The n-Category Café

Well, super division algebras in the submonoid

$$\mathbb{Z}_8 \subset \mathbb{10}$$

happen to be Morita equivalent to the super algebras $Cl_0, \dots, Cl_7$ (real Clifford algebras), while those in the submonoid

$$\mathbb{Z}_2 \subset \mathbb{10}$$

happen to be Morita equivalent to the super algebras $\mathbb{C}\mathrm{l}_0, \mathbb{C}\mathrm{l}_1$ (complex Clifford algebras).

So in the former case $\mathbb{Z}_8$ has a distinguished generator, $Cl_1$ (actually a super division algebra itself), which gives $\mathbb{Z}_8$ a cyclic ordering! But I only know this generator is distinguished using Clifford algebra theory, not from staring at the super Brauer group $\mathbb{Z}_8$ in abstract.

The second case is less interesting: $\mathbb{Z}_2$ has a distinguished generator simply because it has only one possible generator.

I’ve tried things like that but haven’t been able to get it to work. The basic problem is that

$$\mathbb{C}\otimes_\mathbb{R} \mathbb{C} \cong \mathbb{C} \oplus \mathbb{C}$$

whereas the three-fold way and ten-fold way want the tensor product of two $\mathbb{C}$-modules to be another $\mathbb{C}$-module.

Central simple algebras over a field $k$ are closed under tensor product, and every one is Morita equivalent to a division algebra over $k$ whose center is $k$. This gives the usual Brauer group. The problem is that $\mathbb{C}$ is not central simple over $\mathbb{R}$. $\mathbb{C}\otimes_\mathbb{R} \mathbb{C} \cong \mathbb{C} \oplus \mathbb{C}$ is no longer even simple.

If $k$ has characteristic zero, semisimple algebras over $k$ are closed under tensor product (as well as direct sum). So, we can create a rig of Morita equivalence classes of semisimple algebras over $k$. This seems to be the most reasonable thing to do. But this rig consists of finite linear combinations of $\mathbb{R}, \mathbb{C},$ and $\mathbb{H}$ with the following multiplication table:

$$\begin{array}{cccc} \mathbf{\otimes_{\mathbb{R}}} & \mathbf{\mathbb{R}} & \mathbf{\mathbb{C}} & \mathbf{\mathbb{H}} \\ \mathbf{\mathbb{R}} & \mathbb{R} & \mathbb{C} & \mathbb{H} \\ \mathbf{\mathbb{C}} & \mathbb{C} & \mathbb{C} \oplus \mathbb{C} & \mathbb{C} \\ \mathbf{\mathbb{H}} & \mathbb{H} & \mathbb{C} & \mathbb{R} \end{array}$$

I don’t see a systematic way to chop this rig down to the monoid $\mathbb{3} = \{1, 0, -1\}$. The problem is

$$\mathbb{C} \otimes_{\mathbb{R}} \mathbb{C} = \mathbb{C} \oplus \mathbb{C}$$

again.

So, I think the field extensions of our base field $k$ should be treated with some extra respect; this leads to the idea of taking all the Brauer groups and glomming them together into a monoid.

Let me try to explain this structure theorem:

A commutative semigroup has a grading by a semilattice such that the homogeneous components are Archimedean semigroups.

First, a meet-semilattice is a poset $(L,\le)$ where any pair of elements $\alpha, \beta$ has a greatest lower bound, denoted $\alpha \wedge \beta$. Any such semilattice is automatically a commutative semigroup obeying an extra law, the idempotence law:

$$\alpha \wedge \alpha = \alpha$$

Suppose we have a commutative semigroup $(S, \cdot)$ and a homomorphism $f : S \to L$ where $L$ is a semilattice. Then we can partition $S$ into (possibly empty) subsets, one for each element $\alpha \in L$:

$$S_\alpha = f^{-1} \{\alpha\}$$

It’s easy to check that thanks to the idempotence law, each $S_\alpha$ is a semigroup!

So, a homomorphism $f : S \to L$ to a semilattice $L$ is a great way to chop up a commutative semigroup into smaller commutative semigroups.

So, given a commutative semigroup $(S,\cdot)$, how do we get a homomorphism to a semilattice?

We can define a preorder on $S$ by

$$a \le b \iff b \cdot c = a^n \; for \; some \; c \in S, n > 0$$

Roughly speaking, $a \le b$ if you can get from $b$ to some power of $a$ by multiplying $b$ with something. I think this definition is ‘upside down’ compared to how I would visualize things, but I will stick with Grillet’s notation for now to keep from getting completely confused!

