March 26, 2005

This Week's Finds in Mathematical Physics (Week 212)

John Baez

As you may know, theoretical particle physics is highly enamored of "supersymmetry" these days. This is not because there's a shred of experimental evidence for it - there's not - but just because it's such a cool idea from a mathematical point of view. Mathematicians should have gotten this idea and run with it first, but physicists did - and maybe it's turned them into mathematicians.

The unarguable central core of this idea is that everything is made of bosons and fermions. In the Standard Model, most bosons are "force carriers", like photons, which carry the electromagnetic force. Fermions are more like what we'd normally call "matter": leptons and quarks, for example. The one big exception is the Higgs boson, which gives elementary particles their mass and... umm... hasn't been seen yet!

But, at a more fundamental level, the really important thing is that bosons commute:

xy = yx
while fermions anticommute:
xy = -yx
Also, in case you're wondering, bosons commute with fermions.

But already, most mathematicians reading this will be confused and unhappy. What does it mean for two particles to commute, much less anticommute? Does an apple commute with a grape? Here in the suburbs of Los Angeles almost everyone commutes, but that's not what we're talking about.

The whole idea of particles commuting or anticommuting only occurred to people after they invented quantum theory, where the state of any system is described by a unit vector in some Hilbert space. In quantum theory, if you have a system in some state x, and you check to see if it's in the state y, your experiment gives you the answer "yes" with probability

the square of the absolute value of the inner product of x and y.

There! Now you know quantum theory.

Given this setup, when you have a system consisting of two particles, the first in some state x and the second in some state y, it's natural to write the state of the whole system as a kind of product xy. But then you have to figure out what rules you want this product to satisfy!

If you require it to be commutative:

xy = yx
you're saying that there's no difference between the first particle being in state x and the second particle being in state y, and the other way around. In other words, the particles don't have little name tags on them saying who they are.

This seems reasonable, and particles satisfying this rule are called "bosons". But, there's another popular option, called "fermions":

xy = -yx
Here again, the particles don't have name tags, since if we put the whole system in the state xy and check to see if it's in the state yx, we get the same answer as when we check to see if it's in the state xy! See:
|<xy,yx>|2 = |<xy,-xy>|2

           = |<xy,xy>|2
thanks to the absolute value. This means that the states xy and yx are indistinguishable.

Reading what I just said, you'd be forgiven for wondering what's the big difference between fermions and bosons! After all, that absolute value in the formula for probabilities just ignores minus signs.

One difference is the "Pauli exclusion principle". Take a pair of fermions and check to see if they're both in the state x. The probability is always zero, since

xx = -xx  
so xx = 0. So, fermions are antisocial: that's why the electrons in an atom form "shells" with different electrons in different states, instead of all hanging out at the lowest energy state.

Bosons, by contrast, are gregarious: when a store clerk uses a laser scanner to ring up your purchases, that beam of red light is a bunch of photons all in the same state! A laser is a quintessentially quantum-theoretic gadget - we live in a marvelous world, where such things are taken for granted.

After getting used to these ideas for a while - Bose and Einstein worked out the idea of bosons in 1924, Pauli came up with his exclusion principle in 1925, and Dirac systematized the whole business in 1926 - physicists eventually started looking for symmetries that relate bosons and fermions. Supersymmetries! They're not seen in nature, but physicists were looking to see if they're mathematically possible. They turn out not only to be possible, but fascinating.

Formulating supersymmetries in a slick way requires that we take everything we knew about linear algebra and generalize it by letting all our vector spaces have both an "even" or "bosonic" part and an "odd" or "fermionic" part. Mathematically this just amounts to writing our vector space as a direct sum

V = V0 + V1
where V0 is the "even part" and V1 is the "odd part". Such a thing is called a "Z/2-graded vector space", or "super vector space".

So far this is pathetically simple. But then - and this is the really crucial part! - whenever we multiply things, we have to follow this rule:

       even    odd
even|  even    odd
odd |  odd     even
It's a little confusing, since this isn't what happens when you multiply even and odd numbers - it's what happens when you ADD them. But, one quickly adapts.

Also, when we generalize equations involving multiplication, we must remember to stick in an extra minus sign whenever we switch two odd vectors.

So, for example, the usual concept of an algebra gets replaced by that of a "superalgebra". This is a super vector space A equipped with an associative product and unit such that when we multiply even and/or odd vectors, the rules in the above table hold. We say a superalgebra is "supercommutative" if

xy = yx
when at least one of x,y lives in the even part A0, while
xy = -yx
when both x and y live in A1.

Similarly we can define super Lie algebras, super Lie groups, supermanifolds, and so on....

People have done a lot of work on this stuff: it would take me days to explain it all - even longer if I actually knew something about it.

But right now, I just want to zoom in the direction of super division algebras. These are not the most important aspect of "superalgebra" - but they're pretty cool, and Todd Trimble has been explaining them to me lately. Everything interesting I'm about to say is due to him.

