## August 7, 2014

### The Ten-Fold Way (Part 3)

#### Posted by John Baez

My last article on the ten-fold way was a piece of research in progress — it only reached a nice final form in the comments. Since that made it rather hard to follow, let me try to present a more detailed and self-contained treatment here!

But if you’re in a hurry, you can click on this:

and get my poster for next week’s scientific advisory board meeting at the Centre for Quantum Technologies, in Singapore. That’s where I work in the summer, and this poster is supposed to be a terse introduction to the ten-fold way.

First we’ll introduce the ‘Brauer monoid’ of a field. This is a way of assembling all simple algebras over that field into a monoid: a set with an associative product and unit. One reason for doing this is that in quantum physics, physical systems are described by vector spaces that are representations of certain ‘algebras of observables’, which are sometimes simple (in the technical sense). Combining physical systems involves taking the tensor product of their vector spaces and also these simple algebras. This gives the multiplication in the Brauer monoid.

We then turn to a larger structure called the ‘super Brauer monoid’ or ‘Brauer–Wall monoid’. This is the ‘super’ or $\mathbb{Z}_2$-graded version of the same idea, which shows up naturally in physical systems containing both bosons and fermions. For the field $\mathbb{R}$, the super Brauer monoid has 10 elements. This gives a nice encapsulation of the ‘ten-fold way’ introduced in work on condensed matter physics. At the end I’ll talk about this particular example in more detail.

Actually elements of the Brauer monoid of a field are equivalence classes of simple algebras over this field. Thus, I’ll start by reminding you about simple algebras and the notion of equivalence we need, called ‘Morita equivalence’. Briefly, two algebras are Morita equivalent if they have the same category of representations. Since in quantum physics it’s the representations of an algebra that matter, this is sometimes the right concept of equivalence, even though it’s coarser than isomorphism.

## Review of algebra

We begin with some material that algebraists consider well-known.

### Simple algebras and division algebras

Let $k$ be a field.

By an algebra over $k$ we will always mean a finite-dimensional associative unital $k$-algebra: that is, a finite-dimensional vector space $A$ over $k$ with an associative bilinear multiplication and a multiplicative unit $1 \in A$.

An algebra $A$ over $k$ is simple if its only two-sided ideals are $\{0\}$ and $A$ itself.

A division algebra over $k$ is an algebra $A$ such that if $a \ne 0$ there exists $b \in A$ such that $a b= b a = 1$. Using finite-dimensionality the following condition is equivalent: if $a,b \in A$ and $a b = 0$ then either $a = 0$ or $b = 0$.

A division algebra is automatically simple. More interestingly, by a theorem of Wedderburn, every simple algebra $A$ over $k$ is an algebra of $n \times n$ matrices with entries in some division algebra $D$ over $k$. We write this as

$A \cong D[n]$

where $D[n]$ is our shorthand for the algebra of $n \times n$ matrices with entries in $D$.

The center of an algebra over $k$ always includes a copy of $k$, the scalar multiples of $1 \in A$. If $D$ is a division algebra, its center $Z(D)$ is a commutative algebra that’s a division algebra in its own right. So $Z(D)$ is field, and it’s a finite extension of $k$, meaning it contains $k$ as a subfield and is a finite-dimensional algebra over $k$.

If $A$ is a simple algebra over $k$, its center is isomorphic to the center of some $D[n]$, which is just the center of $D$. So, the center of $A$ is a field that’s a finite extension of $k$. We’ll need this fact when defining in the multiplication in the Brauer monoid.

Example. I’m mainly interested in the case $k = \mathbb{R}$. A theorem of Frobenius says the only division algebras over $\mathbb{R}$ are $\mathbb{R}$ itself, the complex numbers $\mathbb{C}$ and the quaternions $\mathbb{H}$. Of these, the first two are fields, while the third is noncommutative. So, the simple algebras over $\mathbb{R}$ are the matrix algebras $\mathbb{R}[n]$, $\mathbb{C}[n]$ and $\mathbb{H}[n]$. The center of $\mathbb{R}[n]$ and $\mathbb{H}[n]$ is $\mathbb{R}$, while the center of $\mathbb{C}[n]$ is $\mathbb{C}$, the only nontrivial finite extension of $\mathbb{R}$.

Example. The case $k = \mathbb{C}$ is more boring, because $\mathbb{C}$ is algebraically closed. Any division algebra $D$ over an algebraically closed field $k$ must be $k$ itself. (To see this, consider $x \in D$ and look at the smallest subring of $D$ containing $k$ and $x$ and closed under taking inverses. This is a finite hence algebraic extension of $k$, so it must be $k$.) So if $k$ is algebraically closed, the only simple algebras over $k$ are the matrix algebras $k[n]$.

Example. The case where $k$ is a finite field has a very different flavor. A theorem of Wedderburn and Dickson implies that any division algebra over a finite field $k$ is a field, indeed a finite extension of $k$. So, the only simple algebras over $k$ are the matrix algebras $F[n]$ where $F$ is a finite extension of $k$. Moreover, we can completely understand these finite extensions, since the finite fields are all of the form $\mathbb{F}_{p^n}$ where $p$ is a prime and $n = 1,2,3,\dots$, and the only finite extensions of $\mathbb{F}_{p^n}$ are the fields $\mathbb{F}_{p^m}$ where $n$ divides $m$.

### Morita equivalence and the Brauer group

Given an algebra $A$ over $k$ we define $Rep(A)$ to be the category of left $A$-modules. We say two algebras $A, B$ over $k$ are Morita equivalent if $Rep(A) \simeq Rep(B)$. In this situation we write $A \simeq B$.

Isomorphic algebras are Morita equivalent, but this equivalence relation is more general; for example we always have $A[n] \simeq A$, where $A[n]$ is the algebra of $n \times n$ matrices with entries in $A$.

We’ve seen that if $A$ is simple, $A \cong D[n]$, and this implies $A \simeq D[n]$. On the other hand, we have $D[n] \simeq D$. So, every simple algebra over $k$ is Morita equivalent to a division algebra over $k$.

As a set, the Brauer monoid of $k$ will simply be the set of Morita equivalence classes of simple algebras over $k$. By what I just said, this is also the set of Morita equivalence classes of division algebras over $k$. The trick will be defining multiplication in the Brauer monoid. For this we need to think about tensor products of algebras.

The tensor product of two algebras $A,B$ over $k$ is another algebra over $k$, which we’ll write as $A \otimes_k B$. This gets along with Morita equivalence:

$A \simeq A' \; and \; B \simeq B' \; \implies \; A \otimes_k A' \simeq B \otimes_k B'$

However, the tensor product of simple algebras need not be simple! And the tensor product of division algebras need not be a division algebra, or even simple. So, we have to be a bit careful if we want a workable multiplication in the Brauer monoid.

