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## Banach algebras, the Gelfand representation, and the commutative Gelfand-Naimark theorem

Banach algebras abstract the properties of closed algebras of operators on Banach spaces. Many basic properties of such operators have elegant proofs in the framework of Banach algebras, and Banach algebras also naturally appear in areas of mathematics like harmonic analysis, where one writes down Banach algebras generalizing the group algebra to study topological groups.

Today we will develop some of the basic theory of Banach algebras, our goal being to discuss the Gelfand representation of a commutative Banach algebra and the fact that, for commutative C*-algebras, this representation is an isometric isomorphism. This implies in particular a spectral theorem for self-adjoint operators on a Hilbert space.

This material can be found in many sources; I am working from Dales, Aiena, Eschmeier, Laursen and Willis’ Introduction to Banach Algebras, Operators, and Harmonic Analysis.

Below all vector spaces are over $\mathbb{C}$, all algebras are unital, and all algebra homomorphisms preserve units unless otherwise stated. In the context of Banach algebras, the last two assumptions are not standard, but in practice non-unital Banach algebras are studied by adjoining units first, so we do not lose much generality.

Definitions and examples

A Banach algebra is a Banach space $A$ which is also an associative algebra (with the same addition) such that the multiplication satisfies $\| ab\| \le \| a \| \| b \|$ and $\| 1 \| = 1$. Abstractly, it is a monoid object in $\text{Ban}_1$. A morphism of Banach algebras is a weak contraction which is also an algebra homomorphism. Note that since we want morphisms to preserve units, any morphism has norm exactly $1$.

(Note also that the zero ring cannot be a Banach algebra with these axioms since $\| 0 \| = 0$ and $\| 1 \| = 1$ is incompatible with $0 = 1$.)

Example. The fundamental example is the algebra $B \Rightarrow B$ bounded linear operators on a Banach space $B$ with the operator norm.

Example. Let $T : B \to B$ be a bounded linear operator on a Banach space. Then the closure of the algebra $\mathbb{C}[T] \subset B \Rightarrow B$ is a commutative Banach algebra. The application of Banach algebra techniques to spectral theory proceeds by studying algebras like this.

Example. Let $\mathcal{S}$ be a collection of bounded linear operators on a Banach space $B$. Their commutant $\mathcal{S}' = \{ T : TS = ST \forall S \in \mathcal{S} \}$ is an intersection of closed subalgebras of $B \Rightarrow B$, hence also a Banach algebra. If $B$ is thought of as a module over the algebra generated by $\mathcal{S}$, then $\mathcal{S}'$ consists of the endomorphisms of this module.

Example. Let $X$ be a topological space and consider again the space $C_b(X, \mathbb{C}) = C_b(X)$ of bounded continuous functions $X \to \mathbb{C}$. This space is a Banach space when equipped with the sup norm, and moreover pointwise multiplication makes it a Banach algebra. If $X$ is discrete, this is just $\ell^{\infty}(X)$. If $X$ is compact, this is just $C(X)$.

Example. Let $X$ be a locally compact Hausdorff space which is not compact. We denote by $C_0(X, \mathbb{C}) = C_0(X)$ the space of continuous functions $f : X \to \mathbb{C}$ which vanish at infinity in the sense that for any $\epsilon > 0$ there exists a compact set $K$ such that $|f(x)| < \epsilon$ for $x \not \in K$; these are precisely the functions vanishing at the point at infinity in the one-point compactification of $X$. This condition is preserved by uniform limits, so $C_0(X)$ is a Banach space when equipped with the sup norm. When equipped with pointwise products, it is a non-unital Banach algebra (the identity cannot vanish at infinity). Adjoining a unit corresponds to passing to the one-point compactification.

Example. Let $C$ be a small category. Recall that the category ring $\mathbb{Z}[C]$ is the free abelian group on the morphisms of $C$, with the product of two morphisms being their composition if defined and zero otherwise (and recall that it is unital if and only if $C$ has finitely many objects.) Since it is in particular a free abelian group on a set, we can tensor it with $\mathbb{C}$ and equip it with the $\ell^1$ norm; the resulting completion $\ell^1(C)$ is non-unital if and only if $C$ has infinitely many objects. Even if $C$ has finitely many objects, in which case the unit of $\ell^1(C)$ is the sum of the identity morphisms in $C$, that unit satisfies $\| 1 \| = |\text{Ob}(C)|$, which is usually not equal to $1$.

Thinking of $\ell^1(C)$ in terms of functions $\text{Mor}(C) \to \mathbb{C}$, the product is the convolution

$\displaystyle (f * g)(c) = \prod_{ab = c} f(a) g(b)$

and so $\ell^1(C)$ is called the convolution algebra of $C$. This definition is usually only stated in the special case that $C$ is a group (that is, a one-object small category with all morphisms invertible), in which case it generalizes the group algebra, although it is sometimes also given when $C$ is a groupoid.

Groupoids naturally arise when some space of interest is quotiented by some equivalence relation or group action. In such situations it often happens that the quotient as a space is very badly behaved, and Alain Connes’ approach to understanding this situation via noncommutative geometry is to replace the quotient by the (noncommutative space corresponding to the) convolution algebra of the corresponding groupoid.

Sub-example. Taking $C$ to be the groupoid of $n$ objects, all of which are isomorphic via a unique isomorphism, we recover $\mathcal{M}_n(\mathbb{C})$ (with norm given by the sum of the absolute values of the entries). This is more or less how Heisenberg discovered matrix mechanics: rather than thinking in terms of vector spaces he thought in terms of transitions between energy levels, which form the morphisms in a groupoid.

