Also available at http://math.ucr.edu/home/baez/week211.html
March 6, 2005
This Week's Finds in Mathematical Physics  Week 211
John Baez
The last time I wrote an issue of this column, the Huyghens probe was
bringing back cool photos of Titan. Now the European "Mars Express"
probe is bringing back cool photos of Mars!
1) Mars Express website, http://www.esa.int/SPECIALS/Mars_Express/index.html
There are some tantalizing pictures of what might be a "frozen sea" 
water ice covered with dust  near the equator in the Elysium Planitia region:
2) Mars Express sees signs of a "frozen sea",
http://www.esa.int/SPECIALS/Mars_Express/SEMCHPYEM4E_0.html
Ice has already been found at the Martian poles  it's easily visible there,
and Mars Express is getting some amazing closeups of it now  here's a
here's a view of some ice on sand at the north pole:
3) Glacial, volcanic and fluvial activity on Mars: latest images,
http://www.esa.int/SPECIALS/Mars_Express/SEMLF6D3M5E_1.html
What's new is the possibility of large amounts of water in warmer parts of
the planet.
Now for some math. It's always great when two subjects you're interested in
turn out to be bits of the same big picture. That's why I've been really
excited lately about Bott periodicity and the "superBrauer group".
I wrote about Bott periodicity in "week105", and about the Brauer group
in "week209", but I should remind you about them before putting them together.
Bott periodicity is all about how math and physics in n+8dimensional space
resemble math and physics in ndimensional space. It's a weird and wonderful
pattern that you'd never guess without doing some calculations. It shows up
in many guises, which turn out to all be related. The simplest one to verify
is the pattern of Clifford algebras.
You're probably used to the complex numbers, where you throw in just *one*
square root of 1, called i. And maybe you've heard of the quaternions, where
you throw in *two* square roots of 1, called i and j, and demand that they
anticommute:
ij = ji
This implies that k = ij is another square root of 1. Try it and see!
In the late 1800s, Clifford realized there's no need to stop here. He
invented what we now call the "Clifford algebras" by starting with the
real numbers and throwing in n square roots of 1, all of which anticommute
with each other. The result is closely related to rotations in n+1
dimensions, as I explained in "week82".
I'm not sure who first worked out all the Clifford algebras  perhaps it was
Cartan  but the interesting fact is that they follow a periodic pattern.
If we use C_n to stand for the Clifford algebra generated by n anticommuting
square roots of 1, they go like this:
C_0 R
C_1 C
C_2 H
C_3 H + H
C_4 H(2)
C_5 C(4)
C_6 R(8)
C_7 R(8) + R(8)
where
R(n) means n x n real matrices,
C(n) means n x n complex matrices, and
H(n) means n x n quaternionic matrices.
All these become algebras with the usual addition and multiplication of
matrices. Finally, if A is an algebra, A + A consists of pairs of guys
in A, with pairwise addition and multiplication.
What happens next? Well, from then on things sort of "repeat" with period 8:
C_{n+8} consists of 16 x 16 matrices whose entries lie in C_n!
So, you can remember all the Clifford algebras with the help of this
eighthour clock:
0
R
7 1
R+R C
6 R H 2
C H+H
5 3
H
4
To use this clock, you have to remember to use matrices of the right size to
get C_n to have dimension 2^n. So, when I write "R + R" next to the "7" on
the clock, I don't mean C_7 is really R + R. To get C_7, you have to take
R + R and beef it up until it becomes an algebra of dimension 2^7 = 128. You
do this by taking R(8) + R(8), since this has dimension 8 x 8 + 8 x 8 = 128.
Similarly, to get C_{10}, you note that 10 is 2 modulo 8, so you look at
"2" on the clock and see "H" next to it, meaning the quaternions. But to get
C_{10}, you have to take H and beef it up until it becomes an algebra of
dimension 2^{10} = 1024. You do this by taking H(16), since this has
dimension 4 x 16 x 16 = 1024.
This "beefing up" process is actually quite interesting. For any associative
algebra A, the algebra A(n) consisting of n x n matrices with entries in A
is a lot like A itself. The reason is that they have equivalent categories
of representations!
