Borcherds seminar notes, uncut
Scott Carnahan
Week 1, 26 Jan 2004, The Wightman axioms and how to fix themSome of you might remember I gave a class a few years ago on the standard
model. It ran into a few technical problems, the main one being the fact that I didn’tknow what I was talking about. I’ve learned a thing or two since then, and I’m going
to try again.
Key insight in studying physics: Physics literature is usually correct, but theyuse mathematical terms in a different way than mathematicians do. The standard
example is that when physicists say “function”, they mean “distribution”. A slightly
less well known example is that when they say “Hilbert space”, they mean “somemodule with a sesquilinear form over a ring of formal power series”.
Physicsfunction
Hilbert space

Mathematics
distribution
some module with a sesquilinear form
over a ring of formal power series

In particular, physicists seem to be completely uninterested in completeness, andbasically never use it. There is a page in Weinberg’s text which gives his definitionof Hilbert space, and mentions that mathematicians like to add this extra axiom forsome reason.
What is a QFT?
Quantum field theory means different things to different people. It’s like the story
about the three blind men examining an elephant. The definition should:
1. be rigorous
2. include QED, QCD, standard model, etc. as examples
Naive definition (that doesn’t work):
We will do the case of single hermitian scalar field φ. Spacetime is R1,3 (withmetric x20 − x21 − x22 − x23 , although people are divided about this). [At this point,
A.J. says something about general relativists vs. particle physicists - one side likes
positive spacelike slices. Borcherds mentions that both sides can give plenty of goodreasons why they are right and the other side is stupid, but they will end up withequivalent theories. Allen makes some obscure comment about getting different Pingroups if spacetime is nonorientable. Borcherds pretends not to hear.]
We want:
1. a space H of states12. For each x ∈ spacetime, a hermitian operator φ(x) on H.
3. The Lorentz group O1,3 (R) (or the connected component of the identity) should
act on H. It should also act on x, and v 7→ φ(x)v should behave well underO1,3 (R).
4. H contains a (possibly unique) vacuum vector VAC fixed by rotations andtranslations.
5. Translations act on H in some nice way. In particular, H should have “positiveenergy”: fix some timelike vector in forward direction - operator is (at least
formally) of the form eiE for some self-adjoint E. E should be a positive
operator (in the sense of its eigenvalues).
Notorious Problem: φ is “too singular” in x for this to work. limx→y φ(x)φ(y)cannot be defined in any sensible way.
Solution: (Gårding, Wightman) Instead of regarding φ(x) as an operator-valued
function, think of it as an operator-valued distribution.R We have operators φ(f ) for
all f ∈ S(R1,n ), the Schwarz space. Think of φ(f ) as R1,3 φ(x)f (x)dx. [n=3 here,
apparently]
Wightman axioms (for a hermitian scalar field)• Standard references:– “CPT, Spin. Statistics, and all that” Streater, Wightman– Some book by Jost
1. φ(f ) are unbounded operators: only defined on some dense subspace D of H.
2. We want to compose them: φ(f )φ(g) so each φ(f ) should map D into D.
3. Regularity: For Φ, Ψ ∈ D, (Φ, φ(f )Ψ) is a tempered distribution (so φ(f ) is anoperator-valued distribution).
4. Lorentz group acts on H with unique fixed vector VAC (up to constants).
f, v 7→ φ(f )v is invariant under Lorentz transforamtions.
5. H is also acted on by translations of spacetime. Action is “positive energy”.
6. [φ(f ), φ(g)] = 0 if the supports of f and g are spacelike separated.
7. D is generated by the action of operators φ(f ) on VAC. So vectors of D are
linear combinations of φ(f1 ) . . . φ(fn )VAC. People use the phrase: “Vacuum iscyclic”
28. Miscellaneous axioms I forgot to write down.
The Wightman axioms work fine for:1. Free field theories (see later)
2. 1 + 1 dimensions (see Glim-Jaffe)
3. 1 isolated example in 3 dimensions. This is some degenerate super renormalizable theory.
There are no nontrivial examples known in 4 dimensions. In particular, QED, QCD,
and the standard model are not known to satisfy them. They are perturbative theories, so they work over formal power series. The Wightman axioms don’t allow forformal power series - they only work over C.
There is a solution, due to Alexander the Great, which is to cheat, and change theproblem to one you can solve. If the main examples of a theory do not satisfy youraxioms, this does not indicate a problem with the examples, but with the axioms.
Thus, we should change the Wightman axioms slightly, to allow perturbative theories.
Advantages:
1. We get rigorous axioms
2. Many theories using Wightman axioms still work for the perturbative case.
3. QED, etc. satisfy these axioms.
Changes to Wightman:
1. Work over a ring C[[λ]] of formal power series instead of over C
2. We can recover H from D as its completion, so we may as well discard theHilbert space, and just work with D. Instead of a Hilbert space, we work withthe module D over C[[λ]]. D should have a sesquilinear form with values inC[[λ]]. [At this point, Marty asks if this should be the space of smooth vectors
or the Gårding subspace of H under a unitary representation of O1,3 . Borcherds
replies that it doesn’t really matter, as long as it satisfies the axioms.]
3. Drop the condition that (, ) on D should be positive (in some sense). We aremainly interested in the positive definite case, but the indefinite case often turns
up during the construction of positive definite examples. Standard example:
Ghosts.
3Other variations (also included in the usual Wightman axioms)
1. Instead of taking φ fixed by O1,3 (R), can take φ in some finite-dimensional real
space acted on by O1,3 (R) e.g. vector fields. We would have φ(f ) for φ ∈ somereal representation Φ, f Schwarz.
2. Can allow “fermions”. Φ and H become “super” vector spaces.
I’m going to be incredibly rude and stop in the middle of my sentence.
Week 2, 2 Feb 2004, Wightman axioms (continued)
Recap and continue extra structures:
1. Action of O1,d−1 (R) on Φ.
2. Fermions. All spaces are graded by Z/2Z, i.e. “superspaces”. These have asign rule: when you exchange the order of A and B, you must add a sign of(−1)degA degB e.g. the rule that [φ(f ), ψ(g)] = 0 if the supports of f and g arespacelike separated becomes:
φ(f )ψ(g) − (−1)degφ degψ ψ(g)φ(f ) = 0
So for fermionic fields φ, ψ, φ(f )ψ(g) + ψ(g)φ(f ) = 0 if supports are spacelike
separated. For fermionic fields, we do not have an action of O1,d−1 (R), butinstead one of the connected, simply connected group Spin1,d−1 (R). This grouphas center of order 2, giving a homomorphism to O1,d−1 (R) which is not onto,
since the latter group has four connected componenets for d > 1 (determinant
can be positive or negative, and the direction of time can be changed: spinor
norm is positive or negative). Note that weak interactions are not invariant
under O1,d−1 (R).
3. We want to allow “ghosts”. The spin-statistics theorem (which can be foundearly in any book on Wightman axioms) says that bosonic fields are exactly
those with integral spin, and fermionic fields are exactly those with half-integral
spin. Ghost fields violate this: They can avoid the spin-statistics theorem,
because the inner product on the space of states is no longer positive definite.
Everything has a ghost grading in Z or Z/2Z, although it doesn’t matter which
one, since we are only concerned with parity. In a Hilbert space, we expect(b, a) = (a, b), but we get (b, a) = (−1)ghost(a)ghost(b) (a, b).
[Note added much later: The ghost grading is in fact unnecessary. It is possible
to formulate equivalent theories with and without this additional grading, and there
is no significant advantage to adding this extra complication.]4

