States, tracial states and faithful normal traces on a von Neumann algebra

Let $K$ be nonzero and finite-dimensional, set $n=\dim K$, and fix an orthonormal basis $e_1,\dots ,e_n$, obtained from a finite basis by Gram–Schmidt. With matrix units $E_{ij}$, define $\sigma (T):=n^{-1}{\sum }_{i=1}^n⟨T{e}_i,e_i⟩$. It is linear and WOT-continuous as a finite sum of matrix coefficients; $\sigma (I_K)=1$. For positive $T$, every $⟨T{e}_i,e_i⟩\ge 0$ by [F13], so $\sigma$ is positive. For $T\in \mathcal{B}(K)$, [F14, F15] give $\sigma (T^{\ast }T)=n^{-1}{\sum }_i\mathrm{\parallel }T{e}_i{\mathrm{\parallel }}^2$, which vanishes only when $T=0$, proving faithfulness. If $T=(t_{ij})$ and $S=(s_{ij})$, then $\mathrm{tr}(TS)={\sum }_{i,j}{t}_{ij}{s}_{ji}={\sum }_{i,j}{s}_{ji}{t}_{ij}=\mathrm{tr}(ST)$, so $\sigma$ is tracial. Hence $\sigma$ is a faithful normal tracial state.

Equip $\mathrm{\Gamma }$ with the discrete topology. Each singleton is open, so distinct points have disjoint singleton neighbourhoods and the space is Hausdorff. The product topology on $\mathrm{\Gamma }\times \mathrm{\Gamma }$ is discrete because each singleton $\{g\}\times \{h\}$ is basic open; hence multiplication and inversion are continuous. Each singleton is a compact neighbourhood, so $\mathrm{\Gamma }$ is locally compact Hausdorff and its topology is a group topology.

By [F1] and [F3], $\mathrm{\#}$ is a Borel measure on $\mathrm{\Gamma }$ and $\mathrm{\#}(\{e\})=1$. Each left or right translation is a bijection and therefore preserves the finite or infinite cardinality of every subset, so $\mathrm{\#}$ is left- and right-invariant. A compact subset is finite because its cover by open singletons has a finite subcover; hence $\mathrm{\#}$ is finite on compact sets. Every Borel set $E$ is open, and $E$ itself is an open superset, so monotonicity makes the infimum in outer regularity equal to $\mathrm{\#}(E)$. For open $U$, compact subsets are finite and every finite subset is compact; thus ${\sup }_{K\subseteq U\text{ compact}}\mathrm{\#}(K)={\sup }_{F\subseteq U\text{ finite}}\mathrm{\mid }F\mathrm{\mid }=\mathrm{\#}(U)$, since every infinite set contains finite subsets of arbitrarily large size by induction. Therefore $\mathrm{\#}$ satisfies the left and right Haar conditions in [F4].

If $\tau$ is any tracial state on $\mathcal{B}(K)$, then for $i\mathrm{\ne }j$, $\tau (E_{ij})=\tau (E_{ii}{E}_{ij})=\tau (E_{ij}{E}_{ii})=0$. Also $\tau (E_{ii})=\tau (E_{ij}{E}_{ji})=\tau (E_{ji}{E}_{ij})=\tau (E_{jj})$. Since ${\sum }_i{E}_{ii}=I_K$, normalization gives $\tau (E_{ii})=1\mathrm{/}n$ for every $i$. The matrix units span $\mathcal{B}(K)$, so $\tau =\sigma$; the normalized matrix trace is the unique tracial state. Applying this uniqueness to the formula from any other orthonormal basis proves that the normalized trace is basis-independent.

