(X)$, $\varphi^{\sharp}(f) = f \circ \varphi$, is a unit-preserving $*$-homomorphism.
2. Conversely, show that every unit-preserving $*$-homomorphism $\Psi : C(Y) \to C(X)$ is of the form $\Psi = \varphi^{\sharp}$ for some continuous map $\varphi : X \to Y$.
3. Show that $\varphi$ is surjective if and only if $\varphi^{\sharp}$ is injective.

<Solution>
**1.** The function $f\circ\varphi$ is continuous, hence an element of $C(X)$, and the identities $(f+\lambda g)\circ\varphi = f\circ\varphi + \lambda(g\circ\varphi)$, $(fg)\circ\varphi = (f\circ\varphi)(g\circ\varphi)$, $\overline{f}\circ\varphi = \overline{f\circ\varphi}$ and $1\circ\varphi = 1$ are all verified pointwise.

**2.** Fix $x \in X$. Then $\mathrm{ev}_{x}\circ\Psi$ is an algebra homomorphism sending $1$ to $1$, hence a character of $C(Y)$. By part 2 of <Ref to="thm-gelfand-naimark" /> we have $\Omega(C(Y)) = \{\mathrm{ev}_{y}\}_{y \in Y}$ with $y \mapsto \mathrm{ev}_{y}$ injective, so there is a unique $\varphi(x) \in Y$ with $\mathrm{ev}_{x}\circ\Psi = \mathrm{ev}_{\varphi(x)}$. That is,
$$
\Psi(f)(x) = f(\varphi(x)) \qquad (f \in C(Y),\ x \in X),
$$
so $\Psi = \varphi^{\sharp}$.

It remains to prove that $\varphi$ is continuous. Let $x_{i} \to x$ be a net. Since $\Psi(f)$ is continuous, $f(\varphi(x_{i})) = \Psi(f)(x_{i}) \to \Psi(f)(x) = f(\varphi(x))$ for every $f$. If $\varphi(x_{i}) \to \varphi(x)$ failed, there would be an open neighbourhood $U$ of $\varphi(x)$ and a subnet $(x_{j})$ with $\varphi(x_{j}) \notin U$. Since $Y$ is normal, Urysohn's lemma provides an $f$ with $f(\varphi(x)) = 1$ and $f = 0$ on $Y \setminus U$, and then $f(\varphi(x_{j})) = 0 \not\to 1$, a contradiction.

**3.** $(\Rightarrow)$ If $\varphi$ is surjective and $f\circ\varphi = 0$, then $f(y) = 0$ for every $y = \varphi(x)$, so $f = 0$ and $\varphi^{\sharp}$ is injective.

$(\Leftarrow)$ We prove the contrapositive. If $\varphi$ is not surjective, then $\varphi(X)$ is compact as a continuous image of a compact set, hence closed since $Y$ is Hausdorff. Pick $y_{0} \in Y \setminus \varphi(X)$; by Urysohn's lemma there is an $f \in C(Y)$ with $f(y_{0}) = 1$ and $f = 0$ on $\varphi(X)$. Then $f \ne 0$ but $\varphi^{\sharp}(f) = 0$, so $\varphi^{\sharp}$ is not injective.

Combining this exercise with <Ref to="thm-gelfand-naimark" />, the category of compact Hausdorff spaces is contravariantly equivalent to the category of unital commutative $C^*$-algebras. Part 3 is a manifestation of that contravariance: surjectivity and injectivity are interchanged.
</Solution>
</Exercise>

## References

- G. J. Murphy, *C\*-Algebras and Operator Theory*, Academic Press, 1990 — Chapter 1 (Banach algebras and spectral theory), Chapter 2 ($C^*$-algebras, Gelfand theory, continuous functional calculus), Chapter 3 (the GNS construction). This is the treatment closest to the present article.
- W. Rudin, *Functional Analysis*, 2nd ed., McGraw-Hill, 1991 — Chapter 10 (Banach algebras), Chapter 11 (commutative Banach algebras, Wiener's theorem). The proof of <Ref to="thm-spectral-radius" /> given here follows this book fairly closely.
- R. V. Kadison and J. R. Ringrose, *Fundamentals of the Theory of Operator Algebras I: Elementary Theory*, Academic Press, 1983 — Chapters 3 and 4.
- J. Dixmier, *C\*-algebras*, North-Holland, 1977 (original: *Les C\*-algèbres et leurs représentations*, Gauthier-Villars, 1964) — Chapters 1 and 2.
- I. M. Gelfand and M. A. Naimark, "On the imbedding of normed rings into the ring of operators in Hilbert space", *Mat. Sbornik* 12 (54) (1943), 197–213. — The original paper containing both the commutative and the noncommutative Gelfand–Naimark theorems.
- A. Connes, *Noncommutative Geometry*, Academic Press, 1994 — Introduction and Chapter II. The point of view in which $C^*$-algebras are used as noncommutative spaces is developed here.

## Appendix: the non-unital case

In the main text we postponed the proof of part 3 of <Ref to="thm-gelfand-naimark" />; we supply it here. The key is to put the correct $C^*$-norm on the unitization of a $C^*$-algebra, since the norm $\|a\| + |\lambda|$ of <Ref to="rem-unitization" /> does not satisfy the $C^*$-identity and is therefore unusable.