A common mistake is to try to identify the ghost grading with the fermionic
grading. The problem is that (a, a) = (−1)ghost(a)ghost(a) (a, a), so (a, a) is imaginary if
a has odd ghost degree. If things of odd fermionic number have imaginary norm, it
messes up attempts to construct a positive definite Hilbert space. This leads to weird(by which I mean unmemorable) conventions, like (a, a) > 0 if a has even degree andi(a, a) > 0 (or < 0) if a has odd degree.
Question:When do you use the fermionic minus sign and when do you use the ghost
minus sign?
Before I answer that, I will explain my greatest discovery. I have solved theFundamental Problem of Hilbert space theory: Should the inner product (, ) be linear
on the left or right?
The key point of the solution is to think of a Hilbert space H as a bimodule
over C, by using complex conjugation as a sort of “antipode” like in a Hopf algebra.
We define the right action of C on H by vλ = λ̄v for all λ ∈ C, v ∈ H.
Similarly, if A is an operator on H, define the right action of A on H by vA = A† v
(where † = adjoint). The inner product now becomes a bilinear map of bimodules
from H × H to C, and the action follows a natural associative law:λ(u, v) = (λu, v) = (uλ̄, v) = (u, λ̄v) = (u, vλ) = (u, v)λSo (u, v) is left linear in u, right linear in v, and the left and right C-module structures
are related by complex conjugation. In particular, for left Hilbert spaces, (, ) should
be linear in the left variable. You can also make right Hilbert spaces if you’re one ofthose people who happens to like things acting on the right for some reason. Similarly,
we have for operators A, (Au, v) = (uA† , v) = (u, A†v). Now define:• λ† := λ for λ ∈ C (view it as a constant-multiplication operator)• A† := usual adjoint for operators• v † := v for v ∈ H. The reason you haven’t seen this one before is that no one
has bothered to come up with a special notation for the identity operator.
Then we have(ab)† = b† a†
for a and b complex numbers, vectors, or operators, whenever it makes sense.
What about ghost grading?
Replace (ab)† = b† a† by (ab)† = (−1)ghost(a)ghost(b) b† a† everywhere.
When do you use:(−1)ghost(a)ghost(b) - when the interchange is done using the adjoint operator5(−1)deg(a)deg(b) - when making things commute or anticommute.
Modified Wightman axioms:
1. Allow ghosts.
2. Work over a formal power series ring instead of C.
Ingredients:
All spaces should be Z/2Z × Z/2Z graded by fermion number and ghost number.
1. A ring R that we work over. R can be a formal power series ring in several
variables, e.g. R[[λ]]. This case covers all interesting examples. Everything also
works if R is an inverse limit of finite-dimensional algebras over R. Alternatively,
R is the dual of some coalgebra over R.
2. A real vector space Φ of hermitian fields. (Note: elements of C ⊗ Φ are called“complex fields”)
3. A space of states D, a real vector space. This is sometimes written D[[λ]]. D[[λ]]is a module over R[[λ]]. Think of D as a dense subspace of some Hilbert space,though it need not be. D should have a sesquilinear inner product (, ) satisfying
λ(u, v) = (λu, v) = (u, λ̄v) and (u, v) = (−1)ghost(u)ghost(v) (v, u). Assume (, ) is
nondegenerate. We do not assume it is positive definite (though we prefer it tobe).
4. An action of the Lie algebra of translations of R1,d−1 on D.
5. An action of Spin1,d−1 on Φ and on D. Technical assumption: φ is a sum offinite-dimensional algebraic representations of Spin1,d−1 (R). D is not usually asum of finite-dimensional representations.
6. A Field map: For φ ∈ Φ a real Schwarz function on R1,d−1 , we have a self-adjoint
operator φ(f ) from D[[λ]] to D[[λ]] such that:• it is linear in φ.

