In the book of Garrett Birkhoff "lattice theory", it is mentioned that there are 28 subspaces that can be obtained from three subspaces in general position in a Hilbert space (using intersections and sums). Apparently this is related to the fact that the free modular lattice on three generators has 28 elements. I am not sure why though, since subspaces in projective geometry should be related to the more restrictive complemented modular lattices. Any hint on that point is welcome.


**Question:** how many subspaces can be obtained from n subspaces in general position in a Hilbert space (or in a projective geometry of dimension N, N big wrt n)? And how does this number is computed?


It is a bit frustrating because the three subspaces problem
is often mentioned in lattice theory books but the general case
is never addressed.



How many subspaces are generated by three or more subspaces in a Hilbert space?
From four subspaces in general position one can generate an infinite number of other subspaces by closing up under joins and meets. This is true even for subspaces of $\mathbb{R}^3$ (any field of characteristic zero in place of $\mathbb{R}$ would do). This is easy to see in the corresponding projective plane picture: take $a = (0: 0 : 1)$, $b = (1: 0: 1)$, $c = (0: 1: 1)$, and $d = (1: 1: 1)$ to be four lines in $\mathbb{R}^3$ written in homogeneous coordinates, so that $a \vee b$, $c \vee d$ are parallel projective lines which meet in the "line at infinity" at $(1: 0: 0)$ and similarly $a \vee c$, $b \vee d$ meet "at infinity" in $(0: 1: 0)$. It's not hard to see from a picture that starting from this initial configuration in $\mathbb{P}^2(\mathbb{R})$, one can generate infinitely many points in the projective plane by drawing lines between points and drawing points of intersection between lines. (A more precise worked-out description is given on pages 150-151 of the very nice book *Combinatorics: The Rota Way* by Joseph P. S. Kung, Gian-Carlo Rota, and Catherine H. Yan ([Google book](https://books.google.com/books?id=nX5Exky4iGQC&pg=PA149&lpg=PA149&dq=28+elements+modular+lattice+subspaces&source=bl&ots=ZKyzE1D1UN&sig=athJ6dfADZC-Ra_A-41oq_jfXPM&hl=en&sa=X&ved=0CE4Q6AEwCWoVChMI5IucppSYxwIVwns-Ch3XQAg_#v=onepage&q=28%20elements%20modular%20lattice%20subspaces&f=false)).) 


I could leave it at that, but here are some further notes: 


- 1. The free modular lattice on four generators is in fact very complicated; it has an undecidable word problem ([reference](http://link.springer.com/article/10.1007%2FBF01455951)). This is perhaps to be expected, since after all the structure of a general linear operator $T: V \to V$ is encoded in four subspaces of $V \oplus V$ (the "$x$-axis" $V \oplus 0$, the "$y$-axis" $0 \oplus V$, the diagonal $\{(x, x): x \in V\}$ which canonically identifies the domain of $T$ with the codomain, and the graph $\{(x, T x): x \in V\}$). Numerous invariants of $T$ can then be extracted as elements in the modular lattice generated by these four; as a fun exercise, you might consider how to define $\text{im}(T^n)$ in terms of meets and joins, starting from these four.



(And that barely scratches the surface in terms of complication: free modular lattices $F(n)$ are much more complicated than free linear lattices, where a "linear lattice" is a lattice of commuting equivalence relations such as any lattice of congruences of a Mal'cev algebraic theory such as the theory of vector spaces $V$. In fact it can be shown that there is no faithful linear representation $F(4) \to \text{Sub}(V)$ for any $V$. This is part of a long story beginning with fundamental papers by Gelfand and Ponomarev, but see for example the paper *Elements of Modular Lattices* by Andrew Cylke, pp. 125-148 from this [Google book](https://books.google.com/books?id=Ok9LPgQOWJMC&pg=PA148&lpg=PA148&dq=free+modular+lattices+faithful+linear+representation&source=bl&ots=VEmrE8BEj_&sig=PmoG63f2GUX8BwRrmCNoTRk_bjg&hl=en&sa=X&ved=0CFkQ6AEwCGoVChMIt4zVtMGXxwIVRSUeCh0bGQ2J#v=onepage&q=free%20modular%20lattices%20faithful%20linear%20representation&f=false), particularly Theorem 3 on page 145, for a specific example.) 
 


- 2. The book by Kung, Rota, and Yan also talks a bit about the 28-element modular lattice freely generated by three elements. This number was determined by Dedekind in 1900 and the structure of the lattice is reproduced in many places; for example page 100 [here](http://www.math.hawaii.edu/~jb/math618/os9uh.pdf). As you can see, the rank of the lattice is $8$, with the generators $x, y, z$ sitting at rank $4$, and so it is not too unexpected that the lattice embeds in the subspace lattice of an $8$-dimensional vector space. In fact, an explicit example in terms of a basis $e\_1, e\_2, \ldots, e\_8$ is



* $x = \text{span}(e\_2, e\_4, e\_5, e\_8)$,
* $y = \text{span}(e\_2, e\_3, e\_6, e\_7)$,
* $z = \text{span}(e\_1, e\_4, e\_6, e\_7 + e\_8)$


although I don't have a short conceptual proof that this example works. In any case, I don't find it very surprising that $F(3)$ embeds in a complemented modular lattice, in roughly the same spirit that the free associative algebra on one element happens to be commutative, or any other similar "coincidental" extra identities which can occur in free structures when the number of generators is small. 
 


- 3. Regarding the number $28$: John Baez [surmises](https://golem.ph.utexas.edu/category/2010/12/solers_theorem.html#c035868) this is connected with the fact that configurations of three subspaces are linear representations of the $D\_4$ quiver, where $D\_4$ also happens to be the Coxeter diagram attached to $SO(8)$ (there's that $8$-dimensional vector space again!), where $SO(8)$ happens also to be a $28$-dimensional Lie group. It seems to me a reasonable guess, although I haven't tried to work out what the connection is exactly. Something Lie-algebraic I would guess. Maybe someone here at MO can shed more light.



