## August 12, 2014

### The Tenfold Way (Part 4)

#### Posted by John Baez Back in 2005, Todd Trimble came out with a short paper on the super Brauer group and super division algebras, which I’d like to TeXify and reprint here.

In it, he gives extremely efficient proofs of several facts I alluded to last time. Namely:

• There are exactly 10 real division superalgebras.

• 8 of them have center $\mathbb{R}$, and these are Morita equivalent to the real Clifford algebras $Cliff_0, \dots, Cliff_7$.

• 2 of them have center $\mathbb{C}$, and these are Morita equivalent to the complex Clifford algebras $\mathbb{C}liff_0$ and $\mathbb{C}liff_1$.

• The real Clifford algebras obey

$Cliff_i \otimes_{\mathbb{R}} Cliff_j \simeq Cliff_{i + j mod 8}$

where $\simeq$ means they’re Morita equivalent as superalgebras.

It easily follows from his calculations that also:

• The complex Clifford algebras obey

$\mathbb{C}liff_i \otimes_{\mathbb{C}} \mathbb{C}liff_j \simeq \mathbb{C}liff_{i + j mod 2}$

These facts lie at the heart of the ten-fold way. So, let’s see why they’re true!

Before we start, two comments are in order. First, Todd uses Deligne’s term ‘super Brauer group’ where I decided to use ‘Brauer–Wall group’. Second, and more importantly, there’s something about Morita equivalence everyone should know.

In my last post I said that two algebras are Morita equivalent if they have equivalent categories of representations. Todd uses another definition which I actually like much better. It’s equivalent, it takes longer to explain, but it reveals more about what’s really going on. For any field $k$, there is a bicategory with

• algebras over $k$ as objects,

$A$-$B$ bimodules as 1-morphisms from the algebra $A$ to the algebra $B$, and

• bimodule homomorphisms as 2-morphisms.

Two algebras $A$ and $B$ over $k$ are Morita equivalent if they are equivalent in this bicategory; that is, if there’s a $A$-$B$ bimodule $M$ and a $B$-$A$ bimodule $N$ such that

$M \otimes_B N \cong A$

as an $A$-$A$ bimodule and

$N \otimes_A M \cong B$

as a $B$-$B$ bimodule. The same kind of definition works for Morita equivalence of superalgebras, and Todd uses that here.

So, with no further ado, here is Todd’s note.

## The super Brauer group and division superalgebras

### The super Brauer group

Let $SuperVect$ be the symmetric monoidal category of finite-dimensional super vector spaces over $\mathbb{R}$. By super algebra I mean a monoid in this category. There’s a bicategory whose objects are super algebras $A$, whose 1-morphisms $M: A \to B$ are left $A$- right $B$-modules in $V$, and whose 2-morphisms are homomorphisms between modules. This is a symmetric monoidal bicategory under the usual tensor product on $SuperVect$.

$A$ and $B$ are Morita equivalent if they are equivalent objects in this bicategory. Equivalence classes $[A]$ form an abelian monoid whose multiplication is given by the monoidal product. The super Brauer group of $\mathbb{R}$ is the subgroup of invertible elements of this monoid.

If $[B]$ is inverse to [A] in this monoid, then in particular $A \otimes (-)$ can be considered left biadjoint to $B \otimes (-)$. On the other hand, in the bicategory above we always have a biadjunction

$\begin{array}{ccl} A \otimes C \to D \\ ------ \\ C \to A^* \otimes D \end{array}$

essentially because left $A$-modules are the same as right $A^*$-modules, where $A^*$ denotes the super algebra opposite to $A$. Since right biadjoints are unique up to equivalence, we see that if an inverse to $[A]$ exists, it must be $[A^*]$.

