How is the free modular lattice on 3 generators related to 8-dimensional space?

Question

Here are three facts which sound potentially related. What are the actual relationships?

1. In 1900, Dedekind constructed the free modular lattice on 3 generators as a sublattice of the lattice of subspaces of an 8-dimensional vector space. If the basis of the vector space is $e_1, \dots, e_8$, he looked at the subspaces $$ X = \langle e_2, e_4, e_5, e_8 \rangle , \; Y = \langle e_2, e_3, e_6, e_7 \rangle, \; Z = \langle e_1, e_4, e_6, e_7 + e_8 \rangle $$ By repeatedly taking intersections and unions, you can build up 28 subspaces starting from these three. This proves the free modular lattice on 3 generators has at least 28 elements. In fact it has exactly 28 elements. I think Dedekind showed this by working out the free modular lattice 'by hand' and noting that it, too, has 28 elements.

2. The dimension of $\mathrm{SO}(8)$ is 28. Its Lie algebra $\mathfrak{so}(8)$, also called $D_4$, has 12 positive roots, and its Cartan algebra has dimension 4. As usual, the Lie algebra is spanned by positive roots, an equal number of negative roots, and the Cartan subalgebra, so we get $$ 28 = 12 + 12 + 4 $$

3. The 3 subspace problem asks us to classify triples of subspaces of a finite-dimensional vector space $V$, up to invertible linear transformations of $V$. There are finitely many possibilities, unlike the situation for the 4 subspace problem. One way to see this is to note that 3 subspaces $X, Y, Z \subseteq V$ give a representation of the $D_4$ quiver. This is nothing profound: a representation of the $D_4$ quiver is just 3 linear maps $X \to V$, $Y \to V$, $Z \to V$, and here we are taking those to be inclusions. The nontrivial part is that indecomposable representations of any Dynkin quiver correspond in a natural one-to-one way with positive roots of the corresponding Lie algebra. So, in particular, the $D_4$ quiver has 12 indecomposable representations. The representation coming from $X, Y, Z \subseteq V$ must be a direct sum of indecomposable representations, so we can classify the possibilities and solve the 3 subspace problem.

Here is a way of making my question more concrete. However, it may be on the wrong track:

Is there a natural 1-1 correspondence between the 28 elements of the free modular lattice on 3 generators and some basis of $\mathfrak{so}(8)$?

Here is another more cautious way:

What is known about the relation between the free modular lattice on 3 generators and the 3 subspace problem?

It would help if one could somehow directly relate the free modular lattice on 3 generators to something built from $D_4$. The free modular lattice on 3 generators looks like this:

For more background information, see:

• John Baez, The free modular lattice on 3 generators, The n-Category Café, September 19, 2015.

Answer 1

The Darmstadt school, particularly Christian Herrmann, studied the Gelfand Ponomarev papers carefully to see what they said about 4-generated modular lattices; see Ch. Herrmann, Rahmen und erzeugende Quadrupel in modularen Verbande, Algebra Universalis, 14(1982) 357-387. Some of the interesting results include:

1. The lattice of subspaces of every $n$-dimensional vector space, $3 \le n < \infty$, over a prime field is $4$-generated. Moreover, the quadruples generating the whole lattice are known. The dimension of each of the subspaces in the generating set can differ from $n/2$ by at most $2$.
2. The lattice of subgroups of $(\mathbb Z/p^k\mathbb Z)^n$ is four generated if $3 \le n < \infty$.

The hope was to classify the modular lattices generated by quadruples and so solve the word problem for $FM(4)$. But is now known that the word problem for free modular lattices is undecidable.

Answer 2

Edited to be more of an answer. But, unfortunately, to be quite wrong. I apologize for having been borne away by my enthusiasm.

The picture of the free modular lattice above shows thirty elements. Comparing with Grätzer's book, it appears that the top and bottom elements should be removed. And that makes sense, as there is clearly no way to generate them from the other elements.

However, I guess your picture could be viewed as the "free modular lattice on three elements with top and bottom element".

The reason I don't want to tell you to change your picture, is that the part of it above the middle rank is exactly the poset of positive roots of $D_4$! I don't know why, though.

Addition starts here: clearly, the most general lattice you can get from three subvectorspaces of a vector space is obtained by adding together one copy of each of the indecomposable representations of $D_4$ (with arrows oriented inwards) such that all the maps are injections, because if you take multiple copies of a representation, it doesn't help you. This is almost what Dedekind did, as described above, but not quite: he is missing two of them: the simple at the central node, and the representation which is one-dimensional at each node (i.e., add $e_9$ in all of $X,Y,Z$, and $e_{10}$ in none of them (just in the ambient space)). Then you get the full 30 elements of the lattice.

Everything from here on is pretty much junk.

For each of element of the lattice, you can ask "which indecomposable representations survive here?" This isn't really enough information, because, for example, the biggest indecomposable only "partly" survives (because only one of its two dimensions at the middle node is hit by the map from $X$). So, for the element X, the answer is "all the indecomposable representations supported over X". For $X \wedge Y$ it is "all indecomposable representations supported over $X$ and $Y$. Etc.

This translates the problem into a question about subsets of indecomposable representations of $D_4$. In fact, it seems to be simpler to think about the subsets of indecomposable representations which are killed at each node. Thus, for the top node, there are none. For the next node down, there is just the simple at the central vertex. Clearly, the set of indecomposables which are killed at a given element of the lattice is closed under subrepresentations. However, more is true. By inspection i.e., blinded by optimism, we see that the set of indecomposables killed by a given lattice element is also closed under extensions. (I thought this was obvious but I no longer think so. However, there is probably a conceptual explanation.) Subcategories closed under extensions and subobjects are called "torsion-free classes". The torsion-free classes in rep($D_4$) are known to form a lattice but not a modular lattice. And in fact, the full lattice (on 30 elements) consists of all the torsion-free classes inside the representations of $D_4$ for which the maps are injective. There are not enough torsion-free classes, and in this case they don't form a modular lattice (it's not even graded).

Answer 3

Here's a nice clean statement that emerged from discussions on The n-Category Café.

A representation of the $D_4$ quiver consists of three linear maps $f_i : L_i \to L$ ($i =1,2,3$) between finite-dimensional vector spaces over your favorite field. We can take direct sums of these representations, and define an indecomposable representation to be one that’s not a direct sum of two others.

Given a representation of the $D_4$ quiver, the images of the maps $f_i$ are subspaces of $L$. These generate a sublattice $\mathcal{L}$ of the lattice of all subspaces of $L$. Clearly $\mathcal{L}$ is a modular lattice with 3 generators.

Theorem. If we take a direct sum of indecomposable representations of the $D_4$ quiver, one from each isomorphism class, we obtain a representation of the $D_4$ quiver whose corresponding modular lattice is the free modular lattice on 3 generators. In this representation the spaces $\mathrm{im} (f_i)$ have dimension 5 and the space $L$ has dimension 10. 10 is the smallest possible dimension for a vector space $L$ containing subspaces that generate a copy of the free modular lattice on 3 generators.

(Here I should emphasize that I'm using lattice to mean a poset for which every finite subset has a least upper bound and greatest lower bound. Such a thing has a top and bottom as well as binary operations $\vee$ and $\wedge$. So, the free modular lattice on 3 generators, in this sense, has 30 elements, including a freely adjoined top and bottom element. This gives a cleaner statement of the result than working with Dedekind's original definition of lattice.)

The proof is lurking in discussions here:

• John Baez, The free modular lattice on 3 generators, The n-Category Café, September 19, 2015.

but I will put a cleaned-up version onto Visual Insight on January 1, 2016.