The category $C$ above is equivalent but not isomorphic to the category with one object and a single morphism. The corresponding algebra is just $\mathbb{C}$, which is similarly Morita equivalent but not isomorphic to $\mathcal{M}_n(\mathbb{C})$.

Example. Let $G$ be a locally compact group (e.g. $\mathbb{R}$). $G$ has a nice left-invariant measure $\mu$ called Haar measure (e.g. Lebesgue measure) which is unique up to scaling; if $G$ is compact we can normalize it so that $\mu(G) = 1$ to get normalized Haar measure. The Banach space $L^1(G)$ can also be equipped with a convolution product, which for groups can be written

$\displaystyle (f \ast g)(t) = \int_X f(s) g(t - s) \, d \mu$.

When $G$ is a group with the discrete topology, counting measure is a Haar measure and we recover $\ell^1(G)$. Like $C_0(X, \mathbb{C})$, typically $L^1(G)$ is not unital (for example when $G = \mathbb{R}$).

Compact Hausdorff spaces

For $X$ compact Hausdorff, the Banach algebras $C(X)$ will be particularly important to keep in mind as examples in this post. Keep in mind the following, which was discussed previously here and here.

Theorem: The functors $X \mapsto C(X, \mathbb{R}), A \mapsto \text{MaxSpec } A$ define a contravariant equivalence of categories from the category $\text{CHaus}$ of compact Hausdorff spaces to a subcategory of the category of $\mathbb{R}$-algebras (where $\text{MaxSpec } A$ is given the Zariski topology).

Basic properties

Banach algebras are an abstraction of closed subalgebras of the algebra of bounded operators on a Banach space in the same way that groups are an abstraction of subgroups of the symmetric group on a set. Accordingly, we have the following result which guarantees that the abstract and concrete notions of Banach algebra coincide.

Cayley’s theorem for Banach algebras: Every Banach algebra $A$ acts faithfully on a Banach space $B$ by bounded linear operators such that the norm on $A$ agrees with the operator norm.

Proof. Consider the action of $A$ on itself by left multiplication; we denote the operator $b \mapsto ab$ by $L_a : A \to A$. This is clearly an injective algebra homomorphism (Cayley’s theorem for algebras). Since $\| ab \| \le \| a \| \| b \|$ we clearly have $\| L_a \| \le \| a \|$; moreover, since $\| a \| = \| a \| \| 1 \|$ we in fact have $\| L_a \| = \| a \|$. $\Box$

Recall that the spectrum $\sigma(a)$ consists of all scalars $\lambda \in \mathbb{C}$ such that $\lambda - a$ is not invertible. This definition turns out to be very fruitful for Banach algebras in particular due to the following results.

Proposition: Let $a$ be an element of a Banach algebra $A$. Then the limit $\lim \| a^n \|^{1/n}$ exists.

Proof. Let $\nu = \liminf \| a^n \|^{1/n}$. Then for any $\epsilon > 0$ we can find $n$ such that $\nu + \epsilon > \| a^n \|^{1/n}$, hence $(\nu + \epsilon)^n > \| a^n \|$. It follows that for $m \ge n$ we have

$\displaystyle \| a^m \| = \| a^{nq + r} \| \le (\nu + \epsilon)^{nq} \| a^r \|$

hence that $\limsup \| a^n \|^{1/n} \le \nu + \epsilon$. Since this is true for every $\epsilon > 0$, we conclude that $\liminf \| a^n \|^{1/n} = \limsup \| a^n \|^{1/n}$. $\Box$

Proposition: Let $a \in A$ be such that $\lim \| a^n \|^{1/n} < 1$. Then $1 - a$ is invertible.

Proof. The geometric series $1 + a + a^2 + ...$ converges absolutely and is a two-sided inverse for $1 - a$. $\Box$

Corollary: Let $\lambda \in \sigma(a)$. Then $|\lambda| \le \lim \| a^n \|^{1/n}$.

Proof. If $\lambda > \lim \| a^n \|^{1/n}$, then $1 - \frac{a}{\lambda}$ is invertible by the above. $\Box$

Accordingly, we define the spectral radius

$\displaystyle \nu(a) = \sup_{\lambda \in \sigma(a)} | \lambda |$.

This is the radius of the smallest circle about the origin containing the spectrum of $a$. The above shows that we always have $\nu(a) \le \lim \| a^n \|^{1/n} \le \| a \|$.

For a Banach algebra $A$ let $\text{GL}(A)$, the general linear group of $A$, denote its group of units.

Proposition: $\text{GL}(A)$ is open. The inversion map $\text{GL}(A) \to \text{GL}(A)$ is continuous (making $\text{GL}(A)$ a topological group).

Proof. By translation it suffices to show that an open neighborhood of the identity in $A$ lies in $\text{GL}(A)$. But this is clear since the open ball of radius, say, $\frac{1}{2}$ about the identity lies in $\text{GL}(A)$ by the above.

Let $a, b \in \text{GL}(A)$. Note the identity $b^{-1} - a^{-1} = b^{-1} (a - b) a^{-1}$, which gives

$\displaystyle \| b^{-1} - a^{-1} \| \le \| b^{-1} \| \| a - b \| \| a^{-1} \| \le (\| a^{-1} \| + \| b^{-1} + a^{-1} \|) \| a - b \| \| a^{-1} \|$.

