## August 17, 2015

### A Wrinkle in the Mathematical Universe

#### Posted by John Baez

Of all the permutation groups, only $S_6$ has an outer automorphism. This puts a kind of ‘wrinkle’ in the fabric of mathematics, which would be nice to explore using category theory.

For starters, let $Bij_n$ be the groupoid of $n$-element sets and bijections between these. Only for $n = 6$ is there an equivalence from this groupoid to itself that isn’t naturally isomorphic to the identity!

This is just another way to say that only $S_6$ has an outer isomorphism.

And here’s another way to play with this idea:

Given any category $X$, let $Aut(X)$ be the category where objects are equivalences $f : X \to X$ and morphisms are natural isomorphisms between these. This is like a group, since composition gives a functor

$\circ : Aut(X) \times Aut(X) \to Aut(X)$

which acts like the multiplication in a group. It’s like the symmetry group of $X$. But it’s not a group: it’s a ‘2-group’, or categorical group. It’s called the automorphism 2-group of $X$.

By calling it a 2-group, I mean that $Aut(X)$ is a monoidal category where all objects have weak inverses with respect to the tensor product, and all morphisms are invertible. Any pointed space has a fundamental 2-group, and this sets up a correspondence between 2-groups and connected pointed homotopy 2-types. So, topologists can have some fun with 2-groups!

Now consider $Bij_n$, the groupoid of $n$-element sets and bijections between them. Up to equivalence, we can describe $Aut(Bij_n)$ as follows. The objects are just automorphisms of $S_n$, while a morphism from an automorphism $f: S_n \to S_n$ to an automorphism $f' : S_n \to S_n$ is an element $g \in S_n$ that conjugates one automorphism to give the other:

$f'(h) = g f(h) g^{-1} \qquad \forall h \in S_n$

So, if all automorphisms of $S_n$ are inner, all objects of $Aut(Bij_n)$ are isomorphic to the unit object, and thus to each other.

Puzzle 1. For $n \ne 6$, all automorphisms of $S_n$ are inner. What are the connected pointed homotopy 2-types corresponding to $Aut(Bij_n)$ in these cases?

Puzzle 2. The permutation group $S_6$ has an outer automorphism of order 2, and indeed $Out(S_6) = \mathbb{Z}_2.$ What is the connected pointed homotopy 2-type corresponding to $Aut(Bij_6)$?

Puzzle 3. Let $Bij$ be the groupoid where objects are finite sets and morphisms are bijections. $Bij$ is the coproduct of all the groupoids $Bij_n$ where $n \ge 0$:

$Bij = \sum_{n = 0}^\infty Bij_n$

Give a concrete description of the 2-group $Aut(Bij)$, up to equivalence. What is the corresponding pointed connected homotopy 2-type?

You can get a bit of intuition for the outer automorphism of $S_6$ using something called the Tutte–Coxeter graph.

Let $S = \{1,2,3,4,5,6\}$. Of course the symmetric group $S_6$ acts on $S$, but James Sylvester found a different action of $S_6$ on a 6-element set, which in turn gives an outer automorphism of $S_6$.

To do this, he made the following definitions:

• A duad is a 2-element subset of $S$. Note that there are ${6 \choose 2} = 15$ duads.

• A syntheme is a set of 3 duads forming a partition of $S$. There are also 15 synthemes.

• A synthematic total is a set of 5 synthemes partitioning the set of 15 duads. There are 6 synthematic totals.

Any permutation of $S$ gives a permutation of the set $T$ of synthematic totals, so we obtain an action of $S_6$ on $T$. Choosing any bijection betweeen $S$ and $T$, this in turn gives an action of $S_6$ on $S$, and thus a homomorphism from $S_6$ to itself. Sylvester showed that this is an outer automorphism!

There’s a way to draw this situation. It’s a bit tricky, but Greg Egan has kindly done it:

Here we see 15 small red blobs: these are the duads. We also see 15 larger blue blobs: these are the synthemes. We draw an edge from a duad to a syntheme whenever that duad lies in that syntheme. The result is a graph called the Tutte–Coxeter graph, with 30 vertices and 45 edges.

The 6 concentric rings around the picture are the 6 synthematic totals. A band of color appears in one of these rings near some syntheme if that syntheme is part of that synthematic total.

If we draw the Tutte–Coxeter graph without all the decorations, it looks like this:

The red vertices come from duads, the blue ones from synthemes. The outer automorphism of $S_6$ gives a symmetry of the Tutte–Coxeter graph that switches the red and blue vertices!

The inner automorphisms, which correspond to elements of $S_6$, also give symmetries: for each element of $S_6$, the Tutte–Coxeter graph has a symmetry that permutes the numbers in the picture. These symmetries map red vertices to red ones and blue vertices to blue ones.

The group $\mathrm{Aut}(S_6)$ has

$2 \times 6! = 1440$

elements, coming from the $6!$ inner automorphisms of $S_6$ and the outer automorphism of order 2. In fact, $\mathrm{Aut}(S_6)$ is the whole symmetry group of the Tutte–Coxeter graph.

For more on the Tutte–Coxeter graph, see my post on the AMS-hosted blog Visual Insight:

Posted at August 17, 2015 9:45 AM UTC

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/2840

### Re: A Wrinkle in the Mathematical Universe

That got me trying to remember why the center of $S_n$ for $n \gt 2$ is trivial. I remember learning something along the lines of the first of the answers given here.

Posted by: David Corfield on August 17, 2015 12:26 PM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

well, the conjugation action is transitive on each conjugacy class (that’s trivial…), and if $n \gt 2$, only one conjugacy class is a singleton (this is slightly less trivial)… but we know what the classes are in each case.

Posted by: Jesse C. McKeown on August 18, 2015 12:04 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

To see that $S_2$ is the only abelian permutation group, I like to visualize a permutation as a bunch of cycles:

Then, conjugating this permutation by an transposition has the effect of ‘snipping’ and ‘reattaching’ edges in this picture.