For example, take $k = \mathbb{R}$. The division algebras over $\mathbb{R}$ are $\mathbb{R}, \mathbb{C}$ and the quaternions $\mathbb{H}$. We have

$\mathbb{H} \otimes_{\mathbb{R}} \mathbb{H} \cong \mathbb{R}[4] \simeq \mathbb{R}$

so this particular tensor product of division algebras over $\mathbb{R}$ is simple and thus Morita equivalent to another division algebra over $\mathbb{R}$. On the other hand,

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

and this is not a division algebra, nor even simple, nor even Morita equivalent to a simple algebra.

What’s the problem with the latter example? The problem turns out to be that the division algebra $\mathbb{C}$ does not have $\mathbb{R}$ as its center: it has a larger field, namely $\mathbb{C}$ itself, as its center.

It turns out that if you tensor two simple algebras over a field $k$ and they both have just $k$ as their center, the result is again simple. So, in Brauer theory, people restrict attention to simple algebras over $k$ having just $k$ as their center. These are called central simple algebras over $k$. The set of Morita equivalence classes of these is closed under tensor product, so it becomes a monoid. And this monoid happens to be be an abelian group: Brauer group of $k$, denoted $Br(k)$. I want to work with all simple algebras over $k$. So I will need to change this recipe a bit. But it will still be good to compute a few Brauer groups.

To do this, it pays to note that element of $Br(k)$ has a representative that is a division algebra over $k$ whose center is $k$. Why? Every simple algebra over $k$ is $D[n]$ for some division algebra $D$ over $k$. $D[n]$ is central simple over $k$ iff the center of $D$ is $k$, and $D[n]$ is Morita equivalent to $D$. Using this, we easily see:

Example. The Brauer group $Br(\mathbb{R})$ is $\mathbb{Z}_2$, the 2-element group consisting of $[\mathbb{R}]$ and $[\mathbb{H}]$. We have

$[\mathbb{R}] \cdot [\mathbb{R}] = [\mathbb{R} \otimes_\mathbb{R} \mathbb{R}] = [\mathbb{R}]$

$[\mathbb{R}] \cdot [\mathbb{H}] = [\mathbb{R} \otimes_\mathbb{R} \mathbb{H}] = [\mathbb{H}]$

$[\mathbb{H}] \cdot [\mathbb{H}] = [\mathbb{H} \otimes_\mathbb{R} \mathbb{H}] = [\mathbb{R}]$

Example. The Brauer group of any algebraically closed field $k$ is trivial, since the only division algebra over $k$ is $k$ itself. Thus $Br(\mathbb{C}) = 1$.

Example. The Brauer group of any finite field $k$ is trivial, since the only division algebras over $k$ are fields that are finite extensions of $k$, and of these only $k$ itself has $k$ as center.

Example. Just so you don’t get the impression that Brauer groups tend to be boring, consider the Brauer group of the rational numbers:

$Br(\mathbb{Q}) = \left\{ (a,x) : \; a \in \{0,\frac{1}{2}\}, \quad x \in \bigoplus_p \mathbb{Q}/\mathbb{Z}, \quad a + \sum_p x_p = 0 \right\}$

where the sum is over all primes. This is a consequence of the Albert–Brauer–Hasse–Noether theorem. The funny-looking $\{0,\frac{1}{2}\}$ is just a way to think about the group $\mathbb{Z}_2$ as a subgroup of $\mathbb{Q}/\mathbb{Z}$. The elements of this correspond to $\mathbb{Q}$ itself and a rational version of the quaternions. The other stuff comes from studying the situation ‘locally’ one prime at a time. However, the two aspects interact.

## The Brauer monoid of a field

Let $k$ be a field and $\overline{k}$ its algebraic completion. Let $L$ be the set of intermediate fields

$k \subseteq F \subseteq \overline{k}$

where $F$ is a finite extension of $k$. This set $L$ is partially ordered by inclusion, and in fact it is a semilattice: any finite subset of $L$ has a least upper bound. We write $F \vee F'$ for the least upper bound of $F,F' \in L$. This is just the smallest subfield of $\overline{k}$ containing both $F$ and $F'$.

We define the Brauer monoid of $k$ to be the disjoint union

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

So, every simple algebra over $k$ shows up in here: if $A$ is a simple algebra over $k$ with center $F$, the Morita equivalence class $[A]$ will appear as an element of $Br(F)$. However, isomorphic copies of the same simple algebra will show up repeatedly in the Brauer monoid, since we may have $F \ne F'$ but still $F \cong F'$.

How do we define multiplication in the Brauer monoid? The key is that the Brauer group is functorial. Suppose we have an inclusion of fields $F \subseteq F'$ in the semilattice $L$. Then we get a homomorphism

$Br_{F', F} : Br(F) \to Br(F')$

as follows. Any element $[A] \in Br(F)$ comes from a central simple algebra $A$ over $F$; the algebra $F' \otimes_F A$ will be central simple over $F'$, and we define

$Br_{F',F} [A] = [F' \otimes_F A]$

Of course we need to check that this is well-defined, but this is well-known. People call $Br_{F',F}$ restriction, since larger fields have smaller Brauer groups, but I’d prefer to call it ‘extension’, since we’re extending an algebra to be defined over a larger field.

It’s easy to see that if $F \subseteq F' \subseteq F''$ then

$Br_{F'', F} = Br_{F'' ,F'} Br_{F', F}$

and this together with

$Br_{F,F} = 1_{Br(F)}$

implies that we have a functor

$Br: L \to AbGp$

So now suppose we have two elements of $BR(k)$ and we want to multiply them. To do this, we simply write them as $[A] \in Br(F)$ and $[A'] \in Br(F')$, map them both into $Br(F \vee F')$, and then multiply them there:

$[A] \cdot [A'] \; := \; Br_{F \vee F', F} [A] \; \cdot Br_{F \vee F', F'} [A']$

This can also be expressed with less jargon as follows:

$[A] \cdot [A'] = [A \otimes_F (F \vee F') \otimes_{F'} A']$

However, the functorial approach gives a nice outlook on this basic result:

Proposition. With the above multiplication, $BR(k)$ is a commutative monoid.

Proof. The multiplicative identity is $[k] \in Br(k)$, and commutativity is obvious, so the only thing to check is associativity. This is easy enough to do directly, but it’s a bit enlightening to notice that it’s a special case of an idea that goes back to A. H. Clifford.

