---
tag: G1P8EW3Q3E
kind: definition
name: Orthomodular lattice
msc: [06C15, 81P10, 03G12]
defines: 6FPKEFX0X4
uses: [J7KN14DYV3]
sources:
  - url: https://en.wikipedia.org/wiki/Complemented_lattice
---

## Definition

Let $L$ be an [[orthocomplemented lattice|J7KN14DYV3]] with orthocomplementation $a \mapsto a^\perp$.
The lattice $L$ is *orthomodular* if for all $a, b \in L$,
$$a \le b \implies b = a \vee (a^\perp \wedge b).$$

## Remarks

An equivalent form is: if $a \le b$ and $a^\perp \wedge b = 0$, then $a = b$.

Example: the lattice of closed subspaces of a [[Hilbert space|HM2HHE3J51]] $H$.
If $M \subseteq N$ are closed subspaces, every $x \in N$ is $x = Px + (x - Px)$, where $P$ is the orthogonal projection onto $M$; then $x - Px \in M^\perp \cap N$.
So $N$ is the closed span of $M$ and $M^\perp \cap N$.
When $H$ is infinite-dimensional, this lattice is orthomodular but not modular.

Non-example: the hexagon ortholattice $O_6$ with elements $0, a, b, b^\perp, a^\perp, 1$ and order generated by $0 < a < b < 1$ and $0 < b^\perp < a^\perp < 1$.
The map exchanging $a$ with $a^\perp$, $b$ with $b^\perp$ and $0$ with $1$ is an orthocomplementation.
Here $a \le b$, but $a^\perp \wedge b = 0$, so $a \vee (a^\perp \wedge b) = a \ne b$.

Every [[Boolean algebra|0XADJKM699]] is orthomodular, and every modular ortholattice is orthomodular.