For example, to conjugate the above permutation with the transposition that switches 9 and 4, we replace the edges going to and from 9 with edges going to and from 4, and replace the edges going to and from 4 with edges going to and from 9. In this case, two cycles become one. Various other things can happen.

This lets you see that only in very few cases does a permutation commute with all transpositions. The permutation needs to be either the identity, or any element of $S_2$. (In the first case there are no edges to switch, in the second switching the edges never has any effect.)

Thanks to the Schur–Weyl duality between $S_n$ and $SU(N)$, this picture has a nice application to 2-dimensional Yang-Mills theory with gauge group $SU(N)$. The edges can be seen as electric field lines wrapping around a circle, and with the passage of time these field lines get ‘snipped’ and ‘reattached’:

A similar but more complicated phenomenon happens with magnetic field lines in a plasma:

It’s called reconnection, and it’s important in phenomena such as solar flares.

Posted by: John Baez on August 18, 2015 1:37 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

I love this recombination idea but it seems to me you’re describing what happens when you multiply a permutation with a transposition, not when you conjugate a permutation with a transposition? If I compute the conjugate of the permutation in your graphic by $(94)$, I get two cycles, with the roles of $9$ and $4$ swapped:

$(9 4) (2) (7) (9 10 11 5 6 8) (1 3 4) (9 4) = (2) (7) (4 10 11 5 6 8) (1 3 9)$

unless I am doing something silly.

Posted by: Bruce Bartlett on September 3, 2015 10:10 PM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

You’re right: ‘reconnecting field lines’ in 2d Yang-Mills theory arise from composing a permutation with a transposition, not conjugating a permutation with a transposition. Obviously conjugating a transposition with something doesn’t change its cycle structure! It’s cleary been too long since I’ve thought about this.

Posted by: John Baez on September 4, 2015 2:11 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Here is a more explicitly categorical way to say essentially the same thing.

Let $\mathcal{F}$ be the category of finite sets and bijections, and let $\mathcal{G}$ be the category of finite graphs and isomorphisms. (We consider only undirected graphs without loops of multiple edges.) We define functors

(1)$\mathcal{F} \xrightarrow{D} \mathcal{G} \xrightarrow{T} \mathcal{G} \xrightarrow{P} \mathcal{F}$

as follows:

• A dyad in a set $X$ is a subset $d\subseteq X$ with $|d|=2$. We let $D X$ be the graph whose vertices are the dyads in $X$, with an edge from $p$ to $p'$ iff $p\cap p'=\emptyset$.
• More generally, a dyad in a graph $G$ is a pair of vertices that are not joined by an edge. This agrees with the previous definition if we regard sets as graphs with no edges.
• A pentad in a graph $G$ is a set $p\subseteq\text{vert}(G)$ such that $|p|=5$ and $G|_p$ is discrete. We let $P G$ denote the set of pentads in $G$.
• A triangle in a graph $G$ is a set $t\subseteq\text{vert}(G)$ such that $|t|=3$ and $G|_t$ is complete. We let $T G$ be the graph whose vertices are the triangles in $G$, with an edge from $t$ to $t'$ iff $t\cap t'\neq\emptyset$. (Note that this is opposite to the rule for dyads.)

Now let $\mathcal{G}_1\subset\mathcal{G}$ denote the category of graphs with the following properties:

• $G$ has $15$ vertices, all of valence $6$.
• If $v$ and $w$ are distinct vertices that are not joined by an edge, then there is a unique pentad $p\in PG$ with $\{v,w\}\subseteq p$.

We also let $\mathcal{F}_1\subset\mathcal{F}$ denote the category of sets of size $6$.

Theorem:

The functors defined above restrict to give equivalences

(2)$\mathcal{F}_1 \xrightarrow{D} \mathcal{G}_1 \xrightarrow{T} \mathcal{G}_1 \xrightarrow{P} \mathcal{F}_1.$

Moreover, in this context we have $T^2\simeq 1\colon\mathcal{G}_1\to\mathcal{G}_1$, and $D$ and $P$ are inverse to each other.

However, $T$ is not equivalent to the identity on $\mathcal{G}_1$, and $P T D$ is not equivalent to the identity on $\mathcal{F}_1$.

Posted by: Neil Strickland on August 17, 2015 1:26 PM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Cool! Did you make this up specially for this comment, or is this to be found somewhere else?

Posted by: John Baez on August 18, 2015 4:12 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

I have some notes that I wrote a few years ago which prove the statement as formulated above, and also make some connections with $PGL_2(\mathbb{F}_5)$ and $PSL_2(\mathbb{F}_9)$ and the symmetries of the icosahedron. But I have not released them anywhere.

Posted by: Neil Strickland on August 18, 2015 7:04 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

In Klein’s Erlangen program, we start with a group $G$. If this acts on some geometrical structure, any ‘figure’ in this structure will be stabilized by some subgroup of $G$, and Klein’s clever idea was to define ‘figures’ as subgroups of $G$.

We can use this idea to define the Tutte–Coxeter graph in a way that makes it obviously have $Aut(S_6)$ as symmetries… though it takes real work to see that this definition gives the picture here:

Let $G$ be a group abstractly isomorphic to $S_6$, but let’s not think of it as a group of permutations of a 6-element set, since the whole point is that we can do this in two inequivalent ways, related by an outer automorphism, and we don’t want to favor either one here!

$G$ has 30 subgroups isomorphic to $S_4 \times S_2$. These will be the vertices of our graph.

Given two such subgroups, $H$ and $H'$, we draw an edge between the corresponding vertices iff $H \cap H'$ is isomorphic to $D_4 \times S_2$, where $D_4$ is the dihedral group that acts as symmetries of the square.

The resulting graph is obviously acted on by $Aut(G)$. What’s not obvious is that it looks like the picture above. For this we need some facts about subgroups of $S_6$.

The 30 subgroups of $G$ isomorphic to $S_4 \times S_2$ come in two classes: 15 are conjugate to each other, and the other 15 are conjugate to each other. If we identify $G$ with a group of permutations of a 6-element set, 15 of these subgroups are stabilizers of duads, while the other 15 are stabilizers of synthemes.