To see what I mean by this, remember that a "representation" of an algebra
is a way for its elements to act as linear transformations of some vector
space. For example, R(n) acts as linear transformations of R^n by matrix
multiplication, so we say R(n) has a representation on R^n. More generally,
for any algebra A, the algebra A(n) has a representation on A^n.
More generally still, if we have any representation of A on a vector space V,
we get a representation of A(n) on V^n. It's less obvious, but true, that
*every* representation of A(n) comes from a representation of A this way.
In short, just as n x n matrices with entries in A form an algebra A(n)
that's a beefedup version of A itself, every representation of A(n) is
a beefedup version of some representation of A.
Even better, the same sort of thing is true for maps between representations
of A(n). This is what we mean by saying that A(n) and A have equivalent
categories of representations. If you just look at the categories of
representations of these two algebras as abstract categories, there's no
way to tell them apart! We say two algebras are "Morita equivalent" when
this happens.
It's fun to study Morita equivalence classes of algebras  say algebras over
the real numbers, for example. The tensor product of algebras gives us a way
to multiply these classes. If we just consider the invertible classes, we get
a *group*. This is called the "Brauer group" of the real numbers.
The Brauer group of the real numbers is just Z/2, consisting of the classes
[R] and [H]. These correspond to the top and bottom of the Clifford clock!
Part of the reason is that
H tensor H = R(4)
so when we take Morita equivalence classes we get
[H] x [H] = [R]
But, you may wonder where the complex numbers went! Alas, the Morita
equivalence class [C] isn't invertible, so it doesn't live in the Brauer
group. In fact, we have this little multiplication table for tensor
product of algebras:
tensor R C H

R  R C H

C  C C+C C(2)

H  H C(2) R(4)
Anyone with an algebraic bone in their body should spend an afternoon
figuring out how this works! But I won't explain it now.
Instead, I'll just note that the complex numbers are very aggressive and
infectious  tensor anything with a C in it and you get more C's. That's
because they're a field in their own right  and that's why they don't
live in the Brauer group of the real numbers.
They do, however, live in the *superBrauer* group of the real numbers,
which is Z/8  the Clifford clock itself!
But before I explain that, I want to show you what the categories of
representations of the Clifford algebras look like:
0
real vector spaces
7 1
split real vector spaces complex vector spaces
6 real vector spaces quaternionic vector spaces 2
complex vector spaces split quaternionic vector spaces
5 3
quaternionic vector spaces
4
You can read this information off the 8hour Clifford clock I showed you
before, at least if you know some stuff:
A real vector space is just something like R^n
A complex vector space is just something like C^n
A quaternionic vector space is just something like H^n
and a "split" vector space is a vector space that's been written as the direct
sum of two subspaces.
Take C_4, for example  the Clifford algebra generated by 4 anticommuting
square roots of 1. The Clifford clock tells us this is H + H. And if you
think about it, a representation of this is just a pair of representations of
H. So, it's two quaternionic vector spaces  or if you prefer, a "split"
quaternionic vector space.
Or take C_7. The Clifford clock says this is R + R... or at least Morita
equivalent to R + R: it's actually R(8) + R(8), but that's just a beefedup
version of R + R, with an equivalent category of representations. So, the
category of representations of C_7 is *equivalent* to the category of split
real vector spaces.
And so on. Note that when we loop all the way around the clock, our
Clifford algebra becomes 16 x 16 matrices of what it was before, but this
is Morita equivalent to what it was. So, we have a truly period8 clock
of categories!
But here's the really cool part: there are also arrows going clockwise and
counterclockwise around this clock! Arrows between categories are called
"functors".
Each Clifford algebra is contained in the next one, since they're built
by throwing in more and more square roots of 1. So, if we have a
representation of C_n, it gives us a representation of C_{n1}. Ditto
for maps between representations. So, we get a functor from the category
of representations of C_n to the category of representations of C_{n1}.
This is called a "forgetful functor", since it "forgets" that we have
representations of C_n and just thinks of them as representations of C_{n1}.