As you know, I'm inordinately fond of the normed division algebras: the real numbers, complex numbers, quaternions and octonions. They're so beautiful, it's a little sad at times that there are only four! Could superalgebra allow for more?

YES! And, they turn out to be related to Bott periodicity.

Nobody seems to have pondered nonassociative super division algebras yet, but Deligne has a nice article about the associative ones, which I mentioned in "week211". I'll give more references later.

So, what's the idea?

I've already told you what a superalgebra is. We say it's a "super division algebra" if every nonzero element that's purely even or purely odd is invertible.

That's pretty easy. What are they like?

Well, I don't completely understand all the options yet, so I'll just list the "central" super division algebras over the real numbers, namely those where the elements that supercommute with everything form a copy of the real numbers. There turn out to be 8, and their beautiful patterns are best displayed in a circular layout:


           R[e] mod e2 - 1    7          1   R[e] mod e2 + 1

C[e] mod e2 - 1, ei + ie   6               2   C[e] mod e2 + 1, ei + ie

           H[e] mod e2 + 1   5           3   H[e] mod e2 - 1

What does this notation mean? Well, as usual R, C, and H stand for the reals, complex numbers, and quaternions. In all but two cases, we start with one of those algebras and throw in an odd element "e" satisfying the relations listed: e is either a square root of +1 or of -1, and in the complex cases it anticommutes with i.

So, for example, super division algebra number 1:

R[e] mod e2 + 1
is just the real numbers with an odd element thrown in that satisfies e2 + 1 = 0. In other words, it's just the complex numbers made into a superalgebra in such a way that i is odd.

The real reason I've arranged these guys in a circle numbered from 0 to 7 is to remind you of the Clifford algebra clock in "week210", where I discussed the super Brauer group of the real numbers, and said it was Z/8.

Indeed, the central super division algebras are a complete set of representatives for this super Brauer group! In particular, the Clifford algebra Cn is super Morita equivalent to the nth algebra on this circle:

 C0 = R           ~ R
 C1 = C           ~ R[e] mod e2 + 1
 C2 = H           ~ C[e] mod e2 + 1, ei + ie
 C3 = H + H       ~ H[e] mod e2 - 1
 C4 = H(2)        ~ H
 C5 = C(4)        ~ H[e] mod e2 + 1
 C6 = R(8)        ~ C[e] mod e2 - 1, ei + ie
 C7 = R(8) + R(8) ~ R[e] mod e2 - 1
where ~ means "super Morita equivalent", and the notation for Clifford algebras was explained last week.

I think this is cool. I'm not quite sure what to do with it yet, though. How much of what people ordinarily do with division algebras can be done with super division algebras? For example, can we define projective spaces over super division algebras? (See "week106" and "week145" for why that would be interesting.)

To read more about this, try:

1) Pierre Deligne, Notes on spinors, in Quantum Fields and Strings: A Course For Mathematicians, volume 1, American Mathematical Society, Providence, 1999. Also available at

A lot of the ideas go back to here:

2) C. T. C. Wall, Graded Brauer groups, J. Reine Angew. Math. 213 (1963/1964), 187-199.

and here's another good reference:

3) Peter Donovan and Max Karoubi, Graded Brauer groups and K-theory with local coefficients, Publications Math. IHES 38 (1970), 5-25. Also available at

I should admit that I have a yearning to classify nonassociative super division algebras. Has anyone ever tried this? It's already plain to see that we have two 16-dimensional nonassociative super division algebras:

O[e] mod e2 + 1
O[e] mod e2 - 1

where e is an odd element that commutes with all the octonions. (I should have mentioned this before, when talking about H[e]: even though the quaternions are noncommutative, we assume that e commutes with all of them.) Maybe one of these algebras deserves to be called the superoctonions. I bet these or something awfully similar are lurking around in string theory.

Hmm... next I wanted to write something about the topology of Bott periodicity and how that fits into what I've been discussing, but I'm running out of energy. Let me say it briefly, without much detail, just in case I never get around to a decent explanation.

Two super algebras are super Morita equivalent precisely when they have equivalent categories of super representations. So, the super Brauer group really consists of 8 different categories: the categories SuperRep(Cn), where Bott periodicity says

SuperRep(Cn+8) ~ SuperRep(Cn)
Moreover these are symmetric monoidal categories, since direct summing lets us "add" objects in these categories in a nice way.

A long time ago, Graeme Segal figured out how to take a symmetric monoidal category and get an infinite loop space from it. I explained this construction in "week199", but for a much more detailed and intense treatment with lots of references to earlier work, try:

4) R. W. Thomason, Symmetric categories model all connective spectra, Theory and Applications of Categories 1 (1995), 78-118. Available at

If we do this to SuperRep(Cn), I think we get something like


that is, the n-fold loop space of something called kO, the "connective K-theory spectrum", which I explained in "week105". The fact that this repeats with period 8:

Ωn+8(kO) ~ Ωn(kO)

is the topological version of Bott periodicity - see "week105" for more. So, we get the topological version of Bott periodicity from the algebraic version by turning symmetric monoidal categories into infinite loop spaces!