• tempered: (Φ, φ(f )Ψ) is a tempered distribution for all Φ, Ψ ∈ D.

[The letter Φ is used for two different things here] Remark: (Φ, φ(f )Ψ) takes
values in R = R[[λ]], so when we say it is a distribution, we just mean thecoefficient of any power of λ is a distribution. ∂(φ(f )v) = φ(∂f )v + φ(f )∂v for∂ an infinitesimal translation.
7. A vacuum vector VAC ∈ D, fixed by translations, rotations, and normalized
such that (VAC, VAC) = 1.
6

Main Axioms:
1. Causality - [φ(f ), ψ(g)] = 0 if f and g are spacelike separated.
2. Positive energy - explanation postponed until next week.
3. Lorentz invariance
4. Vacuum is cyclic - D[[λ]] is generated by the action of φ(f ) of VAC.
Week 3, 9 Feb 2004, Free massive fields
Warning: The sign in (a, b) = (−1)ghost(a)ghost(b) (b̄, ā) might be unnecessary.
About 47% of the time I work it out it goes one way, and about 53% of the timeit goes the other, so I can’t give you an answer until it stabilizes somewhat.
[Note added much later: the ghost grading seems to be unnecessary.]
I said I was going to continue talking about the Wightman axioms, but I got a bit
bored of them, so instead I will describe a basic example of a quantum field theory.
This is given by a free massive hermitian (bosonic) scalar field.
Massive: opposite of massless
Hermitian: this means φ ∈ Φ (not C ⊗ Φ).

Scalar: The Lorentz group acts trivially on Φ.
Examples: Higgs bosons, scalar mesons.
Main idea for constructing free hermitian fields φ:Write φ as a sum of 2 fields, φ+ and φ− . φ+ will be the creation field, and φ− willbe the annihilation field - it “kills particles”. Note that φ+ and φ− are not in Φ.
For historical reasons, φ+ is sometimes called φ− , and φ− is sometimes called φ+ .
At some point during the early days of quantum physics, someone made a sign error,
and it has propogated ever since. It’s not clear why anyone would use a minus sign
to indicate creation.
Commutation relations:• [φ+ (f ), φ+ (g)] = [φ− (f ), φ− (g)] = 0.
• [φ+ (f ), φ− (g)] = some scalar (in C) depending on f and g. Instead of anarbitrary operator, you end up just multiplying by a constant.
• φ− (f )(VAC) = 0.
These are basically the characteristic relations of any free field. We need to choose
the scalar. [φ(f ), φ(g)] is some sort of bilinear form on f and g, and should be
7

