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August 17, 2015

A Wrinkle in the Mathematical Universe

Posted by John Baez

Of all the permutation groups, only S 6S_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 nBij_n be the groupoid of nn-element sets and bijections between these. Only for n=6n = 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 6S_6 has an outer isomorphism.

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

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

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

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

By calling it a 2-group, I mean that Aut(X)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 nBij_n, the groupoid of nn-element sets and bijections between them. Up to equivalence, we can describe Aut(Bij n)Aut(Bij_n) as follows. The objects are just automorphisms of S nS_n, while a morphism from an automorphism f:S nS nf: S_n \to S_n to an automorphism f:S nS nf' : S_n \to S_n is an element gS ng \in S_n that conjugates one automorphism to give the other:

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

So, if all automorphisms of S nS_n are inner, all objects of Aut(Bij n)Aut(Bij_n) are isomorphic to the unit object, and thus to each other.

Puzzle 1. For n6n \ne 6, all automorphisms of S nS_n are inner. What are the connected pointed homotopy 2-types corresponding to Aut(Bij n)Aut(Bij_n) in these cases?

Puzzle 2. The permutation group S 6S_6 has an outer automorphism of order 2, and indeed Out(S 6)= 2.Out(S_6) = \mathbb{Z}_2. What is the connected pointed homotopy 2-type corresponding to Aut(Bij 6)Aut(Bij_6)?

Puzzle 3. Let BijBij be the groupoid where objects are finite sets and morphisms are bijections. BijBij is the coproduct of all the groupoids Bij nBij_n where n0n \ge 0:

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

Give a concrete description of the 2-group Aut(Bij)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 6S_6 using something called the Tutte–Coxeter graph.

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

To do this, he made the following definitions:

• A duad is a 2-element subset of SS. Note that there are 6choose2=15 {6 \choose 2} = 15 duads.

• A syntheme is a set of 3 duads forming a partition of SS. 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 SS gives a permutation of the set TT of synthematic totals, so we obtain an action of S 6S_6 on TT. Choosing any bijection betweeen SS and TT, this in turn gives an action of S 6S_6 on SS, and thus a homomorphism from S 6S_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 6S_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 6S_6, also give symmetries: for each element of S 6S_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 Aut(S 6)\mathrm{Aut}(S_6) has

2×6!=1440 2 \times 6! = 1440

elements, coming from the 6!6! inner automorphisms of S 6S_6 and the outer automorphism of order 2. In fact, Aut(S 6)\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

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Re: A Wrinkle in the Mathematical Universe

That got me trying to remember why the center of S nS_n for n>2n \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>2n \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 2S_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 2S_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 nS_n and SU(N)SU(N), this picture has a nice application to 2-dimensional Yang-Mills theory with gauge group SU(N)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)(94), I get two cycles, with the roles of 99 and 44 swapped:

(94)(2)(7)(91011568)(134)(94)=(2)(7)(41011568)(139) (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 𝒢\mathcal{G} be the category of finite graphs and isomorphisms. (We consider only undirected graphs without loops of multiple edges.) We define functors

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

as follows:

  • A dyad in a set XX is a subset dXd\subseteq X with |d|=2|d|=2. We let DXD X be the graph whose vertices are the dyads in XX, with an edge from pp to pp' iff pp=p\cap p'=\emptyset.
  • More generally, a dyad in a graph GG 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 GG is a set pvert(G)p\subseteq\text{vert}(G) such that |p|=5|p|=5 and G| pG|_p is discrete. We let PGP G denote the set of pentads in GG.
  • A triangle in a graph GG is a set tvert(G)t\subseteq\text{vert}(G) such that |t|=3|t|=3 and G| tG|_t is complete. We let TGT G be the graph whose vertices are the triangles in GG, with an edge from tt to tt' iff ttt\cap t'\neq\emptyset. (Note that this is opposite to the rule for dyads.)

Now let 𝒢 1𝒢\mathcal{G}_1\subset\mathcal{G} denote the category of graphs with the following properties:

  • GG has 1515 vertices, all of valence 66.
  • If vv and ww are distinct vertices that are not joined by an edge, then there is a unique pentad pPGp\in PG with {v,w}p\{v,w\}\subseteq p.

We also let 1\mathcal{F}_1\subset\mathcal{F} denote the category of sets of size 66.

Theorem:

The functors defined above restrict to give equivalences

(2) 1D𝒢 1T𝒢 1P 1. \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 21:𝒢 1𝒢 1T^2\simeq 1\colon\mathcal{G}_1\to\mathcal{G}_1, and DD and PP are inverse to each other.

