## September 19, 2015

### The Free Modular Lattice on 3 Generators

#### Posted by John Baez

The set of subspaces of a vector space, or submodules of some module of a ring, is a lattice. It’s typically not a distributive lattice. But it’s always modular, meaning that the distributive law

$a \vee (b \wedge c) = (a \vee b) \wedge (a \vee c)$

holds when $a \le b$ or $a \le c$. Another way to say it is that a lattice is modular iff whenever you’ve got $a \le a'$, then the existence of an element $x$ with

$a \wedge x = a' \wedge x \; \mathrm{and} \; a \vee x = a' \vee x$

is enough to imply $a = a'$. Yet another way to say it is that there’s an order-preserving map from the interval $[a \wedge b,b]$ to the interval $[a,a \vee b]$ that sends any element $x$ to $x \vee a$, with an order-preserving inverse that sends $y$ to $y \wedge b$:

Dedekind studied modular lattices near the end of the nineteenth century, and in 1900 he published a paper showing that the free modular lattice on 3 generators has 28 elements.

One reason this is interesting is that the free modular lattice on 4 or more generators is infinite. But the other interesting thing is that the free modular lattice on 3 generators has intimate relations with 8-dimensional space. I have some questions about this stuff.

One thing Dedekind did is concretely exhibit the free modular lattice on 3 generators as a sublattice of the lattice of subspaces of $\mathbb{R}^8$. If we pick a basis of this vector space and call it $e_1, \dots, e_8$, he looked at the subspaces

$X = \langle e_2, e_4, e_5, e_8 \rangle , \quad Y = \langle e_2, e_3, e_6, e_7 \rangle, \quad Z = \langle e_1, e_4, e_6, e_7 + e_8 \rangle$

By repeatedly taking intersections and unions, he built 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. It looks like this:

This picture makes it a bit hard to see the $S_3$ symmetry of the lattice, but if you look you can see it. (Can someone please draw a nice 3d picture that makes the symmetry manifest?)

If you look carefully here, as Hugh Thomas did, you will see 30 elements! That’s because the person who drew this picture, like me, defines a lattice to be a poset with upper bounds and lower bounds for all finite subsets. Dedekind defined it to be a poset with upper bounds and lower bounds for all nonempty finite subsets. In other words, Dedekind’s kind of lattice has operations $\vee$ and $\wedge$, while mine also has a top and bottom element. So, Dedekind’s ‘free lattice on 3 generators’ did not include the top and bottom element of the picture here. So, it had just 28 elements.

Now, there’s something funny about how 8-dimensional space and the number 28 are showing up here. After all, the dimension of $\mathrm{SO}(8)$ is 28. This could be just a coincidence, but maybe not. Let me explain why.

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, which is this little category here:

This fact is trivial: 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. The Lie algebra corresponding to $D_4$ is $\mathfrak{so}(8)$, the Lie algebra of the group of rotations in 8 dimensions. This Lie algebra has 12 positive roots. So, the $D_4$ quiver has 12 indecomposable representations. The representation coming from any triple of subspaces $X, Y, Z \subseteq V$ must be a direct sum of these indecomposable representations, so we can classify the possibilities and solve the 3 subspace problem!

What’s going on here? On the one hand, Dedekind the free modular lattice on 3 generators shows up as a lattice of subspaces generated by 3 subspaces of $\mathbb{R}^8$. On the other hand, the 3 subspace problem is closely connected to classifying representations of the $D_4$ quiver, whose corresponding Lie algebra happens to be $\mathfrak{so}(8)$. But what’s the relation between these two facts, if any?

Another way to put the question is this: what’s the relation between the 12 indecomposable representations of the $D_4$ quiver and the 28 elements of the free modular lattice on 3 generators? Or, more numerogically speaking: what relationship between the numbers 12 and 28 is at work in this business?

Here’s one somewhat wacky guess. The Lie algebra of $\mathfrak{so}(8)$ 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

$28 = 12 + 12 + 4$

But I don’t really see how this is connected to anything I’d said previously. In particular, I don’t see why 24 of the 28 elements of the lattice of subspaces generated by

$X = \langle e_2, e_4, e_5, e_8 \rangle , \; Y = \langle e_2, e_3, e_6, e_7 \rangle, \; Z = \langle e_1, e_4, e_6, e_7 + e_8 \rangle$

should be related to roots of $D_4$.

I think a more sane, non-numerological approach to this network of issues is to take the $D_4$ quiver representation corresponding to Dedekind’s choice of $X , Y, Z \subseteq \mathbb{R}^8$, decompose it into indecomposables, and see which positive roots those correspond to. I may try my hand at that in the comments, but I’m really looking for some help here.

Posted at September 19, 2015 9:57 PM UTC

TrackBack URL for this Entry:   https://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/2845

### Re: The Free Modular Lattice on 3 Generators

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

Posted by: John Baez on September 20, 2015 1:03 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

To get subspaces $X, Y, Z \subseteq V$ that generate a free modular lattice on 3 generators, it seems natural to take the direct sum of all the indecomposable representations of the $D_4$ quiver for which the maps involved are injective. The idea is that this should give the ‘generic situation’.

Unfortunately, if we do this, taking the direct sum of all these:

   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


we get three 5-dimensional subspaces of a 10-dimensional space. Dedekind saw it was enough to use three 4-dimensional subspaces of an 8-dimensional space. So, somehow this approach is ‘overkill’.

Posted by: John Baez on September 20, 2015 1:11 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Dear John, very nice post !

A small remark : the first and next-to-last indecomposables in your list amount to add a line to V, not touching X,Y,Z, and add the same line to V,X,Y,Z respectively.

