The Super Brauer Group and Super Division Algebras

Todd Trimble

April 27, 2005

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

A and B are super 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 R is the 
subgroup of invertible elements of this monoid. 

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

                    A tensor C --> D 
                   -------------------
                    C --> A* tensor D

(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 --> A* tensor A       A tensor A* --> 1

are equivalences in the bicategory.  Actually, one is an 
equivalence iff the other is, because both of these 
canonical 1-cells 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* tensor A --> Hom(S, S),    

which is a map of super algebras, is an isomorphism. 

2.  As an example, let  A = C1  be the Clifford algebra 
generated by the 1-dimensional space R with the usual 
quadratic form Q(x) = |x|2,  and Z/2-graded in the usual 
way.  Thus, the homogeneous parts of A are 1-dimensional 
and there is an odd generator i satisfying i2 = -1.  The 
opposite A* is similar except that there is an odd generator 
e satisfying  e2 = 1.  Under the map 

           A* tensor A --> Hom(S, S), 

where we write S as a sum of even and odd parts R + Ri, 
this map has a matrix representation 

                               -1   0
             e tensor i  |-->   0   1

                                0  -1 
             1 tensor i  |-->   1   0

                                0   1
             e tensor 1  |-->   1   0

which makes it clear that this map is surjective and thus 
an isomorphism.  Hence [C1]  is invertible. 

One manifestation of Bott periodicity is that [C1] 
has order 8.  We will soon see a very easy proof of this 
fact.  A theorem of C.T.C. Wall is that [C1] in fact 
generates the super Brauer group; I believe this can 
be shown by classifying super division algebras, as 
discussed below. 

3. That [C1] has order 8 is an easy calculation.  Let 
Cr  denote the r-fold tensor of C1.  C2 for instance 
has two super-commuting odd elements  i, j  satisfying 
i2 = j2 = -1;  it follows that  k := ij  satisfies  
k2 = -1, and we get the usual quaternions, graded so that 
the even part is the span  ⟨1, k⟩  and the odd part is ⟨i, j⟩. 

C3 has three super-commuting odd elements i, j, l, all 
of which are square roots of  -1.  It follows that 
e = ijl is an odd central involution (here "central" 
is taken in the ungraded sense), and also that  
i' = jl,  j' = li,  k' = ij satisfy the Hamiltonian equations 

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

so we have  C3 = H[e] mod (e2 - 1).   Note this is the 
same as 

                 H tensor (C1)*

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

Then we see immediately that  C4 = C3 tensor C1 is 
equivalent to purely bosonic H  (since the C1 cancels 
(C1)*  in the super Brauer group). 

At this point we are done: we know that conjugation on 
(purely bosonic) H gives an isomorphism 

                  H* --> H 

hence  [H]-1 = [H*] = [H],  i.e. [H] = [C4] has order 2!  
Hence  [C1]  has order 8. 

4.  All this generalizes to Clifford algebras: if a real quadratic 
vector space  (V, Q)  has signature (r, s), then the super 
algebra Cliff(V, Q)  is isomorphic to  Ar tensor (A*)s, 
where  Ar  denotes r-fold tensor product of  A = C1. 
By the above calculation we see that  Cliff(V, Q)  is 
equivalent to  Cr-s  where r-s is taken modulo 8. 

For the record, then, here are the hours of the super 
Clifford clock, where e denotes an odd element, and ~ 
denotes equivalence: 

C0 ~ R
C1 ~ R + Re,  e2 = -1
C2 ~ C + Ce,  e2 = -1, ei = -ie
C3 ~ H + He,  e2 = 1,  ei = ie, ej = je, ek = ke
C4 ~ H
C5 ~ H + He,  e2 = -1, ei = ie, ej = je, ek = ke
C6 ~ C + Ce,  e2 = 1,  ei = -ie
C7 ~ R + Re,  e2 = 1

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

To prove Wall's result that [C1] generates the super 
Brauer group, we need a lemma: any element in the super 
Brauer group is the class of a super division algebra. 

For this, we need the classification of super division algebras. 
 
I'll take as known that the only associative division algebras 
over R are R, C, H  -- the even part A of an associative super 
division algebra must be one of these cases.  We can 
express the associativity of a super algebra (with even part 
A) by saying that the odd part M is an A-bimodule equipped 
with a A-bimodule map pairing 
 
     < -, - >:  M tensor_A M --> A
 
such that:

(**)  a<b, c> = <a, b>c  for all a, b, c in M.  
 
If the super algebra is a super division algebra which is not 
wholly concentrated in even degree, then multiplication by a 
nonzero odd element induces an isomorphism 
 
                    A --> 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 
 
                   ae = ea'  
 
and that the A-bimodule structure forces  (ab)' = a'b'.  Hence we 
have an automorphism (fixing the real field) 
 
              (--)': A --> A
 
and we can easily enumerate (up to isomorphism) the possibilities 
for associative super division algebras over R: 
 
(1)  A = R.  Here we can adjust e so that either  e2 := <e, e>  is 
-1 or 1.  The corresponding super division algebras occur at 1 o'clock 
and 7 o'clock on the super Brauer clock. 
 
(2)  A = C.  There are two R-automorphisms  C --> C.  In the case 
where the automorphism is conjugation, condition (**) for super 
associativity gives  <e, e>e = e<e, e>  so that  <e, e>  must be 
real.  Again e can be adjusted so that  <e, e> = -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  <e, e>
is 1.  This gives the super algebra  C[e] mod e2 - 1  (where e 
commutes with elements in C).  This does not occur on the 
super Brauer clock over R.  However, it does generate the super 
Brauer group over C (which is of order two). 
 
(3)  A = H.  Here R-automorphisms  H --> H  are given by 
h |--> xhx-1  for x in H.  In other words 
 
                     he = exhx-1
 
whence  ex  commutes with all elements of H (i.e. we can 
assume wlog that the automorphism is the identity).  The 
properties of the pairing guarantee that  h<e, e> = <e, e>h 
for all h in H, so  <e, e>  is real and again we can adjust 
e so that  <e, e> = 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. 
 


© 2005 Todd H. Trimble

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