However, which is which? That depends on how you identify $G$ with a group of permutations of a 6-element set!

If $H$ is the stabilizer of a duad, and $H'$ is the stabilizer of a syntheme, and the duad lies in the syntheme, then $H \cap H'$ will be isomorphic to $D_4 \times S_2$. So, this is why we put an edge between two vertices in this case.

In fact, $G$ has exactly 45 subgroups isomorphic to $D_4 \times S_2$, so we can identify the edges of our graph with these subgroups.

Posted by: John Baez on August 18, 2015 1:13 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Is it obvious that for each subgroup isomorphic to $D_4 \times S_2$, there is exactly one pair of H, H’ such that the subgroup is the intersection of that pair? Or is that just one of the facts about subgroups of $S_6$?

Posted by: Layra on August 18, 2015 1:38 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

It’s not obvious, but I think it’s not hard once you know that every $S_4 \times S_2$ subgroup is the stabilizer of either a duad or a syntheme, which follows from knowing the total number of such subgroups is 15+15 = 30, something I’m willing to look up in a table. Then you just need to consider the various ways duads and synthemes can be related to each other. I believe that only when you have a duad contained in a syntheme is the intersection of their stabilizers $D_4 \times S_2$. Moreover, this stabilizer lets you reconstruct the duad (the two points interchanged by $S_2$) and the syntheme (consisting of this duad and 4 more points, on which $D_4$ acts in a way that preserves their splitting into 2 more duads).

Posted by: John Baez on August 18, 2015 3:03 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Inspired by Dan Piponi’s comment on G+, I just rewrote the first part of this article. I decided to make it less about $S_n$ and more about the groupoid of $n$-element sets. Of course these are equivalent, but I think the categorical content of this ‘wrinkle in the universe’ becomes a bit clearer if we do this.

Posted by: John Baez on August 18, 2015 3:43 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Nobody seems to be nibbling on the puzzles. I think I’ll do the hardest one, Puzzle 2, before I forget how.

Suppose $\mathbf{G}$ is a 2-group and $B \mathbf{G}$ is the corresponding pointed connected homotopy 2-type. $\pi_1(B \mathbf{G})$ consists of isomorphism classes of objects of $\mathbf{G}$, while $\pi_2(B \mathbf{G})$ is the group of automorphisms of the unit object of $\mathbf{G}$. All the higher homotopy groups vanish.

Suppose $\mathbf{G} = Aut(Bij_n)$. Here $\pi_1(B \mathbf{G})$ consists of natural isomorphism classes of equivalences $f: Bij_n \to Bij_n$. These are the same as automorphisms $f: S_n \to S_n$ modulo inner automorphimsm. So, $\pi_1(B \mathbf{G})$ is the outer automorphism group of $S_n$. For $n = 6$, this is $\mathbf{Z}_2$.

On the other hand, $\pi_2(B \mathbf{G})$ consists of natural isomorphisms from the identity functor $1 : Bij_n \to Bij_n$ to itself. These are the same as elements $g \in S_n$ such that

$h = g h g^{-1} \quad \forall h \in S_n$

There’s only one, $g = 1$. So, $\pi_2(B \mathbf{G})$ is the trivial group.

In short, the 2-group $\mathbf{G} = Aut(Bij_n)$ corresponds to a pointed connected homotopy 2-type with $\pi_1 = \mathbb{Z}_2$ and $\pi_2$ trivial. So, it must be $\mathbb{R}P^\infty$, which is actually a homotopy 1-type.

Summary: if we take the automorphism 2-group of the groupoid of 6-element sets, and turn it into a homotopy 2-type, we get $\mathbb{R}P^\infty$.

Posted by: John Baez on August 18, 2015 4:08 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

I wrote:

There’s only one, $g=1$.

Whoops, that’s not always true! This inserts another wrinkle into the fabric of mathematics, connected to David’s comment.

Posted by: John Baez on August 18, 2015 4:14 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

That’s why I was thinking of centers.

If, as you said, $\pi_1(B \mathbf{G})$ is $\operatorname{Out}(G)$, and $\pi_2(B \mathbf{G})$ is $Z(G)$, the center, then can we reinterpret the exact sequence

$1 \to Z(G) \to G \to \operatorname{Aut}(G) \to \operatorname{Out}(G) \to 1$

as an exact homotopy sequence of a fibration?

Posted by: David Corfield on August 18, 2015 8:36 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

That sounds mighty plausible, but I’m not seeing the right fibration right now.

Posted by: John Baez on August 18, 2015 9:28 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Great! I think “the space of $K(G,1)$’s” is the concept I was missing.

Posted by: John Baez on August 20, 2015 1:09 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Let $EM(G,1) = B Aut(K(G,1))$ and $EM_\ast(G,1) = B Aut_\ast(K(G,1))$, where $Aut_\ast$ means automorphisms of a pointed space (i.e. those preserving the basepoint). Since the space of pointed maps $K(G,1) \to K(H,1)$ is equivalent to the space of group homomorphisms $G\to H$, we have $Aut_\ast(K(G,1)) \simeq Aut(G)$, so that $EM_\ast(G,1) \simeq K(Aut(G),1)$.

The notation is intended to suggest that $EM(G,1)$ is “the space of $K(G,1)$s” and $EM_\ast(G,1)$ is “the space of pointed $K(G,1)$s”. This implies there is a forgetful map $EM_\ast(G,1) \to EM(G,1)$. Moreover, since to make an unpointed $K(G,1)$ into a pointed one means precisely to choose an element of a $K(G,1)$, the fiber of this map should be a $K(G,1)$. Thus, we have a fiber sequence

$K(G,1) \to K(Aut(G),1) \to EM(G,1)$

and hence a long exact sequence

$1 \to \pi_2 EM(G,1) \to \pi_1(K(G,1)) \to \pi_1(Aut(G),1) \to \pi_1(EM(G,1)) \to 1$

i.e.

$1 \to \pi_2 EM(G,1) \to G \to Aut(G) \to \pi_1(EM(G,1)) \to 1.$

Finally, it’s not hard to identify $\pi_1(EM(G,1))$ with $Out(G)$ and $\pi_2 EM(G,1)$ with $Z(G)$. I wrote about this in HoTT here.

