Gelfand Theory and Commutative C Star Algebras — Examples

By [L1] the map $e$ is a homeomorphism for every nonempty compact Hausdorff $K$, so in particular every character of $C(K)$ is ${\mathrm{e}\mathrm{v}}_x$ for a unique $x$, and the topology on $\mathrm{\Delta }(C(K))$ is the transported topology of $K$; no computation beyond [L1] is needed.

For $K=[0,1]$ the hypotheses of [L1] hold by [L2], so the description of the characters of $C([0,1])$ follows; explicitly, ${\mathrm{e}\mathrm{v}}_s={\mathrm{e}\mathrm{v}}_t$ would give $f(s)=f(t)$ for all continuous $f$, and the coordinate function $f(u)=u$ then forces $s=t$.

$A(\mathbb{D})$ is a complex vector space closed under pointwise multiplication, with unit the constant function $1$ of norm one, and the supremum norm is submultiplicative; the constant functions and the coordinate function $z$ show that it is nonzero.

$A(\mathbb{D})$ is complete: if $(f_n)\subseteq A(\mathbb{D})$ is Cauchy in the supremum norm, then it is uniformly Cauchy on $\bar{\mathbb{D}}$, so [L3] supplies a uniform limit $f$, and [L3] also makes $f$ continuous. For every filled triangle $\mathrm{\Delta }\subseteq \mathbb{D}$, [L4] gives ${\int }_{\mathrm{\partial }\mathrm{\Delta }}{f}_n=0$ because $f_n$ is holomorphic. Passing to the uniform limit in the contour integral by [L3] gives ${\int }_{\mathrm{\partial }\mathrm{\Delta }}f=0$, and Morera's direction of [L4] makes $f$ holomorphic on $\mathbb{D}$. Hence $f\in A(\mathbb{D})$ and $\mathrm{\parallel }{f}_n-f{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$.

For every $a\in \bar{\mathbb{D}}$ the evaluation ${\mathrm{e}\mathrm{v}}_a$ is a character: it is nonzero, complex-linear and multiplicative, and ${\mathrm{e}\mathrm{v}}_a(1)=1$.

Let $\chi$ be a character of $A(\mathbb{D})$ and put $a:=\chi (z)$, where $z$ denotes the coordinate function. By [L1] $\mathrm{\mid }a\mathrm{\mid }\le \mathrm{\parallel }z{\mathrm{\parallel }}_{\mathrm{\infty }}=1$, so $a\in \bar{\mathbb{D}}$; and for every polynomial $p$ one has $\chi (p)=p(a)$ by linearity, multiplicativity and $\chi (1)=1$.

Polynomials are uniformly dense in $A(\mathbb{D})$: for $f\in A(\mathbb{D})$ and $0<r<1$ the function $f_r(z):=f(rz)$ is holomorphic on the disc $\mathrm{\mid }z\mathrm{\mid }<1\mathrm{/}r$ and agrees with the sum of its Taylor series there, and the partial sums converge uniformly on the compact set $\bar{\mathbb{D}}$ by [L2]; moreover $f_r\to f$ uniformly on $\bar{\mathbb{D}}$ as $r\uparrow 1$ by uniform continuity of $f$ on the compact disc; hence $f$ is a uniform limit of polynomials.

Consequently $\chi (f)=f(a)$ for every $f\in A(\mathbb{D})$: by [step 1.5] take polynomials $p_n\to f$ uniformly and use continuity of $\chi$ from [L1] together with $\chi (p_n)=p_n(a)$ from [step 1.4]; hence $\chi ={\mathrm{e}\mathrm{v}}_a$ with $a=\chi (z)\in \bar{\mathbb{D}}$, so every character is an evaluation at a point of the disc, and the point is unique because ${\mathrm{e}\mathrm{v}}_a={\mathrm{e}\mathrm{v}}_b$ forces $a={\mathrm{e}\mathrm{v}}_a(z)={\mathrm{e}\mathrm{v}}_b(z)=b$.

The map $a\mapsto {\mathrm{e}\mathrm{v}}_a$ is continuous because every coordinate $a\mapsto {\mathrm{e}\mathrm{v}}_a(f)=f(a)$ is continuous; it is a bijection by [step 2.1] and [step 1.3]. Its domain $\bar{\mathbb{D}}$ is compact and its target is Hausdorff by [L6], so [L6] makes it a homeomorphism.