However, TT is not equivalent to the identity on 𝒢 1\mathcal{G}_1, and PTDP T D is not equivalent to the identity on 1\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(𝔽 5)PGL_2(\mathbb{F}_5) and PSL 2(𝔽 9)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 GG. If this acts on some geometrical structure, any ‘figure’ in this structure will be stabilized by some subgroup of GG, and Klein’s clever idea was to define ‘figures’ as subgroups of GG.

We can use this idea to define the Tutte–Coxeter graph in a way that makes it obviously have Aut(S 6)Aut(S_6) as symmetries… though it takes real work to see that this definition gives the picture here:

Let GG be a group abstractly isomorphic to S 6S_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!

GG has 30 subgroups isomorphic to S 4×S 2S_4 \times S_2. These will be the vertices of our graph.

Given two such subgroups, HH and HH', we draw an edge between the corresponding vertices iff HHH \cap H' is isomorphic to D 4×S 2D_4 \times S_2, where D 4D_4 is the dihedral group that acts as symmetries of the square.

The resulting graph is obviously acted on by Aut(G)Aut(G). What’s not obvious is that it looks like the picture above. For this we need some facts about subgroups of S 6S_6.

The 30 subgroups of GG isomorphic to S 4×S 2S_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 GG 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 GG with a group of permutations of a 6-element set!

If HH is the stabilizer of a duad, and HH' is the stabilizer of a syntheme, and the duad lies in the syntheme, then HHH \cap H' will be isomorphic to D 4×S 2D_4 \times S_2. So, this is why we put an edge between two vertices in this case.

In fact, GG has exactly 45 subgroups isomorphic to D 4×S 2D_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×S 2D_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 6S_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×S 2S_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×S 2D_4 \times S_2. Moreover, this stabilizer lets you reconstruct the duad (the two points interchanged by S 2S_2) and the syntheme (consisting of this duad and 4 more points, on which D 4D_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 nS_n and more about the groupoid of nn-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 G\mathbf{G} is a 2-group and BGB \mathbf{G} is the corresponding pointed connected homotopy 2-type. π 1(BG)\pi_1(B \mathbf{G}) consists of isomorphism classes of objects of G\mathbf{G}, while π 2(BG)\pi_2(B \mathbf{G}) is the group of automorphisms of the unit object of G\mathbf{G}. All the higher homotopy groups vanish.

Suppose G=Aut(Bij n)\mathbf{G} = Aut(Bij_n). Here π 1(BG)\pi_1(B \mathbf{G}) consists of natural isomorphism classes of equivalences f:Bij nBij nf: Bij_n \to Bij_n. These are the same as automorphisms f:S nS nf: S_n \to S_n modulo inner automorphimsm. So, π 1(BG)\pi_1(B \mathbf{G}) is the outer automorphism group of S nS_n. For n=6n = 6, this is Z 2\mathbf{Z}_2.

On the other hand, π 2(BG)\pi_2(B \mathbf{G}) consists of natural isomorphisms from the identity functor 1:Bij nBij n1 : Bij_n \to Bij_n to itself. These are the same as elements gS ng \in S_n such that

h=ghg 1hS n h = g h g^{-1} \quad \forall h \in S_n

There’s only one, g=1g = 1. So, π 2(BG)\pi_2(B \mathbf{G}) is the trivial group.

In short, the 2-group G=Aut(Bij n)\mathbf{G} = Aut(Bij_n) corresponds to a pointed connected homotopy 2-type with π 1= 2\pi_1 = \mathbb{Z}_2 and π 2\pi_2 trivial. So, it must be P \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 P \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=1g=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, π 1(BG)\pi_1(B \mathbf{G}) is Out(G)\operatorname{Out}(G), and π 2(BG)\pi_2(B \mathbf{G}) is Z(G)Z(G), the center, then can we reinterpret the exact sequence

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

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

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

and hence a long exact sequence

1π 2EM(G,1)π 1(K(G,1))π 1(Aut(G),1)π 1(EM(G,1))11 \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.

1π 2EM(G,1)GAut(G)π 1(EM(G,1))1.1 \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 π 1(EM(G,1))\pi_1(EM(G,1)) with Out(G)Out(G) and π 2EM(G,1)\pi_2 EM(G,1) with Z(G)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)pi_1(H)=Out(G) is the outer automorphism group of GG and pi 2(H)=Z(G)pi_2(H)=Z(G) is the centre of GG.

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=2n=2, when you get B 2S 2B^2 S_2.