For the lattice generated by X,Y,Z this doesn’t change anything, and if you sum the rest of indecomposables you get your three 4-dimensional subspaces in an 8-dimensional one, surely Dedekinds…

Posted by: bruno on September 22, 2015 1:46 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Great! I think all this shows that there are two definitions of ‘lattice’: one being a poset with upper bounds and lower bounds for all finite subsets, and another being a poset with upper bounds and lower bounds for all nonempty finite subsets.

With the former definition a lattice must have a top and bottom element; with the latter it doesn’t. With the former definition to form the free modular lattice on some generators we freely throw in a top and bottom element; with the latter we don’t. With the former definition the free modular lattice on 3 generators has 30 elements; with the latter it has just 28.

I think you’re saying that the former version is what naturally pops out when we sum all the indecomposable representations of the $D_4$ quiver.

I’m glad I was confused about this, because if I’d seen the number 30 I would not have been so convinced that $D_4$ was relevant… while 28 is the dimension of the $D_4$ Lie algebra. And this clue was not entirely misleading, since the lattice of positive roots of $D_4$ does show up as the top 5 ranks of this lattice:

That accounts for 12 of the dots. The bottom 5 ranks (or negative roots) account for 12 more. The middle rank, with 6 dots, remains a bit mysterious to me.

Posted by: John Baez on September 22, 2015 6:04 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

I think the mysterious $6$ points might possibly come from the projection of the $D_4$ root system onto the plane perpendicular to the plane of rotation of a triality transformation. If this is true, they would form a $G_2$ root system coming from folding the $D_4$.

I should flesh this out as it can be quite confusing. At least to me.

The roots of $D_4$ form the vertices of a $24$-cell.

The vertices of a $24$-cell can be decomposed into $3$ sets of $8$, each comprising the vertices of a $4$-orthoplex. Any two of these orthoplexes taken together form a hypercube dual to the third—i.e. in $4$ dimensions a demihypercube is an orthoplex. (These three orthoplexes can be seen as the $\mathbb{F}_1$ analogues of the vector and two spinor reps of $so_8$. Each vertex of one of the demihypercubes lies at the centre of one of the tetrahedral facets of the dual $4$-orthoplex, surrounded by $4$ points of that orthoplex, and consequently one can realise triality as a cycling among vectors and the two types of isotropic $4$-space of an $8$-space with an orthogonal structure on it.)

There are $2$ types of triality transformation. One of these fixes each of the $3$ orthoplexes setwise; in each orthoplex it fixes one axis and cycles the other three axes. The fixed axes contain $2$ points each, which correspond to the $6$ fixed points in the rotation picture above.

The other type of triality transformation cycles the three orthoplexes and has no fixed points.

Each type of transformation can be realised as a rotation in $4$-space; in fact, if such a rotation realises one type of transformation on a $24$-cell, then it will realise the other type of transformation on the dual $24$-cell.

The rotation takes place in a single plane and fixes a perpendicular plane. (For the type of triality with no fixed points, the plane of rotation is parallel to certain triangular faces of the $24$-cell, whose vertices are cycled by it). Thus we can project the points of the $24$-cell down to the fixed plane. In the case of the triality with $6$ fixed points, they project to $2$ hexagons (mutually dual up to rescaling). In the case of the triality with no fixed points, the $8$ cycles project to a hexagon and two coincident points at the origin.

Posted by: Tim Silverman on September 23, 2015 3:41 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

That’s a beautiful bunch of erudition, Tim! I thought I was pretty wise about the 24-cell, but I didn’t know all this! Now that I think about, this way to inscribe three 4-orthoplexes in a 24-cell is like a grown-up version of how one can inscribe two tetrahedra in a 3-cube.

The big mystery to me, though, is what all this stuff has to do with the free modular lattice on 3 generators. It seems this lattice is deeply connected to triality — surprise, surprise — and also the octonions. I want to figure this out in full detail. Let’s do it!

Posted by: John Baez on September 23, 2015 5:12 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Yeah, that’s mysterious to me too. 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.

Also, I really ought to do the calculations for my idea about the hexagon. My very shaky intuition for $4$-dimensional rotations is wondering loudly if the plane I’m vaguely imagining I’m projecting onto will actually do the job, because it’s off at some angle to the plane perpendicular to the triality rotation in the picture.

By the way, the erudition is actually the result of trying to understand triality in the very simplest possible case—over $\mathbb{F}_1$—because I was having terrible difficulties getting my head round it in any more complicated case …

This took me a long time to understand too:

this way to inscribe three $4$-orthoplexes in a $24$-cell is like a grown-up version of how one can inscribe two tetrahedra in a $3$-cube.

Yeah, “a $4$-demicube is an orthoplex” is the $\mathbb{F}_1$ version of triality; and “a $3$-demicube is a simplex” is the $\mathbb{F}_1$ version of the Klein correspondence!

Posted by: Tim Silverman on September 23, 2015 9:08 PM | Permalink | Reply to this

### What is the empty list a list of?

Thanks for introducing the idea of a bounded lattice to me. It allowed me to solve the problem of what the empty list is a list of. The post (http://wombatlang.blogspot.com.au/2015/09/mathematics-influences-wombat-again.html) doesn’t require any knowledge of the Wombat language. To make this comment (barely) on-topic, let me say that this shows once again the advantage of keeping degenerate cases, and I would guess that it will work out best if you do the same.

Posted by: Robert Smart on September 25, 2015 11:58 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

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?

Posted by: Theo Johnson-Freyd on September 20, 2015 1:45 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

I know a bunch of ways of thinking about triality. I’m not quite sure what you might be looking for.

One interesting thing is that the 3 inequivalent 8-dimensional irreps of $\mathfrak{so}(8)$ can be gotten by geometrically quantizing different Grassmannians for $Spin(8)$. One of these is the variety of all ‘left-handed’ 4-dimensional subspaces of $\mathbb{C}^8$, one is the variety of all ‘right-handed’ 4-dimensional subspaces of $\mathbb{C}^8$, but the third is the variety of all isotropic lines in $\mathbb{C}^8$.

