# Kaplansky's Parallelogram Law - Why is it called like that?

From Klaplansky's Rings of Operators, p.81, the theorem reads as follows:

For any projections $e$ and $f$ (in a von Neumann algebra), we have $(e \cup f) - f \sim e - e \cap f$.

were $\cap,\cup$ denote the meet and join operations in the lattice of projections and $\sim$ denotes Murray-von Neumann-equivalence.

Why is it called the parallelogram law? Kaplansky himself introduces the name before formulating the theorem, but does not motivate it.

This is the motivation I came up with, seeing user218931's answer: \begin{matrix} && e \cup f & \\ &\huge\diagup & & \huge\diagdown \\ e & & & & f\\ &\huge\diagdown & & \huge\diagup \\ &&e \cap f \end{matrix} which is to be unterstand as a diagram in the lattice of projections.

• I just want to comment that this reminds me of the second isomorphism theorem, also called the diamond/parallelogram rule. A nice graph can be seen in the graph of this answer. Commented Apr 10, 2017 at 7:46
• @FrenzyLi: Interesting observation. Thanks! Commented Apr 10, 2017 at 8:13

$\require{AMScd}$ Just draw a picture $$\begin{CD} e\cup f @>>> e\\ @VVV @VVV\\ f @>>> e\cap f \end{CD}$$ (but without the arrow tips). Then $e\cup f - f \sim e - e\cap f$ can be interpreted as “the lengths of the sides $e\cup f$ to $f$ and $e$ to $e\cap f$ are equal”. Symmetrically $e\cup f - e \sim f - e\cap f$ is interpreted as “the lengths of the sides $e\cup f$ to $e$ and $f$ to $e\cap f$ are equal”. So geometrically, this resembles a parallelogram.