The boundary restriction is isometric: by [L5] applied to the bounded domain $\mathbb{D}$ and the function $f$, continuous on the closure, one has ${\sup }_{\bar{\mathbb{D}}}\mathrm{\mid }f\mathrm{\mid }={\max }_{\mathrm{\partial }\mathbb{D}}\mathrm{\mid }f\mathrm{\mid }$, so $\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}=\mathrm{\parallel }f{\mathrm{\mid }}_{\mathrm{\partial }\mathbb{D}}{\mathrm{\parallel }}_{\mathrm{\infty }}$ and $R$ preserves norms; and the character space is $\mathrm{\Delta }(A(\mathbb{D}))\cong \bar{\mathbb{D}}$, which contains points not on the boundary, so it is not the circle alone.

All integer-index sums are transported by the fixed bijection $b:\mathbb{N}\to \mathbb{Z}$; double-index sums use $b\times b$. For nonnegative families the sum is the supremum of finite subsums, so a bijective reindexing preserves it. For absolutely summable complex families apply [L3] to real and imaginary parts. Thus the $\mathbb{N}$-indexed Tonelli and Fubini statements apply to the displayed integer-index sums. Convolution is well defined and $\mathrm{\parallel }a\ast b{\mathrm{\parallel }}_1\le \mathrm{\parallel }a{\mathrm{\parallel }}_1\mathrm{\parallel }b{\mathrm{\parallel }}_1$: for each $n$ the family $(\mathrm{\mid }{a}_j{b}_{n-j}\mathrm{\mid }{)}_j$ has finite sum at most $\mathrm{\parallel }a{\mathrm{\parallel }}_1\mathrm{\parallel }b{\mathrm{\parallel }}_1$ whenever ${\sum }_j\mathrm{\mid }{a}_j\mathrm{\mid }<\mathrm{\infty }$ and $\mathrm{\mid }{b}_{n-j}\mathrm{\mid }$ is bounded along $j$; summing over $n$ and interchanging the order of summation by Tonelli's theorem [L3] gives ${\sum }_n{\sum }_j\mathrm{\mid }{a}_j{b}_{n-j}\mathrm{\mid }=\mathrm{\parallel }a{\mathrm{\parallel }}_1\mathrm{\parallel }b{\mathrm{\parallel }}_1$, so every convolution coordinate is absolutely convergent and the estimate follows.

Convolution is commutative and associative and ${\delta }_0$ is the identity: commutativity is the change of variable $j\mapsto n-j$; associativity is the regrouping of the absolutely summable triple family $(a_i{b}_j{c}_{n-i-j}{)}_{i,j}$ along the two possible bracketing orders, licensed by Tonelli and Fubini for the real and imaginary parts [L3]; and $(a\ast {\delta }_0{)}_n=a_n=({\delta }_0\ast a{)}_n$.

${\mathrm{\ell }}^1(\mathbb{Z})$ is complete: if $(a^{(k)})$ is Cauchy in $\mathrm{\parallel }\cdot {\mathrm{\parallel }}_1$, then each coordinate sequence $(a_n^{(k)}{)}_k$ is Cauchy in $\mathbb{C}$ and converges by [L4] to some $a_n$; for every finite set $F\subseteq \mathbb{Z}$ one has ${\sum }_{n\in F}\mathrm{\mid }{a}_n\mathrm{\mid }={\lim }_k{\sum }_{n\in F}\mathrm{\mid }{a}_n^{(k)}\mathrm{\mid }\le {\sup }_k\mathrm{\parallel }{a}^{(k)}{\mathrm{\parallel }}_1<\mathrm{\infty }$, so $a\in {\mathrm{\ell }}^1(\mathbb{Z})$ with $\mathrm{\parallel }a{\mathrm{\parallel }}_1\le {\sup }_k\mathrm{\parallel }{a}^{(k)}{\mathrm{\parallel }}_1$, and the same finite-subset estimate applied to $a-a^{(k)}$ gives $\mathrm{\parallel }a-a^{(k)}{\mathrm{\parallel }}_1\le {\sup }_{l\ge k}\mathrm{\parallel }{a}^{(l)}-a^{(k)}{\mathrm{\parallel }}_1\to 0$.