It follows that for fixed $a$ we can make $\| b^{-1} - a^{-1} \|$ arbitrarily small by making $\| a - b \|$ sufficiently small. $\Box$

Corollary: The spectrum $\sigma(a)$ is compact.

Proof. The resolvent $R_a : z \mapsto (z - a)^{-1}$ is defined on the complement of the spectrum (the resolvent set $\rho(a) = \mathbb{C} \setminus \sigma(a)$) and has image in $\text{GL}(A)$. By the above, $R_a$ is continuous, hence $R_a^{-1}$ maps open sets to open sets. Since $\text{GL}(A)$ is open in $A$, it follows that $\rho(a)$ is open in $\mathbb{C}$, so the spectrum is closed. Since the spectrum is also bounded, it is compact by Heine-Borel. $\Box$

Ideals

Proposition: Let $A$ be a Banach algebra and $I$ a (left, right, two-sided) ideal of $A$. Then:

1. The closure $\bar{I}$ of $I$ is also a (left, right, two-sided) ideal.
2. If $I$ is proper, then so is $\bar{I}$.
3. If $I$ is two-sided, closed, and proper, the quotient $A/I$ is also a Banach algebra and the map $A \to A/I$ is a morphism of Banach algebras.
4. If $I$ is maximal, it is already closed.

Proof. $1$: $\bar{I}$ is the closure of a subspace, hence also a subspace, hence closed under addition and scalar multiplication. If $x_i \in I$ is a Cauchy sequence converging to $x \in \bar{I}$, then $\| rx_i - rx_j \| \le \| r \| \| x_i - x_j \|$, hence $rx_i$ is a Cauchy sequence converging to $rx$, so $\bar{I}$ is closed under left multiplication by elements of $B$ if $I$ is and similarly for right multiplication.

$2$: An ideal $I$ is proper if and only if it contains no invertible elements if and only if it does not contain the identity (and this is true for left, right, and two-sided ideals). Since the open ball of radius $\frac{1}{2}$ about the identity consists of invertible elements, it must be disjoint from any proper ideal $I$, so $\bar{I}$ cannot contain the identity and hence must also be proper.

$3$: If $I$ is two-sided, closed, and proper, the quotient $A/I$ is a Banach space and an algebra, and the map $A \to A/I$ is a weak contraction and an algebra homomorphism. It remains to show that the quotient norm on $A/I$ satisfies $\| ab \| \le \| a \| \| b \|$, but this follows from the fact that the cosets $ab + I$ contain the possible products of elements in the cosets $a + I, b + I$, hence the infimum of the norms in the former is at most the infimum of the norms in the latter.

This already shows that the quotient norm on $A/I$ satisfies $\| 1 \| = 1$, since $\| b \| \le \| 1 \| \| b \|$ implies $\| 1 \| \ge 1$ and the quotient map $A \to A/I$ is a weak contraction. A direct argument is to observe that the open ball of radius $1$ about $1$ consists only of invertible elements, hence is disjoint from $I$.

$4$: Follows directly from $2$. $\Box$

Corollary: Any algebra homomorphism $f : A \to \mathbb{C}$ is a morphism of Banach algebras.

This is one of several results in Banach algebra theory to the effect that an algebra homomorphism between certain Banach algebras is automatically continuous; see, for example, Section I.5 of Dales et al.

Basic properties assuming the spectrum is non-empty

The following is fundamental.

Black Box #1: The spectrum $\sigma(a)$ is non-empty.

The idea of the proof is that if $\sigma(a)$ were empty, then the resolvent $R_a : z \mapsto (z - a)^{-1}$ would be a non-constant “bounded holomorphic function,” which contradicts Liouville’s theorem. This proof actually works with no modifications for complex $n \times n$ matrices and shows that they have eigenvalues without going through either characteristic polynomials or the fundamental theorem of algebra, and in fact this is one way to prove the fundamental theorem of algebra. In general we need to develop some complex analysis for Banach space-valued holomorphic functions, which we defer to a later post.

For now, it it is more important to see what can be done with this result.

Corollary (Gelfand-Mazur): Any Banach algebra which is a division algebra is isometrically isomorphic to $\mathbb{C}$.

Proof. Let $A$ be such a Banach algebra and let $a \in A$ be nonzero. $\sigma(a)$ is non-empty, so let $\lambda$ be such that $\lambda - a$ is not invertible. Since $a$ is nonzero, it must be invertible, so $\lambda \neq 0$, hence $1 - \frac{a}{\lambda}$ is not invertible. But then it must be equal to $0$, so $a = \lambda$. $\Box$

Corollary: Let $A$ be a commutative Banach algebra. Then $\lambda \in \sigma(a)$ if and only if there exists a morphism $f : A \to \mathbb{C}$ such that $f(a) = \lambda$.

Proof. $\Leftarrow$: if there exists such an $f$, then $\lambda - a$ lies in a maximal ideal, so cannot be invertible. (This argument works in any algebra over $\mathbb{C}$.)

$\Rightarrow$: if $\lambda \in \sigma(a)$, then $\lambda - a$ is not invertible, hence is contained in a maximal ideal $m$, which is already closed. Since $A/m \cong \mathbb{C}$ by Gelfand-Mazur, the quotient map $f : A \to A/m$ is the desired $f$. $\Box$

(Note that we make essential use of the axiom of choice here.)

A digression into quantum mechanics

The non-emptiness of the spectrum, besides leading to the above results, also has the following corollary, which is apparently due to Wielandt.

Proposition: The Weyl algebra $\mathbb{C}[x, \partial_x] \cong \mathbb{C}[x, y]/(yx - xy - 1)$ does not embed into a Banach algebra.