This can be sharpened: an inverse to $[A]$ exists iff the unit and counit

$1 \to A^* \otimes A \qquad A \otimes A^* \to 1$

are equivalences in the bicategory. Actually, one is an equivalence iff the other is, because both of these canonical 1-morphisms are given by the same $A$-bimodule, namely the one given by $A$ acting on both sides of the underlying superspace of $A$ (call it $S$) by multiplication. Either is an equivalence if the bimodule structure map

$A^* \otimes A \to Hom(S, S),$

which is a map of superalgebras, is an isomorphism.

### $Cliff_1$

As an example, let $A = Cliff_1$ be the Clifford algebra generated by the 1-dimensional space $\mathbb{R}$ with the usual quadratic form $Q(x) = x^2$, and $\mathbb{Z}_2$-graded in the usual way. Thus, the homogeneous parts of $A$ are 1-dimensional and there is an odd generator $i$ satisfying $i^2 = -1$. The opposite $A^*$ is similar except that there is an odd generator $e$ satisfying $e^2 = 1$. Under the map

$A^* \otimes A \to Hom(S, S)$

where we write $S$ as a sum of even and odd parts $\mathbb{R} + \mathbb{R}i$, this map has a matrix representation

$e \otimes i \mapsto \left(\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array} \right)$

$1 \otimes i \mapsto \left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right)$

$e \otimes 1 \mapsto \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$

which makes it clear that this map is surjective and thus an isomorphism. Hence $[Cliff_1]$ is invertible.

One manifestation of Bott periodicity is that $[Cliff_1]$ has order 8. We will soon see a very easy proof of this fact. A theorem of C. T. C. Wall is that $[Cliff_1]$ in fact generates the super Brauer group; I believe this can be shown by classifying super division algebras, as discussed below.

### Bott periodicity

That $[Cliff_1]$ has order 8 is an easy calculation. Let $Cliff_r$ denote the $r$-fold tensor power of $Cliff_1$. $Cliff_2$ for instance has two supercommuting odd elements $i, j$ satisfying $i^2 = j^2 = -1$; it follows that $k \;:= i j$ satisfies $k^2 = -1$, and we get the usual quaternions, graded so that the even part is the span $\langle 1, k\rangle$ and the odd part is $\langle i, j\rangle$.

$Cliff_3$ has three supercommuting odd elements $i, j, l,$ all of which are square roots of $-1$. It follows that $e = i j l$ is an odd central involution (here ‘central’ is taken in the ungraded sense), and also that $i' = j l$, $j' = l i$, $k' = i j$ satisfy the Hamiltonian equations

$(i')^2 = (j')^2 = (k')^2 = i'j'k' = -1,$

so we have $Cliff_3 = \mathbb{H}[e]/\langle e^2 - 1\rangle$. Note this is the same as

$\mathbb{H} \otimes Cliff_1^*$

where the $\mathbb{H}$ here is the quaternions viewed as a super algebra concentrated in degree 0 (i.e. is purely bosonic).

Then we see immediately that $Cliff_4 = Cliff_3 \otimes Cliff_1$ is equivalent to purely bosonic $\mathbb{H}$ (since the $Cliff_1$ cancels $Cliff_1^\ast$ in the super Brauer group).

At this point we are done: we know that conjugation on (purely bosonic) $\mathbb{H}$ gives an isomorphism

$\mathbb{H}^* \cong \mathbb{H}$

hence $[{\mathbb{H}}]^{-1} = [\mathbb{H}^*] = [\mathbb{H}]$, i.e. $[\mathbb{H}] = [Cliff_4]$ has order 2! Hence $[Cliff_1]$ has order 8.

### The super Brauer clock

All this generalizes to arbitrary Clifford algebras: if a real quadratic vector space $(V, Q)$ has signature $(r, s)$, then the superalgebra $Cliff(V, Q)$ is isomorphic to $A^{\otimes r} \otimes {A^*}^{\otimes s}$, where $A^{\otimes r}$ denotes the $r$-fold tensor product of $A = Cliff_1$. By the above calculation we see tha $Cliff(V, Q)$ is equivalent to $Cliff_{r-s}$ where $r-s$ is taken modulo 8.

