The n-Category Café

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.

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.

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.

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).

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?

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 $G$.

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!

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

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.

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

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.

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.

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)?

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.

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.

Nice!