So, we have forgetful functors cycling around counterclockwise!
Even better, all these forgetful functors have "left adjoints" going
back the other way. I talked about left adjoints in "week77",
so I won't say much about them now. I'll just give an example.
Here's a forgetful functor:
forget complex structure
complex vector spaces > real vector spaces
which is one of the counterclockwise arrows on the Clifford clock.
This functor takes a complex vector space and forgets your ability to multiply
vectors by i, thus getting a real vector space. When you do this to C^n,
you get R^{2n}.
This functor has a left adjoint:
complexify
complex vector spaces < real vector spaces
where you take a real vector space and "complexify" it by tensoring it with
the complex numbers. When you do this to R^n, you get C^n.
So, we get a beautiful version of the Clifford clock with forgetful functors
cycling around counterclockwise and their left adjoints cycling around
clockwise! When I realized this, I drew a big picture of it in my math
notebook  I always carry around a notebook for precisely this sort of thing.
Unfortunately, it's a bit hard to draw this chart in ASCII, so I won't
include it here.
Instead, I'll draw something easier. For this, note the following mystical
fact. The Clifford clock is symmetrical under reflection around the
3o'clock/7o'clock axis:
0
real vector spaces
7 1
split real vector spaces complex vector spaces
\
\
\
\
6 real vector spaces \ quaternionic vector spaces 2
\
\
\
\
complex vector spaces split quaternionic vector spaces
5 3
quaternionic vector spaces
4
It seems bizarre at first that it's symmetrical along *this* axis instead
of the more obvious 0o'clock/4o'clock axis. But there's a good reason,
which I already mentioned: the Clifford algebra C_n is related to rotations in
n+1 dimensions.
I would be very happy if you had enough patience to listen to a full
explanation of this fact, along with everything else I want to say. But
I bet you don't... so I'll hasten on to the really cool stuff.
First of all, using this symmetry we can fold the Clifford clock in half...
and the forgetful functors on one side perfectly match their left adjoints
on the other side!
So, we can save space by drawing this "folded" Clifford clock:
split real vector spaces
 ^
forget splitting   double
v 
real vector spaces
 ^
complexify   forget complex structure
v 
complex vector spaces
 ^
quaternionify   forget quaternionic structure
v 
quaternionic vector spaces
 ^
double   forget splitting
v 
split quaternionic vector spaces
The forgetful functors march downwards on the right, and their
left adjoints march back up on the left!
The arrows going between 7 o'clock and 0 o'clock look a bit weird:
split real vector spaces
 ^
forget splitting   double
V 
real vector spaces
Why is "forget splitting" on the left, where the left adjoints belong, when
it's obviously an example of a forgetful functor?
One answer is that this is just how it works. Another answer is that it
happens when we wrap all the way around the clock  it's like how going from
midnight to 1 am counts as going forwards in time even though the number is
getting smaller. A third answer is that the whole situation is so symmetrical
that the functors I've been calling "left adjoints" are also "right adjoints"
of their partners! So, we can change our mind about which one is
"forgetful", without getting in trouble.
But enough of that: I really want to explain how this stuff is related
to the superBrauer group, and then tie it all in to the *topology* of Bott
periodicity. We'll see how far I get before giving up in exhaustion....
What's a superBrauer group? It's just like a Brauer group, but where we
use superalgebras instead of algebras! A "superalgebra" is just physics
jargon for a Z/2graded algebra  that is, an algebra A that's a direct
sum of an "even" or "bosonic" part A_0 and an "odd" or "fermionic" part A_1:
A = A_0 + A_1
such that multiplying a guy in A_i and a guy in A_j gives a guy in A_{i+j},
where we add the subscripts mod 2.
The tensor product of superalgebras is defined differently than for algebras.
If A and B are ordinary algebras, when we form their tensor product, we
decree that everybody in A commutes with everyone in B. For superalgebras
we decree that everybody in A "supercommutes" with everyone in B  meaning
that
ab = ba
if either a or b are even (bosonic) while
ab = ba
if a and b are both odd (fermionic).