For the record, then, here are the hours of the super Brauer clock, where $e$ denotes an odd element, and $\simeq$ denotes Morita equivalence:

$\begin{array}{ccl} Cliff_0 & \simeq & \mathbb{R} \\ Cliff_1 & \simeq & \mathbb{R} + \mathbb{R}e, \quad e^2 = -1 \\ Cliff_2 & \simeq & \mathbb{C} + \mathbb{C}e, \quad e^2 = -1, e i = -i e \\ Cliff_3 & \simeq & \mathbb{H} + \mathbb{H}e, \quad e^2 = 1, e i = i e, e j = j e, e k = k e \\ Cliff_4 & \simeq & \mathbb{H} \\ Cliff_5 & \simeq & \mathbb{H} + \mathbb{H}e, \quad e^2 = -1, e i = i e, e j = j e, e k = k e \\ Cliff_6 & \simeq & \mathbb{C} + \mathbb{C} e, \quad e^2 = 1, e i = -i e \\ Cliff_7 & \simeq & \mathbb{R} + \mathbb{R}e, \quad e^2 = 1 \end{array}$

All the superalgebras on the right are in fact division superalgebras, i.e. superalgebras in which every nonzero homogeneous element is invertible.

To prove Wall’s result that $[Cliff_1]$ generates the super Brauer group, we need a lemma: any element in the super Brauer group is the class of a central division superalgebra: that is, one with $\mathbb{R}$ as its center.

Then, if we classify the division superalgebras over $\mathbb{R}$ and show the central ones are Morita equivalent to $Cliff_0, \dots, Cliff_7$, we’ll be done.

### Classifying real division superalgebras

I’ll take as known that the only associative division algebras over $\mathbb{R}$ are $\mathbb{R}, \mathbb{C}, \mathbb{H}$ — the even part $A$ of an associative division superalgebra must be one of these cases. We can express the associativity of a superalgebra (with even part $A$) by saying that the odd part $M$ is an $A$-bimodule equipped with a $A$-bimodule map pairing

$\langle - , - \rangle : M \otimes_A M \to A$ such that:

$a\langle b, c\rangle = \langle a, b\rangle c \; for \; all \; a, b, c \in M \qquad (\star)$

If the superalgebra is a division superalgebra which is not wholly concentrated in even degree, then multiplication by a nonzero odd element induces an isomorphism

$A \to M$

and so $M$ is 1-dimensional over A; choose a basis element $e$ for $M$.

The key observation is that for any $a \in A$, there exists a unique $a' \in A$ such that

$a e = e a'$

and that the $A$-bimodule structure forces $(a b)' = a'b'$. Hence we have an automorphism (fixing the real field) $(-)': A \to A$

and we can easily enumerate (up to isomorphism) the possibilities for associative division superalgebras over $\mathbb{R}$:

1. $A = \mathbb{R}$. Here we can adjust $e$ so that $e^2 \; := \langle e, e\rangle$ is either $-1$ or $1$. The corresponding division superalgebras occur at 1 o’clock and 7 o’clock on the super Brauer clock.

2. $A = \mathbb{C}$. There are two $\mathbb{R}$-automorphisms $\mathbb{C} \to \mathbb{C}$. In the case where the automorphism is conjugation, condition $(\star)$ for super associativity gives $\langle e, e\rangle e = e\langle e, e\rangle$ so that $\langle e, e\rangle$ must be real. Again $e$ can be adjusted so that $\langle e, e\rangle$ equals $-1$ or $1$. These possibilities occur at 2 o’clock and 6 o’clock on the super Brauer clock.

For the identity automorphism, we can adjust $e$ so that $\langle e, e \rangle$ is 1. This gives the super algebra $\mathbb{C}[e]/\langle e^2 - 1\rangle$ (where $e$ commutes with elements in $\mathbb{C}$). This does not occur on the super Brauer clock over $\mathbb{R}$. However, it does generate the super Brauer group over $\mathbb{C}$ (which is of order two).

