Is the lattice of subspaces of a finite-dimensional scalar product space distributive?

Question

A famous 1936 Birkhoff, Von Neumann paper claimed that the following lattice isn't distributive: the set $L(\mathscr{H})$ of closed linear subspaces of an Hilbert space $\mathscr{H}$, endowed with the set theoretic $\subseteq$ relation.

Examples of failing distributivity for $<L(\mathscr{H}),\subseteq>$ seem to me straightforward to be found, and it seems neither required that $\mathscr{H}$ is infinite-dimensional: Consider a finite-dimensional space $\mathscr{H}$. The join is set theoretic intersection; and the meet is linear spanning, denote them $\sqcap, \sqcup$. So despite $$\mathscr{B}\sqcup\mathscr{B}^{\perp}=\mathscr{H}$$ for a general subspace $\mathscr{B}$ we have $$\mathscr{B}\cup\mathscr{B}^{\perp}\neq\mathscr{H}$$ So I can take a subspace $\mathscr{A}$ in $\overline{\mathscr{B}\cup\mathscr{B}^{\perp}}$: $$\mathscr{A}\sqcap\mathscr{B}=\bigl\{0\bigr\}$$ $$\mathscr{A}\sqcap\mathscr{B}^{\perp}=\bigl\{0\bigr\}$$ Then also $$\bigl(\mathscr{A}\sqcap\mathscr{B}\bigr)\sqcup\bigl(\mathscr{A}\sqcap\mathscr{B}^{\perp}\bigr)=\bigl\{0\bigr\}$$ while $$\mathscr{A}\sqcap\bigl(\mathscr{B}\sqcup\mathscr{B}^{\perp}\bigr)=\mathscr{A}\sqcap\mathscr{H}=\mathscr{A}\neq\bigl\{0\bigr\}$$

Now, Popper on Nature in 1968 sustained that Birkhoff and Von Neumann were wrong! Because any uniquely complemented modular lattice is boolean, and every complemented measurable lattice is uniquely complemented. If this is true, why my demonstration seems to work so safely? Or why Popper argument doesn't apply to my example?

It seems so strange that Birkhoff and Von Neumann could be wrong in such a thing, and it doesn't make sense to me that $<L(\mathscr{H}),\subseteq>$ is uniquely complemented. But where is the fault in Popper argument?

Birkhoff and Von Neumann 1936 https://www.jstor.org/stable/1968621?seq=1/subjects

Popper 1968 https://www.nature.com/articles/219682a0

Answer 1

If you look at the definition of "lattice measure" used in Popper's paper, note that he defines "finite additivity" (axiom R4) as (essentially)

For any $A$ and $B$ (not necessarily orthogonal) $$ \mu ( A \sqcup B) = \mu(A) + \mu(B) - \mu (A \sqcap B) \tag{*}$$

assuming I have understood correctly.

Look at equation (9) in Popper's paper -- the use of this form of the "finite additivity axiom", R4, appears to be essential to the argument.

However, it appears to be known (cf. e.g. equation (5) of these slides, can't find a better reference at the moment) that the above formulation of finite additivity is equivalent to the formulation

For any $A$ and $B$ disjoint/orthogonal (i.e. $A \prec B^{\perp} \iff B \prec A^{\perp}$) $$ \mu(A \sqcup B) = \mu (A) + \mu(B) \tag{**}$$

only when the lattice is distributive. In fact Popper's paper probably amounts to a proof of that fact.

I don't know which version of the finite additivity axiom Birkhoff and von Neumann used in their original paper, but mathematically speaking I guess Popper's paper can amount to saying that if Birkhoff and von Neumann had used ($*$) instead of ($**$), they were wrong because ($*$) would imply their lattice is distributive, which it is not. Along those lines, today if you look at any lecture notes that define a notion of (probability) measure on an orthocomplemented lattice, they always use ($**$) and not ($*$).