This is just one of many preorders people put on semigroups, but this one is particularly nice for our purposes.

We can define an equivalence relation on $S$ by

$$a \sim b \iff a \le b \; and \; b \le a$$

Then $S/\sim$ becomes a semilattice with the operation $\wedge$ coming from the multiplication $\cdot$ in $S$ and a partial order coming from the preorder $\le$ in $S$. The quotient map

$$S \to S/\sim$$

is a homomorphism of semigroups, so we’ve got what we want.

Even better, $\sim$ is the finest equivalence relation such that $S /\sim$ becomes a semilattice!

This is part of Theorem 1.2 in Grillet’s book Commutative Semigroups. It looks quite easy to prove.

Let me just check that $S / \sim$ is a semilattice.

I’ll take it for granted that $\le$ as defined above is a preorder on our semigroup $S$. It’s easy to check that this preorder gives one on $S / \sim$.

Note that the preorder $\le$ on $S$ is not in general a partial order: we don’t have

$$a \le b \; and \; b \le a \implies a = b$$

since for a group we have $a \sim b$ for all $a,b$. However, to form $S/\sim$ we identify two elements whenever $a \le b$ and $b \le a$. So the preorder on $S / \sim$ is actually a partial order.

Why is it a semilattice? We need to show that given $[a], [b] \in S/\sim$ they have a greatest lower bound. The only possible choice is $[a \cdot b]$, so let’s check that it works.

I would visualize $a \cdot b$ as bigger than $a$ and $b$, but I think my visualization is upside down compared to Grillet’s. Indeed, we have

$$a \cdot b \le a$$

since $a$ times something equals some power of $a \cdot b$. Similarly

$$a \cdot b \le b$$

so $a \cdot b$ is a lower bound of $a$ and $b$ in $S$. Thus $[a \cdot b]$ is a lower bound of $[a]$ and $[b]$ in $S / \sim$.

Why is it the greatest lower bound?

Suppose $[c]$ is any lower bound of $[a]$ and $[b]$. Then $c \le a$ and $c \le b$. So,

$$c^m = a \cdot x$$

for some $n, x$ and

$$c^n = b \cdot y$$

for some $m, y$. Following my old nose, I get

$$c^{m + n} = a \cdot b \cdot x \cdot y$$

which implies that $c \le a \cdot b$. So, $[a \cdot b]$ is the greatest lower bound of $[a]$ $[b]$.

Apart from all the inequalities pointing the opposite direction from how I would visualize them, this was very easy.

And now for the final piece of this structure theorem! We’ve already seen that starting from a commutative semigroup $S$, we get a semilattice $S/\sim$ and a homomorphism, the quotient map

$$f: S \to S/\sim$$

It follows that each component

$$S_\alpha = f^{-1} \{ \alpha \}$$

is a semigroup. Moreover, $S$ becomes graded by the semilattice $S / \sim$, meaning

$$S_\alpha \cdot S_\beta \subseteq S_{\alpha \wedge \beta }$$

But the final piece is this: each component $S_\alpha$ is an Archimedean semigroup. This is one for which any pair of elements $a, b$ obeys $a \le b$, where $\le$ is the preorder we’ve seen before:

$$a \le b \iff b \cdot c = a^n \; for \; some \; c \in S, n > 0$$

This last piece is obvious, since each component is defined to be an equivalence class of elements of $S$, where $a \sim b$ iff $a \le b$ and $b \le a$.

You can see how the Archimedean property is a generalization of the one we know in real analysis.

Ah, Theorem 2.1 in Grillet’s Commutative Monoids gives a possible strategy for building the ‘Brauer monoid’ of a field out of the Brauer groups of all its algebraic extensions!

Let me first state the theorem. We say a commutative semigroup $S$ is graded by a semilattice $L$ if we can write $S$ as a disjoint union of ‘components’ $S_\alpha$ for $\alpha \in L$, and

$$S_\alpha \cdot S_\beta \subseteq S_{\alpha \vee \beta}$$

for all $\alpha, \beta \in L$. This implies that each component is a semigroup.

We’ve seen that every commutative semigroup has a canonical grading of this sort.