${\delta }_1$ is invertible with inverse ${\delta }_{-1}$, since ${\delta }_1\ast {\delta }_{-1}={\delta }_0$; if $\chi$ is a character and $z:=\chi ({\delta }_1)$, then $1=\chi ({\delta }_0)=\chi ({\delta }_1)\chi ({\delta }_{-1})$ so $\mathrm{\mid }z\mathrm{\mid }\ge 1$ by [L1], while $\mathrm{\mid }z\mathrm{\mid }\le \mathrm{\parallel }{\delta }_1{\mathrm{\parallel }}_1=1$; hence $\mathrm{\mid }z\mathrm{\mid }=1$, and multiplicativity gives $\chi ({\delta }_n)=z^n$ for all $n\in \mathbb{Z}$.

For $a\in {\mathrm{\ell }}^1(\mathbb{Z})$ define $h_a(z)={\sum }_n{a}_n{z}^n$ on $\mathbb{T}$. This series converges absolutely by [L3]. Given $ϵ>0$, choose a finite initial segment of the fixed enumeration with tail sum less than $ϵ\mathrm{/}4$, and choose $N$ so that $[-N,N]$ contains that segment. Then ${\sum }_{\mathrm{\mid }n\mathrm{\mid }>N}\mathrm{\mid }{a}_n\mathrm{\mid }<ϵ\mathrm{/}4$. Put $M={\sum }_{\mathrm{\mid }n\mathrm{\mid }\le N}\mathrm{\mid }n\mathrm{\mid }\mathrm{\mid }{a}_n\mathrm{\mid }$. For $z,w\in \mathbb{T}$, telescoping positive powers and the identity $\mathrm{\mid }{z}^{-1}-w^{-1}\mathrm{\mid }=\mathrm{\mid }z-w\mathrm{\mid }$ give $\mathrm{\mid }{z}^n-w^n\mathrm{\mid }\le \mathrm{\mid }n\mathrm{\mid }\mathrm{\mid }z-w\mathrm{\mid }$ for every integer $n$. Thus $\mathrm{\mid }{h}_a(z)-h_a(w)\mathrm{\mid }\le M\mathrm{\mid }z-w\mathrm{\mid }+ϵ\mathrm{/}2$. Taking $\mathrm{\mid }z-w\mathrm{\mid }<ϵ\mathrm{/}(2(M+1))$ proves continuity, including $M=0$. Once these maps are shown to be characters, this proves continuity into the evaluation topology [L1]; evaluation at ${\delta }_1$ is continuous in the reverse direction.

For $\mathrm{\mid }z\mathrm{\mid }=1$ the map ${\chi }_z(a):={\sum }_n{a}_n{z}^n$ is a character: it is complex-linear, nonzero (${\chi }_z({\delta }_0)=1$) and multiplicative, because expanding ${\chi }_z(a){\chi }_z(b)={\sum }_j{\sum }_k{a}_j{b}_k{z}^{j+k}$ and regrouping along $n=j+k$ (Tonelli and Fubini on the absolutely summable family $(a_j{b}_k{z}^{j+k})$, [L3]) gives ${\sum }_n(a\ast b{)}_n{z}^n={\chi }_z(a\ast b)$.

For every $a\in {\mathrm{\ell }}^1(\mathbb{Z})$ and every character $\chi$ with $z=\chi ({\delta }_1)$ one has $\chi (a)={\sum }_n{a}_n{z}^n$: let $U(a{)}_k=a_{b(k)}$. The definition of the norm gives $\mathrm{\parallel }Ua{\mathrm{\parallel }}_1=\mathrm{\parallel }a{\mathrm{\parallel }}_1$, and $U$ has inverse $(U^{-1}v{)}_n=v_{b^{-1}(n)}$. Define $Q_N=U^{-1}{P}_{N}U$ using precisely the $\mathbb{N}$-indexed $P_N$ of [L2]. Then $\mathrm{\parallel }a-Q_{N}a{\mathrm{\parallel }}_1=\mathrm{\parallel }Ua-P_{N}Ua{\mathrm{\parallel }}_1\to 0$; $Q_{N}a$ is the finite sum ${\sum }_{k=0}^N{a}_{b(k)}{\delta }_{b(k)}$. Approximate $a$ by these $Q_{N}a$, apply linearity and [step 1.4] to each truncation, and pass to the limit using continuity of $\chi$ from [L1]; the series converges absolutely because $\mathrm{\mid }{a}_n{z}^n\mathrm{\mid }=\mathrm{\mid }{a}_n\mathrm{\mid }$.

