[proofplan] The central projection $z$ splits every element $x\in M$ into its two orthogonal central parts $zx$ and $(1-z)x$. We first verify that the two summands $zM$ and $(1-z)M$ are von Neumann algebras with their natural units. Then we prove that the splitting map is a unital $*$-homomorphism, construct its inverse explicitly by addition, and check weak-operator continuity of both maps from the defining matrix coefficients. [/proofplan] [step:Verify that the two central summands are von Neumann algebras] Define $zM=\{za:a\in M\}$ and $(1-z)M=\{(1-z)a:a\in M\}$. Since $z\in Z(M)$ is a projection, we have $z^2=z=z^*$ and $za=az$ for every $a\in M$. Hence $zM$ is closed under multiplication because $(za)(zb)=zab$, it is closed under the involution because $(za)^*=a^*z=za^*$, and $z$ is its multiplicative identity because $z(za)=za$. The same computation with $1-z$ in place of $z$ shows that $(1-z)M$ is a unital self-adjoint algebra with identity $1-z$. It remains to check weak-operator closedness. Let $I$ be a directed set, and let $(za_i)_{i\in I}$ be a net in $zM$ converging in the weak operator topology to an operator $T\in\mathcal{L}(H)$. Since $M$ is weak-operator closed and every $za_i$ lies in $M$, we have $T\in M$. For all $\xi,\eta\in H$, \begin{align*} (zT\xi,\eta)_H=(T\xi,z\eta)_H=\lim_i (za_i\xi,z\eta)_H=\lim_i (za_i\xi,\eta)_H=(T\xi,\eta)_H. \end{align*} Therefore $zT=T$, so $T\in zM$, and $zM$ is weak-operator closed. Replacing $z$ by $1-z$ gives the same conclusion for $(1-z)M$. Thus both $zM$ and $(1-z)M$ are von Neumann algebras. [/step] [step:Show that the splitting map is a unital $*$-homomorphism] Define the map \begin{align*} \Phi: M \to zM\oplus (1-z)M,\qquad x \mapsto (zx,(1-z)x). \end{align*} The unit of the external direct sum $zM\oplus(1-z)M$ is $(z,1-z)$, and \begin{align*} \Phi(1)=(z,1-z). \end{align*} For $x,y\in M$, centrality of $z$ gives \begin{align*} zxy=(zx)(zy), \end{align*} because $(zx)(zy)=z^2xy=zxy$. Similarly, \begin{align*} (1-z)xy=((1-z)x)((1-z)y). \end{align*} Therefore \begin{align*} \Phi(xy)=\Phi(x)\Phi(y). \end{align*} Linearity follows from distributivity of multiplication by the fixed operators $z$ and $1-z$. Finally, \begin{align*} \Phi(x^*)=(zx^*,(1-z)x^*)=((zx)^*,((1-z)x)^*)=\Phi(x)^*, \end{align*} using $z^*=z$, $(1-z)^*=1-z$, and centrality. Thus $\Phi$ is a unital $*$-homomorphism. [guided] The point of centrality is that multiplication by $z$ respects the algebra product. We define the map \begin{align*} \Phi: M \to zM\oplus (1-z)M,\qquad x \mapsto (zx,(1-z)x). \end{align*} The codomain is the external direct sum, whose multiplication, addition, scalar multiplication, and involution are all coordinatewise. Its unit is $(z,1-z)$, because $z$ is the unit of $zM$ and $1-z$ is the unit of $(1-z)M$. First, \begin{align*} \Phi(1)=(z1,(1-z)1)=(z,1-z), \end{align*} so $\Phi$ is unital. For multiplicativity, take $x,y\in M$. Since $z\in Z(M)$, it commutes with both $x$ and $y$, and since $z$ is a projection, $z^2=z$. Hence \begin{align*} (zx)(zy)=z^2xy=zxy. \end{align*} The same computation applies to the complementary central projection $1-z$: \begin{align*} ((1-z)x)((1-z)y)=(1-z)^2xy=(1-z)xy. \end{align*} Thus \begin{align*} \Phi(x)\Phi(y)=((zx)(zy),((1-z)x)((1-z)y))=(zxy,(1-z)xy)=\Phi(xy). \end{align*} Linearity is coordinatewise distributivity: \begin{align*} \Phi(\alpha x+\beta y)=(z(\alpha x+\beta y),(1-z)(\alpha x+\beta y))=\alpha\Phi(x)+\beta\Phi(y) \end{align*} for all $\alpha,\beta\in\mathbb{C}$. For the involution, centrality and self-adjointness of $z$ give \begin{align*} (zx)^*=x^*z=zx^* \end{align*} and similarly \begin{align*} ((1-z)x)^*=x^*(1-z)=(1-z)x^*. \end{align*} Therefore \begin{align*} \Phi(x^*)=(zx^*,(1-z)x^*)=((zx)^*,((1-z)x)^*)=\Phi(x)^*. \end{align*} So $\Phi$ is a unital $*$-homomorphism. [/guided] [/step] [step:Construct the inverse by adding the two central components] If $\Phi(x)=0$, then $zx=0$ and $(1-z)x=0$. Since $z+(1-z)=1$, we obtain \begin{align*} x=zx+(1-z)x=0. \end{align*} Thus $\Phi$ is injective. Define the map \begin{align*} \Psi: zM\oplus(1-z)M \to M,\qquad (u,v) \mapsto u+v. \end{align*} For $a,b\in M$, an arbitrary element of $zM\oplus(1-z)M$ has the form $(za,(1-z)b)$. Set \begin{align*} x=za+(1-z)b\in M. \end{align*} Using $z(1-z)=0$ and $(1-z)z=0$, we get \begin{align*} zx=z(za)+z(1-z)b=za \end{align*} and \begin{align*} (1-z)x=(1-z)za+(1-z)^2b=(1-z)b. \end{align*} Hence $\Phi(x)=(za,(1-z)b)$. Therefore $\Phi$ is surjective and $\Psi=\Phi^{-1}$. [/step] [step:Check weak-operator continuity of the isomorphism and its inverse] Let $I$ be a directed set, and let $(x_i)_{i\in I}$ be a net in $M$ converging to $x\in M$ in the weak operator topology. For every $\xi,\eta\in H$, \begin{align*} (zx_i\xi,\eta)_H=(x_i\xi,z\eta)_H\to (x\xi,z\eta)_H=(zx\xi,\eta)_H. \end{align*} Thus $zx_i\to zx$ in the weak operator topology. The same calculation with $1-z$ in place of $z$ gives $(1-z)x_i\to(1-z)x$ weakly. Hence $\Phi$ is weak-operator continuous. Conversely, let $(u_i,v_i)_{i\in I}$ be a net in $zM\oplus(1-z)M$ converging weakly coordinatewise to $(u,v)$. For every $\xi,\eta\in H$, \begin{align*} ((u_i+v_i)\xi,\eta)_H=(u_i\xi,\eta)_H+(v_i\xi,\eta)_H\to (u\xi,\eta)_H+(v\xi,\eta)_H=((u+v)\xi,\eta)_H. \end{align*} Therefore $\Psi(u_i,v_i)=u_i+v_i$ converges weakly to $\Psi(u,v)=u+v$, so $\Psi$ is weak-operator continuous. The map $\Phi$ is positive because it is a unital $*$-homomorphism. We justify the passage from the displayed weak-operator computations to normality. By [citetheorem:9274], every normal functional on $\mathcal{L}(H)$ has the form \begin{align*} \omega_A(T)=\operatorname{Tr}(AT) \end{align*} for some trace-class operator $A\in\mathcal{T}(H)$ and every $T\in\mathcal{L}(H)$. Trace-class operators are norm limits in $\mathcal{T}(H)$ of finite-rank trace-class operators. On a bounded subset of $\mathcal{L}(H)$, convergence of all vector functionals therefore implies convergence against every trace-class functional: finite-rank trace-class functionals are finite sums of vector functionals, and the trace-class norm approximation controls the remaining error uniformly on bounded sets. The weak-operator convergent nets used above are bounded by the [uniform boundedness principle](/theorems/549), so this bounded-subset argument applies to them. Applying the same reasoning coordinatewise to $zM\oplus(1-z)M\subseteq \mathcal{L}(H)\oplus\mathcal{L}(H)$ shows that the displayed matrix-coefficient computations give ultraweak continuity of $\Phi$ on bounded subsets. Since $\Phi:M\to zM\oplus(1-z)M$ is a positive [linear map](/page/Linear%20Map) between von Neumann algebras, [citetheorem:9275] applies and gives that $\Phi$ is normal. We have shown that $\Phi$ is a normal bijective unital $*$-homomorphism whose inverse is also weak-operator continuous. Therefore $\Phi$ is an isomorphism of von Neumann algebras, and \begin{align*} M\cong zM\oplus(1-z)M. \end{align*} [/step]