The n-Category Café

Here are the 12 indecomposable representations of the $D_4$ quiver. I’ll just write the dimensions of the vector spaces involved, and let you guess the maps between them — they’re not hard to guess with a few hints.

First there are 3 representations where the maps involved are not injective. These cannot show up inside a representation obtained from subspaces $X, Y, Z \subseteq V$:

   1
   0
 0   0

   0
   0
 1   0

   0
   0
 0   1

For the other representations the maps are injective:

   0
   1
 0   0

   1
   1
 0   0

   0
   1
 1   0

   0
   1
 0   1

   1 
   1
 0   1

   0
   1
 1   1

   1
   1
 1   0

   1
   1
 1   1

   1
   2
 1   1

Unfortunately I cannot (without much work) read Dedekind’s paper. It wouldn’t surprise me if in the subspaces-of-$\mathbb{R}^8$ version, the $S_3$ symmetry is hidden. I say this because, if my memory is correct, triality is not that obvious in the $\mathfrak{so}(8)$-action on $\mathbb{R}^8$ (I mean, in the usual matrix representation of $\mathfrak{so}(8)$). I thought I had a favorite presentation of this symmetry, but https://golem.ph.utexas.edu/~distler/blog/archives/002492.html doesn’t have it — perhaps I read it in your blog somewhere?

I’m working on adding some stuff to my answer at MO, but for now, let me point out that some of the representations above are not indecomposable. The representations that have a 2 in the middle should have a 1 in the middle, except for the very last one. Dedekind uses all but two of the ones where the maps are injective, exactly once; if you add in the two he misses, then you get the 30 element poset not the 28 element poset.

Over on MathOverflow, Hugh Thomas found a nice clue. The part of this lattice:

above the middle rank is isomorphic to the poset of roots of $D_4$!

In other words, this portion here:

We see the 4 simple roots at bottom, and then various linear combinations of them.

But why?

It’s curious that the free modular lattice on 3 generators is finite but the free modular lattice on 4 generators is infinite. Is there any way to deduce those facts more easily than by explicitly constructing both? E.g. is there some sort of “3-4-ness” of the algebraic theory of modular lattices? I notice that in all your characterizations of the axiom that makes a lattice modular, there are 3 or 4 elements appearing. And I learned from Sketches of an Elephant (A3.2.2(a)) that any lattice generated by 2 elements is modular, so no identity in fewer than 3 variables can characterize modularity.

Yet another way to say that a lattice is modular is that it doesn’t contain the following sublattice:

This lattice is called $N_5$, and the fact that this condition is equivalent to modularity is one half of the so-called $M_3$-$N_5$ theorem.

Isn’t that amazing? $N_5$ is the simplest example of a non-modular lattice, so it’s clear that if your lattice contains a copy of $N_5$ as a sublattice then it can’t be modular. But the theorem gives us the converse too.

It’s as if there was a theorem saying “a group is abelian if and only if it contains no subgroup isomorphic to $S_3$”. But that’s false!

The other half of the $M_3$-$N_5$ theorem states that a lattice is distributive if and only if contains no sublattice isomorphic to either $N_5$ or this lattice, called $M_3$:

Tim Silverman wrote:

I would really like to have an explicit description of this lattice in terms of its generators, because I don’t have much intuition for what it would look like.

Okay, let’s try to figure out how to take the free modular lattice on 3 generators:

and label each dot by an expression built from the 3 generators $X,Y,Z$ using $\wedge$ and $\vee$.

For starters, the top dot is $\top$ and the bottom dot is $\bot$. These rather degenerate guys were not in Dedekind’s original 28-element lattice.

I believe $X \vee Y \vee Z$ is the topmost (or ‘truest’) proposition short of $\top$ that one can build from $X,Y$ and $Z$ using $\wedge$ and $\vee$. So, it should be the dot right below the top dot.

Similarly, $X \wedge Y \wedge Z$ should be the dot right above the bottom dot.

But the key question is: where are $X, Y$ and $Z$?

Since the generators $X, Y$ and $Z$ play symmetrical roles, the whole picture should have $S_3$ symmetry, and it obviously does. But more importantly, we know $X,Y$ and $Z$ must lie on some horizontal equilateral triangle centered at the $z$ axis.

Since the definition of ‘modular lattice’ has a completely symmetry between top and bottom, and between $\wedge$ and $\vee$, the free modular lattice on 3 generators must inherit this symmetry — and we can see it does. More importantly, I believe this symmetry implies that the 3 generators $X, Y, and Z$ must appear at the middle rank of this lattice, the rank with 6 dots in it. (This sounds intuitive; turning that intuition into a proof is a bit subtle but I think I can do it.)