I think for Puzzle 3 you get the product of the first two puzzles: B 2S 2×ℝℙ 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=2n=2, when you get B 2S 2B^2 S_2.

Right! Great!

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

Puzzle 4. What does K( 2,2)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(,2)=P 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×P 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(U()/(/2))B\left(U(\mathcal{H})/(\mathbb{Z}/2)\right), since U()/(/2)U(\mathcal{H})/(\mathbb{Z}/2) is a B/2B\mathbb{Z}/2. One can almost combine this with this MO answer giving a ‘natural’ BPU()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)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×ℝℙ 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 n=0 Bij n Bij \simeq \sum_{n = 0}^\infty Bij_n

where Bij nBij_n is the groupoid of nn-element sets and bijections between these.

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

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

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

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

We’ve seen that

|Aut(Bij n)|={* n2,6 K(/2,2) n=2 K(/2,1) n=6 |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(/2,2)×K(/2,1) |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 0Bij 11Bij_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 BijBij that switch them.

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

E=1+1+12!+13!+ 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, GG, and one looks for a kind of wreath product of the permutation group of the collection with Aut(G)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

I haven’t thought about it….

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 6S 15S_6 \to S_{15}, is the outer automorphism given by an inner automorphism of S 15S_{15} that preserves S 6S_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 6S 15D:S_6 \to S_{15} and S:S 6S 15S:S_6 \to S_{15}.

We also have the homomorphism TC:S 6Symm(Tutte-Coxeter graph)TC:S_6 \to Symm(\text{Tutte-Coxeter graph}), where TC(S 6)TC(S_6) consists of all 720 symmetries of the Tutte-Coxeter graph that leave duads as duads and synthemes as synthemes.

Now, suppose GG is any symmetry of the Tutte-Coxeter graph that exchanges duads and synthemes, and define G XG_X, G YG_Y as the elements of S 15S_{15} such that:

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

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

Given any permutation pp in S 6S_6, since GG exchanges duads and synthemes and TC(p)TC(p) does not, the graph symmetry:

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

does not exchange duads and synthemes, so there exists some qq in S 6S_6 such that:

TC(q)=G 1TC(p)GTC(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 1S(p)G XD(q) = G_X^{-1} S(p) G_X

S(q)=G Y 1D(p)G YS(q) = G_Y^{-1} D(p) G_Y

In other words, since pp is arbitrary:

D(S 6)=G X 1S(S 6)G XD(S_6) = G_X^{-1} S(S_6) G_X

So G XG_X gives an inner automorphism of S 15S_{15} that maps one inclusion of S 6S_6 in S 15S_{15}, S(S 6)S(S_6), into the other one, D(S 6)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 6S 15D:S_6 \to S_{15} and S:S 6S 15S:S_6 \to S_{15} are injective, so you can construct an outer automorphism F:S 6S 6F: S_6 \to S_6 as:

F(p)=D 1(G X 1S(p)G X)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,𝔽 2)Sp(4,\mathbb{F}_2) (which are isomorphic to S 6S_6) sitting inside GL(4,𝔽 2)=PGL(4,𝔽 2)GL(4,\mathbb{F}_2)=PGL(4,\mathbb{F}_2) which at the least is a subgroup of S 15S_{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 6S_6 is also isomorphic to O(8,𝔽 2)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,𝔽 2)Sp(4,\mathbb{F}_2) means the group of transformations preserving a nondegenerate alternating form on an 8-dimensional vector space over 𝔽 2\mathbb{F}_2, not a 4-dimensional one. Then Sp(4,𝔽 2)Sp(4,\mathbb{F}_2) would be just another name for O(8,𝔽 2)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 𝔽 2\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)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 6S_6 is isomorphic to PSp(4,𝔽 2)PSp(4, \mathbb{F}_2).

Take a 66-dimensional vector space over 𝔽 2\mathbb{F}_2: say sextuples of ones and zeros. Take the subspace orthogonal to (1,1,1,1,1,1)\left(1,1,1,1,1,1\right), getting a 55-dimensional space, consisting of sextuples with an even number of ones. Then take the quotient by addition of (1,1,1,1,1,1)\left(1,1,1,1,1,1\right), getting a 44-dimensional space.

Now think of this in another way: the first space is the set of subsets of a set SS of 66 elements corresponding to coordinates (for each each vector, take the subset of coordinates consisting of those with a component value of 11). The 55-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 44-dimensional space: the identity, consisting of the equivalence class {(0,0,0,0,0,0),(1,1,1,1,1,1)}\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 SS; and the other elements, all of which are two-element equivalence classes containing a 22-element subset of SS together with its 44-element complement.