Proof. Suppose to the contrary that $x, y$ are elements of a Banach algebra satisfying $yx - xy = 1$. We make the elementary observation that $\sigma(a + 1) = \sigma(a) + 1 = \{ \lambda + 1 : \lambda \in \sigma(a) \}$, from which it follows that

$\sigma(yx) = \sigma(xy) + 1$.

But we know that $\sigma(yx) = \sigma(xy)$, so $\sigma(xy)$ must be invariant under addition by $1$. Since $\sigma(xy)$ is non-empty, it must contain arbitrarily large elements, and this contradicts compactness. $\Box$

This result is not needed for what follows, but it does imply that it is not possible to describe the position and momentum operators as elements of a Banach algebra. Equivalently, it is not possible to represent position and momentum as bounded operators on a Banach space. In their usual representation on $L^2(\mathbb{R})$, both position and momentum are unbounded (discontinuous) and densely defined (can only be applied to certain dense subspaces of $L^2$).

Fortunately, there is a way out: instead of dealing with position and momentum directly, we can deal with the strongly continuous one-parameter unitary groups they generate via Stone’s theorem (see also the Stone-von Neumann theorem). Indeed, these unitary groups are defined on $L^2(\mathbb{R})$ by

$\displaystyle f(x) \mapsto e^{itx} f(x)$

for position and (up to normalization)

$\displaystyle f(x) \mapsto f(x + t)$

for momentum, both of which are defined everywhere on $L^2(\mathbb{R})$.

The Gelfand representation

Recall that for any commutative ring $A$ we can define a homomorphism

$\displaystyle A \ni a \mapsto \prod_m (a \bmod m) \in \prod_m A/m$

where the product runs over all maximal ideals of $A$ and $a \bmod m$ is the image of $r$ in $A/m$. This represents $A$ as a “ring of functions” on $\text{MaxSpec } A$, although ideally all of the residue fields $A/m$ should be the same for this interpretation to really make sense, and the kernel of this homomorphism is the Jacobson radical $J(A)$.

We now know that if $A$ is a Banach algebra, we always have $A/m \cong \mathbb{C}$, so the above homomorphism lands in the space of functions $\text{MaxSpec } A \to \mathbb{C}$. It’s natural to want to equip $\text{MaxSpec } A$, the spectrum, Gelfand spectrum, or Gelfand space of $A$, with a topology such that the corresponding functions are all continuous. This is usually done by identifying an element of $\text{MaxSpec } A$ with the corresponding map $A \to \mathbb{C}$, which is in particular a continuous linear functional, hence $\text{MaxSpec } A$ naturally sits as a subspace of $A^{\ast}$. The initial topology on $A^{\ast}$ making all evaluations against elements of $A$ continuous is the weak-* topology, and the topology we want on $\text{MaxSpec } A$ is the subspace topology with respect to this topology.

(An equivalent way to define the weak-* topology, which makes sense for $A$ replaced by any topological vector space $V$, is to think of $V^{\ast}$ as a subspace of the space of functions $V \to \mathbb{C}$, or equivalently a subspace of the product $\mathbb{C}^V$ of $|V|$ copies of $\mathbb{C}$; then the weak-* topology is the subspace topology with respect to the product topology on this product.)

Since morphisms $A \to \mathbb{C}$ have norm $1$, $\text{MaxSpec } A$ in fact lies in the unit ball of $A^{\ast}$. We now need the following result, which shows that the weak-* topology is very different from the norm topology on $A^{\ast}$.

Theorem (Banach-Alaoglu): let $V$ be a normed space. Then the unit ball of $V^{\ast}$ is compact in the weak-* topology.

Proof. Thinking of $V^{\ast}$ as a subspace of $\mathbb{C}^V$ with the product topology, the unit ball consists of linear functionals which send the unit ball $B$ of $V$ to the unit ball (disk) $D = \{ z \in \mathbb{C} : |z| \le 1 \}$, hence can be thought of as a subspace of $D^B$ with the product topology. This is compact by Tychonoff’s theorem. The unit ball of $V^{\ast}$ is a closed subspace of $D^B$ (since it is defined by the additional requirement that it consists of linear functions, which impose a collection of closed conditions), hence is also compact. $\Box$

(Note that $D$ is compact Hausdorff, so we only need Tychonoff’s theorem for compact Hausdorff spaces here, hence we only need the ultrafilter lemma and not the full strength of the axiom of choice. In fact Banach-Alaoglu is equivalent to the ultrafilter lemma.)

Corollary: For $A$ a commutative Banach algebra, $\text{MaxSpec } A$ is compact Hausdorff in the weak-* topology.

It follows that the homomorphism $A \to \prod_m A/m$ above, applied to a commutative Banach algebra, gives us its Gelfand representation

$A \ni a \mapsto \hat{a} \in C(\text{MaxSpec } A)$.

Since $\text{MaxSpec } A$ is compact Hausdorff, the space of continuous functions on it is itself a Banach algebra with the sup norm. We record the following properties of the Gelfand representation to summarize what we have already proven.

1. $\hat{a}$ has range the spectrum $\sigma(a)$ of $a$.
2. Consequently $\| \hat{a} \|$ is the spectral radius $\nu(a)$.
3. Consequently the Gelfand representation is a weak contraction (hence a morphism of Banach algebras).
4. The kernel of the Gelfand representation is $J(A)$. It is also the set of elements of $A$ with spectral radius zero (the quasi-nilpotent elements).
5. Consequently the Gelfand representation is faithful if and only if $A$ is semiprimitive, if and only if $A$ has no nonzero quasi-nilpotents.