Lorentz-invariant and translation invariant, so we try:Zfˆ(p)ĝ(−p) × (some Lorentz-invariant measure)d4 p
where fˆ and ĝ are Fourier transforms:fˆ(p) :=

Z

e2πixp f (x)dx

The measure is almost the simplest possible. [some drawing of light cones andhyperbolic sheets goes here] Choose a positive real m. This is the mass. Take the set ofvectors in momentum space (R1,3 ) satisfying p2 = m2 and p > 0 (in particular, ignore
the bottom component of the hyperboloid), and take a Lorentz-invariant measure
supported on that set. This measure is unique up to scalars (One could take othermeasures, e.g. supported on more than one hypersurface - this gives “generalized freefield theories” which don’t offer significant advantages). The statement that “mass =m” is equivalent to the assertion that the measure has support on (the positive partof) the hypersurface p2 = m2 .
Main problem: need to check causality condition - that [φ(f ), φ(g)] = 0 if f and g
are spacelike separated.

[φ+ (f ) + φ− (f ), φ+ (g) + φ− (g)] = [φ+ (f ), φ− (g)] − [φ+ (g), φ− (f )]Z= fˆ(p)ĝ(−p) − ĝ(p)fˆ(−p) × (measure)d4 pZ= e2πixp f (x)e−2πiyp g(y) × M(p)dpdxdy
Vital: M changes sign under rotations changing the direction of time. Take theinvariant measure supported on the positive part of p2 = m2 , and subtract the corresponding measure supported on the negative part.
The condition that the commutator vanishes for x − y spacelike is equivalent tothe condition that the Fourier transform of M(p) vanishes on spacelike values. So
causality for bosonic scalar fields reduces to the following property of the measure M:M̂ (x) = 0 if x is spacelike
M has the following properties:
1. Invariant under time-preserving rotations82. Changes sign under time-reversing rotations
The formula for M can be given explicitly using Bessel functions, but properties 1
and 2 are sufficient to show that M̂ = 0 where we want it to.
1. If M transforms as above, so does M̂ . This is a trivial fact about invarianceunder group actions.
2. If a measure M̂ has these transformation properties, then M̂ vanishes on spacelike vectors.
Proof: If M is a function, the result is trivial: For any spacelike vector x, there isa time-reversing rotation σ fixing x. We then have −M(x) = M(σx) = M(x), so
M(x) = 0. For distributions, one needs to work slightly harder, so this is left as
an exercise. The main point is that any rotation-invariant
distribution on spacelikepvectors depends only on the “radius” r = −(x, x)
Define D1 to be the space of Schwarz functions on {p|p2 = m2 , p > 0}. This is
the space of 1-particle states. Let D be the symmetric algebra of D1 . We identify D1with the space of states of the form φ+ (f )VAC (look at fˆ on {p2 = m2 , p > 0}). By
the commutation relations, φ+ (f1 )φ+ (f2 ) . . . φ+ (fn )(VAC) can be identified with thesymmetric algebra product of f1 , . . . , fn .
How does φ− (g) act? It is uniquely determined by:
1. φ− (g)(VAC) = 0
2. [φ+ (f ), φ− (g)] is known
For example:
φ− (g)φ+ (f1 )φ+ (f2 )(VAC) = [φ− (g), φ+ (f1 )]φ+ (f2 )(VAC)+ φ+ (f1 )[φ− (g), φ+(f2 )](VAC)+ φ+ (f1 )φ+ (f2 )φ− (g)(VAC)
and the last term necessarily vanishes.
D is positive-definite, because the inner product (, ) is determined by the condition
that φ+ is adjoint to φ− . When you take the Fourier transform, something funny
happens, but since we’re out of time, I won’t explain that.
Week 4, 16 Feb 2004, Presidents’ Day - no class
The following people were seen trying the (locked) door of room 939 to see if theseminar would happen:
1. Nicolai Reshetikhin
9