So, two have some nice relation to 4d spaces in $\mathbb{C}^8$, while the third does not.

Posted by: John Baez on September 20, 2015 2:33 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

A very symmetric view of triality is to take the $F_4$, pick out an embedded $D_4$, and then break the adjoint representation of $F_4$ into the adjoint representation of $D_4$ plus three 8-dimensional subspaces invariant under the $D_4$ (28 + 8 + 8 + 8 = 52). The three 8-dimensional spaces are the standard and two spinor representations of $D_4$, and usual Lie algebra 3-form, restricted to taking a vector from each subspace, gives the triality 3-form. The outer automorphisms of $D_4$ are just conjugation by elements of $F_4$.

Posted by: Layra on September 20, 2015 5:42 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

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.

Posted by: Hugh Thomas on September 20, 2015 2:48 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Thanks for catching that mistake! I’m going to take the liberty of fixing the comments, here and here, that had mistakes in the lists of quiver representations. Just to reduce the amount of confusion in the universe.

Posted by: John Baez on September 20, 2015 2:57 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

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?

Posted by: John Baez on September 20, 2015 2:52 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

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.

Posted by: Mike Shulman on September 20, 2015 3:23 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

It’s a nice question, but don’t know anything about it. Does anyone out there know general theorems saying that if a variety of algebras has some property, then the algebra on $n$ generators is finite? Or infinite?

Obviously some relations are more ‘coercive’ than others. Compare commutative monoids to commutative idempotent monoids.

Posted by: John Baez on September 20, 2015 6:31 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

As to the 3 or more finite/infinite question, a conceptual explanation is given in work by Gelfand and Ponomarev. For any $n$ they construct cubicles of $2^n$ elements which have a representation theoretic interpretation, see for example this paper by Claus Ringel or this one by Dlab and Ringel. Gelfand and Ponmarev show that there are countably many such cubicles if $n \geq 4$.

As to the numerology. Can it be that 28 = 2x8 + 3x4 first the two cubicles and then three times the 4 elements of the 2 generator lattice?

Posted by: lieven le bruyn on September 20, 2015 9:41 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

It’s not too hard to deduce that an infinite modular lattice can be generated from four elements. If you work in the projective plane $\mathbb{P}^2(\mathbb{R})$, based of course on the modular lattice of vector subspaces of $\mathbb{R}^3$, and start with the four points $(0, 0), (1, 0), (0, 1), (1, 1)$ (or in homogeneous coordinate notation, $0, 0, 1), (1, 0, 1), (0, 1, 1), (1, 1, 1)$), you can get to work right away, joining pairs of points in lines and meeting pairs of lines in points, and pretty quickly generate an infinite collection of points. If you’re feeling lazy, there’s a picture of this on page 151 from a nice book by Kung, Rota, and Yan. (Courtesy of Google books, hope everyone can see it.)

It was from some writing of Rota that I first learned that fundamental invariants of a general linear operator $T: V \to V$ on a vector space are encoded in the free modular lattice on four elements. The idea is that we can recover $T$ from four subspaces of $V \oplus V$: the “$x$-axis” $V \oplus 0$, the “$y$-axis” $0 \oplus V$, the diagonal “$x = y$” given by $\{(v, v): v \in V\}$, and the graph of “$y = T x$”: $\{(v, T v): v \in V\}$. For example, as an exercise, one can construct $Im(T^n)$ by taking suitable meets and joins. From this point of view, too, it should be no wonder that the free modular lattice on four elements is infinite.

More discussion at MathOverflow here.

Posted by: Todd Trimble on September 23, 2015 2:04 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

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 $M_3$:

Posted by: Tom Leinster on September 20, 2015 12:42 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Nice! I wrote down three definitions of ‘modular lattice’ because I can never remember the usual one:

x ≤ b implies x ∨ (a ∧ b) = (x ∨ a) ∧ b

probably because it’s trying to be ‘as efficient as possible’ in a stupid way.

It’s better to relax a bit and openly admit that this definition says ∨ distributes over ∧ when the guy being ∨ed is ≤ one of the guys being ∧ed.

But even that sentence hard for me to remember, since by flipping symbols I can get a total of 4 similar sentences of this sort.

I’m told the definition is secretly self-dual, but I don’t know if that means 2 of the similar sentences are equivalent or all 4.

I could find out, but right now I like this definition better: there’s an isomorphism of posets

$[a \vee b, b] \cong [a, a \wedge b]$

sending $x$ to $x \wedge a$, with inverse sending $y$ to $y \vee a$.

I might like it even better if one of you guys said it more conceptually using more category theory. Something about an adjunction?

Posted by: John Baez on September 20, 2015 4:01 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Déjà vu: John’s comment here with a comment from me immediately after.

In particular, the adjunction you want is that for the pair of poset maps

$[a \wedge b, b] \stackrel{\overset{- \wedge b}{\leftarrow}}{\underset{a \vee -}{\longrightarrow}} [a, a \vee b]$

we have $(a \vee -) \dashv (- \wedge b)$. This is true for any lattice. There’s nothing particularly profound here: given $a \wedge b \leq x \leq b$ and $a \leq y \leq a \vee b$, we have

$\array{ a \vee x \leq y & \Leftrightarrow & a \leq y \; and \; x \leq y & by\; universal\; property\; of\; \vee \\ & \Leftrightarrow & x \leq y & since\; a \leq y\; is\; already\; given \\ & \Leftrightarrow & x \leq y \; and\; x \leq b & since\; x \leq b\; is\; already\; given \\ & \Leftrightarrow & x \leq y \wedge b & by\; universal\; property\; of\; \wedge }$

As John recalled, a lattice is modular if this adjoint pair is an adjoint equivalence, for all elements $a, b$.