Apart from these modifications, the superBrauer group works almost like the
Brauer group. We start with superalgebras over our favorite field  here
let's use the real numbers. We say two superalgebras are "Morita equivalent"
if they have equivalent categories of representations. We can multiply
these Morita equivalence classes by taking tensor products, and if we just
keep the invertible classes we get a group: the superBrauer group.
As I've hinted already, the superBrauer group of the real numbers is Z/8 
just the Clifford algebra clock in disguise!
Here's why:
The Clifford algebras all become superalgebras if we decree that all the
square roots of 1 that we throw in are "odd" elements. And if we do this,
we get something great:
C_n tensor C_m = C_{n + m}
The point is that all the square roots of 1 we threw in to get C_n
*anticommute* with those we threw in to get C_m.
Taking Morita equivalence classes, this mean
[C_n] [C_m] = [C_{n+m}]
but we already know that
[C_{n+8}] = [C_n]
so we get the group Z/8. It's not obvious that this is *all* the superBrauer
group, but it actually is  that's the hard part.
Now let's think about what we've got. We've got the superBrauer group,
Z/8, which looks like an 8hour clock. But before that, we had the categories
of representations of Clifford algebras, which formed an 8hour clock with
functors cycling around in both directions.
In fact these are two sides of the same coin  or clock, actually. The
superBrauer group consists of Morita equivalence classes of Clifford
algebras, where Morita equivalence means "having equivalent categories
of representations". But, our previous clock just shows their categories
of representations!
This suggests that the functors cycling around in both directions are secretly
an aspect of the superBrauer group. And indeed they are! The functors going
clockwise are just "tensoring with C_1", since you can tensor a representation
of C_n with C_1 and get a representation of C_{n+1}. And the functors going
counterclockwise are "tensoring with C_{1}"... or C_7 if you insist, since
C_{1} doesn't strictly make sense, but 7 equals 1 mod 8, so it does the
same job.
Hmm, I think I'm tired out. I didn't even get to the topology yet! Maybe
that'll be good as a separate little story someday. If you can't wait,
just read this:
4) John Milnor, Morse Theory, Princeton U. Press, Princeton, New Jersey, 1963.
You'll see here that a representation of C_n is just the same as a vector
space with n different anticommuting ways to "rotate vector by 90 degrees",
and that this is the same as a real inner product space equipped with a map
from the nsphere into its rotation group, with the property that the north
pole of the nsphere gets mapped to the identity, and each great circle
through the north pole gives some action of the circle as rotations. Using
this, and stuff about Clifford algebras, and some Morse theory, Milnor gives a
beautiful proof that
Omega^8(SO(infinity)) ~ SO(infinity)
or in English: the 8fold loop space of the infinitedimensional rotation
group is homotopy equivalent to the infinitedimensional rotation group!
The thing I really like, though, is that Milnor relates the forgetful functors
I was talking about to the process of "looping" the rotation group. That's
what these maps from spheres into the rotation group are all about... but I
want to really explain it all someday!
I learned about the superBrauer group here:
5) V. S. Varadarajan, Supersymmetry for Mathematicians: An Introduction,
American Mathematical Society, Providence, Rhode Island, 2004.
though the material here on this topic is actually a summary of some
lectures by Deligne in another book I own:
6) P. Deligne, P. Etingof, D.S. Freed, L. Jeffrey, D. Kazhdan, J. Morgan,
D.R. Morrison and E. Witten, Quantum Fields and Strings: A Course For
Mathematicians 2 vols., American Mathematical Society, Providence, 1999.
Notes also available at http://www.math.ias.edu/QFT/
Varadarajan's book doesn't go as far, but it's much easier to read, so I
recommend it as a way to get started on "super" stuff.

Previous issues of "This Week's Finds" and other expository articles on
mathematics and physics, as well as some of my research papers, can be
obtained at
http://math.ucr.edu/home/baez/
For a table of contents of all the issues of This Week's Finds, try
http://math.ucr.edu/home/baez/twf.html
A simple jumpingoff point to the old issues is available at
http://math.ucr.edu/home/baez/twfshort.html
If you just want the latest issue, go to
http://math.ucr.edu/home/baez/this.week.html