(Some sources use “semisimple” to mean semiprimitive. The people who use this convention use “Artinian semisimple” to mean semisimple.)

Commutative semiprimitive Banach algebras therefore have a relatively concrete description as algebras of continuous functions on compact Hausdorff spaces; note, however, that even if $A$ is semiprimitive we are not guaranteed that the spectral radius coincides with the norm on $A$. For an important class of commutative Banach algebras which we now introduce, we will be able to show that the Gelfand representation is an isometric isomorphism.

$^{\dagger}$-algebras

For $H$ a Hilbert space, the Banach algebra $H \Rightarrow H$ of bounded linear operators on $H$ crucially has extra structure not present in a general Banach algebra given by the adjoint. In abstracting this structure, we are led to the following sequence of definitions.

A *-ring (or involutive ring, or ring with involution) is a ring $R$ equipped with an anti-homomorphism $r \mapsto r^{\ast}$ which squares to the identity. Explicitly, this is a map satisfying

1. $1^{\ast} = 1$,
2. $(r + s)^{\ast} = r^{\ast} + s^{\ast}$,
3. $(rs)^{\ast} = s^{\ast} r^{\ast}$,
4. $r^{\ast \ast} = r$.

Equivalently, a *-ring is a small $\text{Ab}$-enriched dagger category, so I am going to call them $^{\dagger}$-rings and use $^{\dagger}$ for the involution even though “dagger ring” appears to have some unrelated meaning in mathematics. A morphism of $^{\dagger}$-rings is a ring homomorphism preserving the involution.

Example. Any commutative ring is a $^{\dagger}$-ring with trivial involution.

Example. $\mathbb{C}$ is a $^{\dagger}$-ring with the involution being complex conjugation.

Example. The quaternions $\mathbb{H}$ are a $^{\dagger}$-ring with the involution sending $i, j, k$ to $-i, -j, -k$.

Example. Let $D$ be a small dagger category. The category ring $\mathbb{Z}[D]$ is a $^{\dagger}$-ring with the involution extending the involution on $D$. In particular, a groupoid is a dagger category with the involution being inverse, so we get an example for any groupoid and in particular any group.

Example. Let $R$ be a commutative ring. The ring $\mathcal{M}_n(R)$ of $n \times n$ matrices over $R$ is a $^{\dagger}$-ring with the involution being transpose. (This is more or less a special case of the above applied to the groupoid of $n$ equivalent objects after tensoring with $R$.)

A *-algebra over a commutative *-ring $R$ is an algebra over $R$ in the ordinary sense which is also equipped with an involution making it a *-ring compatible with the one on $R$. Equivalently, it is a *-ring $S$ together with a morphism $R \to Z(S)$ of *-rings. Again, I will call these $^{\dagger}$-algebras. A morphism of $^{\dagger}$-algebras is an $R$-linear morphism of $^{\dagger}$-rings.

Example. Let $A$ be a $^{\dagger}$-ring which is also an algebra over $\mathbb{R}$ (with the involution restricted to $\mathbb{R}$ being trivial). Then the tensor product $\mathbb{C} \otimes_R A$ is a complex $^{\dagger}$-algebra (that is, a $^{\dagger}$-algebra over $\mathbb{C}$ as a $^{\dagger}$-ring) with the involution being complex conjugation on $\mathbb{C}$ and the existing involution on $A$.

Letting $A = \mathbb{Z}[D]$ be the category ring of a dagger category we get the complex $^{\dagger}$-algebras $\mathbb{C}[D]$.

Example. The algebra $H \Rightarrow H$ is a complex $^{\dagger}$-algebra with the involution being adjoint, as is any $^{\dagger}$-closed subalgebra.

We will primarily be interested in complex $^{\dagger}$-algebras, which are $^{\dagger}$-algebras over $\mathbb{C}$ regarded as a $^{\dagger}$-ring using complex conjugation. In the commutative case, this is more or less a more convenient way to study real algebras.

Theorem: The functor sending a commutative complex $^{\dagger}$-algebra to its subalgebra of self-adjoint elements is an equivalence of categories to the category of commutative $\mathbb{R}$-algebras.

Proof. The inverse of this functor sends an algebra $R$ to the tensor product $\mathbb{C} \otimes_{\mathbb{R}} R$ with involution given by complex conjugation. That this is the inverse follows from the fact that for any element $a$ of a commutative complex $^{\dagger}$-algebra we can write

$\displaystyle a = \text{Re}(a) + \text{Im}(a) = \frac{a + a^{\dagger}}{2} + i \frac{a - a^{\dagger}}{2i}$

which is precisely an element of $\mathbb{C}$ tensor the self-adjoint subalgebra of $A$; moreover, morphisms out of $A$ are uniquely and freely determined by what they do to self-adjoint elements by the above. $\Box$

Example. Let $A = \mathbb{C}[z, z^{-1}]$ be the algebra of Laurent polynomials with involution given by $z^{\dagger} = z^{-1}$. Identifying $A$ with the space of Fourier polynomials $\mathbb{C}[e^{i\theta}, e^{-i\theta}]$, we may identify the self-adjoint subalgebra with the space of trigonometric polynomials $\mathbb{R}[\cos \theta, \sin \theta]$.