Posted by: Mike Shulman on August 19, 2015 8:44 PM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

This is all linked to the automorphism 2-group, isn’t it? A webpage on what we call 2-groups points to an example

The homomorphism j:G—>Aut(G) which sends elements of G to the corresponding inner automorphism can be viewed as a cat1-group (H,s,t) where H is the semi-direct product Gx|Aut(G) and s(g,a)=(1,a), t(g,a)=(1,j(g)a). Here $pi_1(H)=Out(G)$ is the outer automorphism group of $G$ and $pi_2(H)=Z(G)$ is the centre of $G$.

Posted by: David Corfield on August 18, 2015 8:56 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Puzzle 1: only the trivial homotopy type, unless $n=2$, when you get $B^2 S_2$.

I think for Puzzle 3 you get the product of the first two puzzles: $B^2S_2 \times \mathbb{RP}^\infty$.

Posted by: David Roberts on August 18, 2015 6:04 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

David wrote:

Puzzle 1: only the trivial homotopy type, unless $n=2$, when you get $B^2 S_2$.

Right! Great!

For some reason homotopy theorists never say $S_2$ here; they’d call this space $B^2 \mathbb{Z}_2$ — or even more commonly, since $\mathbb{Z}_2$ is being treated as a discrete group, $K(\mathbb{Z}_2, 2)$.

Puzzle 4. What does $K(\mathbb{Z}_2, 2)$ ‘look like’? In other words, is there a fairly pleasant concrete description of this space?

Of course we have $K(\mathbb{Z},2) = \mathbb{C}P^\infty$ — that may help.

I think for Puzzle 3 you get the product of the first two puzzles: $B^2S_2 \times \mathbb{R}P^\infty$.

Not quite!

Posted by: John Baez on August 18, 2015 9:22 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

One could take a model for $B\left(U(\mathcal{H})/(\mathbb{Z}/2)\right)$, since $U(\mathcal{H})/(\mathbb{Z}/2)$ is a $B\mathbb{Z}/2$. One can almost combine this with this MO answer giving a ‘natural’ $BPU(\mathcal{H})$, but I’m not so sure it all goes through.

Posted by: David Roberts on August 19, 2015 2:05 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Well, the only other option then is the truncation of the sphere spectrum to a 2-type!

Posted by: David Roberts on August 18, 2015 1:27 PM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

John wrote:

Puzzle 3. Give a concrete description of the 2-group $Aut(Bij)$, up to equivalence. What is the corresponding pointed connected homotopy 2-type?

David replied, almost correctly:

I think for Puzzle 3 you get the product of the first two puzzles: $B^2S_2 \times \mathbb{RP}^\infty$.

John replied:

Not quite!

Let me say a bit about why. The groupoid of finite sets and bijections is the coproduct

$Bij \simeq \sum_{n = 0}^\infty Bij_n$

where $Bij_n$ is the groupoid of $n$-element sets and bijections between these.

Assume that every equivalence of groupoids $f: Bij \to Bij$ maps each $Bij_n$ to itself. If this were true, we’d have

$Aut(Bij) \simeq \prod_{n = 0}^\infty Aut(Bij_n)$

and turning each of these 2-groups $Aut(Bij)$ into a connected pointed homotopy 2-type, say $|Aut(Bij)|$, we’d get

$|Aut(Bij)| \simeq \prod_{n = 0}^\infty |Aut(Bij_n)|$

We’ve seen that

$|Aut(Bij_n)| = \left\{ \begin{array}{cl} \ast & n \ne 2,6 \\ K(\mathbb{Z}/2,2) & n = 2 \\ K(\mathbb{Z}/2,1) & n = 6 \end{array} \right.$

So, given that assumption, we’d get

$|Aut(Bij)| = K(\mathbb{Z}/2,2) \times \K(\mathbb{Z}/2,1)$

just as David suggested (though he expressed in it different symbols).

However, I think that assumption isn’t true! So we need to correct this calculation a bit. There’s one more wrinkle in the mathematical universe… or at least in the groupoid of finite sets.

Posted by: John Baez on August 19, 2015 3:40 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Oh, is is due to the fact $Bij_0 \simeq Bij_1 \simeq 1$? That will mess things up a little.

Posted by: David Roberts on August 19, 2015 6:54 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Yes, the groupoid of 0-element sets is equivalent to the groupoid of 1-element sets, so there are autoequivalences of $Bij$ that switch them.

We can see this clearly if we write $E$ for the groupoid of finite sets and $\frac{1}{n!}$ for the groupoid of $n$-element sets. Then

$E = 1 + 1 + \frac{1}{2!} + \frac{1}{3!} + \cdots$

and there’s a symmetry that switches the first two terms. Funny, eh?

Posted by: John Baez on August 19, 2015 7:09 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

This extra symmetry puts me in mind of something I was aiming at in the old Klein 2-geometry days, where there is a collection of identical groups, $G$, and one looks for a kind of wreath product of the permutation group of the collection with $Aut(G)$.

Here it was. There’s John speculating about a “more categorified wreathe product”. Is that any clearer how to do it 8 years later?

Posted by: David Corfield on August 23, 2015 10:14 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Posted by: John Baez on August 26, 2015 2:47 PM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

David Roberts wrote on G+:

“Given that there are two homomorphisms $S_6 \to S_{15}$, is the outer automorphism given by an inner automorphism of $S_{15}$ that preserves $S_6$?”

To make things concrete, suppose we number all the duads 1 to 15, and all the synthemes 1 to 15, in some arbitrary fashion. We then have two specific homomorphisms $D:S_6 \to S_{15}$ and $S:S_6 \to S_{15}$.

We also have the homomorphism $TC:S_6 \to Symm(\text{Tutte-Coxeter graph})$, where $TC(S_6)$ consists of all 720 symmetries of the Tutte-Coxeter graph that leave duads as duads and synthemes as synthemes.