The map $z\mapsto {\chi }_z$ is a bijection $\mathbb{T}\to \mathrm{\Delta }({\mathrm{\ell }}^1(\mathbb{Z}))$: it is injective because ${\chi }_z={\chi }_{z^{\mathrm{\prime }}}$ forces $z={\chi }_z({\delta }_1)={\chi }_{z^{\mathrm{\prime }}}({\delta }_1)=z^{\mathrm{\prime }}$, and it is surjective by [step 2.1] applied to the character $\chi$ and its value $z=\chi ({\delta }_1)\in \mathbb{T}$ from [step 1.4].

By [step 1.5] and [step 3.1] the assignment is a continuous bijection with continuous inverse between the compact Hausdorff space $\mathbb{T}$ and the Hausdorff character space $\mathrm{\Delta }({\mathrm{\ell }}^1(\mathbb{Z}))$, hence a homeomorphism; and by [step 2.1] the Gelfand transform of $a$ is the Laurent series $\widehat{a}({\chi }_z)={\sum }_n{a}_n{z}^n$.

The multiplication $(a+b\epsilon )(c+d\epsilon )=ac+(ad+bc)\epsilon$ is associative, commutative and complex-bilinear, and the norm is submultiplicative: $\mathrm{\parallel }xy\mathrm{\parallel }=\mathrm{\mid }ac\mathrm{\mid }+\mathrm{\mid }ad+bc\mathrm{\mid }\le \mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }c\mathrm{\mid }+\mathrm{\mid }a\mathrm{\mid }\mathrm{\mid }d\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid }\mathrm{\mid }c\mathrm{\mid }\le (\mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }b\mathrm{\mid })(\mathrm{\mid }c\mathrm{\mid }+\mathrm{\mid }d\mathrm{\mid })=\mathrm{\parallel }x\mathrm{\parallel }\mathrm{\parallel }y\mathrm{\parallel }$; the unit is $1=1+0\epsilon$ with $\mathrm{\parallel }1\mathrm{\parallel }=1$, and the algebra is complete because its two coordinates are controlled by the norm ($\mathrm{\mid }a\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }$, $\mathrm{\mid }b\mathrm{\mid }\le \mathrm{\parallel }x\mathrm{\parallel }$) and conversely $\mathrm{\parallel }x\mathrm{\parallel }\le 2\max (\mathrm{\mid }a\mathrm{\mid },\mathrm{\mid }b\mathrm{\mid })$, so the norm is equivalent to the Euclidean norm on ${\mathbb{C}}^2$; hence $A$ is a nonzero commutative unital complex Banach algebra.

Every character $\chi$ of $A$ satisfies $\chi (\epsilon {)}^2=\chi ({\epsilon }^2)=\chi (0)=0$ by multiplicativity [L1], so $\chi (\epsilon )=0$ since $\mathbb{C}$ is a field; hence $\chi (a+b\epsilon )=a\chi (1)+b\chi (\epsilon )=a$ by linearity and unitality [L1]; in particular $\mathrm{\Delta }(A)$ consists of the single character ${\chi }_0(a+b\epsilon )=a$.

The element $\epsilon =0+1\epsilon$ has norm $\mathrm{\parallel }\epsilon \mathrm{\parallel }=1\mathrm{\ne }0$, while its Gelfand transform vanishes identically: $\widehat{\epsilon }({\chi }_0)={\chi }_0(\epsilon )=0$ by [step 1.2]; hence $\mathrm{\parallel }\widehat{\epsilon }{\mathrm{\parallel }}_{\mathrm{\infty }}=0\mathrm{\ne }1=\mathrm{\parallel }\epsilon \mathrm{\parallel }$, so the Gelfand transform of $A$ is not isometric (and not injective, since $\epsilon \mathrm{\ne }0$ has zero transform).

$h$ is a homeomorphism with $h^{-1}=h$, since $h(h(t))=t$, and $u$ is continuous with $\mathrm{\mid }u(t)\mathrm{\mid }=1$ by [L2]; hence by [L1] the map $Tf(t)=u(t)f(h(t))$ is a surjective linear isometry.

$T$ is not unital: $T\mathbf{1}=u$, and $u(t)=e^{it}$ is not the constant function $1$ because at the endpoint $t=1$, [L2] and [L3] give $\mathrm{Im}u(1)=\sin 1\ge 5\mathrm{/}6>0$.