In modern language: suppose we have any semilattice $L$ and any functor $B: L \to AbGp$. This gives an abelian group $B(x)$ for any $x \in L$, and a homomorphism

$B_{x', x} : B(x) \to B(x')$

whenever $a \le a'$. Then the disjoint union

$\coprod_{x \in L} B(x)$

becomes a commutative monoid if we define the product of $a \in B(x)$ and $a' \in B(x')$ by

$a \cdot a' = B_{x \vee x',x} (a) \; \cdot \; B_{x \vee x', x'}(a')$

Checking associativity is an easy fun calculation, so I won’t deprive you of the pleasure. Moreover, there’s nothing special about abelian groups here: a functor $B$ from $L$ to commutative monoids would work just as well. ∎

Let’s see a couple of examples:

Example. The Brauer monoid of the real numbers is the disjoint union

$BR(\mathbb{R}) = Br(\mathbb{R}) \sqcup Br(\mathbb{C})$

This has three elements: $[\mathbb{R}]$, $[\mathbb{C}]$ and $[\mathbb{H}]$. Leaving out the brackets, the multiplication table is

$\begin{array}{lrrr} \cdot & \mathbf{\mathbb{R}} & \mathbf{\mathbb{C}} & \mathbf{\mathbb{H}} \\ \mathbf{\mathbb{R}} & \mathbb{R} & \mathbb{C} &\mathbb{H} \\ \mathbf{\mathbb{C}} & \mathbb{C} & \mathbb{C} & \mathbb{C} \\ \mathbf{\mathbb{H}} & \mathbb{H} & \mathbb{C} & \mathbb{R} \end{array}$

So, this monoid is isomorphic to the multiplicative monoid $\mathbb{3} = \{1, 0, -1\}$. This formalizes the multiplicative aspect of Dyson’s ‘threefold way’, which I started grappling with in my paper Division algebras and quantum theory. If you read that paper you can see why I care: Hilbert spaces over the real numbers, complex numbers and quaternions are all important in quantum theory, so they must fit into a single structure. The Brauer monoid is a nice way to describe this structure.

Example. The Brauer monoid of a finite field $k$ is the disjoint union

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

where $L$ is the lattice of subfields of the algebraic closure $\overline{k}$ that are finite extensions of $k$. However, we’ve seen that $Br(F)$ is always the trivial group. Thus

$Br(k) \cong L$

with the monoid structure being the operation $\vee$ in the lattice $L$.

Example. The Brauer monoid of $\mathbb{Q}$ seems quite complicated to me, since it’s the disjoint union of $Br(F)$ for all $F \subset \overline{\mathbb{Q}}$ that are finite extensions of $\mathbb{Q}$. Such fields $F$ are called algebraic number fields, and their Brauer groups can, I believe, be computed using the Albert–Brauer–Hasse–Noether theorem. However, here we are doing this for all algebraic number fields, and also keeping track of how they ‘fit together’ using the so-called restriction maps $Br_{F', F} : Br(F) \to Br(F')$ whenever $F\subseteq F'$. The absolute Galois group of a field always acts on its Brauer monoid, so the rather fearsome absolute Galois group of $\mathbb{Q}$ acts on $Br(\mathbb{Q})$, for whatever that’s worth.

Fleeing the siren song of number theory, let us move on to my main topic of interest, which is the ‘super’ or $\mathbb{Z}_2$-graded version of this whole story.

## Review of superalgebra

We now want to repeat everything we just did, systematically replacing the category of vector spaces over $k$ by the category of super vector spaces over $k$, which are $\mathbb{Z}_2$ graded vector spaces:

$V = V_0 \oplus V_1$

We call the elements of $V_0$ even and the elements of $V_1$ odd. Elements of either $V_0$ or $V_1$ are called homogeneous, and we say an element $a \in V_i$ has degree $i$. A morphism in the category of super vector spaces is a linear map that preserves the degree of homogeneous elements.

The category of super vector spaces is symmetric monoidal in a way where we introduce a minus sign when we switch two odd elements.

### Simple superalgebras and division superalgebras

A superalgebra is a monoid in the category of super vector spaces. In other words, it is a super vector space $A = A_0 \oplus A_1$ where the vector space $A$ is an algebra in the usual sense and

$a \in A_i, \; b \in A_j \quad \implies \quad a \cdot b \in A_{i + j}$

where we do our addition mod 2. There is a tensor product of superalgebras, where

$(A \otimes B)_i = \bigoplus_{i = j + k} A_j \otimes B_k$

and multiplication is defined on homogeneous elements by:

$(a \otimes b)(a' \otimes b') = (-1)^{i+j} \; a a' \otimes b b'$

where $b \in B_i, a' \in A_j$ are the elements getting switched.

An ideal $I$ of a superalgebra is homogeneous if it is of the form

$I = I_0 \oplus I_1$

where $I_i \subseteq A_i$. We can take the quotient of a superalgebra by a homogeneous two-sided ideal and get another superalgebra. So, we say a superalgebra $A$ over $F$ is simple if its only two-sided homogeneous ideals are $\{0\}$ and $A$ itself.

A division superalgebra over $k$ is a superalgebra $A$ such that if $a \ne 0$ is homogeneous then there exists $b \in A$ such that $a b= b a = 1$.

At this point it is clear what we aim to do: generalize Brauer groups to this ‘super’ context by replacing division algebras with division superalgebras. Luckily this was already done a long time ago, by Wall:

• C. T. C. Wall, Graded Brauer groups, Journal für die reine und angewandte Mathematik 213 (1963–1964), 187–199.

He showed there are 10 division superalgebras over $\mathbb{R}$ and showed how 8 of these become elements of a kind of super Brauer group for $\mathbb{R}$, now called the ‘Brauer–Wall’ group. The other 2 become elements of the Brauer–Wall group of $\mathbb{C}$. A more up-to-date treatment of some of this material can be found here:

• Pierre Deligne, Notes on spinors, in Quantum Fields and Strings: a Course for Mathematicians, vol. 1, AMS. Providence, RI, 1999, pp. 99–135.

Nontrivial results that I state without proof will come from these sources.

Every division superalgebra is simple. Conversely, we want a super-Wedderburn theorem describing simple superalgebras in terms of division superalgebras. However, this must be more complicated than the ordinary Wedderburn theorem saying every simple algebra is a matrix algebra $D[n]$ with $D$ a division algebra.

After all, besides matrix algebras, we have ‘matrix superalgebras’ to contend with. For any $p,q \ge 0$ let $k^{p|q}$ be the super vector space with even part $k^p$ and odd part $k^q$. Then its endomorphism algebra

$k[p|q] = End(k^{p|q})$

becomes a superalgebra in a standard way, called a matrix superalgebra. Matrix superalgebras are always simple.

Deligne gives a classification of ‘central simple’ superalgebras, and from this we can derive a super-Wedderburn theorem. But what does ‘central simple’ mean in this context?