Now, suppose $G$ is any symmetry of the Tutte-Coxeter graph that exchanges duads and synthemes, and define $G_X$, $G_Y$ as the elements of $S_{15}$ such that:

$G_X(i) =\text{the syntheme number of} \, G(duad \, i)$

$G_Y(i) = \text{the duad number of} \, G(syntheme \, i)$

Given any permutation $p$ in $S_6$, since $G$ exchanges duads and synthemes and $TC(p)$ does not, the graph symmetry:

$G^{-1} TC(p) G$

does not exchange duads and synthemes, so there exists some $q$ in $S_6$ such that:

$TC(q) = G^{-1} TC(p) G$

If we project out the individual effects of the two sides of this equation on duad numbers and syntheme numbers respectively, it tells us that:

$D(q) = G_X^{-1} S(p) G_X$

$S(q) = G_Y^{-1} D(p) G_Y$

In other words, since $p$ is arbitrary:

$D(S_6) = G_X^{-1} S(S_6) G_X$

So $G_X$ gives an inner automorphism of $S_{15}$ that maps one inclusion of $S_6$ in $S_{15}$, $S(S_6)$, into the other one, $D(S_6)$.

Posted by: Greg Egan on August 18, 2015 8:04 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

I should add that the homomorphisms $D:S_6 \to S_{15}$ and $S:S_6 \to S_{15}$ are injective, so you can construct an outer automorphism $F: S_6 \to S_6$ as:

$F(p) = D^{-1}(G_X^{-1} S(p) G_X)$

Posted by: Greg Egan on August 18, 2015 8:42 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Thanks, Greg! We could also ask how this looks from the point of view of the two copies of $Sp(4,\mathbb{F}_2)$ (which are isomorphic to $S_6$) sitting inside $GL(4,\mathbb{F}_2)=PGL(4,\mathbb{F}_2)$ which at the least is a subgroup of $S_{15}$, if not iso to it (on my phone and can’t look up what I wrote earlier!)

Posted by: David Roberts on August 18, 2015 10:40 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

I believe that $S_6$ is also isomorphic to $O(8,\mathbb{F}_2)$. This is presumably because the difference between a symplectic structure and an orthogonal structure evaporates in characteristic 2. I also hope David is using the convention where $Sp(4,\mathbb{F}_2)$ means the group of transformations preserving a nondegenerate alternating form on an 8-dimensional vector space over $\mathbb{F}_2$, not a 4-dimensional one. Then $Sp(4,\mathbb{F}_2)$ would be just another name for $O(8,\mathbb{F}_2)$.

(There are a lot of subtleties in characteristic 2: for example, there’s a difference between antisymmetric and alternating bilinear forms, and I believe there are also 2 kinds of orthogonal structure on an even-dimensional vector space over $\mathbb{F}_2$. So, I could be mixed up here.)

Posted by: John Baez on August 18, 2015 11:59 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Sadly it has to be a 4-dimensional vector space, as it’s a subgroup of $GL(4,F_2)$. I got it from here: http://www.savbb.sk/~karabas/edu/sschool14/material/PermGpsNotes.pdf

Posted by: David Roberts on August 18, 2015 12:54 PM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Here is how $S_6$ is isomorphic to $PSp(4, \mathbb{F}_2)$.

Take a $6$-dimensional vector space over $\mathbb{F}_2$: say sextuples of ones and zeros. Take the subspace orthogonal to $\left(1,1,1,1,1,1\right)$, getting a $5$-dimensional space, consisting of sextuples with an even number of ones. Then take the quotient by addition of $\left(1,1,1,1,1,1\right)$, getting a $4$-dimensional space.

Now think of this in another way: the first space is the set of subsets of a set $S$ of $6$ elements corresponding to coordinates (for each each vector, take the subset of coordinates consisting of those with a component value of $1$). The $5$-dimensional subspace consists of subsets of even cardinality. The quotient identifies subsets with their complements. Sum of vectors is symmetric difference of sets.

There are two types of element in the $4$-dimensional space: the identity, consisting of the equivalence class $\left\{\left(0,0,0,0,0,0\right),\left(1,1,1,1,1,1\right)\right\}$, or, in the alternative formulation the empty set together the whole of $S$; and the other elements, all of which are two-element equivalence classes containing a $2$-element subset of $S$ together with its $4$-element complement.

If we go over to projective space, all we do is throw away the identity (because $\mathbb{F}_2$ only has one non-zero scalar, making everything very simple). This gives us a space whose points are duads.

The vector sum of two duads depends on their overlap. Letting a, b, c, d, e and f be the elements of $S$ in some order:

$(ab)+(ab)$ is the identity, e.g.$\left(1,1,0,0,0,0\right)+\left(1,1,0,0,0,0\right)=\left(0,0,0,0,0,0\right)$.

$(ab)+(bc)=0$, e.g. $\left(1,1,0,0,0,0\right)+\left(0,1,1,0,0,0\right)=\left(1,0,1,0,0,0\right)$. (EDIT: should be $(ab) + (bc) = (ac)$)

$(ab)+(cd)=(ef)$, e.g. $\left(1,1,0,0,0,0\right)+\left(0,0,1,1,0,0\right)=\left(1,1,1,1,0,0\right)\sim\left(0,0,0,0,1,1\right)$.

Then, in the projective space there are two kinds of line, of the form $\left\{(ab),(bc),(ac)\right\}$ and $\left\{(ab),(cd),(ef)\right\}$ respectively.

Now there is a symplectic structure on the duads given by parity of overlap

$<(ab),(ab)>=0$
$<(ab),(bc)>=1$
$<(ab),(cd)>=0$

An example symplectic basis would be $\left\{(ab),(bc),(de),(ef)\right\}$.

Since both the vector space (or projective space) structure and the symplectic structure depend only on the overlap between pairs, it’s not hard to see that the symplectic group must be $S_6$. The two types of lines are the non-isotropic and isotropic lines respectively. (That is, lines of the form $\left\{(ab),(bc),(ac)\right\}$ have all their points mutually non-orthogonal, and lines of the form $\left\{(ab),(cd),(ef)\right\}$ have all their points mutually orthogonal.)