Posted by: Todd Trimble on September 23, 2015 1:35 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

There’s also the handy dandy nLab which records various proofs of equivalence between definitions of modular lattice, if anyone is interested. The nLab prefers (and gives as first definition) the definition that John gave, involving isomorphisms between intervals.

Posted by: Todd Trimble on September 23, 2015 1:42 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

The advantage of senility is that you get to experience the joy of discovery repeatedly. In that comment you linked to, I was wondering how the free modular lattice on 3 generators was related to $D_4$, and how the 28 elements of the former are related to the 24 roots of the latter. I’d completely forgotten that I wondered about that. I’d also forgotten that in your reply, you’d explained modularity in terms of an adjoint equivalence.

And yet somehow these ‘forgotten’ things continued to lurk inside me, so that I half-knew that the best explanation of modularity would be in terms of an adjunction… and now, I’ve almost figured out the answer to my old puzzle about the free modular lattice on 3 generators vs. $D_4$.

Do the rest of you out there also experience this phenomenon, that you keep forgetting and re-learning various mathematical facts, yet make a bit of progress in understanding them each time around? It’s almost as if I need to ‘forget’ things to let them stew and ferment and make new connections in the subconscious mind. Or maybe that’s just a way to make it seem less pathetic.

Posted by: John Baez on September 23, 2015 5:00 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

I remembered that I’d stated the $M_3$-$N_5$ theorem on some other Café thread a few years ago, but I couldn’t remember where and it wasn’t worth my while to look it up. But the nLab entry that Todd just mentioned links to it, and I was slightly spooked to see just how similar that old comment was to my comment the other day.

Another feature of getting older is that “a few years ago” turns out to be 5.

(Incidentally, that nLab entry makes the point that in my statement of the $M_3$-$N_5$ theorem, you have to interpret “sublattice” in the sense that you would if you didn’t insist on lattices having a top or a bottom. In other words, it’s a subposet closed under binary joins and meets.)

Posted by: Tom Leinster on September 23, 2015 5:46 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

By the way, John: modularity is manifestly self-dual if you use your currently preferred definition!

Posted by: Todd Trimble on September 23, 2015 8:28 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Yes, that’s nice! By the way: since modularity is a weakened version of distributivity, is there some way to restate the distributive law for lattices that makes it look like a strengthening of the law saying that

$[a \wedge b, b] \stackrel{\overset{- \wedge b}{\leftarrow}}{\underset{a \vee -}{\longrightarrow}} [a, a \vee b]$

is an adjoint equivalence for all $a,b$? This may be a dumb question; I don’t have a good idea of how to turn distributivity into something about adjunctions.

Posted by: John Baez on September 23, 2015 9:00 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

I don’t know offhand, but it would be nice. The self-duality of distributive lattices is not something that looks manifestly obvious to me, and that’s an annoyance.

The forbidden subgraph characterization of modular lattices and of distributive lattices, if taken as a definition, does make the self-duality look obvious in both cases. But I haven’t thought about it long enough to decide whether it would be a good idea to use that as an actual definition (sort of like the Möbius band characterization of orientable surfaces Tom mentioned a little while ago!).

Posted by: Todd Trimble on September 23, 2015 11:01 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

For what it’s worth, one nice manifestly self-dual way to axiomatize distributive lattices is as those lattices where $a \wedge b \leq c$ and $a \leq b \vee c$ entails $a \leq c$.

Posted by: Sridhar Ramesh on September 24, 2015 5:36 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

And, I realize now, a modular lattice is one where this last property holds so long as the extra condition $c \leq a$ is assumed.

Posted by: Sridhar Ramesh on September 24, 2015 5:58 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

A self-dual equational axiom for distributive lattices is $(x\wedge y)\vee(x\wedge z)\vee (y\wedge z) = (x\vee y)\wedge (x\vee z)\wedge(y\vee z).$ This is an exercise in many books on lattice theory. Notice the two sides are the top and bottom of the $M_3$ in $FM(3)$.

I have an applet for drawing lattices: http://math.hawaii.edu/~ralph/LatDraw/. (Chrome no longer supports applets but it should still work with Firefox, perhaps after giving it permission.)

This discussion mentioned Christian Herrmann’s result that $FM(4)$ has an undecidable word problem and gave some heuristics on why it should be true. But the result is much deeper than that. If we take the class of all sublattices of the lattices of all subgroups of an Abelian group, the free lattice on 4 generators for this class has a solvable word problem. (Of course, this class includes all lattices of subspaces of a vector space.)

Posted by: Ralph Freese on September 27, 2015 10:02 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

This discussion mentioned Christian Herrmann’s result that $FM(4)$ has an undecidable word problem and gave some heuristics on why it should be true.

Professor Freese, you are quite right that that part of the discussion was somewhat superficial (and your bringing it up jarred me into recognition of your name; if I recall correctly, you also had a hand in establishing the undecidability of the word problem for $FM(n)$, for $n \geq 4$).

Anyway, very glad to see you joined this discussion; I’m hoping you have some answers to questions I have.

The solvability or decidability result you mentioned sounds very interesting; can you tell me where that is proven?

This reminds me of some intriguing work by Rota and coworkers on so-called “linear lattices” or lattices of commuting equivalence relations; I think I first learned of this from Rota’s essay here. I don’t know what the state of the art is in this area, but Rota mentions a conjecture of Gel’fand that the (word problem for the) free linear lattice in four generators is decidable. Can you say anything about that?

Rota also mentioned some interesting work by himself, Finberg, and Mainetti on a kind of logical and graphical calculus for linear lattices. A Google books link is here. They illustrate this calculus by showing how to efficiently derive equations such as the Desarguesian identity and some other identities unearthed by Haiman.