The supercommutator of two homogeneous elements $a \in A_i$, $b \in A_j$ of a superalgebra $A$ is

$[a,b] = a b - (-1)^{i+j} b a$

We can extend this by bilinearity to all elements of $A$. We say $a,b \in A$ supercommute if $[a,b]= 0$. The supercenter of $A$ is the set of elements in $A$ that supercommute with every element of $A$. If all elements of $A$ supercommute, or equivalently if the supercenter of $A$ is all of $A$, we say $A$ is supercommutative.

I believe a superalgebra $A$ over $k$ is central simple if $A$ is simple and its supercenter is just $k \subseteq A_0$, the scalar multiples of the identity. Deligne gives a more complicated definition of ‘central simple’, but then in Remark 3.5 proves it is equivalent to being semisimple with supercenter just $k$. I believe this is equivalent to the more reasonable-sounding condition I just gave, but have not carefully checked.

In Remark 3.5, Deligne says that by copying an argument in Chapter 8 of Bourbaki’s Algebra one can show:

Proposition. Any central simple superalgebra over $k$ is of the form $D[p|q]$ for some division superalgebra $D$ whose supercenter is $k$. Conversely, any superalgebra of this form is central simple.

Starting from this, Guo Chuan Thiang showed me how to prove the:

Super-Wedderburn Theorem. Suppose $A$ is a simple superalgebra over $k$, where $k$ is a field not of characteristic 2. Its supercenter $Z(A)$ is purely even, and $Z(A)$ is a field extending $k$. $A$ is isomorphic to $D[p|q]$ where $D$ is some division superalgebra $D$ over $Z(A)$.

It follows that any simple superalgebra over $k$ is of the form $D[p|q]$ where $D$ is a division superalgebra over $k$. Conversely, if $D$ is any division algebra over $k$, then $D[p,q]$ is a simple superalgebra over $k$.

Proof. Suppose $A$ is a simple superalgebra over $k$, and let $Z(A)$ be its supercenter. Suppose $a$ is a nonzero homogeneous element of $Z(A)$. Then $a A$ is a graded two-sided ideal of $A$. Since this ideal contains $a$ it is nonzero. Thus, this must be $A$ itself. So, there exists $b \in A$ such that $a b = 1$.

If $a$ is even, $b$ must be as well, and we obtain $b a = a b = 1$, so $a$ has an inverse. Thus, the even part of $Z(A)$ is a field.

If $a$ is odd, it satisfies $a^2=-a^2$. Multiplying on the left by $b$, then it follows that $a = -a$, so $a = 0$, since $k$ is not of characteristic 2.

In short, nonzero elements of $Z(A)$ must be even and invertible. It follows that $Z(A)$ is purely even, and is a field extending $k$. $A$ is central over this field $Z(A)$, so by the previous proposition we see $A \cong D[p|q]$ for some division superalgebra $D$ over $Z(A)$. $D$ will automatically be a division superalgebra over the smaller field $k$ as well.

Conversely, suppose $D$ is a division algebra over $k$. Since $D$ is simple, its supercenter will be a field $F$ extending $k$. By the previous proposition $D[p|q]$ will be a central simple superalgebra over $F$. It follows that $D[p|q]$ is simple as a superalgebra over $k$. ∎

Here is an all-important example:

Example. Let $k[\sqrt{-1}]$ be the free superalgebra over $k$ on an odd generator whose square is -1. This superalgebra has a 1-dimensional even part and a 1-dimensional odd part. It is a division superalgebra. It is not supercommutative, since $\sqrt{-1}$ does not supercommute with itself. It is central simple: its supercenter is just $k$. Over an algebraically closed field $k$ of characteristic other than 2, the only division superalgebras are $k$ itself and $k[\sqrt{-1}]$.

I don’t understand what happens in characteristic 2.

### Morita equivalence and the Brauer–Wall group

The Brauer–Wall group consists of Morita equivalence classes of central simple superalgebras, or equivalently, Morita equivalence classes of division superalgebras. For this to make sense, first we need to define Morita equivalence.

Given a superalgebra $A$ over $k$ we define a left module to be a super vector space $V$ over $k$ equipped with a morphism (that is, a grade-preserving linear map)

$A \otimes V \to V$

obeying the usual axioms of a left module. We define a morphism of left $A$-modules in the obvious way, and let $Rep(A)$ be the category of left $A$-modules.

We say two algebras $A, B$ over $k$ are Morita equivalent if $Rep(A) \simeq Rep(B)$. In this situation we write $A \simeq B$.

Example. Every matrix superalgebra $k[p|q]$ is Morita equivalent to $k$.

Example. If $A \simeq A'$ and $B \simeq B'$ then $A \otimes_k A' \simeq B \otimes_k B'$.

Example. Since every central simple superalgebra over $k$ is of the form $D[p|q] = D \otimes k[p|q]$ for some division superalgebra $D$ whose supercenter is just $k$, the previous two examples imply that every central simple superalgebra over $k$ is Morita equivalent to a division superalgebra whose center is just $k$.

We define the Brauer–Wall group $Bw(k)$ of the field $k$ to be the set of Morita equivalence classes of central simple superalgebras over $k$, given the following multiplication:

$[A] \otimes [B] \; := \; [A \otimes B]$

This is well-defined because the tensor product of central simple superalgebras is again central simple. Given that, $Bw(k)$ is clearly a commutative monoid. But in fact it’s an abelian group.

Since every central simple superalgebra over $k$ is Morita equivalent to a division superalgebra whose center is just $k$, we can compute Brauer–Wall groups by focusing on these division superalgebras.

Example. For any algebraically closed field $k$, the Brauer–Wall group $Bw(k)$ is $\mathbb{Z}_2$, where the two elements are $[k]$ and $[k[\sqrt{-1}]]$. In particular, $Bw(\mathbb{C})$ is $\mathbb{Z}_2$. Wall showed that this $\mathbb{Z}_2$ is related to the period-2 phenomenon in complex K-theory and the theory of complex Clifford algebras. The point is that

$\mathbb{C}[\sqrt{-1}]^{\otimes n} \cong \mathbb{C}liff_n$

where $\mathbb{C}liff_n$ is the complex Clifford algebra on $n$ square roots of -1, made into a superalgebra in the usual way. It is well-known that

$\mathbb{C}liff_2 \simeq \mathbb{C}liff_0$

and this gives the period-2 phenomenon.

Example. $Bw(\mathbb{R})$ is much more interesting: by a theorem of Wall, this is $\mathbb{Z}_8$. This is generated by $[\mathbb{R}[\sqrt{-1}]]$. Wall showed that this $\mathbb{Z}_8$ is related to the period-8 phenomenon in real K-theory and the theory of real Clifford algebras. The point is that

$\mathbb{R}[\sqrt{-1}]^{\otimes n} \cong Cliff_{n}$

where $Cliff_{n}$ is the real Clifford algebra on $n$ square roots of -1, made into a superalgebra in the usual way. It is well-known that

$Cliff_{8} \simeq Cliff_0$

and this gives the period-8 phenomenon.