Thus (necessarily isotropic) points are duads, and isotropic lines are synthemes, explaining why the duads and synthemes with the obvious incidence relation form a generalised quadrangle.

It’s a bit tedious to work out, but planes consist of a distinguished duad, together with the $6$ duads formed from the $4$ elements of $S$ not lying in the distinguished duad. The distinguished duad is then the pole of the plane (i.e. the unique point orthogonal to everything in the plane).

These planes are, of course, Fano planes.

Posted by: Tim Silverman on August 18, 2015 11:25 PM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Due to the Dynkin diagram isomorphism $B_2=C_2$, there is a generic exceptional isomorphism $PSp(4,q)\simeq P\Omega(5,q)$ where $P\Omega$ is the simple orthogonal group (which may or may not be $PSO(5,q)$.)

This can be seen as a folding of the Klein Correspondence—that is, of the Dynkin diagram isomorphism $A_3=D_3$—as follows (at lightning speed):

$A_3$ corresponds to $PSL(4,q)$. Take the $4$-dimensional vector space on which this naturally acts (e.g. from the left). The bivectors of this space form a $6$-dimensional vector space and $PSL(4,q)$ preserves an orthogonal structure on this space given by wedging bivectors together. In fact, $PSL(4,q)\simeq P\Omega^+(6,q)$. (Where the $+$ picks out one of the two possible types of orthogonal structure.)

The bivectors give bilinear forms on $\mathbb{F}_q^4$ by wedging vectors on left and right, which (except for the zero bivector) are either:

i) degenerate of rank 2, in which case the bivector is the wedge of two vectors and hence corresponds to a plane (or, projectively, a line);

or

ii) non-degenerate, in which case the bilinear form is a symplectic form.

Up in $6$-dimensional orthogonal space, these two possibilities turn out to correspond to isotropic and non-isotropic vectors (or, projectively, points). There are lots of other correspondences: for our purposes, the most important are:

(a) two mutually orthogonal isotropic points in projective $5$-space correspond to intersecting lines in projective $3$-space; hence an isotropic line (consisting entirely of mutually orthogonal isotropic points) corresponds to a plane pencil of projective $3$-space: all the lines lying in a particular plane and passing through a particular point.

(b) a non-isotropic and an isotropic point which are mutually orthogonal in projective $5$-space correspond to a symplectic structure and one of its isotropic lines in projective $3$-space.

Now, we get from an orthogonal projective $5$-space to an orthogonal projective $4$-space by fixing a non-isotropic point and taking its perpendicular space. As a non-isotropic point corresponds to a symplectic structure in projective $3$-space, this “explains” why we have $PSp(4,q)\simeq P\Omega(5,q)$.

Now, let us look at orthogonal projective $5$-space and take the perpendicular space to one of its non-isotropic points, and look at the isotropic points and lines of this perpendicular space.

By the Klein correspondence and remark (b) above, the isotropic points of orthogonal projective $4$-space correspond to isotropic lines of the symplectic projective $3$-space.

From remark (a) the isotropic lines of orthogonal projective $4$-space then correspond to a plane pencil of isotropic lines of symplectic projective $3$-space. But then their point of intersection is just the pole of the plane they lie in (because, since the lines are isotropic, the point is orthogonal to all the points of all the lines, i.e. all the points of the plane). Each pole is the pole of just one plane, so we get a bijection between isotropic lines of orthogonal projective $4$-space and all points (which are all isotropic) of symplectic projective $3$-space.

So the folded Klein correspondence matches orthogonal isotropic planes to symplective isotropic lines and vice versa. So the generalised quadrangles coming from the orthogonal and symplectic sides of the isomorphism are dual to each other.

Posted by: Tim Silverman on August 19, 2015 12:11 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

These days I teeter on the brink between loving finite simple groups and finite geometries and thinking “oh, there’s so much to learn, and so much is already known, that there’s no point in bothering—I’ll never get anywhere near the frontier”. I think it all depends on whether I meet new information at a rate I can handle, or too fast.

Your first comment on this thread made me love the relation between $S_6$ and $PSp(2,\mathbb{F}_2)$: all those lovely duads and synthemes show up so nicely starting from linear algebra over $\mathbb{F}_2$, and at the end we bump into the Fano plane again!

This comment, alas, provoked the other reaction. Maybe it shouldn’t have, since a big ingredient seems to be an $\mathbb{F}_q$ version of the Klein correspondence, which I already love in its guise

$Spin(6) \cong SU(4)$

But somehow there was a bit too much for me, too fast. I think I should just learn to avert my eyes at certain moments.

Posted by: John Baez on August 19, 2015 4:55 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Very nice, Tim!

One thing I don’t follow:

(ab)+(bc)=0, e.g. (1,1,0,0,0,0)+(0,1,1,0,0,0)=(1,0,1,0,0,0).

Why is the RHS of the first equation here 0, rather than (ac)?

Posted by: Greg Egan on August 19, 2015 12:22 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Oops, that was a typo caused by doing things too fast late at night. The sum is, indeed, $(ac)$.

Posted by: Tim Silverman on August 19, 2015 1:03 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

This is terrific stuff! I’ve been starting to read your notes to me again.

Posted by: John Baez on August 19, 2015 3:45 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Thanks!

Be warned—I think there are a few dreadful typos in my notes too.

I used to assiduously proofread everything I wrote, including blog comments, but I’ve got into a hurry in my old age.

Posted by: Tim Silverman on August 19, 2015 9:16 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Now in characteristic $2$, we have an additional, completely different correspondence between $PSp(4,q)\simeq P\Omega(5,q)$.

In fact we have a general correspondence between $PSp(2n,q)\simeq P\Omega(2n+1,q)$, corresponding to an isomorphism between the Dynkin diagrams $C_n$ and $B_n$ (which is more familiar in characteristic $1$).

Very crudely: since, in characteristic $2$, $1=-1$, it is also true that symmetric and antisymmetric bilinear forms become the same thing.

Less crudely: fix a quadratic form $Q$. We then get a bilinear form $B$ via

$B(u,v)=Q(u+v)-Q(u)-Q(v)$

Generally, $B$ will obviously be symmetric.