We say $S$ is a Clifford semigroup if it is equipped with a grading by a semilattice $L$ (not necessarily the canonical one) with the additional property that each component is a group. The components will obviously then be abelian groups.

The theorem says that from a Clifford semigroup we can build a functor from $L$ (viewed as a category with a unique morphism from $\alpha$ to $\beta$ when $\beta \le \alpha$) to the category of abelian groups. Conversely, given a functor from a semilattice to the category of abelian groups, we can build a Clifford semigroup.

(I believe these two processes are ‘inverses’: that is, they form an equivalence between the category of $L$-graded semigroups whose components are groups, and the category of functors from $L$ to $AbGp$. But Grillet doesn’t say this.)

The idea seems pretty simple. Given a Clifford semigroup, we define a functor from $L$ to $AbGp$ as follows. To each $\alpha \in L$ we assign the group $S_\alpha$. And whenever $\beta \le \alpha$, we have a homomorphism $S_\alpha \to S_\beta$ sending $x \in S_\alpha$ to $x 1_\beta$, where $\beta$ is the identity of $S_\beta$. This indeed maps $S_\alpha$ to $S_\beta$ since

$$x \in S_\alpha, 1_\beta \in S_\beta \; \implies \; x 1_\beta \in S_{\alpha \wedge \beta} = S_\beta$$

And it’s clearly a homomorphism!

It’s also pretty easy to see that these homomorphisms fit together to define a functor from $L$ to $AbGp$.

(Again Grillet seems to be doing things ‘upside down’ by treating $L$ as a category with a unique morphism from $\alpha$ to $\beta$ when $\beta \le \alpha$. But this is just a matter of convention. I won’t change conventions while I’m still reading his book!)

So here’s my latest idea for defining the ‘Brauer monoid’ of a field $k$. We let $K$ be a finite extension of $k$, and let $L$ be the lattice of subfields of $K$ containing $k$.

For each field $F \in L$, let $Br(F)$ be the Brauer group of $F$. If $F$ is contained in a bigger field $F' \in L$, there’s a homomorphism

$$Br(F) \to Br(F')$$

sending each Morita equivalence class $[A]$ of central simple algebras over $F$ to the class $[F' \otimes_F A]$ of central simple algebras over $F'$. (Apparently people call this homomorphism a ‘restriction map’.)

I believe this should give a functor from the lattice $L$ to AbGp — and thus, by the theorem I mentioned in my last comment, an $L$-graded commutative semigroup!

(Now our conventions have flipped, so we’re thinking of $L$ as a category with a unique morphism from $F$ to $F'$ when $F \subseteq F'$. But that’s okay.)

This commutative semigroup should be a commutative monoid: the Brauer monoid of the extension $k \subseteq K$.

This seems like a more spiffy version of my ‘collage’ idea.

It may seem odd that I described a Brauer monoid not for a field but for a field $k$ and a finite extension $K$. This gives a bit of extra flexibility, but we may not want that flexibility. I restricted myself to finite extensions of the field $k$, that is those where the larger field $K$ is a finite-dimensional vector space over $k$, because I only want my Brauer monoid to know about finite-dimensional division algebras over $k$.

If the algebraic closure of $k$ is a finite extension of $k$, we can use that as our choice of $K$ and speak simply of the Brauer monoid of $k$.

This condition holds for $k = \mathbb{R}$, but not $k = \mathbb{Q}$.

Now I see that we can easily avoid this condition: let $K$ be the algebraic closure of $k$ but let $L$ be the lattice of fields $k \subseteq F \subseteq K$ that are finite extensions of $k$. Define a functor from $L$ to $AbGp$ sending each such field $F$ to the Brauer group $Br(F)$, and sending each inclusion $F \hookrightarrow F'$ to the canonical map $Br(F) \to Br(F')$. Use Theorem 2.1 in Grillet’s book to create a commutative monoid from this functor. This is the Brauer monoid of $k$.

It agrees with the previously mentioned one when the algebraic closure of $k$ is a finite extension of $K$.

Next I’d like to describe the Brauer monoid a bit more directly, without mentioning that theorem in Grillet’s book.

How do we explicitly describe the commutative semigroup coming from a functor

$$F : L \to AbGp$$

where $L$ is a lattice? Grillet explains this in the proof of his Theorem 2.1, a theorem that goes back to here:

• A. H. Clifford, Semigroups admitting relative inverses, Annals of Math. 42 (1941) 1037–1049.