2. Noah Snyder
Week 5, 23 Feb 2004, Spinors and Clifford algebras
Today and possibly next week, I will be talking about everything worth knowing
about spinors and Clifford algebras. For our purposes, we only need to describe
Clifford algebras for R0,n and R1,n , but out of some kind of bloody-mindedness, wewill define them for all fields. [AJ: even characteristic 2?] ... including characteristic
2, which is the case that scares people. It turns out that the reason people are scaredis that they are using the wrong definitions. We will be doing the characteristic 2
case in order to force ourselves to use the best definition.
Let V be a finite-dimensional vector space over a field F , with quadratic form N.
This is a degree 2 map from V to F such that N(a + b) − N(a) − N(b) is bilinear ina and b.
Example: If (, ) is a symmetric bilinear form, then N(a) = (a, a) is a quadratic
form. If char(F ) 6= 2, then quadratic forms are equivalent to symmetric blinear forms.
In characteristic 2, quadratic forms are the right thing to use, and bilinear forms arewrong.
Warning: If N(a) = (a, a), then the bilinear form of N is two times the originalone. (You can make quadratic forms from bilinear forms in characteristic 2 also, butthey tend to be rather degenerate.) We say V is nondegenerate if the bilinear formof N is nondegenerate.
The Clifford algebra CV (F ) of V is the algebra generated by the vector space V
with relations Q(v) = v 2 for v ∈ V . I’m going to skip the details of the construction,
since you know how it goes.
At this point, most people spend time proving that they have dimension 2n , but
I’m not going to bother. Instead, I will just tell what they are for the cases we careabout.
Let’s calculate CV (R) for all non-degenerate quadratic forms over R. We write
Cm,n (R) for the Clifford algebra of x21 + . . . + x2m − x2m+1 − . . . − x2m+n . The first few
cases are easy to compute and are as follows:
C1,0 ∼
C0,1 ∼
C0,0 ∼
=R⊕R
=C
=R
∼
∼
C0,2 (R) = H C1,1 (R) = C2,0 (R) ∼= M2 (R)
After these, it becomes a bit tiresome to figure them out directly, so we use alemma:
Cm+2,n (R) ∼= Cn,m (R) ⊗ M2 (R)∼Cm+1,n+1 (R) = Cm,n (R) ⊗ M2 (R)Cm,n+2 (R) ∼= Cn,m (R) ⊗ H
10

Note the order reversal in the first and third formulas.
More generally: Suppose dim U is even, U non-degenerate. ThenCU ⊕V ∼= CU (F ) ⊗ CV (−disc U ) (F )where V (−disc U) is the vector space V , with the norms of the vectors given bytaking the usual norms and multiplying by minus the discriminant of U.
Sketch of proof: We can assume char F 6= 2 since the characteristic 2 case istrivial. Then we can choose orthogonal bases γ1 , . . . , γm for U and γm+1 , . . . for V .
(this always exists in characteristic not 2, and is never possible in characteristic 2 if0
the form is nondegenerate). Consider elements γm+k:= γ1 γ2 . . . γm γm+k . An easy
calculation shows:0
1. γm+k
commutes with γi (i ≤ m)
0
2
2. (γm+k)2 = −disc(U)γm+k20
and −disc(U) turns out to be γ12 γ22 . . . γm
, so γ1 , . . . , γm , γm+1
, . . . satisfy the relations
for CU ⊗ CV (−disc U ) .

Now, to calculate all Clifford algebras over R, we just need to recall how to taketensor products of matrix algebras:R ⊗ anything ∼C⊗C∼= anything
= C⊕C
∼
∼
C ⊗ H = M2 (C)H ⊗ H = M4 (R)∼Ma (X) ⊗ Mb (Y ) ∼
M
(X
⊗
Y
)
M
(M
= ab
a
b (X)) = Mab (X)[Actually, we can do away with the C ⊗ C, but might want Ma (X ⊕ Y ) ∼= Ma (X) ⊕Ma (Y )] So from the lemma, we have for Cm,n (R):
m

n

0
1
2
3
4
5
6
7
8

0
1 2
3
4
5
6
7
8
R
C H H ⊕ H M2 (H) M4 (C) M8 (R) M8 (R ⊕ R) M16 (R)R⊕R
M2 (R)
M2 (C)
M2 (H)
M2 (H ⊕ H)
M4 (H)
M8 (C)
M16 (R)

The rest of the table is filled in by multiplying the dimension of the matrix algebraby 2 each time you go down and to the right. Each time you go down by 8 or to11the right by 8, you multiply the dimension of the matrix algebra by 16. That is acomplete description of all useful Clifford algebras.
Now, we write CV (F ) = C0 (F ) ⊕ C1 (F ), where C0 (F ) is spanned by products ofeven numbers of γi (basis vectors for V ), and C1 (F ) is spanned by products of odd
numbers of γi . This splitting is possible because the defining relations of the Clifford
algebra only had terms of even degree, so degree is defined mod 2. C0 (F ) is the objectof interest for us, since it gives us spinors.
Lemma: CU0 ⊕V ∼= CV (disc(U )) for U 1-dimensional and nondegenerate.
Proof: Let γ be a basis for U, and γ1 , . . . , γn a basis for V . Then γγ1 , . . . , γγngenerate CU0 ⊕V , and (γγi )2 = −γi2 γ 2 = γi2 (−disc(U)), etc. The relations are satisfied.