$T$ is not multiplicative: $(T\mathbf{1})(T\mathbf{1})=u^2$ while $T(\mathbf{1}\cdot \mathbf{1})=u$, and $u^2\mathrm{\ne }u$ because $u(t)=e^{it}\mathrm{\ne }0$ and $u(t)\mathrm{\ne }1$ for some $t$; at $t=1$, equality $u(1{)}^2=u(1)$ would, by division by the nonzero $u(1)$, force $u(1)=1$, contradicting [step 1.2].

In the real case $u\equiv -1$ has $\mathrm{\mid }u\mathrm{\mid }=1$ and $h=h^{-1}$; directly, $\mathrm{\parallel }Tf{\mathrm{\parallel }}_{\mathrm{\infty }}={\sup }_{t\in [0,1]}\mathrm{\mid }f(1-t)\mathrm{\mid }=\mathrm{\parallel }f{\mathrm{\parallel }}_{\mathrm{\infty }}$ and $T^2=I$, so $T$ is a surjective real-linear isometry. Moreover, $T\mathbf{1}=-1\mathrm{\ne }1$ and $T(\mathbf{1}\cdot \mathbf{1})=-1\mathrm{\ne }(-1)(-1)=1$, so $T$ is neither unital nor multiplicative.

The cofinite filter $\mathcal{F}=\{A\subseteq \mathbb{N}:\mathbb{N}\setminus A\text{ finite}\}$ is a proper filter: it contains $\mathbb{N}$, omits $\mathrm{\varnothing }$ (whose complement is infinite), and is closed under finite intersections and upward inclusion.

By [L3] extend $\mathcal{F}$ to an ultrafilter $\mathcal{U}$; since $\mathcal{F}\subseteq \mathcal{U}$, no singleton belongs to $\mathcal{U}$ (a singleton has infinite complement, so it is not in the cofinite filter, and its complement is in $\mathcal{U}$, so the singleton is not), that is, $\mathcal{U}$ is free.

$M_{\mathcal{U}}=\{f:Z(f)\in \mathcal{U}\}$ is a maximal ideal of ${\mathbb{R}}^{\mathbb{N}}$ by [L2], and it is not fixed: if $M_{\mathcal{U}}=M_n=\{f:f(n)=0\}$ for some $n$, then for the characteristic function $f:={\mathbf{1}}_{\{n\}}$ one has $Z(f)=\mathbb{N}\setminus \{n\}\in \mathcal{U}$ by freeness, so $f\in M_{\mathcal{U}}$ while $f(n)=1\mathrm{\ne }0$, so $f\mathrm{\notin }{M}_n$, a contradiction.

By [L1] the maximal ideals of $C(\mathbb{N},\mathbb{R})$ are the $M_p$ with $p\in \beta \mathbb{N}$ uniquely determined, and by [step 2.1] there is a maximal ideal that is not fixed. By [L1] its point $p$ lies outside $\mathbb{N}$, so $\beta \mathbb{N}\setminus \mathbb{N}\mathrm{\ne }\mathrm{\varnothing }$. This proves the point-set parametrisation claimed in the example; [L1] supplies no topology on the maximal-ideal set, and none is inferred.

Equivalently, $\beta \mathbb{N}\mathrm{\ne }\mathbb{N}$ directly: $\mathbb{N}$ is dense in the compact space $\beta \mathbb{N}$ by [L4], so if $\beta \mathbb{N}=\mathbb{N}$ then $\mathbb{N}$ would be compact, contradicting [L4].

The ultrafilters on $\mathcal{P}(\mathbb{N})$ are exactly the maximal filters of subsets of $\mathbb{N}$ ("set ultrafilters") by the complement dichotomy [L2]; the principal ultrafilters $U_n$ are pairwise distinct and $\{U_n\}=[\{n\}]$ is open, so the map $n\mapsto U_n$ is an injective continuous map from discrete $\mathbb{N}$ onto a discrete subspace.

The image $\{U_n:n\in \mathbb{N}\}$ is dense in the ultrafilter space: a nonempty basic clopen $[A]$ with $A\mathrm{\ne }\mathrm{\varnothing }$ contains $U_n$ for every $n\in A$.

