Introduction E9F5FC

Questions FFFFC0

• View
• Edit
• History
• Print

Bott periodicity, Super division algebras

I'm studying

• Todd Trimble. The Super Brauer Group and Super Division Algebras

Super Brauer Group

Let {$\textrm{SuperVect}$} be the symmetric monoidal category of finite-dimensional super vector spaces over {$\mathbb{R}$}.

• A super vector space {$V=V_0+V_1$} is a {$\mathbb{Z}_2$}-graded vector space.
• The tensor product is given by {$(V\otimes W)_0 = (V_0\otimes W_0)\oplus(V_1\otimes W_1)$} and {$(V\otimes W)_1 = (V_0\otimes W_1)\oplus(V_1\otimes W_0)$}.
• A symmetric monoidal category has a tensor product {$A\otimes B$} which is symmetric in that {$A\otimes B$} is naturally isomorphic to {$B\otimes A$} for all objects {$A,B$}.
• Naturally ismomorphic means...

By super algebra I mean a monoid in this category. There's a bicategory whose objects are super algebras , whose 1-morphisms are left - right-modules in , and whose 2-morphisms are homomorphisms between modules. This is a symmetric monoidal bicategory under the usual tensor product on .

 and 
 are Morita equivalent if they are equivalent objects in this bicategory. Equivalence classes 
 form an abelian monoid whose multiplication is given by the monoidal product. The super Brauer group of 
 is the subgroup of invertible elements of this monoid.

If

 is inverse to [A] in this monoid, then in particular 
 can be considered left biadjoint to 

. On the other hand, in the bicategory above we always have a biadjunction

essentially because left -modules are the same as right -modules, where

 denotes the super algebra opposite to 

. Since right biadjoints are unique up to equivalence, we see that if an inverse to

 exists, it must be 

. This can be sharpened: an inverse to

 exists iff the unit and counit

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 -bimodule, namely the one given by

 acting on both sides of the underlying superspace of 
 (call it 

) by multiplication. Either is an equivalence if the bimodule structure map which is a map of superalgebras, is an isomorphism.

As an example, let

 be the Clifford algebra generated by the 1-dimensional space 
 with the usual quadratic form 

, and -graded in the usual way. Thus, the homogeneous parts of

 are 1-dimensional and there is an odd generator 
 satisfying 

. The opposite

 is similar except that there is an odd generator 
 satisfying 

. Under the map where we write

 as a sum of even and odd parts 

, this map has a matrix representation

which makes it clear that this map is surjective and thus an isomorphism. Hence

 is invertible.

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

Bott periodicity That

 has order 8 is an easy calculation. Let 
 denote the 

-fold tensor power of .

 for instance has two supercommuting odd elements 
 satisfying 

; it follows that

 satisfies 

, and we get the usual quaternions, graded so that the even part is the span

 and the odd part is 

.

 has three supercommuting odd elements 
 all of which are square roots of 

. It follows that

 is an odd central involution (here 'central' is taken in the ungraded sense), and also that 

, ,

 satisfy the Hamiltonian equations

so we have . Note this is the same as where the

 here is the quaternions viewed as a super algebra concentrated in degree 0 (i.e. is purely bosonic).

Then we see immediately that

 is equivalent to purely bosonic 
 (since the 
 cancels 
 in the super Brauer group).

At this point we are done: we know that conjugation on (purely bosonic)

 gives an isomorphism

hence , i.e.

 has order 2! Hence 
 has order 8.

The super Brauer clock All this generalizes to arbitrary Clifford algebras: if a real quadratic vector space

 has signature 

, then the superalgebra

 is isomorphic to 

, where

 denotes the 

-fold tensor product of . By the above calculation we see that

 is equivalent to 
 where 
 is taken modulo 8.

For the record, then, here are the hours of the super Brauer clock, where

 denotes an odd element, and 
 denotes Morita equivalence:

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 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

 as its center.

Then, if we classify the division superalgebras over

 and show the central ones are Morita equivalent to 

, we'll be done.