Define $U:{\mathrm{\ell }}^2(\mathrm{\Gamma },\mathbb{C})\to L^2(\mathrm{\Gamma },\mathrm{\#};\mathbb{C})$ by sending a family to its pointwise function class. By [F1] every function is measurable, and by [F3] the only counting-null set is empty, so each $L^2$ class has a unique pointwise representative. Since $\mathrm{\#}$ is a left Haar measure by step 2.1, [F5] applies with $c=\mathrm{\#}(\{e\})=1$: ${\int }_{\mathrm{\Gamma }}\mathrm{\mid }a(g){\mathrm{\mid }}^{2}d\mathrm{\#}(g)={\sum }_g\mathrm{\mid }a(g){\mathrm{\mid }}^2$. Hence a function represents an $L^2$ class exactly when its family is square-summable, so $U$ is onto and preserves norms. For $a,b\in {\mathrm{\ell }}^2$, [F7] and Cauchy–Schwarz make $a\bar{b}$ absolutely summable; [F5] gives ${\int }_{\mathrm{\Gamma }}\mathrm{\mid }a\bar{b}\mathrm{\mid }d\mathrm{\#}={\sum }_g\mathrm{\mid }a(g)\overline{b(g)}\mathrm{\mid }<\mathrm{\infty }$, so [F6] makes it integrable and [F5] gives $⟨Ua,Ub{⟩}_{L^2}={\int }_{\mathrm{\Gamma }}a\bar{b}d\mathrm{\#}={\sum }_{g}a(g)\overline{b(g)}=⟨a,b{⟩}_{{\mathrm{\ell }}^2}$. By [F9], $L^2(\mathrm{\Gamma },\mathrm{\#};\mathbb{C})$ is a Hilbert space under AC; thus $U$ transports its complete Hilbert structure to ${\mathrm{\ell }}^2(\mathrm{\Gamma },\mathbb{C})$, and the coordinate vectors have dense span by [F8].

For $g\in \mathrm{\Gamma }$, the left and right regular formulas reindex coordinates by bijections. Each bijection induces a bijection of finite subsets and preserves the corresponding finite sums by [F18]; taking their suprema preserves the square sum. Hence $\mathrm{\parallel }{\lambda }_{\mathrm{\Gamma }}(g)f{\mathrm{\parallel }}_2=\mathrm{\parallel }f{\mathrm{\parallel }}_2=\mathrm{\parallel }{\rho }_{\mathrm{\Gamma }}(g)f{\mathrm{\parallel }}_2$. They are bounded linear isometries with inverses ${\lambda }_{\mathrm{\Gamma }}(g^{-1})$ and ${\rho }_{\mathrm{\Gamma }}(g^{-1})$, respectively; thus they are unitary. Direct substitution gives ${\lambda }_{\mathrm{\Gamma }}(g){\lambda }_{\mathrm{\Gamma }}(h)={\lambda }_{\mathrm{\Gamma }}(gh)$, ${\rho }_{\mathrm{\Gamma }}(g){\rho }_{\mathrm{\Gamma }}(h)={\rho }_{\mathrm{\Gamma }}(gh)$, and ${\lambda }_{\mathrm{\Gamma }}(g){\rho }_{\mathrm{\Gamma }}(h)={\rho }_{\mathrm{\Gamma }}(h){\lambda }_{\mathrm{\Gamma }}(g)$.

The linear span ${\mathcal{A}}_{\mathrm{\Gamma }}$ of $\{{\lambda }_{\mathrm{\Gamma }}(g):g\in \mathrm{\Gamma }\}$ is a unital $\ast$-algebra by step 4.1 and ${\lambda }_{\mathrm{\Gamma }}(g{)}^{\ast }={\lambda }_{\mathrm{\Gamma }}(g^{-1})$ by [F14]. Thus $L(\mathrm{\Gamma })=W^{\ast }({\lambda }_{\mathrm{\Gamma }}(\mathrm{\Gamma }))$ is its WOT closure by [F17]. Each ${\rho }_{\mathrm{\Gamma }}(h)$ commutes with ${\mathcal{A}}_{\mathrm{\Gamma }}$ by step 4.1; [F12] makes its commutant WOT-closed, so every $T\in L(\mathrm{\Gamma })$ commutes with every right regular operator.