Example. More generally, Wall showed that as long as $k$ doesn’t have characteristic 2, $Bw(k)$ is an iterated extension of $\mathbb{Z}_2$ by $k^*/(k^*)^2$ by $Br(k)$. For a quick modern proof, see Lemma 3.7 in Deligne’s paper. In the case $k = \mathbb{R}$ all three of these groups are $\mathbb{Z}_2$ and the iterated extension gives $\mathbb{Z}_8$.

## The Brauer–Wall monoid

And now the rest practically writes itself. Let $k$ be a field and $\overline{k}$ its algebraic completion. As before, let $L$ be the semilattice of intermediate fields

$k \subseteq F \subseteq \overline{k}$

where $F$ is a finite extension of $k$.

We define the underlying set of Brauer–Wall monoid of $k$ to be the disjoint union

$BW(k) = \coprod_{F \in L} Bw(F)$

To make this into a commutative monoid, we use the functoriality of the Brauer–Wall group. Suppose we have an inclusion of fields $F \subseteq F'$ in the semilattice $L$. Then we get a homomorphism

$Bw_{F', F} : Bw(F) \to Bw(F')$

as follows:

$Bw_{F',F} [A] = [F' \otimes_F A]$

and this gives a functor

$Bw: L \to AbGp$

Using this, we multiply two elements in the Brauer–Wall monoid as follows. Given $[A] \in Bw(F)$ and $[A'] \in Bw(F')$, their product is

$[A] \cdot [A'] \; := \; Bw_{F \vee F', F} [A] \; \cdot Bw_{F \vee F', F'} [A']$

or in other words

$[A] \cdot [A'] = [A \otimes_F (F \vee F') \otimes_{F'} A']$

Proposition. With the above multiplication, $BR(k)$ is a commutative monoid.

Proof. The same argument that let us show associativity for multiplication in the Brauer monoid works again here. ∎

Example. As a set, the Brauer–Wall monoid of the real numbers is the disjoint union

$BW(\mathbb{R}) = Bw(\mathbb{R}) \sqcup Bw(\mathbb{C}) \cong \mathbb{Z}_8 \sqcup \mathbb{Z}_2$

The monoid operation — let’s call it addition now — is the usual addition on $\mathbb{Z}_8$ when applied to two elements of that group, and the usual addition on $\mathbb{Z}_2$ when applied to two elements of that group. The only interesting part is when we add an element $a \in \mathbb{Z}_8$ and an element $b \in \mathbb{Z}_2$. For this we need to convert $a$ into an element of $\mathbb{Z}_2$. For that we use the homomorphism

$Bw_{\mathbb{C}, \mathbb{R}} : Bw(\mathbb{R}) \to Bw(\mathbb{C})$

which sends $[\mathbb{R}[\sqrt{-1}]]$ to $[\mathbb{C}[\sqrt{-1}]]$. More concretely,

$Bw_{\mathbb{C}, \mathbb{R}} : \mathbb{Z}_8 \to \mathbb{Z}_2$

takes an integer mod 8 and gives the corresponding integer mod 2.

So, very concretely,

$BW(\mathbb{R}) \cong \mathbb{10} = \{0,1,2,3,4,5,6,7,\mathbf{0}, \mathbf{1}\}$

where the monoid operation in $\mathbb{10}$ is addition mod 8 for two lightface numbers, but addition mod 2 for two boldface numbers or a boldface and a lightface one.

## Conclusion

I had meant to include a section explaining in detail how the 10 elements of this monoid $\mathbb{10}$ correspond to 10 kinds of matter, but this post is getting too long. So for now, at least, you can click on this picture to get an explanation of that!

### References

Besides what I’ve already mentioned about the classification of simple superalgebras, here are some other links. Wall proved a kind of super-Wedderburn theorem starting in the section of his paper called Elementary properties. Natalia Zhukavets has an Introduction to superalgebras which in Theorem 1.5 proves that in an algebraically closed field of characteristic different than 2, any simple superalgebra is of the form $k[p|q]$ or $D[n]$ where $D = k[u]$, $u$ being an odd square root of 1. Over an algebraically closed field, this superdivision algebra $D$ is isomorphic to the division algebra that I called $k[\sqrt{-1}]$. Over a field that is not algebraically closed, they can be different, and there can be many nonisomorphic division algebras obtained by adjoining to $k$ an odd square root of $a \in k$ where $a \ne 0$.

Jinkui Wan and Weiqiang Wang have a paper with a Digression on superalgebras which summarizes Wall’s results in more modern language. Benjamin Gammage has an expository paper with a Classification of finite-dimensional simple superalgebras. This only classifies the ‘central’ ones — but as we’ve seen, that’s the key case.

Posted at August 7, 2014 4:00 AM UTC

TrackBack URL for this Entry:   http://golem.ph.utexas.edu/cgi-bin/MT-3.0/dxy-tb.fcgi/2760

### Re: The Ten-Fold Way (Part 3)

Posted by: Colin Backhurst on August 7, 2014 1:13 PM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

Whoops, I wrote the extension as .png instead of .pdf. It’s fixed now — thanks!

Posted by: John Baez on August 7, 2014 1:49 PM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

I wonder if there’s any way to make sense of the Brauer monoid construction from the Brauer groups of field extensions in light of the fact that the Brauer group of a ring relates to its second etale cohomology group with coefficients in the corresponding multiplicative group.

Posted by: David Corfield on August 7, 2014 1:42 PM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

If as Wikipedia says

$Br(K) \cong H^2(Gal(K^s/K), K^{s}^{\ast}),$

for $K^s$ the separable closure, which for perfect fields is the algebraic closure, then there’s some sum over the second cohomology groups of Galois groups of finite extensions.

Hmm, perhaps the explanation Wikipedia then gives about what this has to do with division algebras over $K$ helps.

Posted by: David Corfield on August 7, 2014 2:11 PM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

Maybe. Everything I’m doing with Brauer groups seems very elementary, so the connections with étale cohomology and the like seems like overkill. Part of this is that I’m mainly concerned with the Brauer groups of $\mathbb{R}$ and $\mathbb{C}$, which are very dull until you liven them up with a bit of ‘superness’ — and even then, pretty simple. As I was writing this up I wondered what the Brauer group of $\mathbb{Q}$ is, and that seems a lot more interesting: we start seeing class field theory in action.