3. $A = \mathbb{H}$. Here $\mathbb{R}$-automorphisms $\mathbb{H} \to \mathbb{H}$ are given by $h \mapsto x h x^{-1}$ for $x \in \mathbb{H}$. In other words

$h e = e x h x^{-1}$

whence $e x$ commutes with all elements of $\mathbb{H}$ (i.e. we can assume wlog that the automorphism is the identity). The properties of the pairing guarantee that $h\langle e, e\rangle = \langle e, e\rangle h$ for all $h in \mathbb{H}$, so $\langle e, e \rangle$ is real and again we can adjust $e$ so that $\langle e, e\rangle$ equals $1$ or $-1$. These cases occur at 3 o’clock and 5 o’clock on the super Brauer clock.

This appears to be a complete (even if a bit pedestrian) analysis.

Posted at August 12, 2014 1:46 AM UTC

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### Re: The Ten-Fold Way (Part 4)

While Todd humbly calls his analysis ‘pedestrian’, I am a bit slow at elementary algebraic tricks, so let me expand this a bit more:

Here $\mathbb{R}$-automorphisms $\mathbb{H} \to \mathbb{H}$ are given by $h \mapsto x h x^{-1}$ for $x \in \mathbb{H}$. In other words

$h e = e x h x^{-1}$

whence $e x$ commutes with all elements of $\mathbb{H}$ (i.e. we can assume wlog that the automorphism is the identity).

He’s starting with the known fact that all automorphisms of $\mathbb{H}$ are inner and the fact that $h e = e h'$ where $h'$ is the result of applying an automorphism to $h$, obtaining

$h e = e x h x^{-1}$

then multiplying both sides to get

$h e x = e x h$

But this means we can replace $e$ by $f = e x$ and get a new odd element $f$ with

$h f = f h$

reducing to the case where the automorphism is trivial.

Posted by: John Baez on August 12, 2014 8:51 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 4)

If you take the Complex Clifford Algebra CCl(2) of periodicity 2 and then take the completion of the union of all its tensor products you get the hyperfinite II1 factor von Neumann algebra (see your week 175) that is “… the smallest von Neumann algebra containing the creation and annihilation operators on a fermionic Fock space of countably infinite dimension …”.

Could you comment on what you would get if you looked at the Real Clifford Algebra Cl(8) of periodicity 8 in the same way, by making the completion of the union of all its tensor products, and how such an algebra might be interpreted physically as an Algebraic Quantum Field Theory ?

Posted by: Tony Smith on August 17, 2014 8:55 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 4)

I don’t have anything great to say about this, except that somebody studying real C*-algebras and von Neumann algebras should pursue this analogy! Maybe someone already has; at least there exist people who study those subjects. For example:

Li Bingren, Real operator algebras, 1996.

has some basic results on the classification of real von Neumann algebras (near the end), but nothing about a hyperfinite II1 factor.

Posted by: John Baez on August 19, 2014 12:24 PM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 4)

I am not an expert, but my impression is that the hyperfinite II_1 factor shows up all over the place. There’s a reason for the word “the” in its name. (And a non-reason: different copies of it tend not to be canonically isomorphic.)

Posted by: Theo on September 2, 2014 3:00 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 4)

I hope I do not sound too unpolite, if I say that I mostly skipped through the last two posts. In particular I needed to skip thinking about questions which popped up, when skipping, like here:

where we write S as a sum of even and odd parts $\mathbb{R} + \mathbb{R}i$, this map has a matrix representation …

the question: “why would he come up with such a representation?”

needed to be skipped.