So: where are the generators $X, Y$ and $Z$? Are they the corners of the big equilateral triangle in the middle rank, or the little one?

Since $X \vee Y \vee Z$ is the least upper bound of $X, Y$, and $Z$, and it’s the dot directly below the topmost dot, we can answer this question. $X,Y$ and $Z$ must be the corners of the big equilateral triangle in the middle rank.

Now we can easily see where $X \vee Y$, $Y \vee Z$ and $Z \vee X$ are.

For example, $X \vee Y$ must be the lowest dot that you can get to by climbing up along edges from either $X$ or $Y$. So, it must be one of the dots directly below the dot directly below the top dot. If the damned picture would stop rotating, I could tell you which one.

So, we have

1. $\top$ on top.

2. $X \vee Y \vee Z$ in the next rank down: in Dedekind’s representation, this is the 8-dimensional subspace of $\mathbb{R}^8$.

3. $X \vee Y$, $Y \vee Z$ and $Z \vee X$ in the rank below that: these are 7-dimensional subspaces of $\mathbb{R}^8$.

4. Three things in the rank below that: these are 6-dimensional subspaces of $\mathbb{R}^8$.

5. Four things in the rank below that: these are 5-dimensional subspaces of $\mathbb{R}^8$.

6. $X, Y,$ and $Z$ and three other things in the rank below that — the middle rank. These are 4-dimensional subspaces of $\mathbb{R}^8$.

7. Four things in the rank below that: these are 3-dimensional subspaces of $\mathbb{R}^8$.

8. Three things in the rank below that: these are 2-dimensional subspaces of $\mathbb{R}^8$.

9. $X \wedge Y$, $Y \wedge Z$ and $Z \wedge Y$ in the rank below that: these are 1-dimensional subspaces of $\mathbb{R}^8$.

10. $X \wedge Y \wedge Z$ in the rank below that: the 0-dimensional subspace of $\mathbb{R}^8$.

11. $\bot$ in the bottom rank.

At least that’s what I think. But what about the rest?

From what you figured out about the labelling, the rest of it follows quite easily. I have written it out slightly sketchily below.

If we look at the top cube, you have determined the labels of its top element, and the next three elements. The next three are $\wedge$s of pairs from the rank above and the bottom is the $\wedge$ of all three. Similarly we can fill in the bottom cube. Now it is easy to fill in the rest of the diagram using the bottom element of the top cube, the top element of the bottom cube, and $X$, $Y$, $Z$.

Now I want to set up the machinery to explicitly identify the 28 elements in the free modular lattice on 3 generators (without $\top$ and $\bot$) with subspaces of $\mathbb{R}^8$. The idea is to take a direct sum of certain indecomposable representations of the $D_4$ quiver.

As explained by Bruno in a comment above, we need to use the following seven representations of the $D_4$ quiver:

$A$:

   1
   1
 0   0

$B$:

   0
   1
 0   1

$C$:

   0
   1
 1   0

$D$:

   0
   1
 1   1

$E$:

   1
   1
 1   0

$F$:

   1
   1
 0   1

$G$:

   1
   2
 1   1

I’ve written the dimensions of the vector spaces assigned to the dots in the $D_4$ quiver; the maps are determined up to isomorphism by knowing that these representations are indecomposable and the maps are injective. I’ll think of these maps as inclusions.

Taking the direct sum of all seven quiver representations, we get one whose middle dot gets assigned an 8-dimensional vector space. Abusing notation a little, I’ll call this vector space

$$A \oplus B \oplus C \oplus D \oplus E \oplus F \oplus G$$

where now $A, \dots, G$ denote the vector spaces assigned to the middle dot by the quiver representations of the same name.

The vector spaces $A, B, C, E, E, F$ are 1-dimensional, since there’s a 1 on the middle dot of the corresponding quiver representations. The vector space $G$ is 2-dimensional, because the corresponding quiver representation has a 2 labelling the middle dot:

   1
   2
 1   1

In the direct sum of all six quiver representations, the top dot gets assigned a vector space I’ll call $X$. The bottom right dot gets assigned a vector space $Y$, and the bottom left dot gets assigned a vector space $Z$. Since all the maps in our quiver representations are inclusions, we have

$$X, Y, Z \subset A \oplus B \oplus C \oplus D \oplus E \oplus F \oplus G \cong \mathbb{R}^8$$

The cool fact is that these 3 subspaces of $\mathbb{R}^8$ generate a free modular lattice. The next step is to determine exactly what these 3 subspaces are. I’ll do that in the next comment.