If we go over to projective space, all we do is throw away the identity (because 𝔽 2\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 SS in some order:

(ab)+(ab)(ab)+(ab) is the identity, e.g.(1,1,0,0,0,0)+(1,1,0,0,0,0)=(0,0,0,0,0,0)\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(ab)+(bc)=0, e.g. (1,1,0,0,0,0)+(0,1,1,0,0,0)=(1,0,1,0,0,0)\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) + (bc) = (ac))

(ab)+(cd)=(ef)(ab)+(cd)=(ef), e.g. (1,1,0,0,0,0)+(0,0,1,1,0,0)=(1,1,1,1,0,0)(0,0,0,0,1,1)\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 {(ab),(bc),(ac)}\left\{(ab),(bc),(ac)\right\} and {(ab),(cd),(ef)}\left\{(ab),(cd),(ef)\right\} respectively.

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

<(ab),(ab)>=0&lt;(ab),(ab)&gt;=0
<(ab),(bc)>=1&lt;(ab),(bc)&gt;=1
<(ab),(cd)>=0&lt;(ab),(cd)&gt;=0

An example symplectic basis would be {(ab),(bc),(de),(ef)}\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 6S_6. The two types of lines are the non-isotropic and isotropic lines respectively. (That is, lines of the form {(ab),(bc),(ac)}\left\{(ab),(bc),(ac)\right\} have all their points mutually non-orthogonal, and lines of the form {(ab),(cd),(ef)}\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 66 duads formed from the 44 elements of SS 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 2B_2=C_2, there is a generic exceptional isomorphism PSp(4,q)PΩ(5,q)PSp(4,q)\simeq P\Omega(5,q) where PΩP\Omega is the simple orthogonal group (which may or may not be PSO(5,q)PSO(5,q).)

This can be seen as a folding of the Klein Correspondence—that is, of the Dynkin diagram isomorphism A 3=D 3A_3=D_3—as follows (at lightning speed):

A 3A_3 corresponds to PSL(4,q)PSL(4,q). Take the 44-dimensional vector space on which this naturally acts (e.g. from the left). The bivectors of this space form a 66-dimensional vector space and PSL(4,q)PSL(4,q) preserves an orthogonal structure on this space given by wedging bivectors together. In fact, PSL(4,q)PΩ +(6,q)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 𝔽 q 4\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 66-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 55-space correspond to intersecting lines in projective 33-space; hence an isotropic line (consisting entirely of mutually orthogonal isotropic points) corresponds to a plane pencil of projective 33-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 55-space correspond to a symplectic structure and one of its isotropic lines in projective 33-space.

Now, we get from an orthogonal projective 55-space to an orthogonal projective 44-space by fixing a non-isotropic point and taking its perpendicular space. As a non-isotropic point corresponds to a symplectic structure in projective 33-space, this “explains” why we have PSp(4,q)PΩ(5,q)PSp(4,q)\simeq P\Omega(5,q).

Now, let us look at orthogonal projective 55-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 44-space correspond to isotropic lines of the symplectic projective 33-space.

From remark (a) the isotropic lines of orthogonal projective 44-space then correspond to a plane pencil of isotropic lines of symplectic projective 33-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 44-space and all points (which are all isotropic) of symplectic projective 33-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 6S_6 and PSp(2,𝔽 2)PSp(2,\mathbb{F}_2): all those lovely duads and synthemes show up so nicely starting from linear algebra over 𝔽 2\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 𝔽 q\mathbb{F}_q version of the Klein correspondence, which I already love in its guise

Spin(6)SU(4) 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)(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.

I’ll add a note to your previous comment, to warn people about that typo.

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 22, we have an additional, completely different correspondence between PSp(4,q)PΩ(5,q)PSp(4,q)\simeq P\Omega(5,q).

In fact we have a general correspondence between PSp(2n,q)PΩ(2n+1,q)PSp(2n,q)\simeq P\Omega(2n+1,q), corresponding to an isomorphism between the Dynkin diagrams C nC_n and B nB_n (which is more familiar in characteristic 11).

Very crudely: since, in characteristic 22, 1=11=-1, it is also true that symmetric and antisymmetric bilinear forms become the same thing.

Less crudely: fix a quadratic form QQ. We then get a bilinear form BB via

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

Generally, BB will obviously be symmetric.