To get our commutative semigroup, we start by taking the disjoint union

$$S = \coprod_{\alpha \in L} F(\alpha)$$

Then we put a multiplication on it. Say we are given $a \in F(\alpha)$ and $b \in F(\beta)$ and we want to multiply them. Since we have

$$\alpha, \beta \ge \alpha \wedge \beta$$

our functor $F$ gives group homomorphisms from $F(\alpha)$ and $F(\beta)$ to $F(\alpha \wedge \beta)$ (using Grillet’s upside-down conventions). We use these to map $a$ and $b$ into the semigroup $F(\alpha \wedge \beta)$, and then we multiply them in there.

It’s pretty easy to see that this recipe makes $S$ into a commutative semigroup: the associative law is the fun thing to check here.

And if our semilattice $L$ has a top element $\top$, then the identity $1 \in F(\top)$ will serve as an identity for $S$, so $S$ will be a commutative monoid.

(There should be some quick name for a meet-semilattice with a top element, just as there’s a quick name for a semigroup with an identity element.)

So, let me unravel all the constructions and describe my ‘Brauer monoid of a field’ more directly.

We start with a field $k$ and pick an algebraic closure of it, say $K$. We let $L$ be the collection of all intermediate fields $k \subseteq F \subseteq K$ that are finite extensions of $k$. We form the disjoint union

$$\mathbf{BR}(k) = \coprod_{F \in L} Br(F)$$

where $Br(F)$ is the usual Brauer group of $F$. An element of $Br(F)$ is a Morita equivalence class $[A]$ of central simple algebras over $F$.

Now we make $\mathbf{BR}(k)$ into a commutative monoid. Suppose we have two elements of $\mathbf{BR}(k)$:

$$[A] \in Br(F), \quad [A'] \in Br(F')$$

How do we multiply them? We take the smallest field $E \in L$ that contains both $F$ and $F'$. We extend our algebras $A$ and $A'$ to algebras over $E$, getting

$$\tilde{A} = E \otimes_F A , \qquad \tilde{A'} = E \otimes_{F'} A$$

If I know what I’m doing, these are both central simple algebras over $E$. So, we can tensor them over $E$ and get another central simple algebra $\tilde{A} \otimes_E \tilde{A'}$. This gives an element

$$[\tilde{A} \otimes_E \tilde{A'}] \in Br(E) \subseteq \mathbf{BR}(k)$$

And that’s how we multiply two guys in $\mathbf{BR}(k)$.

It’s pretty simple when all is said and done. The detour through commutative semigroup theory could be avoided, but I needed it to reach this idea… and it was fun to learn that stuff: commutative semigroups don’t get the respect they deserve. (Or at least commutative monoids don’t: semigroups that aren’t monoids deserve to be spat on and kicked, or else mercifully provided with an identity element.)

Now I can just generalize this stuff to the ‘super’ case and define the super Brauer monoid of any field. When that field is $\mathbb{R}$, this should be the commutative monoid $\mathbb{10}$, and we’ll get a ‘$\mathbb{10}$-graded symmetric monoidal category’

$$\text{syntax error at token }}$$

describing all 10 kinds of matter and how you can tensor them.

I guess Mike’s paper on constructing symmetric monoidal bicategories includes a proof that this is really true.

Yep. You probably also know about Niles Johnson’s work using this bicategory (and related ones) for Morita theory.

So, something like Mike’s idea should also hold at the level where we allow $R$ to vary.

One way to assemble this sort of data is what I called in the paper a “2x1-category”: a (pseudo) 2-category internal to 1-categories, just as a double category (or “1x1-category”) is a a 1-category internal to 1-categories. You have a category of objects (commutative rings and ring homomorphisms), a category of 1-cells (two-sided algebras and algebra homomorphisms), and a category of 2-cells (bimodules and bimodule maps). There are also other ways of presenting these data, e.g. you could talk about a functor $CRing \to DblCat$. Neither of these includes bimodules between commutative rings explicitly, but you can see them by regarding a commutative ring as an algebra over itself in the tautological way. I guess you could maybe try to include them explicitly in a more triple-categorical sort of structure.

I’m can’t say based on my own experience whether or not this buys us anything, but I remember Peter May caring about structures of this sort at some point — I think towards the end of Parametrized homotopy theory they had to look at an analogous thing where for every space $B$ there was a bicategory (actually a fibrant double category) of spaces-over-$B$ and parametrized-spectra-over-spaces-over-$B$, varying as $B$ varies. And IIRC Chris Douglas and his collaborators have used related structures to talk about “conformal nets”, whatever those are.