But, the interesting puzzle here is: what process can we do to SuperRep(Cn) to get SuperRep(Cn+1), which is the algebraic version of looping? And I think the answer is: "taking super representations of C1 in it". You see,

C1 ⊗ Cn = Cn+1

where I'm using the super tensor product of superalgebras, and this means that the category of representations of C1 in SuperRep(Cn) is SuperRep(Cn+1).

And, if I were trying to really explain this instead of merely scribbling notes about it, I would try to explain why this is because C1 is the complex numbers, and the unit circle in the complex numbers is related to loops.

But, sigh, that will have to wait.

One more thing before I quit for today...

I just saw a cool paper by Dror Bar-Natan, Thang Le and Dylan Thurston about the "Duflo isomorphism". This is a cousin of the Poincare-Birkhoff-Witt theorem, which in its best form says that the universal enveloping algebra UL of a Lie algebra L is isomorphic as a coalgebra to the symmetric algebra SL. You'll often see worse versions of the PBW theorem in textbooks, and ugly proofs, but James Dolan showed me the nice version and proof a while back.

The kernel of the idea is this: if L is the Lie algebra of a group G, UL consists of left-invariant differential operators on G, and there's a map UL → SL sending any differential operator to its "symbol". This is an isomorphism of vector spaces and even of coalgebras, but not of algebras.

Anyway, there's something vaguely similar relating the invariant subalgebras of UL and SL. By "invariant" here, I mean that since L acts as derivations of UL and SL, we can look at the subalgebra of either one consisting of guys who are killed by these derivations; such guys are called "invariant". Physicists call invariant elements of UL "Casimirs", after the first physicist to think about this stuff. They commute with everything else in UL. Invariant elements of SL are like classical Casimirs: there's a Poisson bracket on SL, and these are the guys whose Poisson bracket with everyone vanishes.

The Duflo map is an algebra isomorphism from the invariant subalgebra of SL to the invariant subalgebra of UL. So, it's like a very nice way to quantize Casimirs, one that gets along with multiplication. It's called the "Duflo map" because it was invented by Harish-Chandra for semisimple Lie algebras and for Kirillov in general. Kirillov conjectured that it was always an isomorphism; what Duflo did is prove it:

5) Michel Duflo, Operateurs differentiels bi-invariants sur un groupe de Lie, Ann. Sci. Ecole Norm. Sup. 10 (1977), 265-288.

Apparently all known proofs are sort of hard! According to Bar-Natan, Le and Thurston:

In the book of Dixmier, the proof is given only in the last chapter and it utilizes most of the results developed in the whole book, including many classification results (a situation Godement called "scandalous"). As discussed below, there have been several recent proofs that do not use classification results, but they all use tools from well outside the natural domain of the problem.
The proof by Bar-Natan, Le and Thurston uses the connection between knot theory and Lie algebras - namely, the theory of Vassiliev invariants. I think there's still something slightly scandalous about this, but it's awfully interesting. Anyway, take a look:

6) Dror Bar-Natan, Thang T. Q. Le and Dylan P. Thurston, Two applications of elementary knot theory to Lie algebras and Vassiliev invariants, Geometry and Topology 7 (2003), 1-31. Available at and also as math.QG/0204311.

For more, try Thurston's thesis:

7) Dylan P. Thurston, Wheeling: a diagrammatic analogue of the Duflo isomorphism, math.QG/0006083.

and, just for fun, Deligne's handwritten letter to Bar-Natan:

8) Pierre Deligne, letter to Dror Bar-Natan about the Duflo map, available at

Addendum: Todd Trimble has kindly allowed me to append some rough notes in which he outlines proofs of some results above.
From: Todd Trimble 
Subject: notes on super Brauer 
To: John Baez 
Cc: James Dolan 
Date: Sun, 27 Mar 2005 19:30:00 -0500 


These are some notes on some of the super Brauer discussion.  

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. 

[To be continued.  I had wanted to show that every 
element in the super Brauer group must be of the form 
[A] where A is a super division algebra, and then classify 
super (associative) division algebras, showing on a case 
by case basis that those not in the super Clifford clock 
above are seen not to belong to the super Brauer group.] 

Todd finished off the job later... though by this point we had formulated the grander goal of classifying not-necessarily-associative super division algebras!
From: Todd Trimble
Subject: super division algebras 
To: John Baez
Date: Wed, 27 Apr 2005 22:17:12 EDT 

This is a warm-up to classifying super division algebras 
over R, where I'll consider just the associative case.  
Nothing I say will be deep, but I found it somewhat fun 
and diverting, and there may be echoes of things to come. 
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 
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) 

© 2005 John Baez