But maybe the connection to étale cohomology would make more sense if I understood why people got interested in Brauer groups in the first place. I’m interested in them now because noncommutative algebras show up as algebras of operators in quantum mechanics, and simple algebras are fundamental to classifying physical systems. But it seems people got interested in Brauer groups as part of thinking about commutative algebra, like algebraic number fields (and later, schemes). So it feels like there’s some weird relation between geometry and noncommutative geometry at work here.

More precisely: if a field is like a space, a field extension $F \hookrightarrow F'$ is like a covering space or ramified covering (with the direction of the arrow turned around). But a simple algebra over $F$, say $F \hookrightarrow A$, is like a ‘noncommutative bundle’ over the space for $F$, since $A$ is like a ‘noncommutative space’. And the process whereby any simple algebra $A$ becomes a boring old matrix algebra when we extend it to an algebra over $F'$ for a big enough extension $F'$ (by forming $F' \hookrightarrow F' \otimes_F A$) is like the process whereby a bundle over a space trivializes when we lift it to a big enough cover. Taking $F'$ to be the algebraic closure of $F$ always does the trick; this is a bit like lifting a bundle from a space to its universal cover, insofar as the absolute Galois group of $F$ is like a fundamental group.

But the stuff on the nLab pages suggests another line of thinking, more related to topological quantum field theory. So the overall picture is quite murky in my mind.

More history would help. Why did Grothendieck get interested in generalizing Brauer theory from fields to commutative rings? I know why he wanted to generalize everything from fields to commutative rings, but why did he care about Brauer groups? What did he think they were about?

Posted by: John Baez on August 7, 2014 3:48 PM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

I don’t know anything about the history, but I think thinking in terms of noncommutative geometry is misleading. Rather, think in terms of ordinary geometry, but one categorical dimension up: a simple algebra up to Morita equivalence over a field is a bundle of module categories over its spectrum, so should be regarded as something like a 2-vector bundle. Taking algebras which are invertible up to Morita equivalence is like taking 2-line bundles, as the nLab says.

The classification problem for invertible algebras up to Morita equivalence defines a stack of 2-groupoids on the category of algebraic extensions of a field, sending an extension L to the 2-groupoid whose objects are invertible L-algebras, whose morphisms are invertible L-bimodules between these, and whose 2-morphisms are isomorphisms of L-bimodules. In particular, taking L to be the algebraic closure gives us a single isomorphism class of object (namely the algebraic closure) and a single isomorphism class of bimodule (namely the algebraic closure) with a group of automorphisms given by the group of units of the algebraic closure. In other words, you get $B^2 G_m$ as a local system on $Spec k$, which you should think of in terms of its etale homotopy type $B Gal(\overline{k}/k)$. So that’s the relationship to etale cohomology in a nutshell as I currently understand it.

Posted by: Qiaochu Yuan on August 8, 2014 12:35 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

When you say the connection to noncommutative geometry is “misleading” it makes me feel rebellious… what were Albert and Brauer and Hasse and Noether doing, lumping fields and division algebras together in the same chunk of math, if not some sort of noncommutative algebraic geometry?

But you are definitely on a better line of thought. The part that makes me happiest is this… I’m going to rewrite a little to make me even happier:

The classification problem for invertible algebras up to Morita equivalence defines a stack of 2-groupoids on the category of algebraic extensions of a field, sending an extension $F$ to the 3-group whose objects are invertible $F$-algebras, whose morphisms are invertible $F$-bimodules between these, and whose 2-morphisms are isomorphisms of $F$-bimodules.

It seems like my ‘Brauer monoid’ arises as a decategorification of this! Instead of the category of algebraic extensions of a field, I was using the mere lattice of algebraic extensions of a field, thinking of them as intermediate fields $k \subseteq F \subseteq \overline{k}$. And instead of a stack assigning to each $F$ the 3-group of invertible $F$-algebras, I’m using a mere functor that assigns to each $F$ the group of (Morita equivalence classes of) invertible $F$-algebras.

(I said “3-group” where you’d said “2-groupoid” because when you restrict attention to invertible $F$-algebras you really get a 3-group, and this is why we get a group of Morita equivalence classes of invertible $F$-algebras.)

This is great, because now we can run this process backward and see the 10-fold way as a ‘super’ version of this whole story. There should be a stack of 3-groups on the category of algebraic extensions of a field $k$, sending an extension $F$ to the 3-group whose objects are invertible $F$-superalgebras, whose morphisms are invertible $F$-superbimodules between these, and whose 2-morphisms are isomorphisms of $F$-superbimodules. And applying this to $k = \mathbb{R}$ we get a nice view of the 10-fold way!

People would have so much more fun if they did condensed matter physics over $\mathbb{Q}$

Posted by: John Baez on August 8, 2014 1:32 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

As we learn from Kapranov, the step to super-algebra is the removal of one level of truncation from the full story.

Posted by: David Corfield on August 8, 2014 8:18 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

Looking back at the discussion of the super algebra entry, I see I was confused by the numbering.

We have Urs saying

the degree=1 entry is labeled “boson/fermion superdegree” because the looping of the free abelian 3-group on a single generator sits inside the category of $\mathbb{Z}_2$-graded vector spaces equipped with the symmetric braiding that introduces a minus sign whenever an odd-graded element is commuted with an odd graded element. Hence it’s the category of super-vector spaces and under this identification the first $\mathbb{Z}_2$ in that table is identified with the grading group of super vector spaces.

But in the entry it says

Generally then super-grading and hence super-algebra arises from the 2-truncation (3-coskeleton) of the free abelian ∞-group on a single generator, which is the sphere spectrum $\mathbb{S}$. So the $\mathbb{Z}_2$-grading of superalgebra comes from the stable homotopy groups of spheres $\pi_n(\mathbb{S})$ in degree 1 and 2.

Which leads to abelian 2-group being listed under $n=1$super-degree/ and abelian 3-group being listed under $n=2$/spin.

But then

degree 0 and 1 together encode $\mathbb{Z}$-grading and the Koszul-sign rule on that;

degree 1 and 2 together encode $\mathbb{Z}_2$-grading (of super vector spaces, say) and the Koszul-sign rule on that;

degree 1 and 2 and 3 together match to the grading on (global sections of) super 2-vector spaces; (one Koszul sign rule on super 2-vector spaces, another one on their super 2-line automorphisms)

Posted by: David Corfield on August 8, 2014 9:40 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

By the way, last time Allen Knutson asked about the ‘periodicity’ aspect of the Brauer–Wall group $Bw(\mathbb{R}) \cong \mathbb{Z}_8$, namely the choice of a special ‘best’ generator.

Here I sort of answered that by realizing that for any field $k$ there’s a specially important division superalgebra, namely $k[\sqrt{-1}]$, where we adjoin an odd square root of -1. When $k$ is algebraically closed this is the only division superalgebra over $k$ besides $k$ itself. But when $k = \mathbb{R}$ this is the ‘best’ generator of $Bw(\mathbb{R}) \cong \mathbb{Z}_8$. It’s also the real Clifford algebra $Cliff_1$.

