The n-Category Café

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.