Anyways, as said already I just wanted to get a glimpse on whats going on here and at least I think I have now understood that you want to assign this topological insulator classification to the “product table.” That is the dimension in this classification table seems to be thought to refer to the respective Clifford algebra (like (up to cyclic ambiguity) let $dim j \simeq Cliff_{j-1}$). Moreover following your matter clock one can replace the algebra notations with the respective Clifford algebra (like A1 becomes $Cliff_0$ etc.) and following Todd Trimbles formula for the tensor product of Clifford algebras perform the product (where the dimensioanlity had been replaced by the Clifford algebra), which is again a Clifford algebra however with some strange constraints due to those $T$ and $C$ operators. The “homotopy classes” (or something else) of those Clifford Algebras with strange constraints are thus (up to cyclic ambiguity) : $Cliff_0 \simeq 0$,$Cliff_1 \simeq 0$,$Cliff_2 \simeq \mathbb{Z}$, $Cliff_3 \simeq 0$,$Cliff_4 \simeq \mathbb{Z}_2$,$Cliff_5 \simeq \mathbb{Z}_2$, $Cliff_6 \simeq \mathbb{Z}$, $Cliff_7 \simeq 0$.

Is that right?

### Re: The Ten-Fold Way (Part 4)

I wrote:

…at least I think I have now understood that you want to assign this topological insulator classification to the “product table.”That is the dimension in this classification table seems to be thought to refer to the respective Clifford algebra (like (up to cyclic ambiguity) let $dim j \simeq Cliff_{j−1}$

John - If you didn’t intended to make this assignment then I am aware that this assignment may sound eventually a bit (mildly stated) “unconventional”….

However in order to decide “how unconventional” one would probably need to understand that “Bogoliubov deGennes formalism” which seems not even listed on Google (at least I couldn’t find it in my Google search engine universe). It seems you know more about this, that is in particular you came up with that “topological order” table, where I don’t know where you took that from (it doesn’t seem to be in the Zirnbauer et al paper upon first glance). Where is it from? So in short: whatever you know about that formalism this assignment may eventually sound weird to you. If it does - it may be nice to hear that you do so and eventually a few words why. Likewise if this assignment is a well-known fact than it would be nice to hear that it is.

### Re: The Ten-Fold Way (Part 4)

In particular I needed to skip thinking about questions which popped up, when skipping, like here:

where we write S as a sum of even and odd $\mathbb{R} + \mathbb{R}i$, this map has a matrix representation…

the question: “why would he come up with such a representation?”

I know this is what not what you wanted to talk about, but I can’t resist answering your question. This matrix representation is something you’d find in any book on Clifford algebras. The idea is this:

The Clifford algebra Todd calls $A$ is built from the real numbers together with a square root of $-1$. This is the complex numbers, and it’s pretty well-known that you can think of complex numbers as special $2 \times 2$ real matrices, using the map

$i \mapsto \left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right)$

The Clifford algebra Todd calls $A^*$ consists of the real numbers together with a square root of $+1$ — an extra one, besides $\pm 1$. This algebra is much less famous than the complex numbers, but it’s still well-known enough to have a Wikipedia entry: it’s called the split complex numbers. If we we denote this extra square root of $+1$ by $e$, you can think of the split complex numbers as special $2 \times 2$ real matrices using the map

$e \mapsto \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$

Todd’s goal was to understand the algebra $A^* \otimes A$. This consists of the real numbers together with a square root of $-1$, namely $1 \otimes i$, and a square root of $+1$, namely $e \otimes 1$. These anticommute, since we’ve defined $i$ and $e$ to be odd (that is, fermionic).

You can think of this algebra as consisting of $2 \times 2$ matrices using the tricks we’ve already seen:

$1 \otimes i \mapsto \left(\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array} \right)$

$e \otimes 1 \mapsto \left(\begin{array}{cc} 0 & 1 \\ 1 & 0 \end{array} \right)$

You can see that these two matrices anticommute. We also get

$e \otimes i \mapsto \left(\begin{array}{cc} -1 & 0 \\ 0 & 1 \end{array} \right)$

and of course

$1 \otimes 1 \mapsto \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)$

But these 4 matrices span the space of all $2 \times 2$ real matrices1 So $A^* \otimes A$ is isomorphic to the algebra of $2 \times 2$ real matrices. And that’s what Todd wanted to show.

