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  "body": "\n## Definition\n\nLet $L$ be an [[orthocomplemented lattice|J7KN14DYV3]] with orthocomplementation $a \\mapsto a^\\perp$.\nThe lattice $L$ is *orthomodular* if for all $a, b \\in L$,\n$$a \\le b \\implies b = a \\vee (a^\\perp \\wedge b).$$\n\n## Remarks\n\nAn equivalent form is: if $a \\le b$ and $a^\\perp \\wedge b = 0$, then $a = b$.\n\nExample: the lattice of closed subspaces of a [[Hilbert space|HM2HHE3J51]] $H$.\nIf $M \\subseteq N$ are closed subspaces, every $x \\in N$ is $x = Px + (x - Px)$, where $P$ is the orthogonal projection onto $M$; then $x - Px \\in M^\\perp \\cap N$.\nSo $N$ is the closed span of $M$ and $M^\\perp \\cap N$.\nWhen $H$ is infinite-dimensional, this lattice is orthomodular but not modular.\n\nNon-example: the hexagon ortholattice $O_6$ with elements $0, a, b, b^\\perp, a^\\perp, 1$ and order generated by $0 \u003c a \u003c b \u003c 1$ and $0 \u003c b^\\perp \u003c a^\\perp \u003c 1$.\nThe map exchanging $a$ with $a^\\perp$, $b$ with $b^\\perp$ and $0$ with $1$ is an orthocomplementation.\nHere $a \\le b$, but $a^\\perp \\wedge b = 0$, so $a \\vee (a^\\perp \\wedge b) = a \\ne b$.\n\nEvery [[Boolean algebra|0XADJKM699]] is orthomodular, and every modular ortholattice is orthomodular.\n",
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