## The Super Brauer Group and Super Division Algebras

### The super Brauer group

Let $\mathrm{SuperVect}$ 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 $\mathrm{SuperVect}$.

$A$ 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 $$A^* \otimes A \to \mathrm{Hom}(S, S),$$ which is a map of superalgebras, is an isomorphism.

### $\mathrm{Cliff}_1$

As an example, let $A = \mathrm{Cliff}_1$ 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 $$A^* \otimes A \to \mathrm{Hom}(S, S)$$ 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 $[\mathrm{Cliff}_1]$ is invertible.

One manifestation of Bott periodicity is that $[\mathrm{Cliff}_1]$ has order 8. We will soon see a very easy proof of this fact. A theorem of C. T. C. Wall is that $[\mathrm{Cliff}_1]$ in fact generates the super Brauer group; I believe this can be shown by classifying super division algebras, as discussed below.

### Bott periodicity

That $[\mathrm{Cliff}_1]$ has order 8 is an easy calculation. Let $\mathrm{Cliff}_r$ denote the $r$-fold tensor power of $\mathrm{Cliff}_1$. $\mathrm{Cliff}_2$ 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 $\langle i, j\rangle$.

$\mathrm{Cliff}_3$ 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 $\mathrm{Cliff}_3 = \mathbb{H}[e]/\langle e^2 - 1\rangle$. Note this is the same as $$\mathbb{H} \otimes \mathrm{Cliff}_1^*$$ 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 $\mathrm{Cliff}_4 = \mathrm{Cliff}_3 \otimes \mathrm{Cliff}_1$ is equivalent to purely bosonic $\mathbb{H}$ (since the $\mathrm{Cliff}_1$ cancels $\mathrm{Cliff}_1^\ast$ 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. $[\mathbb{H}] = [\mathrm{Cliff}_4]$ has order 2! Hence $[\mathrm{Cliff}_1]$ 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 $\mathrm{Cliff}(V, Q)$ is isomorphic to $A^{\otimes r} \otimes {A^*}^{\otimes s}$, where $A^{\otimes r}$ denotes the $r$-fold tensor product of $A = \mathrm{Cliff}_1$. By the above calculation we see that $\mathrm{Cliff}(V, Q)$ is equivalent to $\mathrm{Cliff}_{r-s}$ 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: $$\begin{array}{ccl} \mathrm{Cliff}_0 & \simeq & \mathbb{R} \\ \mathrm{Cliff}_1 & \simeq & \mathbb{R} + \mathbb{R}e, \quad e^2 = -1 \\ \mathrm{Cliff}_2 & \simeq & \mathbb{C} + \mathbb{C}e, \quad e^2 = -1, e i = -i e \\ \mathrm{Cliff}_3 & \simeq & \mathbb{H} + \mathbb{H}e, \quad e^2 = 1, e i = i e, e j = j e, e k = k e \\ \mathrm{Cliff}_4 & \simeq & \mathbb{H} \\ \mathrm{Cliff}_5 & \simeq & \mathbb{H} + \mathbb{H}e, \quad e^2 = -1, e i = i e, e j = j e, e k = k e \\ \mathrm{Cliff}_6 & \simeq & \mathbb{C} + \mathbb{C} e, \quad e^2 = 1, e i = -i e \\ \mathrm{Cliff}_7 & \simeq & \mathbb{R} + \mathbb{R}e, \quad e^2 = 1 \end{array}$$ 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 $[\mathrm{Cliff}_1]$ 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 $\mathrm{Cliff}_0, \dots, \mathrm{Cliff}_7$, 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: $$a\langle b, c\rangle = \langle a, b\rangle c \;\; \textrm{for all} \; a, b, c \in M \qquad (\star)$$ 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.