**Unitization by the left regular representation.** Let $A$ be a $C^*$-algebra without a unit, and equip the $*$-algebra $\tilde{A} = A \oplus \mathbb{C}$ with the product of <Ref to="rem-unitization" /> and the involution $(a,\lambda)^{*} = (a^{*}, \bar{\lambda})$. For $x = (a,\lambda)$ consider the bounded operator $L_{x}(b) = ab + \lambda b$ on the Banach space $A$, and put $\|x\|_{\mathrm{op}} := \|L_{x}\|_{B(A)}$. The relation $L_{xy} = L_{x}L_{y}$ is immediate from the definition of the product, so $\|\cdot\|_{\mathrm{op}}$ is a submultiplicative seminorm.

**Injectivity of $L$.** Suppose $L_{x} = 0$. If $\lambda = 0$, then $ab = 0$ for all $b$; taking $b = a^{*}$ gives $\|a\|^{2} = \|aa^{*}\| = 0$ (part 1 of <Ref to="prop-cstar-norm" /> and the $C^*$-identity), so $a = 0$. If $\lambda \ne 0$, then $e := -a/\lambda$ satisfies $eb = b$ for all $b$. Taking adjoints gives $ce^{*} = c$ for all $c$; substituting $b = e^{*}$ in the former and $c = e$ in the latter yields $ee^{*} = e^{*}$ and $ee^{*} = e$, so $e = e^{*}$, and hence $be = be^{*} = b$ as well, making $e$ a unit for $A$. This contradicts the hypothesis.

**Agreement with the original norm on $A$.** The inequality $\|L_{a}\| \le \|a\|$ follows from submultiplicativity. Conversely, if $a \ne 0$ then $b = a^{*}/\|a\|$ has $\|b\| = 1$ and $\|L_{a}(b)\| = \|aa^{*}\|/\|a\| = \|a\|$, so $\|L_{a}\| \ge \|a\|$. Hence $A \subset \tilde{A}$ is an isometric embedding.

**The $C^*$-identity.** For $x \in \tilde{A}$ and $b \in A$ with $\|b\| \le 1$ we have $xb \in A$, so the $C^*$-identity of $A$ applies:
$$
\|xb\|^{2} = \|(xb)^{*}(xb)\| = \|b^{*}(x^{*}x)b\| \le \|b^{*}\|\,\|(x^{*}x)b\| \le \|x^{*}x\|_{\mathrm{op}} .
$$
Taking the supremum gives $\|x\|_{\mathrm{op}}^{2} \le \|x^{*}x\|_{\mathrm{op}}$. Combined with submultiplicativity this yields $\|x\|_{\mathrm{op}}^{2} \le \|x^{*}\|_{\mathrm{op}}\|x\|_{\mathrm{op}}$, that is $\|x\|_{\mathrm{op}} \le \|x^{*}\|_{\mathrm{op}}$; applying this to $x^{*}$ gives the reverse inequality, so $\|x^{*}\|_{\mathrm{op}} = \|x\|_{\mathrm{op}}$. Therefore
$$
\|x\|_{\mathrm{op}}^{2} \le \|x^{*}x\|_{\mathrm{op}} \le \|x^{*}\|_{\mathrm{op}}\|x\|_{\mathrm{op}} = \|x\|_{\mathrm{op}}^{2}
$$
consists of equalities throughout. Completeness follows from the fact that $A$ is complete as a closed subspace and $\tilde{A} = A \oplus \mathbb{C}1$ adds only finitely many dimensions. Hence $\tilde{A}$ is a unital $C^*$-algebra.

**One-point compactification of the character space.** Let $A$ be a commutative $C^*$-algebra without a unit. The map $\chi_{\infty}(a + \lambda 1) := \lambda$ is a character of $\tilde{A}$ with kernel $A$. If $\chi \ne \chi_{\infty}$ then $\chi|_{A} \in \Omega(A)$, and conversely, as in the proof of part 2 of <Ref to="lem-character" />, every element of $\Omega(A)$ extends uniquely to a character of $\tilde{A}$. This correspondence $\Omega(\tilde{A})\setminus\{\chi_{\infty}\} \to \Omega(A)$ is a bijection, and both sides carry the weak-$*$ topology, so it is a homeomorphism. Hence $\Omega(\tilde{A})$ is the one-point compactification of $\Omega(A)$, and $\Omega(A)$, being an open subset of a compact Hausdorff space, is locally compact Hausdorff.

**Completion of the proof.** Applying part 1 of <Ref to="thm-gelfand-naimark" /> to $\tilde{A}$, the map $\Gamma_{\tilde{A}} : \tilde{A} \to C(\Omega(\tilde{A}))$ is an isometric $*$-isomorphism, and under it
$$
A = \ker\chi_{\infty} \ \longleftrightarrow\ \{F : F(\chi_{\infty}) = 0\} = C_{0}\bigl(\Omega(\tilde{A})\setminus\{\chi_{\infty}\}\bigr) = C_{0}(\Omega(A))
$$
(on the open subset obtained by deleting one point from a compact space, the continuous functions vanishing at infinity are exactly the restrictions of the continuous functions vanishing at that point). The restriction $\Gamma_{\tilde{A}}|_{A}$ is precisely $\Gamma_{A}$, so $\Gamma_{A} : A \to C_{0}(\Omega(A))$ is an isometric $*$-isomorphism. The equivalence of unitality and compactness follows from $A \cong C_{0}(\Omega(A))$ and the last claim of <Ref to="ex-c0x" />.