$z(M)M$ is atomic

An *orthocomplementation* on a bounded lattice is a function that maps each element a to an *orthocomplement* $a^\perp$ in such a way that the following axioms are satisfied:

– Complement law: $a\vee a^\perp=1$ and $a\wedge a^\perp =0$;
– Involution law: $(a^\perp)^\perp=a$;
– Order-reversing law: $a\leq b$ $\Rightarrow$ $a^\perp \geq b^\perp$;
– Orthomodular law: $a\leq b$ $\Rightarrow$ $b=a\vee (b\wedge a^\perp).$

a pair of elements $a, b$ is *orthogonal* if $a\leq b^\perp$.
An *orthomodular lattice* is a bounded lattice which is equipped with an orthocomplementation.
The natural homomorphism between orthomodular lattices is a map presevers all the operations ($\vee, \wedge, \perp$).
A map between orthomodular lattices that preserves the join and the orthogonality of any two orthogonal elements, called *ortho-morphism*, is weaker than the natural homomorphism.

**Theorem**. (a)All projections in a von Neumann algebra forms a complete orthomodular lattice.

(b) If $z$ is the supremum of all minimal peojections in a von Neumann algebra $M$, then $zM$ is atomic.

**Theorem**. For any element $l$ in an atomic complete lattice $L$, we have $$l=\sup\{a\in \atom(L)|a\leq l\},$$ where $\atom(L)$ is the set of all atoms of $L$.

*Proof*. Denote $\sup\{a\in \mathscr \atom(L)|a\leq l\}$ by $l_0$, then $l_0\leq l$ and $l_0\in L$. If $l_0\neq l$, then $l\wedge l_0^\perp\neq 0$ according to the Orthomodular law. Thus $l\wedge l_0^\perp$ dominates an atom, say $a$. So, $a\leq l_0^\perp$ and $a\leq l$. But by definition of $l_0$, the later one implies $a\leq l_0$, a contradiction to the former one by the complement law.


Old Version

Suppose $M$ is a W*-algebra, $z$ is the supremum of all minimal projections in $M$. It is well-known that $z$ is a central porjection in $M$. Let $M_0=zM$, then $M_0$ is also a W*-algebra. Denote the collection of all minimal projections in a W*-algebra $M$ by $\mathscr P_m^M$.

**Lemma**. $\mathscr P_m^M=\mathscr P_m^{M_0}$.

*Proof*. Since $m=zm$ for any minimal projection $m$ in $M$, $\mathscr P_m^M\subset\mathscr P_m^{M_0}$. Conversely, suppose $m_0$ is a minimal projection in $M_0$, $p\in M$ and $p\leq m_0$, then $zp\leq zm=m$, hence $zp=0$ or $zp=m_0$. Therefore, $m_0p=m_0zp$ equals to $0$ or $m_0$, i.e., $m_0$ is a minimal projection in $M$.

Now, we can denote $\mathscr P_m^M$ and $\mathscr P_m^{M_0}$ by the same notation $\mathscr P_m$.

**Theorem**. $M_0$ is atomic, that is, each projection in $M_0$ dominates a minimal projection.

*Proof*. Suppose $p$ is a non-zero projection in $M_0$, then $p\leq z$, so, there is a minimal projection $m$ such that $pm\neq 0$( otherwise, $z=\sup\{m\in \mathscr P_m\}\leq 1-p$, $pz=0$ ), which is equivalent to $pmp\neq 0$. Since $$pmpMpmp\subset pmMmp=p(\mathbb{C}m) p=\mathbb{C}pmp,$$ $(pmp)^2$ is a scalar multiple of $pmp$. Let $m_p=pmp/\|pmp\|$, then $m_p$ is a projection and $m_pMm_p= \mathbb C m_p$, that is, $m_p$ is a minimal projection. What’s more, $m_p\leq p$ for $m_pp=pm_p=m_p$.

**Theorem**. For any projection $q$ in $M_0$, we have $q=\sup\{q\in \mathscr P_m^{M_0}|m\leq a\}$.

*Proof*. Denote $\sup\{m\in \mathscr P_m^{M_0}|m\leq q\}$ by $q_0$, then $q_0\leq q$ and $q_0\in q$. If $q_0\neq q$, then $q-q_0$ dominates an atom, say $m$. So, $m\leq q$, and thus $m\leq q_0$ by the definition of $q_0$. Therefore, $m(q-q_0)=mq-mq_0=m-m=0$, a contradiction to $m\leq q-q_0$.

Leave Your Thoughts

Your email address will not be published. Required fields are marked *