Let $K$ be compact Hausdorff and $f:\mathbb{N}\to K$ continuous (that is, arbitrary). For an ultrafilter $U$ on $\mathbb{N}$ let $f_{\ast }U:=\{B\subseteq K:f^{-1}(B)\in U\}$, an ultrafilter on $K$, which converges to a unique point by [L3]; define $F(U)$ to be that limit.

The map $F$ extends $f$: for the principal ultrafilter $U_n$ the pushforward $f_{\ast }{U}_n$ is the principal ultrafilter at $f(n)$, which converges to $f(n)$, so $F(U_n)=f(n)$.

The map $F$ is continuous. Let $F(U)\in V$ with $V\subseteq K$ open. By [L5] choose an open $W$ with $F(U)\in W$ and $\bar{W}\subseteq V$. Since $f_{\ast }U$ converges to $F(U)$, one has $W\in f_{\ast }U$, equivalently $f^{-1}(W)\in U$, so the basic open $[f^{-1}(W)]$ contains $U$. If $U^{\mathrm{\prime }}$ lies in this basic open, then $W\in f_{\ast }{U}^{\mathrm{\prime }}$; because $f_{\ast }{U}^{\mathrm{\prime }}$ converges to $F(U^{\mathrm{\prime }})$, its limit lies in $\bar{W}$ (otherwise the open complement of $\bar{W}$ would also belong to the ultrafilter). Thus $F(U^{\mathrm{\prime }})\in \bar{W}\subseteq V$, proving $[f^{-1}(W)]\subseteq F^{-1}(V)$ and hence continuity.

Uniqueness: $F$ is determined on the dense subset $\{U_n:n\in \mathbb{N}\}$ by [step 1.2] and [step 1.4], and $K$ is Hausdorff, so two continuous extensions agree.

By [step 1.1], [step 1.2], [step 2.1] and [step 2.2] the pair (ultrafilter space, $n\mapsto U_n$) is a Hausdorff compactification of $\mathbb{N}$ satisfying the universal property [L4], so it is a Stone–Čech compactification; by [L1] the basic clopens are the $[A]$ and the algebra of clopens is canonically $\mathcal{P}(\mathbb{N})$.

Every ultrafilter $U$ of $\mathcal{P}(\{1,2,3\})$ is principal: the join $\{1\}\cup \{2\}\cup \{3\}=\{1,2,3\}$ belongs to $U$, so one of the singletons belongs to $U$ by the dichotomy [L1], and then $U=\uparrow \text{\hspace{0.17em}⁣}\{i\}$ for that $i$; conversely each $\uparrow \text{\hspace{0.17em}⁣}\{i\}$ is an ultrafilter by [L1].

Consequently $\mathrm{Ult}(B)=\{\uparrow \text{\hspace{0.17em}⁣}\{1\},\uparrow \text{\hspace{0.17em}⁣}\{2\},\uparrow \text{\hspace{0.17em}⁣}\{3\}\}$ has three points, and the basic open sets $[S]=\{U:S\in U\}$ are in bijection with the subsets $S\subseteq \{1,2,3\}$ through $S\leftrightarrow \{i:i\in S\}$; in particular every subset of the three-point space is basic open, so the topology is discrete and $\mathrm{Clop}(\mathrm{Ult}(B))=\mathcal{P}(\mathrm{Ult}(B))$ has eight elements, matching $\mathrm{\mid }B\mathrm{\mid }=8$.

Let $B$ be a general finite Boolean algebra. If $B$ is trivial, then $\mathrm{A}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}(B)=\mathrm{\varnothing }$, the unique map $B\to \mathcal{P}(\mathrm{\varnothing })$ is an isomorphism, and both $B$ and $\mathcal{P}(\mathrm{\varnothing })$ have no ultrafilters; their Stone space is the empty discrete space. If $B$ is nontrivial, [L2] says that distinct atoms have meet $0$ and every $b\in B$ is the join of the atoms below it. Thus $b\mapsto \{a\text{ atom}:a\le b\}$ is a bijection onto the power set of the finite atom set and preserves joins, meets and complements. Hence $B\cong \mathcal{P}(\mathrm{A}\mathrm{t}\mathrm{o}\mathrm{m}\mathrm{s}(B))$; the argument of [step 1.1], with the finite nonempty atom set in place of $\{1,2,3\}$, says its ultrafilters are the principal ones at atoms, so its Stone space is finite and discrete.