(There’s actually another equally canonical generator, namely the superdivision algebra we get from $k$ by adjoining an odd square root of $+1$. This is the one Deligne uses. But in applications of real Clifford algebras to rotation groups I’ve decided that square roots of $-1$ are more important than square roots of $+1$: geometrically, getting a bunch of linear isometries of $\mathbb{R}^n$ that act as anticommuting square roots of $-1$ is a way to get a bunch of linearly independent vector fields on the corresponding unit sphere.)

Posted by: John Baez on August 7, 2014 4:23 PM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

Actually there’s stuff about a super-Wedderburn theorem in Wall’s paper…

Posted by: John Baez on August 7, 2014 4:40 PM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

As an alternative/complement to Deligne’s notes (which I found difficult to follow), there is Chapter 6 of Varadarajan’s Supersymmetry for Mathematicians. It contains a pedagogical account of super-Brauer theory, with fairly detailed proofs.

http://www.ams.org/bookstore-getitem/item=cln-11

Posted by: Guo Chuan Thiang on August 7, 2014 9:53 PM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

That book seemed quite readable when I looked at it once. Does he present a classification theorem for simple superalgebras? That’s what I’d really like to see.

Posted by: John Baez on August 8, 2014 5:43 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

As I recall, he deals mainly with central simple superalgebras, and the more general (non-central) case might be implicit. In any case, there is a graded Artin-Wedderburn structure theorem, which works for more generally gradings, and also for graded rings:

Section 9.5 of Rings and Their Modules (Bland)

Methods of Graded Rings (Nastasescu, van Oystaeyen)

As I understand it, the simple graded algebras are, as ungraded algebras, matrix algebras over some graded division algebra. Then it needs to be given an appropriate grading.

Posted by: Guo Chuan Thiang on August 8, 2014 1:37 PM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

Guo wrote:

As I recall, he deals mainly with central simple superalgebras, and the more general (non-central) case might be implicit.

I wish I could see that it were. In the ordinary ungraded case, the center of a simple algebra $A$ over a field $k$ is not only commutative, it’s a field in its own right, some extension $F$ of $k$. Then we can reinterpret $A$ as a central simple algebra over $F$. So, the classification of simple algebras can be reduced to the classification of central simple algebras.

(Unfortunately, I don’t know how to show the center of $A$ is a field except through appealing to the classification of simple algebras! This says $A$ is isomorphic to the algebra of matrices over some division algebra $D$. Using this I can see the center of $A$ is the center of $D$, which is a commutative division algebra and thus a field. But this failure of mine could easily be due to my ignorance of some standard algebra tricks.)

In the $\mathbb{Z}_2$-graded case, the supercenter of a simple superalgebra $A$ is supercommutative. But why should it be commutative, much less a field? I don’t know, so I don’t see how to reduce the classification of simple superalgebras to the classification of central simple superalgebras.

(Again, this could easily be just my ignorance. I haven’t even read Wall’s and Deligne’s papers carefully yet. I don’t seem to know an example of a simple superalgebra whose supercenter is not a field.)

Thanks for the references! I want to get this stuff straightened out in my mind.

Posted by: John Baez on August 8, 2014 4:20 PM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

A possible argument goes like this. If $a$ is a non-zero homogeneous super-central element of a simple super-algebra $A$, then $A a$ is a graded 2-sided ideal, which must be $A$ itself. So $a$ has an inverse (which is also super-central). If $a$ is odd, it satisfies $a^2=-a^2$, then it follows that $a=0$. So non-zero super-central elements must be even (hence central) and invertible. I think this should work.

Posted by: Guo Chuan Thiang on August 8, 2014 7:30 PM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

Wow, that’s nice and quick! Thanks. So now I know: a simple superalgebra $A$ has a field as its center, and it’s a central simple superalgebra over this field.

Posted by: John Baez on August 9, 2014 2:37 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

I’ve updated my post to include a ‘super-Wedderburn theorem’ and a fairly careful and pedantic proof, based on your ideas and Deligne’s claimed classification of central simple superalgebras.

I still need to compare Jinkui Wan and Weiqiang Wang’s Digression on superalgebras and Benjamin Gammage’s Classification of finite-dimensional simple superalgebras, both of which look a bit more complicated than what I wrote. Are they consistent with it?

Posted by: John Baez on August 10, 2014 10:39 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

I’ve not checked in detail, but they look the same as Deligne’s classification. One of the Lemmas in Varadarajan’s book says that if $D$ is a super-division algebra over $\mathbb{R}$ which is not purely even, then a matrix superalgebra over $D$ (which I think is your $D[p|q]$), is actually expressible as $M_{p+q}(\mathbb{R})\otimes D$, which I think are the Type I superalgebras in Gammage’s terminology.

Deligne’s classification looks more similar to the general Wedderburn theorem for more general gradings. But it doesn’t distinguish between the cases where $D$ is purely even, from the other cases. This distinction seems to be interesting as well, e.g. for the super-Schur’s lemma. Also, Greg Moore did mention that 10=3+7 is another natural way of thinking about the superdivision algebras over the reals.

Your proposed unification of the real and complex super-Brauer groups, and the role of the super-Brauer groups in “twisting” $K$-theory, makes me wonder whether there is a correspondingly nice way to see how the complex and real (and quaternionic) $K$-theories are related!

Finally, did you mean $\mathbb{C}liff(1)\cong\mathbb{C}\oplus\mathbb{C}$ in your chart?

Posted by: Guo Chuan Thiang on August 12, 2014 1:07 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

Guo wrote:

Your proposed unification of the real and complex super-Brauer groups, and the role of the super-Brauer groups in “twisting” $K$-theory, makes me wonder whether there is a correspondingly nice way to see how the complex and real (and quaternionic) $K$-theories are related!

There should be! The ‘multiplication table’ I described

$\begin{array}{lrrr} \cdot & \mathbf{\mathbb{R}} & \mathbf{\mathbb{C}} & \mathbf{\mathbb{H}} \\ \mathbf{\mathbb{R}} & \mathbb{R} & \mathbb{C} &\mathbb{H} \\ \mathbf{\mathbb{C}} & \mathbb{C} & \mathbb{C} & \mathbb{C} \\ \mathbf{\mathbb{H}} & \mathbb{H} & \mathbb{C} & \mathbb{R} \end{array}$

implies that there’s a nice way to tensor two quaternionic vector bundles and get a real one, etc.