We then have

B(u,u) =Q(u+u)Q(u)Q(u) =Q(2u)2Q(u) =4Q(u)2Q(u) =2Q(u)\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 22, we can use this to go the other way:

Q(u)=12B(u,u)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 22, we have, from the above,

B(u,u)=0B(u,u)=0

So, in this case, BB, if non-degenerate, is a symplectic form. Also, since we can’t go back from BB to QQ, 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+12n+1, and in characteristic 22, we must have that the bilinear form corresponding to a quadratic form is degenerate: there is a 11-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)PΩ(5,q)PSp(4,q)\simeq P\Omega(5,q) in the characteristic 22 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)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 6S_6 is also isomorphic to O(8,𝔽 2)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,𝔽 2)Sp(4,\mathbb{F}_2) means the group of transformations preserving a nondegenerate alternating form on an 8-dimensional vector space over 𝔽 2\mathbb{F}_2, not a 4-dimensional one. Then Sp(4,𝔽 2)Sp(4,\mathbb{F}_2) would be just another name for O(8,𝔽 2)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 6O(5,𝔽 2) S_6 \cong O(5,\mathbb{F}_2)

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

PSp(4,𝔽 2)S 6 PSp(4,\mathbb{F}_2) \cong S_6

and

PSp(4,𝔽 2)PΩ(5,𝔽 2). PSp(4,\mathbb{F}_2) \cong P\Omega(5,\mathbb{F}_2).

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

But this is something even I understand:

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

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

:Λ 2 4Λ 2 4 \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 Λ 2 4\Lambda^2\mathbb{C}^4 by

g(μ,ν)vol=μν g(\mu, \nu) vol = \mu \wedge \star \nu

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

PSp(4,)SO(5,) 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 4×44\times 4 matrices MM over 𝔽 2\mathbb{F}_2 such that:

M TΩM=ΩM^T \Omega M = \Omega

where:

Ω=(0 0 1 0 0 0 0 1 1 0 0 0 0 1 0 0)\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 MM.

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 GG is complete exactly if Center(G)\mathrm{Center}(G) and Out(G)\mathrm{Out}(G) are trivial, i.e., if the space of self-homotopy-equivalences of BGBG 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 nS_ns.

Here is one neat fact about complete groups (see this). Consider the map GAut(G)G\to \mathrm{Aut}(G). We can iterate this, obtaining a sequence GAut(G)Aut(Aut)(G)\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 GG has trivial center, then GAut(G)G\to \mathrm{Aut}(G) is injective and Aut(G)\mathrm{Aut}(G) turns out to have trivial center as well. Therefore, every centerless group embeds in a complete group. (If GG 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

Thanks, Charles! I vaguely remembered reading about this construction:

GAut(G)Aut(Aut(G)) 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 GG, Aut(G)Aut(G) is also centerless, so we have inclusions

GAut(G)Aut(Aut(G)) G \hookrightarrow Aut(G) \hookrightarrow Aut(Aut(G)) \hookrightarrow \cdots

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

It’s also nice to see the example of G=D 4G = 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 GAut(G)G \to Aut(G) is not an isomorphism, yet we have Aut(G)GAut(G) \cong G. This is a case where we need to go ‘forever’ (more precisely, to step ω+1\omega+1) for this process of repeatedly taking AutAut 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 A_\infty spaces. Personally I’m more interested in juicy examples of what happens when you iterate the AutAut 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 XBAut(X)\coprod_X B Aut(X), if GG is complete then the component corresponding to BGB 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 BijBij be the groupoid of finite sets and bijections. Let Aut(Bij)Aut(Bij) be the 2-group of autoequivalences f:BijBijf: Bij \to Bij and natural isomorphisms between these. Let |Aut(Bij)||Aut(Bij)| be the corresponding connected pointed homotopy 2-type: that is, the one whose fundamental 2-group is Aut(Bij)Aut(Bij).

Then:

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

S 0S 1S_0 \cong S_1

The second factor of K(/2,1)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)/2 Out(S_6) \cong \mathbb{Z}/2

The factor of K(/2,2)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)/2 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)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 𝔽 2\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)(ab). Its neighbours should share different synthemes with it. This means that they must be

a) disjoint from (ab)(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)(cd) and (ce)(ce).

The bottom right duad, opposite (ab)(ab), also has these as neighbours, and so must be disjoint from them. That leaves aa, bb and ff 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 aa, the duad must be (af)(af).

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

Conversely, given a syntheme, we have 33 possibilities as to which duad defines the diagonals, and then 22 possibilites as to which of the other two duads corresponds to which diagonal. I guess that gives 15×6=9015\times6=90 cycles.

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

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