We then have

$\array{\arrayopts{\collayout{left}} B(u,u)&=Q(u+u)-Q(u)-Q(u)\\ &=Q(2u)-2Q(u)\\ &=4Q(u)-2Q(u)\\ &=2Q(u)}$

If we are not in characteristic $2$, we can use this to go the other way:

$Q(u)=\frac{1}{2}B(u,u)$.

We turn out to get a bijection between non-degenerate quadratic forms and non-degenerate symmetric bilinear forms.

But in characteristic $2$, we have, from the above,

$B(u,u)=0$

So, in this case, $B$, if non-degenerate, is a symplectic form. Also, since we can’t go back from $B$ to $Q$, our bijection can’t be constructed and a symplectic form may (and will) correspond to multiple different quadratic forms.

Now in a vector space of odd dimensions $2n+1$, and in characteristic $2$, we must have that the bilinear form corresponding to a quadratic form is degenerate: there is a $1$-dimensional subspace of vectors orthogonal to everything. But the quadratic form need not be zero on this subspace; in fact, we don’t want it to be (the quadratic form should be “degenerate but not singular”), otherwise we just have a boring degenerate case.

Now, orthogonal transformations of the full odd-dimensional space must preserve the bilinear form, and hence must preserve the subspace. But taking the quotient by this subspace reduces the bilinear form to a perfectly good non-degenerate symplectic form on a space one dimension lower. And it turns out that symplectic transformations of this space lift uniquely to orthogonal transformations of the full space.

However, since the quadratic form is non-zero on the special subspace, the isotropic points and lines of the quadratic form correspond precisely to the isotropic points and lines of the symplectic form on the quotient space. So we get, for instance, a new isomorphism between $PSp(4,q)\simeq P\Omega(5,q)$ in the characteristic $2$ case, that gives not dual generalised quadrangles but the same generalised quadrangles.

So, we can go up one isomorphism and down the other to find that the generalised quadrangle is self-dual, with an outer automorphism of $PSp(4,q)$ that corresponds to an interchange of the points and lines of the generalised quadrangle.

Posted by: Tim Silverman on August 19, 2015 1:01 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

John wrote:

I believe that $S_6$ is also isomorphic to $O(8,\mathbb{F}_2)$. This is presumably because the difference between a symplectic structure and an orthogonal structure evaporates in characteristic 2. I also hope David is using the convention where $Sp(4,\mathbb{F}_2)$ means the group of transformations preserving a nondegenerate alternating form on an 8-dimensional vector space over $\mathbb{F}_2$, not a 4-dimensional one. Then $Sp(4,\mathbb{F}_2)$ would be just another name for $O(8,\mathbb{F}_2)$.

David wrote:

Sadly it has to be a 4-dimensional vector space.

Oh! I got my information from something Tim Silverman wrote, and I misread it. He actually said

$S_6 \cong O(5,\mathbb{F}_2)$

So, presumably this related to the things he said here:

$PSp(4,\mathbb{F}_2) \cong S_6$

and

$PSp(4,\mathbb{F}_2) \cong P\Omega(5,\mathbb{F}_2).$

The $\Omega$ is apparently some sort of sophisticated version of the letter $O$ that only experts on finite fields know how to use.

But this is something even I understand:

$PSp(4,\mathbb{C}) \cong SO(5,\mathbb{C})$

It comes from taking $\mathbb{C}^4$ with a symplectic structure $\omega$ on it, using this to create a volume form $vol$ and a Hodge star operator

$\star : \Lambda^2\mathbb{C}^4 \to \Lambda^2\mathbb{C}^4$

in the way we more ordinarily do with a metric, and then creating an inner product on $\Lambda^2\mathbb{C}^4$ by

$g(\mu, \nu) vol = \mu \wedge \star \nu$

The symplectic transformations of $\mathbb{C}^4$ obviously preserve this inner product on $\Lambda^2\mathbb{C}^4$, and they preserve the (dualized) symplectic structure $\omega \in \Lambda^2\mathbb{C}^4$, so they preserve the 5-dimensional subspace of $\Lambda^2\mathbb{C}^4$ that’s orthogonal to $\omega$. This gives a homomorphism from $Sp(4,\mathbb{C})$ to $O(5,\mathbb{C})$, which is two-to-one and half-onto. So, we get an isomorphism

$PSp(4,\mathbb{C}) \cong SO(5,\mathbb{C})$

Something roughly like this should work over any field, but characteristic 2 still gives me the creeps.

Posted by: John Baez on August 19, 2015 6:50 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

I got as far as a brute-force computerised enumeration of $4\times 4$ matrices $M$ over $\mathbb{F}_2$ such that:

$M^T \Omega M = \Omega$

where:

$\Omega = \left(\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \end{array}\right)$

There are 720 such $M$.

Posted by: Greg Egan on August 18, 2015 12:31 PM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Nice!

Posted by: John Baez on August 19, 2015 3:53 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

People might be interested in the notion of a complete group (which I learned about from my student Nima Rasekh). A group $G$ is complete exactly if $\mathrm{Center}(G)$ and $\mathrm{Out}(G)$ are trivial, i.e., if the space of self-homotopy-equivalences of $BG$ is contractible.

There is a fair amount of literature on complete groups. Apparently, there are quite a few finite complete groups around, in addition to (most of) the $S_n$s.

Here is one neat fact about complete groups (see this). Consider the map $G\to \mathrm{Aut}(G)$. We can iterate this, obtaining a sequence $\G\to \mathrm{Aut}(G) \to \mathrm{Aut}(\mathrm{Aut})(G)\to \dots$. The theorem is that this sequence eventually stabilizes (possibly after transfinite iterations), and therefore arrives at a complete group.

In particular, if $G$ has trivial center, then $G\to \mathrm{Aut}(G)$ is injective and $\mathrm{Aut}(G)$ turns out to have trivial center as well. Therefore, every centerless group embeds in a complete group. (If $G$ is a finite centerless group the sequence will stabilize after a finite number of steps, so every finite centerless group is contained in a finite complete group.)

