Let $L$ be an orthocomplemented lattice and $a,b\in L$. $a$ is said to be orthogonal to $b$ if $a\leq b^{{\perp}}$, denoted by $a\perp b$. If $a\leq b^{{\perp}}$, then $b=b^{{\perp\perp}}\leq a^{{\perp}}$, so $\perp$ is a symmetric relation on $L$. It is easy to see that, for any $a,b\in L$, $a\perp b$ implies $a\wedge b=0$, and $a\perp a^{{\perp}}$.

For any $a\in L$, define $M(a):=\{c\in L\mid c\perp a\mbox{ and }1=c\vee a\}$. An element of $M(a)$ is called an orthogonal complement of $a$. We have $a^{{\perp}}\in M(a)$, and any orthogonal complement of $a$ is a complement of $a$.

If we replace the $1$ in $M(a)$ by an arbitrary element $b\geq a$, then we have the set

An element of $M(a,b)$ is called an orthogonal complement of $a$ relative to $b$. Clearly, $M(a)=M(a,1)$. Also, for $a,c\leq b$, $c\in M(a,b)$ iff $a\in M(c,b)$. As a result, we can define a symmetric binary operator $\oplus$ on $[0,b]$, given by $b=a\oplus c$ iff $c\in M(a,b)$. Note that $b=b\oplus 0$.

Before the main definition, we define one more operation: $b-a:=b\wedge a^{{\perp}}$. Some properties: (1) $a-a=0$, $a-0=a$, $0-a=0$, $a-1=0$, and $1-a=a^{{\perp}}$; (2) $b-a=a^{{\perp}}-b^{{\perp}}$; and (3) if $a\leq b$, then $a\perp(b-a)$ and $a\oplus(b-a)\leq b$.

A lattice $L$ is called an orthomodular lattice if

1. 1.

   $L$ is orthocomplemented, and

2. 2.

   (orthomodular law) if $x\leq y$, then $y=x\oplus(y-x)$.

The orthomodular law can be restated as follows: if $x\leq y$, then $y=x\vee(y\wedge x^{{\perp}})$. Equivalently, $x\leq y$ implies $y=(y\wedge x)\vee(y\wedge x^{{\perp}})$. Note that the equation is automatically true in an arbitrary distributive lattice, even without the assumption that $x\leq y$.

For example, the lattice $\mathbb{C}(H)$ of closed subspaces of a hilbert space $H$ is orthomodular. $\mathbb{C}(H)$ is modular iff $H$ is finite dimensional. In addition, if we give the set $\mathbb{P}(H)$ of (bounded) projection operators on $H$ an ordering structure by defining $P\leq Q$ iff $P(H)\leq Q(H)$, then $\mathbb{P}(H)$ is lattice isomorphic to $\mathbb{C}(H)$, and hence orthomodular.

A simple example of an orthocomplemented lattice that is not orthomodular is the benzene:

Note that $a\leq b$, but $a\vee(b\wedge a^{{\perp}})=a\vee 0=a\neq b$.

An nice example of an orthomodular lattice that is not modular can be found in the reference below.

Remarks.

• Orthomodular lattices were first studied by John von Neumann and Garett Birkhoff, when they were trying to develop the logic of quantum mechanics by studying the structure of the lattice $\mathbb{P}(H)$ of projection operators on a Hilbert space $H$. However, the term was coined by Irving Kaplansky, when it was realized that $\mathbb{P}(H)$, while orthocomplemented, is not modular. Rather, it satisfies a variant of the modular law as indicated above.

• More generally, an orthomodular poset $P$ is an orthocomplemented poset such that

  1. (a)

     given any pair of orthogonal elements $x,y\in P$ ($x\leq y^{{\perp}}$), their greatest lower bound exists ($x\vee y$ exists). Simply put, $x\perp y$ implies $x\vee y\in P$.

  2. (b)

     for any $x,y\in P$ such that $x\leq y$, the orthomodular law holds (the right hand side of the orthomodular law exists via the first condition).

  From this definition, we see that an orthomodular lattice is just an orthomodular poset that is also a lattice.