I think one of their hopes for future work was to use this calculus as an aid to establishing decidability in free linear lattices, and so I was wondering if you were aware of any further research on that front.

(For what it’s worth, and speaking speculatively here, the work by Finberg, Mainetti, and Rota looks somewhat similar to work of Freyd and Scedrov mentioned in section 2.158 of their book Categories, Allegories; it’s in section 2.158 if you care to look; a Google books link is here. Again a graphical calculus of circuits is employed, and again they establish the Desarguesian identity, and moreover they claim a decidability result for deciding which equations in the language of allegories hold in the allegory of sets – to me it sounds suspiciously similar to asking which equations in the language of lattices hold in lattices of commuting equivalence relations. But as I say, this is pretty speculative.)

Posted by: Todd Trimble on September 28, 2015 2:13 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

I did prove the word problem for free modular lattices on at least 5 generators is undecidable; thanks for remembering. The decidability result mentioned in my previous post is proved in C. Herrmann, On the equational theory of submodule lattices, Proc. Houston Lattice Theory Conference, 1973, 105-118. These proceedings are available here. The main idea is that if a lattice equation fails in some submodule lattice, it fails in the submodule lattice of $(Z_{p^k})^n$, for some $p$, $k$ and $n$.

On linear lattices, Haiman went on to prove the nice result that this class of lattices is not finitely axiomatizable. But there are still open problems such as: is the homomorphic image of a linear lattice linear?

Thanks for the link in Freyd and Scedrov’s book. It certainly seems related to Haiman’s result (although I can’t say I’ve mastered what is going on there).

Posted by: Ralph Freese on September 29, 2015 3:24 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Dear Todd, adding to Ralph’s comments: if one speaks about finitely presented modular lattices, unsolvability occurs already in the vector space setting for 5 generators (Hutchinson and a paper in JSL - just dont recall the authors name, both late 70ties), for 4 generators in the abelian group setting. In contrast, speaking about equationsm that is free lattices in some variety, only proof of unsolvability fails already for Arguesian lattices (since the gluing constructions dont work), proof of solvability fails already for normal subgroup lattices since no decent set of generators for the variety is in sight or since there is nothing like Smith normal form for matrices (as hsed by Hutchinson and Czedli for submodule lattices).

Posted by: Christian Herrmann on October 14, 2017 8:54 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Thanks, Ramesh! That’s nice.

Posted by: John Baez on September 24, 2015 7:00 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

I just realized another nice manifestly self-dual way to axiomatize distributive lattices is as those lattices on which, for every $x$, the operation $y \mapsto (x \wedge y, x \vee y)$ is injective. [Which is just the (manifestly self-dual) second definition* of modularity in John Baez’s original post with the comparability precondition dropped].

I also just realized how these particular formalizations of the distributivity and modularity laws relate directly to the $M_3$-$N_5$ characterization (if $y_1$ and $y_2$ have the same meets and joins with $x$ without being equal, then the (non-bounded) lattice generated by them and $x$ comprises $N_5$ or $M_3$ according as to whether $y_1$ and $y_2$ are comparable or not).

Perhaps this had been obvious to everyone all along!

[*: This very clean definition is not among the many equivalent axiomatizations noted on the nLab article on modular lattices. Someone (perhaps even me) should add it when they have time!]

Posted by: Sridhar Ramesh on September 27, 2015 6:21 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Oh, no, that second paragraph is slightly glib: it works when $y_1$ and $y_2$ are comparable, but if they are not comparable, we may not get $M_3$ directly, as we are not guaranteed that $(y_1 \wedge y_2, y_1 \vee y_2)$ matches $(y_1 \wedge x, y_1 \vee x)$. Hm…

Posted by: Sridhar Ramesh on September 27, 2015 6:39 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Ah, rather, the correct thing is to now consider violations of distributivity in specifically modular lattices, and note that these induces $M_3$ like so:

Take $x$, $y_1$, and $y_2$ like before (that is, such that $x$ produces the same meets and joins with both $y$s), and, being in a modular lattice, let $z$ be the “modular projection” of $x$ into the lattice generated by $y_1$ and $y_2$ [that is, $(x \vee (y_1 \wedge y_2)) \wedge (y_1 \vee y_2)$, or equivalently, the dual of this]. Then, I claim, the lattice generated by $y_1$, $y_2$, and $z$ will be $M_3$.

Proof: We just need to show that $z \wedge y_1 \leq y_2$ (everything else nontrivial will then follow by symmetry and duality). To show this, note that $z \wedge y_1 = (x \vee (y_1 \wedge y_2)) \wedge y_1$. By modularity, this is the same as $(y_1 \wedge y_2) \vee (x \wedge y_1)$. Both disjuncts here entail $y_2$, and thus we are done.

Posted by: Sridhar Ramesh on September 27, 2015 7:27 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Another manifestly self-dual axiom for modularity can be found by taking one of the middle-ranked elements in the free modular lattice on 3 generators and swapping all the join and meet operations:

$(a \wedge b) \vee ((a \vee b) \wedge c) = (a \vee b) \wedge ((a \wedge b) \vee c)$

This equation follows from the axiom for modularity:

$a \le b \Rightarrow a \vee (b \wedge c) = (a \vee b) \wedge (a \vee c)$

since $a \wedge b \le a \vee b$. And we can derive that axiom from the self-dual one, since if $a \le b$ we have $a \vee b = b$ and $a \wedge b = a$, and substituting into the self-dual axiom gives:

$a \vee (b \wedge c) = (a \vee b) \wedge (a \vee c)$

Posted by: Greg Egan on September 28, 2015 12:24 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

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!

On the other hand, it’s quite reminiscent of Kuratowski’s (true) theorem about planar graphs.