Posted by: Charles Rezk on August 19, 2015 4:50 PM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Do you think there is a similar theorem for spaces or oo-groupoids?

Posted by: Egbert Rijke on August 19, 2015 7:12 PM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

$G \to Aut(G) \to Aut(Aut(G)) \to \cdots$

but I’d forgotten all the details. Joel David Hamkins’ paper is full of fun stuff. Since I’m into finite groups on this thread, it’s nice to know that for a centerless group $G$, $Aut(G)$ is also centerless, so we have inclusions

$G \hookrightarrow Aut(G) \hookrightarrow Aut(Aut(G)) \hookrightarrow \cdots$

and in the case where $G$ is finite, these stabilize after finitely many steps.

It’s also nice to see the example of $G = D_4$ is the 8-element dihedral group: the symmetry group of a square, which has already featured in this thread. This does have a nontrivial center, so the natural map $G \to Aut(G)$ is not an isomorphism, yet we have $Aut(G) \cong G$. This is a case where we need to go ‘forever’ (more precisely, to step $\omega+1$) for this process of repeatedly taking $Aut$ to stabilize.

For a certain kind of person, it might be a lot of fun to consider the analogous questions for topological groups, or ‘$\infty$-groups’, or $A_\infty$ spaces. Personally I’m more interested in juicy examples of what happens when you iterate the $Aut$ construction than the general theory of when it stabilizes.

Posted by: John Baez on August 20, 2015 3:36 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

We ran into “complete groups” (what a terrible name) on the HoTT list when wondering about what can be embedded into the universe of types. Since classically the universe of types is $\coprod_X B Aut(X)$, if $G$ is complete then the component corresponding to $B G$ is contractible. Thus, however many complete groups you can find, you can embed that size of discrete set into the universe.

Posted by: Mike Shulman on August 20, 2015 9:46 PM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Since we came very close to answering Puzzle 3, but didn’t quite make it, let me record the answer here, for posterity.

Let $Bij$ be the groupoid of finite sets and bijections. Let $Aut(Bij)$ be the 2-group of autoequivalences $f: Bij \to Bij$ and natural isomorphisms between these. Let $|Aut(Bij)|$ be the corresponding connected pointed homotopy 2-type: that is, the one whose fundamental 2-group is $Aut(Bij)$.

Then:

$Aut(Bij) \simeq K(\mathbb{Z}/2,1) \times K(\mathbb{Z}/2,1) \times K(\mathbb{Z}/2,2)$

The first factor of $K(\mathbb{Z}/2,1)$ comes from the autoequivalence $f: Bij \to Bij$ that switches the groupoid of 0-element sets and the groupoid of 1-element sets:

$S_0 \cong S_1$

The second factor of $K(\mathbb{Z}/2,1)$ comes from the autoequivalence of the groupoid of 6-element sets that is not naturally isomorphic to the identity:

$Out(S_6) \cong \mathbb{Z}/2$

The factor of $K(\mathbb{Z}/2,2)$ comes from the nontrivial natural automorphism of the identity functor on the groupoid of 2-element sets:

$Z(S_2) \cong \mathbb{Z}/2$

Posted by: John Baez on August 20, 2015 4:15 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

Here are some more puzzles, some of which I don’t know the answers to. I suspect Tim Silverman could help out, since they might be related to ‘generalized quadrangles’:

Puzzle 1: How many cycles of length 8 does the Tutte–Coxeter graph contain?

Puzzle 2: What do these cycles correspond to in the Cremona–Richmond configuation?

Puzzle 3: What do these cycles correspond to in terms of duads, synthemes or other features of the 6-element set?

I think I can answer Puzzle 2. The red and blue vertices of the Tutte–Coxeter graph correspond to the points and lines of the Cremona–Richmond configuration, which is also called the generalized quadrangle $GQ(2,2)$:

The shortest cycles in the Tutte–Coxeter graph have length 8, touring four red and four blue vertices in an alternating red/blue/red/blue/red/blue/red/blue way. These should correspond to quadrangles in the Cremona–Richmond configuration.

And specially for Tim:

Puzzle 4: What do these cycles of length 8 in the Tutte–Coxeter graph correspond to in terms of a 4-dimensional symplectic vector space over $\mathbb{F}_2$?

Posted by: John Baez on August 20, 2015 5:35 AM | Permalink | Reply to this

### Re: A Wrinkle in the Mathematical Universe

OK, these cycles should indeed correspond to ordinary (non-generalised) quadrangles in GQ(2,2). I.e. we have squares whose vertices are duads and whose edges are synthemes with the obvious incidence relation.

Suppose the top left duad is $(ab)$. Its neighbours should share different synthemes with it. This means that they must be

a) disjoint from $(ab)$

but

b) not disjoint from each other

(because otherwise the top left duad together with its two neighbours would all lie in a single syntheme).

So there is a point shared between these diagonally opposite duads. For the sake of concreteness, let’s say they are $(cd)$ and $(ce)$.

The bottom right duad, opposite $(ab)$, also has these as neighbours, and so must be disjoint from them. That leaves $a$, $b$ and $f$ for it to be drawn from, so it must share a point with the top left duad. Fixing this shared point then forces the other member of the duad. E.g. if the shared point is $a$, the duad must be $(af)$.

To summarise, each diagonal defines a point, shared between the two duads on the diagonal. Here, the points are $a$ and $c$. Then each diagonal defines a pair of points, giving the other members of the corresponding duads. So here, we have a pair $(bf)$ corresponding to the $a$ diagonal, and $(de)$ corresponding to the $c$ diagonal. These must be disjoint because neighbouring duads are disjoint. So generally, each quadrangle gives rise to a syntheme, here $(ac)(bf)(de)$.

Conversely, given a syntheme, we have $3$ possibilities as to which duad defines the diagonals, and then $2$ possibilites as to which of the other two duads corresponds to which diagonal. I guess that gives $15\times6=90$ cycles.

Posted by: Tim Silverman on August 22, 2015 9:39 PM | Permalink | Reply to this

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