One of the Lemmas in Varadarajan’s book says that if $D$ is a super-division algebra over $\mathbb{R}$ which is not purely even, then a matrix superalgebra over $D$ (which I think is your $D[p|q]$), is actually expressible as $M_{p+q}(\mathbb{R}) \otimes D$.

My $D[p|q]$ is $\mathbb{R}[p|q] \otimes D$, where $\mathbb{R}[p|q]$ is the ‘real matrix superalgebra’ consisting of endomorphisms of the super vector space with $p$ even dimensions and $q$ odd ones. Note that $\mathbb{R}[p|0]$ is the ordinary ‘purely even’ matrix algebra you’re calling $M_p(\mathbb{R})$.

So, you seem to be saying that if $D$ is not purely even,

$D \otimes \mathbb{R}[p|q] \cong D \otimes \mathbb{R}[p+q|0]$

If so, there’s a kind of redundancy in the classification I gave: $D \otimes \mathbb{R}[p|q]$ only depends, up to isomorphism, on $p+q$.

That does seem to be what some of my references were saying, and it’s good to know.

Finally, did you mean $\mathbb{C}liff(1)\cong\mathbb{C}\oplus\mathbb{C}$ in your chart?

Whoops, yes! Maybe too late to fix in my actual poster (they’ll print it out and I was hoping to keep it around for years), but I’ve fixed it here.

Posted by: John Baez on August 12, 2014 6:46 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

Sorry to bombard you with references, but you may also be interested in a point of view which relates the “super-representation theory” of the super-division algebras (equivalently the Clifford algebras) with the real-complex $K$-theory (semi-?)ring of a point.

This is one way of understanding the groups which appear in the $d=0$ column of various versions of the Periodic Table. As far as I see, and having asked some people involved in this work, there are some subtle issues in interpreting that column which have not been fully appreciated or recognised. This is mainly due to the fact that $\pi_0$ is somewhat different from the homotopy groups $\pi_d$ for $d\geq 1$.

Posted by: Guo Chuan Thiang on August 12, 2014 10:02 PM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

Your poster seems to be alluding to the electron shell model of atoms - L=2, K=8, L+K=10. Can we expect further developments with M=18 and a 28 fold way?

Posted by: RodMcGuire on August 7, 2014 6:32 PM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

Heh, since how we’re talking about wacky connections that don’t make sense, I might as well mention that my poster was actually alluding to this:

These are the 8 trigrams and yin-yang symbol as you often see them in Taoist stuff.

I don’t think there’s any deep connection, but I was amused to discover that in Wall’s approach, the 8-element Brauer-Wall group of $\mathbb{R}$ shows up as an extension of $\mathbb{Z}_2$ by $\mathbb{Z}_2$ by $\mathbb{Z}_2$. So, the 8 central superdivision algebras over $\mathbb{R}$ really are classified by the answers to 3 yes/no questions, and you could use trigrams to label them.

Posted by: John Baez on August 8, 2014 12:52 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

Dear John, Is there any way to extract some non-abelian Lie group of symmetries in your approach. I am thinking of the Standard Model group or some bigger group. Louis Crane proposed a couple of years ago a scheme based on the McKay correspondence between a finite group A4 and the Lie group E6. E_6 is a good grand unification symmetry, and the goal of Crane’s approach was to include discrete symmetries in particle physics.

Posted by: Aleksandar Mikovic on August 8, 2014 9:00 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

I don’t know how any of the stuff I’m doing here would be related to the Standard Model, grand unification, or $E_6$. It’s part of a line of thought that’s trying to understand charge conjugation and time reversal symmetry in condensed matter physics. A great introduction is here:

or for a more mathematical approach:

Posted by: John Baez on August 8, 2014 11:04 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

I’d just like to point out a typo on line two of the poster: “They can have an antiunitary T operator with C^2 = ±1 or…” should have been “…with T^2 or…”.

Posted by: Stefan on August 8, 2014 9:55 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

Whoops! I’ve already sent it off to the printer, but I’ll see if I can get it fixed in time. Anyway, I’ll fix the version here.

Posted by: John Baez on August 8, 2014 10:53 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

This Secret Blogging Seminar – The Brauer Groupoid - is very relevant. It points you to Fusion categories and homotopy theory which has

Proposition 4.9. $BrPic(Vec_k)$ is isomorphic to the classical Brauer group $Br(k)$.

And back to the blog post

the points in the Brauer-Picard groupoid are exactly the field extensions of k.

Posted by: David Corfield on August 8, 2014 10:03 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

Pp. 34-35 of these slides by Noah Snyder give the Brauer-Picard groupoid packaging of simple algebras over $\mathbb{R}$. You can see the three-ness of the three-fold way there.

Presumably this is straightforwardly generalizable to superalgebras.

Posted by: David Corfield on August 8, 2014 1:46 PM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

Thanks, that’s cool! I’m happy to avoid the use of fusion categories in what I’m doing here, but it provides a nice generalization, and it should be useful in relating the 10-fold way to ‘topological insulators’.

Posted by: John Baez on August 10, 2014 10:43 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

Tiny nitpick: For finite fields, $\mathbb{F}_{p^m}$ is an extension of $\mathbb{F}_{p^n}$ iff $n | m$. (Not $m \ge n$.)

Great article!

Posted by: Oisin McGuinness on August 9, 2014 5:25 PM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

Whoops, I’ll fix that. Thanks! Glad you liked the article.

Posted by: John Baez on August 10, 2014 9:20 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

I’m wondering whether there is a simpler way to define the [super] BW monoid. If $A$ and $B$ and simple [super] algebras over $k$, then $A\otimes B \cong \bigoplus_i D_i[n_i]$ where the $D_i$ are division algebras over $k$. Is it possible that $D_i \cong D_j$ for all $i$ and $j$? (Call this condition (X); I wanted to call it star/asterisk, but markdown/itex/whatever is being extremely annoying.)

This is true when $k = R$, for example.
Are there examples of $k$ for which (X) is known to fail?

If (X) is true for $k$, then we can define $[A] \bullet [B] = [D]$ where $D$ is the unique division algebra for which $A \otimes B \cong \bigoplus_i D[n_i]$

I apologize if I’m missing something obvious. I read the article several days ago (great article!), but I didn’t get around to asking this question until now.

Posted by: Kevin Walker on August 19, 2014 3:49 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 3)

Thanks! I don’t know if condition X always holds, but I fear that taking $k = \mathbb{R}$ as an example is a dangerous way to generate conjectures, because $\mathbb{R}$ has such a pathetically small collection of algebraic extensions. I suspect that something like $k = \mathbb{Q}$ would be much better to look at, but that veers over toward the other direction, of being unmanageably complicated for anyone except a professional algebraic number theorist.

Posted by: John Baez on August 19, 2014 12:30 PM | Permalink | Reply to this

Post a New Comment