Posted by: Mark Meckes on September 20, 2015 4:41 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

And a surface is orientable if and only if it contains no copy of the Möbius band as a subspace. A friend liked to use that as the definition of orientability when teaching the classification of orientable topological surfaces to undergraduates.

When he first told me that, I laughed at his brazenness, but soon afterwards I came over to the point of view that it wasn’t a bad idea.

Posted by: Tom Leinster on September 23, 2015 6:01 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Posted by: Jesse C. McKeown on September 20, 2015 6:47 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Wow — that’s GREAT!

That’s exactly the image I was dreaming of, including the rotation.

Would it be okay if I feature this on my blog Visual Insight? I’d like to do a post about the free modular lattice on 3 generators and its relation to $D_4$.

Posted by: John Baez on September 20, 2015 7:18 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Yes, of course.

Ulrik, that’s awesome!

Posted by: Jesse C. McKeown on September 20, 2015 11:06 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

By the way, it’s amusing to compare the lattice of Dyck words, better known to category theorists as vertices of the Stasheff polytope, or parenthesized strings of letters — but made into a lattice in a certain way which the picture (or the link) explains:

These are the 14 parenthesized strings of 5 letters, also known as the vertices of the associahedron. See how it looks just a bit like Jesse’s picture above?

Posted by: John Baez on September 20, 2015 7:38 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Before I saw Jesse’s version, I made an interactive visualization (by just adapting a javascript molecule viewer). Feel free to just wherever and whenever.

Posted by: Ulrik Buchholtz on September 20, 2015 8:10 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Wow, Ulrik — that’s great too! I can link to it from Visual Insight, but I don’t think I can include it in a Wordpress blog post. (At least I’ve never succeeded in doing that kind of thing.)

Posted by: John Baez on September 20, 2015 10:42 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

This visualization is wonderful! It’s interesting to note that it comprises a bistable optical illusion (a la the infamous “Spinning Dancer” animated silhouette).

I like to imagine the two Necker cubes spinning oppositely, taunting my brain to deal with the fallout of how to perceive the middle.

Posted by: Sridhar Ramesh on September 25, 2015 7:00 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Yes, today it seems to be rotating counterclockwise. At first I was shocked when it seemed to be turning the opposite way from how I remembered it. I’m unable to get it to reverse directions by sheer mental effort.

Posted by: John Baez on September 25, 2015 8:32 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Alas, it’s not great to try to describe the two possible spin directions as “clockwise” and “counterclockwise”, since either one is naturally described as rotating counterclockwise (follow the 2d path of any particular dot within the flat image; it will be counterclockwise).

Anyway, I find it easiest to switch between the two perspectives by focusing on a cube to start; I can switch my perception of a Necker cube at will, and the rest of the image follows when I do so.

Posted by: Sridhar Ramesh on September 25, 2015 9:34 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Would it be possible to have access to the source code or tool you used to generate this image, Jesse?

I’m trying to explain the ambiguity of this image to others, but finding it difficult to do so with words, and thus would like to produce an augmented version of this visual (specifically, I feel it would be easy to do by coloring all the edges of the wireframe distinctly and then rendering it with edges closer to the viewer occluding those further back, using the two mirror-image natural interpretations of depth information to produce two separate renderings).

(This is all just about the image qua bistable optical illusion rather than qua Hasse diagram; I’m afraid that’s what the friends I’m showing this to are more interested in understanding)

Posted by: Sridhar Ramesh on September 25, 2015 10:39 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

The geometry is done in kseg, from which 24 pngs were exported; these were then imported as layers into the gimp, which is happy to export gif animations. Let me see if I can share the kseg construction in g++good or some such nowhere.

(one could wish that kseg also exported geogebra files, or maybe I should just have adopted geogebra… )

Posted by: Jesse C. McKeown on September 26, 2015 1:14 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

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?

Posted by: John Baez on September 23, 2015 10:29 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

The lower triple of points in the cube at the top consists of

$\array{ (X\vee Y)\wedge(X\vee Z)\\ (X\vee Y)\wedge(Y\vee Z)\\ (X\vee Z)\wedge(Y\vee Z) }$

The point at the bottom of the cube is

$(X\vee Y)\wedge(Y\vee Z)\wedge(X\vee Z)$

and similarly in the bottom cube.

Posted by: Tim Silverman on September 23, 2015 11:38 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Great!

Posted by: John Baez on September 24, 2015 4:52 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

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 $X$, $Y$, $Z$.

Posted by: Hugh Thomas on September 23, 2015 11:00 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Yes, that sounds right! It would be nice to have a 3d, non-rotating picture of this lattice correctly labelled by these expressions.

Posted by: John Baez on September 24, 2015 4:56 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Posted by: Greg Egan on September 24, 2015 9:18 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Wow — thanks, Greg! These two diagrams will make it a lot easier to keep track of what’s going on here. I’ll write a Visual Insight article on this stuff.

Posted by: John Baez on September 24, 2015 5:46 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Posted by: Greg Egan on September 24, 2015 11:59 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

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.