Put ${\tau }_{\mathrm{\Gamma }}(T)=⟨T{\delta }_e,{\delta }_e⟩$. This is a linear WOT-continuous matrix coefficient by [F10]. Also ${\tau }_{\mathrm{\Gamma }}(I)=1$, and if $P\in L(\mathrm{\Gamma })$ is positive then ${\tau }_{\mathrm{\Gamma }}(P)=⟨P{\delta }_e,{\delta }_e⟩\ge 0$ by [F13]; thus ${\tau }_{\mathrm{\Gamma }}$ is positive. Its WOT continuity gives continuity on the unit ball, which [F16] identifies with order-normality for a positive functional.

For $S\in L(\mathrm{\Gamma })$, $S^{\ast }S$ is positive because $⟨S^{\ast }S\xi ,\xi ⟩=\mathrm{\parallel }S\xi {\mathrm{\parallel }}^2$ by [F14, F15]; thus [F13] makes ${\tau }_{\mathrm{\Gamma }}(S^{\ast }S)$ real. The adjoint identity and conjugate symmetry give $\overline{{\tau }_{\mathrm{\Gamma }}(S^{\ast }S)}=⟨S{\delta }_e,S{\delta }_e⟩=\mathrm{\parallel }S{\delta }_e{\mathrm{\parallel }}^2$, so ${\tau }_{\mathrm{\Gamma }}(S^{\ast }S)=\mathrm{\parallel }S{\delta }_e{\mathrm{\parallel }}^2$. If this is zero, then $S{\delta }_e=0$. For each $g\in \mathrm{\Gamma }$, step 4.1 gives ${\delta }_g={\rho }_{\mathrm{\Gamma }}(g^{-1}){\delta }_e$, and step 5.1 gives $S{\delta }_g=S{\rho }_{\mathrm{\Gamma }}(g^{-1}){\delta }_e={\rho }_{\mathrm{\Gamma }}(g^{-1})S{\delta }_e=0$. The span of these coordinate vectors is dense by step 3.1, so boundedness of $S$ implies $S=0$. Therefore ${\tau }_{\mathrm{\Gamma }}$ is faithful.

On generators, ${\tau }_{\mathrm{\Gamma }}({\lambda }_{\mathrm{\Gamma }}(g))=1$ when $g=e$ and $0$ otherwise. Hence for $g,h\in \mathrm{\Gamma }$, ${\tau }_{\mathrm{\Gamma }}(\lambda (g)\lambda (h))={\mathbf{1}}_{gh=e}={\mathbf{1}}_{hg=e}={\tau }_{\mathrm{\Gamma }}(\lambda (h)\lambda (g))$. Bilinearity proves ${\tau }_{\mathrm{\Gamma }}(AB)={\tau }_{\mathrm{\Gamma }}(BA)$ for all $A,B\in {\mathcal{A}}_{\mathrm{\Gamma }}$. For fixed $A\in {\mathcal{A}}_{\mathrm{\Gamma }}$, [F10, F11] make both maps $T\mapsto {\tau }_{\mathrm{\Gamma }}(AT)$ and $T\mapsto {\tau }_{\mathrm{\Gamma }}(TA)$ WOT-continuous, so their equality extends from the WOT-dense algebra ${\mathcal{A}}_{\mathrm{\Gamma }}$ to every $T\in L(\mathrm{\Gamma })$. Now fix such a $T$; the same continuity in the first variable extends the equality from $A\in {\mathcal{A}}_{\mathrm{\Gamma }}$ to all $A\in L(\mathrm{\Gamma })$. Thus ${\tau }_{\mathrm{\Gamma }}$ is tracial on $L(\mathrm{\Gamma })$.

Steps 6.1, 7.1 and 7.2 show that ${\tau }_{\mathrm{\Gamma }}$ is positive, normalized, normal, faithful and tracial, hence a faithful normal tracial state; steps 1.1 and 2.2 show the corresponding existence and uniqueness claim for the normalized matrix trace. The zero-Hilbert-space case has no state because $I_H=0$, as stated in the definition.