I knew that you were not trying to discover a huge intricate sunken cathedral, but being concrete (and real-world relevant); I’m just in a bimodule/algebra frame of mind at the moment.

One reason that people care about such things is that this iterated algebra construction is one way we can access $n$-vector spaces. Also, Urs is currently thinking about how to globalise the arithmetic part of algebraic topology, instead of working ‘prime-by-prime’, in his cohesive setup, and this feels very much like it. All these things pasted together should say something about twists of differential algebraic K-theory, but perhaps that’s a bit of wishful thinking.

Nad wrote:

If I understood you correctly then with the threefold way you have some way to “cook up” higher dimensional Hilbert spaces, via some tensor product construction.

In quantum mechanics, as you know, when we combine two systems we take the tensor product of their Hilbert spaces. This is fine if we’re working only with complex Hilbert spaces, and it’s also fine if we’re working only with real Hilbert spaces, but it doesn’t work with quaternionic Hilbert spaces!

The reason is that the quaternions are noncommutative. A quaternionic vector space is a left module over the quaternions, but you can’t tensor two left modules over a noncommutative algebra and get another left module. Stephen Adler wrote a book on quaternionic quantum mechanics where he got rather confused by this issue (in my humble opinion).

You might wonder why we should care about real or quaternionic Hilbert spaces. The answer is that even if you take complex Hilbert spaces as fundamental, the study of conjugate-linear symmetries, like time reversal and charge conjugation, gives rise to real and quaternionic Hilbert spaces. I explained this here:

• John Baez, Division algebras and quantum theory.

and this paper probably provides some of the motivation — the ‘why should we do this?’ stuff — that you’re missing.

But very roughly:

• quantum systems that don’t have time reversal symmetry are described by complex Hilbert spaces;

• quantum systems that do have time reversal symmetry are described by complex Hilbert spaces with an operator $T$ that is norm-preserving, conjugate-linear ($i T = -T i$) and obeys $T^2 = \pm 1$. When $T^2 = 1$ we can obtain from this data a real Hilbert space; when $T^2 = -1$ we can obtain from this data a quaternionic Hilbert space. Conversely, from a real or quaternionic Hilbert space we can get a complex Hilbert space with this extra data!

We can then think about combining these systems. So, we’re tensoring two complex Hilbert spaces equipped with extra data — or no extra data. But it turns out this problem is the same as the problem of tensoring real, complex and quaternionic Hilbert spaces!

For example, if we have two complex Hilbert spaces $H$ and $H'$ with time reversal operators $T$ and $T'$ obeying $T^2 = -1$, ${T'}^2 = -1$, the Hilbert space $H \otimes H$ gets an operator $T \otimes T'$ with $(T \otimes T')^2 = 1$. This corresponds to how we can tensor two quaternionic Hilbert spaces and get a real one!

Further analysis (done in this paper) reveals the multiplication table here:

$$\begin{array}{cccc} \mathbf{\otimes} & \mathbf{real} & \mathbf{complex} & \mathbf{quaternionic} \\ \mathbf{real} & real & complex & quaternionic \\ \mathbf{complex} & complex & complex & complex \\ \mathbf{quaternionic} & quaternionic & complex & real \end{array}$$

This ultimately arises from the fact that if we tensor the algebras $\mathbb{R}, \mathbb{C}, \mathbb{H}$, thinking of them as algebras over the real numbers, we get other algebras as follows:

$$\begin{array}{lrrr} \mathbf{\otimes} & \mathbf{\mathbb{R}} & \mathbf{\mathbb{C}} & \mathbf{\mathbb{H}} \\ \mathbf{\mathbb{R}} & \mathbb{R} & \mathbb{C} & \mathbb{H} \\ \mathbf{\mathbb{C}} & \mathbb{C} & \mathbb{C} \oplus \mathbb{C} & \mathbb{C}[2] \\ \mathbf{\mathbb{H}} & \mathbb{H} & \mathbb{C}[2] & \mathbb{R}[4] \end{array}$$

But how, exactly, does it arise?

My post here, and especially my comments on it, work out exactly how, in a way that generalizes to the 10-fold way, which is the ‘super’ version of the same story… arising naturally when you consider charge conjugation as well as time reversal.