We may think of the above as being essentially about the Galois theory of the Galois extension $\mathbb{R} \to \mathbb{C}$; more generally, if $k \to L$ is any Galois extension, then there is a functor

$\displaystyle L \otimes_k - : k\text{-Alg} \to L\text{-Alg}$

and we may promote this to an equivalence of categories by attaching to $L \otimes_k A$ the data of the action of the Galois group $\text{Gal}(L/k)$ on the left factor; taking the $\text{Gal}(L/k)$-invariant subalgebra then gives the inverse functor. None of this requires that the $k$-algebras or $L$-algebras in question are commutative.

However, noncommutative complex $^{\dagger}$-algebras do not fit into this formalism since the involution reverses the order of multiplication. In particular, the self-adjoint elements no longer form a subalgebra; if $a, b$ are self-adjoint but don’t commute then

$\displaystyle (ab)^{\dagger} = b^{\dagger} a^{\dagger} = ba \neq ab$.

The self-adjoint elements are instead closed under the anticommutator

$\displaystyle a \circ b = \frac{ab + ba}{2}$

and axiomatizing this observation leads to the theory of Jordan algebras. However, we won’t be concerned much with anticommutators; if $a, b$ are observables in a quantum system, we generally prefer to look at the observable $- i [a, b]$, for example when $a$ is position and $b$ is momentum, in view of Noether’s theorem.

C*-algebras

We are now ready for the following definition axiomatizing the properties of closed $^{\dagger}$-closed subalgebras of $H \Rightarrow H$. A C*-algebra is a Banach algebra $A$ equipped with an involution $^{\dagger} : A \to A$ making it a complex $^{\dagger}$-algebra and also satisfying the B*-identity

$\displaystyle \| a^{\dagger} a \| = \| a \|^2$.

(See Wikipedia for the history here.) A morphism of C*-algebras is a morphism of Banach algebras which respects the involution.

(It is not obvious that the B*-identity is either useful or enough to capture everything we want. Happily, both are true.)

Proposition: Let $H$ be a Hilbert space. Then the algebra $H \Rightarrow H$ of bounded linear operators $H \to H$ is a C*-algebra with involution the adjoint.

Proof. Let $T : H \to H$ be such an operator. Then

$\displaystyle \| T^{\dagger} T \| = \sup_{\| v \| = 1, \| w \| = 1} |\langle T^{\dagger} T v, w \rangle| = \sup_{\| v \| = 1, \| w \| = 1} |\langle Tv, Tw \rangle| = \| T \|^2$

and the conclusion follows. $\Box$

Corollary: Any closed $^{\dagger}$-closed subalgebra of $H \Rightarrow H$ is also a C*-algebra.

Example. Let $T : H \Rightarrow H$ be a bounded linear operator. Then $T, T^{\dagger}$ generate a $^{\dagger}$-closed subalgebra of $H \Rightarrow H$ whose closure is also $^{\dagger}$-closed (this follows from the fact that taking adjoints preserves norms, hence is itself continuous with norm $1$), hence gives a C*-algebra. This C*-algebra is commutative if and only if $T$ is normal.

Example. Let $S$ be a collection of bounded linear operators on a Hilbert space $H$ which is $^{\dagger}$-closed. Then their commutant $S'$ is also $^{\dagger}$-closed, hence is a C*-algebra. These C*-algebras are precisely the von Neumann algebras.

Example. If $X$ is a compact Hausdorff space then $C(X)$ is a commutative C*-algebra with the involution being complex conjugation. It may be regarded as a space of operators acting by multiplication on $L^2(X)$ if $X$ is equipped with a suitably nice Borel measure, although I am not sure of the precise hypotheses we need for the operator norm to match the sup norm. At a minimum the measure of every open set must be positive.

We now need the following.

Black Box #2 (Beurling-Gelfand spectral radius formula): Let $a$ be an element of a Banach algebra $A$. Then the spectral radius satisfies

$\displaystyle \nu(a) = \lim_{n \to \infty} \| a^n \|^{1/n}$.

This is a surprising result in light of the fact that the LHS depends only on the algebra structure of $A$ while the RHS is defined in terms of the norm on $A$. The crux of the proof is the following version of the Cauchy integral formula, which we will not prove here:

$\displaystyle a^n = \frac{1}{2\pi i} \oint_{|z| = r} \frac{z^n}{z - a} \, dz = \frac{1}{2\pi i} \oint_{|z| = r} z^n R_a(z) \, dz$

for all $r > \nu(a)$. If $M_r = \sup_{|z| = r} \| R_a(z) \|$, then it follows that

$\displaystyle \| a^n \| \le r^{n+1} M_r$

hence that $\lim_{n \to \infty} \| a^n \|^{1/n} \le r$ for all $r > \nu(a)$. Since we already know that $\nu(a) \le \lim_{n \to \infty} \| a^n |^{1/n}$, the conclusion follows. Again, we defer discussion of the relevant complex analysis to a later post.

Corollary: Let $a$ be a self-adjoint element of a C*-algebra. Then $\| a \| = \nu(a)$.

Proof. The B*-identity gives $\| a^2 \| = \| a \|^2$, and iterating this gives

$\displaystyle \| a^{2^n} \| = \| a \|^{2^n}$

for all $n \in \mathbb{N}$. Taking $2^n$-th roots of both sides and taking limits, the conclusion follows by the spectral radius formula. $\Box$

Corollary: Let $a$ be an element of a C*-algebra $A$. Then $\| a \| = \sqrt{\nu(a^{\dagger} a)}$. Consequently, $\| a^{\dagger} \| = \| a \|$.