Posted by: John Baez on September 24, 2015 6:15 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

We need a bit more notation to name all the subspaces of $\mathbb{R}^8$ associated to the quiver representation I’m calling $G$:

   1
2
1   1


Let’s number the outer dots $1,2,3$ starting from the top and going clockwise. So, if we call the vector spaces in this quiver representation $G_1, G_2, G_3$ and $G$, they go here:

   G_1
G
G_3  G_2


The vector space $G$ is 2-dimensional, while $G_1, G_2, G_3$ are 1-dimensional and distinct, so that

$G_1 \vee G_2 \vee G_3 = G$

and

$G_1 \wedge G_2 = G_2 \wedge G_3 = G_3 \wedge G_1 = \{0\}$

Now we’re ready to work out the space $X$. It’s the vector space assigned to the top dot by the direct sum of all seven quiver representations. So, using all the notation we’ve set up,

$X = A \oplus E \oplus F \oplus G_1$

Similarly $Y$ is the vector space assigned to the bottom right dot by the direct sum of all seven quiver representations:

$Y = B \oplus D \oplus F \oplus G_2$

and $Z$ is the vector space that this direct sum assigns to the bottom left dot:

$Z = C \oplus D \oplus E \oplus G_3$

Using these facts we can work out the subspace of

$A \oplus B \oplus C \oplus D \oplus E \oplus F \oplus G \cong \mathbb{R}^8$

corresponding to any element of the free modular lattice on $X,Y,Z$. For example we have

$X \wedge Y = ( A \oplus E \oplus F \oplus G_1) \wedge (B \oplus D \oplus F \oplus G_2) = F$

where I used the fact that $G_1 \wedge G_2 = \{0\}$.

Someday when I’m feeling energetic I’ll work out all 28 distinct subspaces built from $X,Y,Z$ using $\wedge$ and $\vee$. Given everything we’ve done so far, this is now routine.

It will be especially fun to see all six 4-dimensional subspaces — the ones that correspond to dots in the middle rank of this picture:

The three outer dots are the 4-dimensional subspaces $X,Y,Z$, but there are three more. We can see if Tim Silverman’s guess about them is correct. For this it will help to think of $G_1, G_2, G_3 \subset G$ as three lines in the plane, going through the origin at 120$^\circ$ angles to each other. That choice makes the ‘triality’ symmetry into a nice geometrical symmetry of this plane, and thus of

$A \oplus B \oplus C \oplus D \oplus E \oplus F \oplus G \cong \mathbb{R}^8$

Posted by: John Baez on September 24, 2015 6:37 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

John wrote

$G_1\vee G_2\vee G_3=G$

In fact, the join of any two of these is $G$, so they are playing the roles of $e_7$, $e_8$ and $e_7+e_8$ in Dedekind’s realisation.

Posted by: Tim Silverman on September 24, 2015 10:06 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Right — I woke up in the middle of last night and realized I should have said

$G_1 \vee G_2 = G_2 \vee G_3 = G_3 \vee G_1 = G$

I found Dedekind’s $e_7, e_8$ and $e_7 + e_8$ to be upsettingly asymmetrical but of course it’s really not. Over $\mathbb{R}$ we should use 3 lines in the plane that are all at 120${}^\circ$ angles to each other, since then the triality symmetries will act as isometries and presumably your hoped-for connection to the Lie group $G_2$ will show up. But Dedekind’s formula is nice because it works over the rationals.

(It’s just some sort of bizarre coincidence that we’re using $G_2$ to mean something completely different here.)

Posted by: John Baez on September 24, 2015 5:14 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Also, $Z$ should include $C$, not $A$.

Posted by: Tim Silverman on September 24, 2015 11:01 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Thanks — I’ll go back and fix that, since I may want to refer to it.

Posted by: John Baez on September 24, 2015 5:34 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

$\array{ ((X\vee Y)\vee Z)&A\oplus B\oplus C\oplus D\oplus E\oplus F\oplus G\\ (X\vee Y)&A\oplus B\oplus D\oplus E\oplus F\oplus G\\ (X\vee Z)&A\oplus C\oplus D\oplus E\oplus F\oplus G\\ (Y\vee Z)&B\oplus C\oplus D\oplus E\oplus F\oplus G\\ ((X\vee Y)\wedge (X\vee Z))&A\oplus D\oplus E\oplus F\oplus G\\ ((X\vee Y)\wedge (Y\vee Z))&B\oplus D\oplus E\oplus F\oplus G\\ ((X\vee Z)\wedge (Y\vee Z))&C\oplus D\oplus E\oplus F\oplus G\\ (((X\vee Y)\wedge (X\vee Z))\wedge (Y\vee Z))&D\oplus E\oplus F\oplus G\\ (X\vee (Y\wedge Z))&A\oplus D\oplus E\oplus F\oplus G_1\\ (Y\vee (X\wedge Z))&B\oplus D\oplus E\oplus F\oplus G_2\\ (Z\vee (X\wedge Y))&C\oplus D\oplus E\oplus F\oplus G_3\\ ((X\vee (Y\wedge Z))\wedge (Y\vee Z))&D\oplus E\oplus F\oplus G_1\\ ((Z\vee (X\wedge Y))\wedge (X\vee Y))&D\oplus E\oplus F\oplus G_3\\ ((Y\vee (X\wedge Z))\wedge (X\vee Z))&D\oplus E\oplus F\oplus G_2\\ X&A\oplus E\oplus F\oplus G_1\\ Y&B\oplus D\oplus F\oplus G_2\\ Z&C\oplus D\oplus E\oplus G_3\\ (X\wedge (Y\vee Z))&E\oplus F\oplus G_1\\ (Y\wedge (X\vee Z))&D\oplus F\oplus G_2\\ (Z\wedge (X\vee Y))&D\oplus E\oplus G_3\\ (((X\wedge Y)\vee (X\wedge Z))\vee (Y\wedge Z))&D\oplus E\oplus F\\ ((X\wedge Y)\vee (X\wedge Z))&E\oplus F\\ ((X\wedge Y)\vee (Y\wedge Z))&D\oplus F\\ ((X\wedge Z)\vee (Y\wedge Z))&D\oplus E\\ (X\wedge Y)&F\\ (X\wedge Z)&E\\ (Y\wedge Z)&D\\ ((X\wedge Y)\wedge Z)& \{0\} }$

Posted by: Tim Silverman on September 24, 2015 12:15 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

Thanks! I’ll be using this…

Posted by: John Baez on September 24, 2015 6:04 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

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.

Posted by: John Baez on September 26, 2015 5:54 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

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.)