This gives you everything you want to know about the Clifford algebra. Now wedefine the Clifford group ΓV . We will find:
1 → F × → ΓV (F ) → OV (F ) → 1∪∪
k
1 → ±1 → Pin(F ) → OV (F )k∪∪
1 → ±1 → Spin(F ) → SOV (F )Where OV (F ) is the orthogonal group of V . Note that the last map on the second
row is usually not surjective.
We are interested in the Spin group, but it is easier to define the Clifford groupand look for Spin inside it.
Definition of ΓV (F ): Let α(v) = v for v ∈ CV0 (F ), and α(v) = −v for v ∈ CV1 (F ),so α is the unique automorphism of CV (F ) induced by v 7→ −v. (If you are interested
in categories, you can think of Clifford algebras as a sort of functor, and α as theimage of the inversion automorphism)
ΓV (F ) is the set of invertible elements x ∈ CV (F ) such that xV α(x)−1 ⊂ V .
(note that V ⊂ CV (F ).

A common “error” among authors: using xV x−1 ⊂ V instead of xV α(x)−1 ⊂ V .
It’s not really an error since the theory can still be made to work, but it makes things
much more difficult. Old books tend to use xV x−1 and newer books tend to use
xV α(x)−1 so people seem to have caught on to this.

(v,x)
First reason: If x ∈ V , v 7→ xvα(x)−1 is the reflection in x: v 7→ v − N
x. The(x)old way gives negative reflections, and you should use refletions whenever possible,
since they fix almost everything. They are the simplest rotations.

Warning: In characteristic 2, these are often called “orthogonal transvections”. The
only apparent reason for this is that authors want to make characteristic 2 sounddifficult.
12[Marco: What kind of action do you get from mixed-type elements in ΓV (F )?
Borcherds: There turn out not to be any.]
Lemma: If x ∈ ΓV (F ) acts trivially on V , then x ∈ F × ⊂ CV (F ).

This is an easy exercise, and is false if we use the action v 7→ xvx−1 .
Now, we want to show ΓV (F ) acts as rotations on V .

Define N α (x) = α(x)T x, where T is the transpose: (γ1 . . . γn )T = (γn . . . γ1 ). T
is the antiautomorphism of CV (F ) fixing all elements of V . As it happens, T is not
all that important - we will just use it in one or two proofs. α is also not terribly
important - we will eventually throw it away and replace it with something better.
An easy calculation shows: N α (x) ∈ F × if x ∈ ΓV (F ). This is because N α (x)acts trivially on V . Next:
N α (xy) = α(y)T α(x)T xy= α(y)T N α (x)y= α(y)T yN α (x)= N α (y)N α (x)So N α is a homomorphism from ΓV (F ) to F × . Note that this is not a homomorphismfrom the Clifford algebra, and doesn’t map its elements to F .
Since N α (v) = −N(v) for v ∈ V , we see that ΓV (F ) preserves (−)N on V , soΓV (F ) → OV (F ). OV (F ) is generated by reflections if char F 6= 2, and there is aunique counterexample in characteristic 2: F = F2 , V = A ⊕ B, where both A andB have dimension 2 and quadratic form x2 + y 2 + xy. Reflections generate an index
2 subgroup in this case. However, in all cases, the map ΓV (F ) → OV (F ) is onto, so
we get:1 → F × → ΓV (F ) → OV (F ) → 1
In particular, the Clifford group is a central extension of OV (F ). Before we goon, we ought to tidy up notation a bit: We can split ΓV (F ) = Γ0V (F ) ∪ Γ1V (F ),where ΓiV (F ) = ΓV (F ) ∩ CVi (F ) This can be checked in all but one case by looking
at reflections.
We define the Dickson invariant: a homomorphism ΓV (F ) → Z/2Z given by
ΓiV (F ) → i. In characteristic not 2, det(x) = (−1)Dickson invariant , while in characteristic
2, the determinant is always 1, so it carries strictly less information.
The spinor norm is defined as: N(x) = N α (x)(−1)Dickson invariant . This is a homomorphism ΓV (F ) → F × . Notice that is coincides with the norm N on {v ∈ V |N(v) 6=0}. It extends the homomorphism, and is uniquely defined by this property.
Week 6, 1 March 2004, Spinors and Clifford algebras, continued
Recall from last time:131. V , a vector space over a field F , with a quadratic form N (or Q, if I’m notpaying attention). This gives rise to a corresponding bilinear form (u, v) 7→
N(u + v) − N(u) − N(v).
2. A Clifford algebra CV (F ) generated by V with relations a2 = N(a).
3. The Clifford group ΓV (F ) of invertible elements x of the Clifford algebra satisfying xV α(x)−1 ⊂ V , where if we have the splitting CV (F ) = CV0 (F ) ⊕ CV1 (F ),α is the antiautomorphism which acts as −1 on CV1 (F ) and as 1 on CV0 (F ).
Note: Some authors leave out the α, which leads to problems later on.
T is the antiautomorphism of CV (F ) which acts as identity on V , and such that(v1 v2 . . . vn )T = vn vn−1 . . . v1 .
We define the twisted spinor norm N α be N α (x) = α(x)T x. This is not toointeresting, but it is convenient for proofs. We define the spinor norm N by N(x) =xT x.
We showed N α is a homomorphism ΓV (F ) → F × (not a homomorphism on
CV (F )), and that ΓV (F ) = Γ0V (F ) ∪ Γ1V (F ), where ΓiV (F ) = ΓV (F ) ∩ CVi (F ). ΓV (F )is generated by V if dim(V ) > 0 (except for one case over F = F2 . This is just incase there are any Gabbers out there).
Recall N(x) = N α (x)(−1)Dickson invariant(x) , where the Dickson invariant is 0 if
x ∈ Γ0V (F ) and 1 if x ∈ Γ1V (F ). The spinor norm is defined uniquely (for V 6= 0) by
N(v) = N(v) for all v ∈ V with N(v) 6= 0. Here, the first N is spinor norm, andthe last two are the quadratic form. Since these two things actually coincide here, we
may as well use the same letter for them.
Note: Many authors define N α to be the spinor norm, so that spinor norm =−norm on V, which introduces a pointless, unmemorable sign.
We get an exact sequence which can be uniquely extended to a diagram:
1 → F × → ΓV (F ) →OV (F )
→ 1
& ↓N
↓N
F×
→ F × /(F × )2 → 1