Proof. The B*-identity and the above gives $\| a \|^2 = \| a^{\dagger} a \| = \nu(a^{\dagger} a)$ since $a^{\dagger} a$ is self-adjoint. Now, $\| a^{\dagger} \| = \sqrt{\nu(a a^{\dagger})}$, and since $\lambda - b$ is invertible if and only if $(\lambda - b)^{\dagger} = \overline{\lambda} - b^{\dagger}$ is invertible, it follows that $\nu(b) = \nu(b^{\dagger})$, hence $\nu(a^{\dagger} a) = \nu(a a^{\dagger})$, and the conclusion follows. $\Box$

So the norm on a C*-algebra is already uniquely determined by its complex $^{\dagger}$-structure. Moreover:

Corollary: Let $A, B$ be C*-algebras. Any complex $^{\dagger}$-algebra morphism $\phi : A \to B$ is a weak contraction; consequently, it is already a morphism of C*-algebras. Equivalently, the category $C^{\ast}\text{-Alg}$ of C*-algebras is a full subcategory of the category of complex $^{\dagger}$-algebras.

Proof. If $\lambda - a \in A$ is invertible in $A$ then $\phi(\lambda - a) = \lambda - \phi(a)$ is invertible in $B$. Consequently, $\sigma(\phi(a)) \subseteq \sigma(a)$, so $\nu(\phi(a)) \le \nu(a)$. By the above, it then follows that $\| \phi(a) \| \le \| a \|$. $\Box$

Before we proceed, we will also need the following.

Lemma: Let $a$ be a self-adjoint element of a C*-algebra $A$. Then $a$ has real spectrum.

Proof. Consider

$\displaystyle u_t = \exp(ita) = \sum_{n=0}^{\infty} \frac{(ita)^n}{n!}$.

This sum converges absolutely for all $a$ and we have $\| u_t \| \le \exp(\|a\|)$. Standard analytic arguments show that

$\displaystyle u_t^{\dagger} u_t = u_{-t} u_t = \exp(-ita) \exp(ita) = 1$

and we know that $\| u_t \| = \| u_t^{\dagger} \| = \sqrt{ \nu(1) } = 1$ by the above. It follows that if $\lambda : A \to \mathbb{C}$ is any morphism of Banach algebras, it is in particular a weak contraction, so $|\lambda(u_t)|, |\lambda(u_{-t})| \le 1$. But $\lambda(u_t) = e^{it \lambda(a)}$ and $\lambda(u_t) = e^{-it \lambda(a)}$; combining these two conditions we conclude that $\lambda(a)$ has zero imaginary part. $\Box$

Gelfand duality

We now have enough to prove the following remarkable result.

Theorem (commutative Gelfand-Naimark): Let $A$ be a commutative C*-algebra. Then the Gelfand representation $A \to C(\text{MaxSpec } A)$ is an isomorphism of C*-algebras.

In particular, commutative C*-algebras are semiprimitive.

Proof. It remains to show that 1) the Gelfand representation respects $^{\dagger}$, 2) the Gelfand representation is an isometry (hence $\| a \| = \nu(a)$) and in particular injective, 3) the Gelfand representation is surjective.

1): Let $a + bi \in A$ where $a, b$ are self-adjoint (recall that every element of $A$ is uniquely of this form). If $\lambda : A \to \mathbb{C}$ is a morphism, then $\lambda(a), \lambda(b)$ are real by the lemma above, hence

$\displaystyle \lambda((a + bi)^{\dagger}) = \lambda(a - bi) = \lambda(a) - \lambda(b) i = \overline{\lambda(a) + \lambda(b) i}$.

2): This follows from $\| a \| = \sqrt{\nu(a^{\dagger} a)}$ and 1), since the Gelfand representation preserves spectral radius. If $\nu(a) = 0$ it follows now that $\| a \| = 0$, hence $a = 0$, hence the Gelfand representation is injective.

3): By 2), $A$ may be identified with a closed subalgebra of $C(\text{MaxSpec } A)$ with norm the induced norm which is closed under complex conjugation, and by definition of the maximal spectrum it separates points. By the complex Stone-Weierstrass theorem, it then follows that $A = C(\text{MaxSpec } A)$. $\Box$

Remember that when we exhibited a contravariant equivalence between compact Hausdorff spaces and certain $\mathbb{R}$-algebras we did not specify what algebras arose in this way. The Gelfand-Naimark theorem supplies the answer: they are precisely the self-adjoint subalgebras of commutative C*-algebras. Combined with results we already know about this equivalence, we conclude the following.

Gelfand duality: The functors $A \mapsto \text{MaxSpec } A, X \mapsto C(X)$ form a contravariant equivalence between commutative C*-algebras and compact Hausdorff spaces.

Corollary: Let $X$ be a topological space. Then $\text{MaxSpec } C_b(X)$ is the Stoneâ€“Cech compactification $\beta X$ of $X$.

Gelfand duality is a very suggestive theorem: it suggests that doing topology with compact Hausdorff spaces is equivalent to doing algebra with commutative C*-algebras. The idea that noncommutative C*-algebras therefore constitute a generalization of compact Hausdorff spaces is the basis of the noncommutative topology program.

But even without getting into such issues, Gelfand duality also has relatively concrete applications. For example, it implies a weak form of the spectral theorem: if $T : H \to H$ is a normal operator on a Hilbert space $H$, then considering the commutative C*-algebra $A_T$ generated by $T$ it follows that $T$ is isomorphic, in an appropriate sense, to a multiplication operator

$\displaystyle M_f(g) = fg$

where $f$ is a continuous complex-valued function on a compact Hausdorff space $X$ (namely the image of $T$ in $C(\text{MaxSpec } A_T)$). Ideally we would want $g$ to be an element of $L^2(X)$ and we would want the identification between $T$ and $M_f$ to be unitary, and it seems that we need to find an appropriate Borel measure on $X$ to say this. However, even without such a measure we still see that $T$ is diagonalizable in a suitable sense.