Posted by: John Baez on September 27, 2015 1:11 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

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.

Posted by: John Baez on September 29, 2015 5:26 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

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

Posted by: John Baez on January 3, 2016 10:09 PM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

The news has been so unrelentingly bad these past few weeks that I’m taking momentary refuge in good old numerology. It’s a fun story about the numbers 3, 8, 24 and 28, so I figured I might as well write about it, even if the 28 in this story might not be the same 28 from the free modular lattice on 3 generators.

First, the 3. Consider the Pauli matrices:

(1)$\sigma_x = \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array}\right),\qquad \sigma_y = \left(\begin{array}{cc} 0 & -i \\ i & 0 \end{array}\right),\qquad \sigma_z = \left(\begin{array}{cc} 1 & 0 \\ 0 & -1 \end{array}\right).$

Note that of these three matrices, only $\sigma_y$ is antisymmetric, and also note that we have

(2)$\sigma_z \sigma_x = -\sigma_x\sigma_z = i\sigma_y.$

This much is familiar, though that minus sign gets around. For example, it is the fuel that makes the GHZ thought-experiment go, because it means that

(3)$\sigma_x \otimes \sigma_x \otimes \sigma_x = -(\sigma_x \otimes \sigma_z \otimes \sigma_z)(\sigma_z \otimes \sigma_x \otimes \sigma_z)(\sigma_z \otimes \sigma_z \otimes \sigma_x).$

And this leads us to where the 8 comes into the story. Let’s consider the finite-dimensional Hilbert space made by composing three qubits. This state space is eight-dimensional, and we build the three-qubit Pauli group by taking tensor products of the Pauli matrices, considering the $2 \times 2$ identity matrix to be the zeroth Pauli operator. There are 64 matrices in the three-qubit Pauli group, and we can label them by six bits. The notation

(4)$\left(\begin{array}{ccc} m_1 & m_3 & m_5 \\ m_2 & m_4 & m_6\end{array}\right)$

means to take the tensor product

(5)$(-i)^{m_1m_2} \sigma_x^{m_1} \sigma_z^{m_2} \otimes (-i)^{m_3m_4} \sigma_x^{m_3} \sigma_z^{m_4} \otimes (-i)^{m_5m_6} \sigma_x^{m_5} \sigma_z^{m_6}.$

Now, we ask: Of these 64 matrices, how many are symmetric and how many are antisymmetric? We can only get antisymmetry from $\sigma_y$, and (speaking heuristically) if we include too much antisymmetry, it will cancel out. More carefully put: We need an odd number of factors of $\sigma_y$ in the tensor product to have the result be an antisymmetric matrix. Otherwise, it will come out symmetric. Consider the case where the first factor in the triple tensor product is $\sigma_y$. Then we have $(4-1)^2 = 9$ possibilities for the other two slots. The same holds true if we put the $\sigma_y$ in the second or the third position. Finally, $\sigma_y \otimes \sigma_y \otimes \sigma_y$ is antisymmetric, meaning that we have

(6)$9 \cdot 3 + 1 = 28$

antisymmetric matrices in the three-qubit Pauli group. In the notation established above, they are the elements for which

(7)$m_1 m_2 + m_3 m_4 + m_5 m_6 = 1 \mod 2.$

Puzzle: This has a secret geometrical meaning in terms of the Fano plane. What is it?

Hint: $28 = 7 \cdot 4 = 7 \cdot (7 - 3).$

We have a 3 (the nontrivial elements of the single-qubit Pauli group), an 8 (the dimension of the three-qubit Hilbert space), and a 28 (the number of antisymmetric three-qubit Pauli operators). Does this story have a 24 as well? Sneakily, yes—because the same combinatorial notation that we used to enumerate the 28 antisymmetric Pauli operators also enumerates the 28 bitangents to a quartic curve. This is a lovely piece of nineteenth-century geometry, in the genre of the 27 lines on a cubic surface. It connects to Galois theory, as this paper explains:

I didn’t think that 24 entered into this numerology until today, when I read that a generic plane quartic has 24 flex points.

I could keep going—rolling on into quantum information theory, combinatorial designs, integral octonions—but for now, I’ll just say I found it rather cute that an 8-dimensional space also yields an interesting collection of 28 things built up from 3 things in this other way.

Posted by: Blake Stacey on December 8, 2016 12:13 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

On a tangentially (ha!) related note, the numerical ‘reason’ one can construct an isomorphism $PGL(3,2) \simeq PSL(2,7)$ is that $168=2^3\cdot 3\cdot 7$ (no factor of 5 (!)) is divisible by $\binom{8}{k}$ for $k=0,1,2,3,5,6,7,8$, and not for $k=4$ (in this latter case, the factor of 4 is not cancelled out). This means that $PSL(2,7)$ cannot act transitively on the set of 4-tuples in the projective line over $\mathbb{F}_7$, which has $\binom{8}{4} = 28$ elements. This 28-element set has an involution $\tau$ with no fixed points (send a 4-tuple to its complement), and the action of $PSL(2,7)$ on the set of 4-tuples descends to the quotient by $\tau$. There is an orbit of size 7 (coming from an orbit of size 14 on the set of 4-uples), and this orbit “is” the Fano plane.

Or rather, this is the story I’m telling myself, amalgamating the two answers here on math.stackexchange by Mariano Suárez-Álvarez and Grigory M. It’s actually a little bit more subtle than that, but the fact about the binomial coefficient (which I think is also responsible for an exceptional isomorphism or two involving $A_8$) is where it comes from.

Posted by: David Roberts on December 8, 2016 5:38 AM | Permalink | Reply to this

### Re: The Free Modular Lattice on 3 Generators

That’s a nice tangent!

Posted by: Blake Stacey on December 8, 2016 6:21 PM | Permalink | Reply to this

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