for all F and all V , where the diagonal arrow is the squaring map x 7→ x2 , andF × /(F × )2 is the image if F × under N.
So we get a spinor norm homomorphism OV (F ) → F × /(F × )2 uniquely definedby N(reflection of v ∈ V, N(v) 6= 0) is N(v).

For people who like weird and irritating exceptions, you may recall there is one
example where Γ is not generated by V . However, the spinor norm homomorphism
is still uniquely defined even for this irritating example, since F × is trivial.
14

Example: Suppose V is positive or negative definite over R. In this case, we haveF × /(F × )2 ∼= {±1}. If V is positive definite, N(v) = 1, since the norm of any reflecting
vector is positive. Thus spinor norm is identically 1. If V is negative definite, thenN(reflection) = −1 ∈ R× /(R× )2 , so spinor norm = determinant.
Now, suppose V has vectors of positive greater than 0 and of norm less than 0.

For reflections of norm > 0, det = −1 and N = 1

For reflections of norm < 0, det = −1 and N = −1

so det × N maps OV (R) onto {−1, 1} × {−1, 1} if V is indefinite. Thus, OV (R) has
at least 4 components (in fact exactly four). OV (R) has 2 components if V is definite(and if dim V > 0 for all the Gabbers in the audience. Gabber was well-known
for pointing out cases involving the zero dimensional vector space or the empty set,
and people found it rather annoying, but someone’s proof of the Atiyah-Singer index
theorem fell apart when Gabber pointed out that the union of two vector spaces is
not a vector space.)
For Lorentzian vector spaces, where V = Rn,1 , this is easy to see. A rotation
has spinor norm > 0/ < 0 if it fixes/exchanges the two light cones. [drawing of light
cones]
Definition: PinV (F ) = kernel of the map N : ΓV (F ) → F × . x ∈ F × and N(x) = 1 if
and only if x = ±1, so we get and exact sequence: 1 → {±1} → PinV (F ) → OV (F ).
The last map is not onto in general, for example if V = Rn,1 . (Note that some authors
define PinV (F ) to be the elements with N(x) = ±1. This makes PinV (R) → OV (R)onto over the reals, but messes things up completely over other fields.)
Subtle point: PinV and OV are algebraic groups over F . This basically means they
are functors from the category of F -algebras to the category of groups. The map
PinV → OV is surjective as a map of algebraic groups, but PinV (F ) → OV (F ) neednot be surjective.
If you’re into Galois cohomology, I’ll mention you get a Galois cohomology exact
sequence:N