If $H$ is finite-dimensional, then so is $A_T$, and this is possible if and only if $\text{MaxSpec } A_T$ is finite with the discrete topology; call its points $1, 2, ... n$. Note that $n$ must be the number of distinct eigenvalues of $T$ and the points $1, 2, ... n$ may be identified with the set (not multiset) of eigenvalues of $T$. Moreover, $C(\text{MaxSpec } A_T)$ contains $n$ distinguished idempotents $e_1, e_2, ... e_n$, the indicator functions of the points $1, 2, ... n$. These elements are self-adjoint idempotents in $H \Rightarrow H$, so they must in fact be projections, and they are precisely the projections onto the eigenspaces of $T$. So the spectral theorem provided by Gelfand duality really does recover the ordinary finite-dimensional spectral theorem.

Another application is as follows. Let $b$ be a normal element of a C*-algebra $A$. The closure of $\mathbb{C}[b, b^{\dagger}]$ defines a commutative C*-subalgebra $B$ of $A$ (the minimal such thing). By Gelfand duality, $B \cong C(\text{MaxSpec } B)$. Now, the latter space is closed under composition by any continuous function $\mathbb{C} \to \mathbb{C}$, Hence if $f$ is any such function, we can define a continuous functional calculus $f \mapsto f(b)$ which allows us to apply continuous functions $\mathbb{C} \to \mathbb{C}$ to normal elements of $A$ and to get other elements of $A$. For example, we can define the absolute value $|b| \in B \subseteq A$ (not to be confused with the norm $\| b \|$) by taking the pointwise absolute value in $C(\text{MaxSpec } B)$.

$f$ only needs to be defined on the spectrum of $b$ in order for $f(b)$ to be well-defined. Thus if $b$ is self-adjoint then we can make sense of $f(b)$ for any continuous $f : \mathbb{R} \to \mathbb{C}$. If $b$ is self-adjoint and has positive spectrum then we can make sense of, for example, $\sqrt{b}$ and $\log b$.

As a more specific application (although I do not think I need the full power of the continuous functional calculus to conclude this), recall that a one-parameter subgroup of a topological group $G$ is a continuous group homomorphism $\phi : \mathbb{R} \to G$.

Theorem: The map assigning an element $a$ of a C*-algebra $A$ the function $u_t = \exp(ita)$ is a bijection from self-adjoint elements of $A$ to norm-continuous one-parameter subgroups of the unitary group $\text{U}(A) = \{ u \in \text{GL}(A) : u^{\dagger} = u^{-1} \}$ of $A$.

Proof. If $a$ is self-adjoint, $\exp(ita)$ lands in the commutative C*-subalgebra generated by $a$, and any one-parameter subgroup of $\text{U}(A)$ takes values in a commutative C*-subalgebra, so we may assume without loss of generality that $A$ is commutative. If $a$ is self-adjoint, then $\exp(ita)$, thought of as a one-parameter family of elements of $C(\text{MaxSpec } A)$, is a one-parameter subgroup by the boundedness of $a$, and taking adjoints shows that it is unitary. Conversely, if $u_t$ is a one-parameter subgroup of $\text{U}(A)$, then for $t$ sufficiently small $u_t$, as a function on $\text{MaxSpec } A$, will only take values in a ball of small radius around $1$ by norm-continuity. On a sufficiently small such ball we can define a branch of the logarithm, and then $\log u_t$ is a one-parameter family of functions $\text{MaxSpec } A \to \mathbb{C}$ such that $\log u_{t+s} = \log u_t u_s = \log u_t + \log u_s$. Since $\log u_t$ is continuous in $t$, it follows that $\log u_t = ita$ for some $a \in A$, and by unitarity $a$ is self-adjoint. $\Box$

The above theorem describes the exponential map from the unitary Lie algebra

$\displaystyle \mathfrak{u}(A) = \{ a \in A : a^{\dagger} = -a \}$

of skew-adjoint elements of $A$ (which are precisely the elements $ia$ where $a$ is self-adjoint) to the unitary group $\text{U}(A)$. However, it is stated for self-adjoint elements so that it becomes a version of Noether’s theorem for C*-algebras, with self-adjoint elements of a C*-algebra $A$ taking the role of observables of some quantum system (the noncommutative space associated to $A$) and one-parameter subgroups of $\text{U}(A)$ taking the role of one-parameter groups of symmetries of the system.

Note that already to talk about position and momentum we needed unbounded operators which can’t embed into a Banach algebra or strongly continuous (not norm-continuous) one-parameter subgroups, so this is an incomplete discussion.

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### 4 Responses

1. Reblogged this on Observer.

2. […] (This post was originally intended to go up immediately after the sequence on Gelfand duality.) […]

3. I’ve only had time to skim this very briefly, but it seems you pass over what to me is the crux of the Gelfand rep, namely that it is left adjoint to the natural contravariant functor from compact Hausdorff spaces to unital Banach algebras. Unlike certain topologists from Berkeley I really do think it is a theorem about the category of commutative unital Banach algebras and not just one about the subcategory of unital commutative C*-algebras đŸ™‚

• Point. I think a proof of this can be assembled from the ingredients in the post, though.