(But again, this is not Todd’s invention; this is standard Clifford algebra theory.)

Posted by: John Baez on August 29, 2014 8:41 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 4)

Is that right?

Yes, that’s right. I don’t know much about ‘topological insulators’, so I prefer to think in a more mundane way about quantum systems whose Hamiltonians either don’t have or have time reversal symmetry (that can square to 1 or -1), and either don’t have or have charge conjugation symmetry (that can square to 1 or -1). Then tensoring such systems corresponds to tensoring Clifford algebras.

If I had time I would write another post on this subject, which would make this aspect more clear! I didn’t really get around to explaining it yet, and I didn’t even get around to fully understanding it. (I need to explain things to understand them, and vice versa.)

Posted by: John Baez on August 29, 2014 8:48 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 4)

But these 4 matrices span the space of all $2 \times 2$ real matrices1 So $A^* \otimes A$ is isomorphic to the algebra of $2 \times 2$ real matrices. And that’s what Todd wanted to show.

(But again, this is not Todd’s invention; this is standard Clifford algebra theory.)

Aha. Sorry I don’t know much about Clifford Algebras. Is there a standard textbook where this representation is mentioned?

Anyways do I get this right - the multiplication formula:

$Cliff_i \otimes_{\mathbb{R}} Cliff_j \simeq Cliff_{i + j mod 8}$

is then also standard Clifford algebra theory?

Furthermore you wrote:

Yes, that’s right. I don’t know much about ‘topological insulators’, so I prefer to think in a more mundane way about quantum systems whose Hamiltonians either don’t have or have time reversal symmetry (that can square to 1 or -1), and either don’t have or have charge conjugation symmetry (that can square to 1 or -1). Then tensoring such systems corresponds to tensoring Clifford algebras.

Yes a quick glance into Moore (page 237) reveals:

Now we will give an elegant description of how the 10 classical symmetric spaces arise directly from the representations of Clifford algebras

I also don’t know next to nothing about topological insulators and in particular it seems that my pommeranian farmer background prevents me even from appreciating the elegance in a physics course book, that is I don’t understand next to nothing from that explanation in Moore. I guess one needs some lectures on homogenous spaces, Clifford algebras and the like in order to understand this. But as said I just want to get a rough understanding of the involved structural features.

So one question: If I would choose a matrix representation for the elements of the homogenous space and a matrix representation for the Clifford algebra’s together with those $T$ and $C$ constraints, would those representations be isomorphic?

I meanwhile saw that the first table in here is also from the paper Ryu et al.

On a first glance it looks as if the entries in the table (like for the real Clifford algebras) are $\pi_d(HomSpace),$ where d is the dimension of the boundary of a topological insulator.

That is they write (p. 10):

A problem of Anderson localization is in general described by a random Hamiltonian (i.e., one that lacks translational symmetry). That Hamiltonian will be in one of the ten symmetry classes listed in table 1 and we are currently focusing on Hamiltonians describing the boundary of the topological insulator (superconductor). Now, as it turns out, at long length scales (much larger thanthe “mean free path”) a description in terms of a “non-linear-sigma-model” (NL$\sigma$M) emerges.

There are no literature reference or further explanations (at least not which I have found) why this is so.