For anyone who has trouble figuring out what we did in this discussion, here’s a summary.

A representation of the $D_4$ quiver is simply a triple of linear maps $f_i : L_i \to L$ ($i = 1,2,3$) between finite-dimensional vector spaces over your favorite field of characteristic zero. 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.

We say a representation of the $D_4$ quiver is injective if all the linear maps $f_i : L_i \to L$ are injective.

Given a representation of the $D_4$ quiver, the images of the maps $f_i$ generate a sublattice of the lattice of all subspaces of $L$. This is a modular lattice with 3 generators.

Theorem. If we take a direct sum of injective indecomposable representations of the $D_4$ quiver, one from each isomorphism class, we obtain a representation whose corresponding modular lattice is the free modular lattice on 3 generators, with 30 elements.

In this representation the maps $f_i : L_i \to L$ are injective, the spaces $L_i$ have dimension 5, and the space $L$ has dimension 10. These are the smallest possible dimensions possible for a representation of the $D_4$ quiver that gives the free modular lattice on 3 generators.

Here I’m defining a lattice to be one equipped with a top and bottom element. If we drop this requirement we can omit two isomorphism classes of injective indecomposable representations from the direct sum, and the direct sum will be a representation whose corresponding modular lattice is the free modular-lattice-without-top-and-bottom on 3 generators. This lattice has just 28 elements, since it’s missing the top and bottom of the previous one. In this variant the maps $f_i : L_i \to L$ are still injective, but the spaces $L_i$ have dimension 4, and the space $L$ has dimension 8.

Let me state the result in another way, which is equivalent to the earlier statement, but shorter:

A representation of the $D_4$ quiver is a triple of 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 $f_i$ generate a sublattice $\mathcal{L}$ of the lattice of all subspaces of $L$. $\mathcal{L}$ 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 $f_i : L_i \to L$ 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. These are the smallest possible dimensions possible for a representation of the $D_4$ quiver that gives the free modular lattice on 3 generators.

(Details: there are 12 isomorphism classes of indecomposable representations of the $\mathrm{D}_4$ quiver. 3 of them give a 0-dimensional contribution to the spaces $\mathrm{im}(f_i)$ and $L$ and can be omitted without changing the resulting modular lattice. In the remaining 9, the maps $f_i$ are injective. Above I’m defining a lattice to be one equipped with a top and bottom element. If instead we work with modular lattices having only $\wedge$ and $\vee$ operations we can omit 2 more isomorphism classes of indecomposable representations from the direct sum, and obtain a representation whose corresponding modular lattice is the free modular-lattice-without-top-and-bottom on 3 generators. This lattice has 2 fewer elements, since it’s missing the top and bottom of the previous one. In this variant the spaces $\mathrm{im}(f_i)$ have dimension 4, and the space $L$ has dimension 8.)

Here’s a related idea already in the literature:

• R. B. Stekolʹshchik, Perfect elements in a modular lattice associated with the extended Dynkin diagram $\tilde{E}_6$ (in Russian) Funktsional. Anal. i Prilozhen. 23 (1989), 90–92; translation in Funct. Anal. Appl. 23 (1990), 251–254.

A representation of a modular lattice $\mathcal{L}$ is a lattice homomorphism from $\mathcal{L}$ into the lattice of subspaces of an $k$-dimensional vector space over a field $k$. Stekolʹshchik considers the modular lattice $L_6$ generated by six elements $x_1, x_2, x_3, y_1, y_2 and y_3$ subject only to the relations $x_i \le y_i$ for $i=1,2,3$. He notes that indecomposable representations of $L_6$ bijectively correspond to indecomposable representations of the quiver given by taking the extended Dynkin diagram $\tilde{E}_6$ and putting arrows on the edges that point to the middle node:

It’s pretty obvious that this $\tilde{E}_6$ quiver is related to the lattice $L_6$ in the same way that the $D_4$ quiver is related to the free modular lattice on 3 generators. This larger quiver will not have finitely many indecomposable representations, but it will still have a ‘tame’ representation theory.

There is now a page summarizing and I hope clarifying the discussions in this thread:

• John Baez, The free modular lattice on 3 generators, Visual Insight, 1 January 2016.

I have written a browser-based web application that can generate and visualize some free modular lattices:

Online demo: www.source-code.biz/lattice
Source code: github.com/chdh/lattice-generator

I just stumbled across FeMoco the iron and molybdenum cofactor which is used in the only way that lifeforms fix atmospheric nitrogen. I looks like a reduced 3 generator free modular lattice.

I don’t know what this means or how to state what the reduction is.