They further write …

It is the homotopy groups of the NL$\sigma$M target spaces $G/H$ which determine whether it is possible to add such a topological term to a given NL$\sigma$M action (see table 2). Specifically, a $theta$-term (Pruisken term) can appear when $\pi_d(G/H)= \mathbb{Z},$.. and a WZW term can be included when $\pi_{d+1}(G/H)= \mathbb{Z}.$ (see also p. 12)

Those extra terms are necessary for evading Anderson localization (which I guess is relevant for superconductivity :) ):

Now, the NL$\sigma$M on the (d-1) dimensional boundary of the d dimensional topological insulator (superconductor) completely evades Anderson localization if a certain extra term of topological origin can be added to the action of the NL$\sigma$M.

I haven’t understood why they think so, it seems ? at least partially this is motivated by the discussion of the properties of (massless) Dirac Hamiltonians.

But concluding - their interpretation of the table seems a bit different from the one which refers to a product table.

### Re: The Ten-Fold Way (Part 4)

Sorry I don’t know much about Clifford algebras. Is there a standard textbook where this representation is mentioned?

There are a number of textbooks on Clifford algebras, which take different approaches. Two of my favorites are:

• Paul Budinich and Andrzej Trautman, The Spinorial Chessboard.

and

The first focuses on Clifford algebras and their representations (called ‘spinors’ and ‘pinors’). The second is mainly a detailed study of the Dirac operator and its applications to differential geometry and topology… but it starts with a self-contained introduction to Clifford algebras and spinors. If you look at section 4, you’ll see all the facts about $Cl_{r,s}$, the Clifford algebra generated by $r$ square roots of $-1$ and $s$ square roots of $+1$, all anticommuting.

Anyways do I get this right - the multiplication formula:

$Cliff_i \otimes_{\mathbb{R}} Cliff_j \simeq Cliff_{i + j mod 8}$

is then also standard Clifford algebra theory?

Yes. Here $Cliff_i = Cl_{i,0}$ is the Clifford algebra generated by $i$ anticommuting square roots of $-1$, $\otimes$ is the tensor product of superalgebras, and $\simeq$ means Morita equivalence. If you prefer thinking about isomorphism of superalgebras, you should remember

$Cliff_i \otimes_{\mathbb{R}} Cliff_j \cong Cliff_{i + j}$

where $\otimes$ is the tensor product of superalgebras and $\cong$ is isomorphism, and also

$Cliff_{i+8} \cong Cliff_i \otimes Cliff_8$

$Cliff_8$ is the algebra of $16 \times 16$ real matrices.

More generally we have

$Cl_{r+8,s} \cong Cl_{r,s+8} \cong Cl_{r,s} \otimes Cliff_8$

This gives rise to a nice $8 \times 8$ square pattern of Clifford algebras which you can find in Michelson and Lawson’s book, and which is the reason for the phrase ‘spinorial chessboard’. All this stuff about the number 8 is also the reason the octonions exist.

I taught a course about this stuff, and you can read my course notes. This subject is so beautiful that I’ve long wanted to write a textbook about it… but there are already plenty of textbooks about it, so I’ve decided it would be a bad use of time, even though it would be fun. Maybe when I retire.

Posted by: John Baez on August 30, 2014 8:47 AM | Permalink | Reply to this

### Re: The Ten-Fold Way (Part 4)

John, many thanks for the references to textbooks on Clifford algebras! It is quite remarkable that the one by H.B. Lawson and M. Michelsohn is fully openly available.

And of course if you feel like making further comments like regarding the remarks to my questions to the Ryu et al article that would be appreciated too.

As a matter of fact on a first glance it seems that the Bogoliubov de Gennes formalism or parts of it seems at least controversial, like I had found a critizising website at http://www.superconductivity.us/index.html of someone who is according to the Vita displayed there: http://www.superconductivity.us/credentials.html a former Harvard postdoc and currently a professor for physics in Puerto Rico. He critisizes the de deGennes equations on that page: http://www.superconductivity.us/why-new-theory.html

Frankly, I don’t understand a word from what he writes - apparently there is some pairing problem, but I have already problems to understand what’s supposed to be combined or paired here. I just wanted to let you know. In particular I can’t really say wether this link is useful for you or not.