1 → ±1 → PinV (F ) → OV (F ) → H 1 (Gal(F /F ), ±1) → H 1 (Gal(F /F ), PinV (F )) →where H 1 (Gal(F̄ /F ), ±1) ∼= F × /(F × )2 if F is perfect [actually unnecessary] andcharacteristic not 2.
In general, when you have an exact sequence of Galois modules, you get a longexact sequence of Galois cohomology, except this one isn’t very long, since the groups
are nonabelian.
We define SpinV (F ) to be the kernel of the Dickson invariant PinV (F ) → {0, 1}. Ifdetchar(F ) 6= 2, then this is the same as the kernel of det : PinV (F ) → OV (F ) → {±1}.
15(Note: SOV (F ) should be defined as elements of Dickson invariant 0, as this also
works if char(F ) = 2. If you are not working in characteristic 2, it doesn’t reallymatter, but since in characteristic 2 it is just as easy to get the definition wrong as it
is to get it right, you might as well use the correct definition.)
We have inclusions: SpinV (F ) ⊂ Γ0V (F ) ⊂ CV0 (F ) ⊂ CV (F ), and homomorphism
SpinV (F ) → SOV (F ) which is not always onto. Representations of CV0 (R) give representations of SpinV (R), called spinor representations. I never quite figured out when
they are called spin and when they are called half-spin.
0
Examples: Spin1,3 (R) ⊂ C1,3
(R). Last lecture, we showed how to work out every
possible Clifford algebra you could conceivably be interested in, and you will seethat C1,3 (R) ∼= M2 (C), so Spin1,3 (R) has a 4 dimensional real representation, witha complex structure. We get a homomorphism Spin1,3 (R) → GL2 (C). A little morework gives an isomorphism Spin1,3 (R) ∼= SL2 (C)The last one is widely used in physics textbooks, and I think it is a bad idea,
because if you rely on this accidental isomorphism, it means you can’t work out
Quantum Field Theory in any dimension other than 4.
Exercises:

Spin1,2 (R) ∼= SL2 (R)Spin (R) ∼= SL2 (H)1,5

Spin2,2 (R) ∼= SL2 (R) × SL2 (R)Spin (R) ∼= SL4 (R)3,3

4 dimensional orthogonal groups often split like on the third line above, but not always. I want to finish this lecture mainly by giving warnings about standard mistakes.
Warning:

Om,n (R) ∼= On,m(R)SOm,n (R) ∼= SOn,m(R)∼Spinm,n (R) = Spinn,m (R)Pinm,n (R) ∼6= Pinn,m (R)The easiest example of this is Pin1,0 (R) ∼= Z/4Z.
= (Z/2Z)2 and Pin0,1 (R) ∼Why is OV (F ) not a simple group?
It turns out it is essentially simple, but it has a large amount of rubbish attached
to it. We’ll just list the possible reasons for a group to be not simple, and if none ofthem apply, we have a simple group.
1. Dickson invariant OV (F ) → Z/2Z (sometimes called the quasi-determinant or
pseudo-determinant)
2. Spinor norm N : OV (F ) → F × /(F × )2 (not necessarily onto)16

3. OV (F ) may have center ±1 (may not be in the kernel of Dickson or N)
4. If dim V ≤ 4, you have lots of special exceptions (especially if F is small, finite)
SpinV (F )/(center) is simple if dim V ≥ 5 and V is nondegenerate. The center canhave order 1, 2, or 4, and all of them can occur, if you remember that F can havecharacteristic 2.
I will now make a table of the most common mistakes, and their consequences.
You’ll notice that none of them are particularly serious, and that’s why people continue to make them. It’s like some sort of genetic disease that isn’t quite severe
enough to keep itself from propagating to the next generation.
Most common mistakes
1) Use of (, ) instead of N todefine CV (F )2) Use of xV x−1 instead of xV α(x)−1
to define the Clifford group3) Defining spinor norm to be N α ,not N (very common, even amongpeople who get everything else right)
4) Saying the groups Spinm,n (R)are simply connected5) Defining SO using determinant
instead of Dickson invariant

Consequence
Things go wrong if char(F ) = 2. Of
course, no one in their right mind
would study orthogonal groups incharacteristic 2, anyway
Several minor irritations: reflection
is not always in ΓV (F ), andΓV (F ) → OV (F ) is not always onto
Spurious minus sign:spinor norm = −norm
You’ll look kind of stupid, because
they are not. Spinn,0, Spin0,n , Spinm,1 ,and Spin1,m are simply connected
for m ≥ 3. Spinm,n (R) is not
simply connected for general m, n:
usually, π1 (Spinm,n (R)) ∼= Z/2Z
Warning: the algebraic group Spinm,nis simply connected for m + n ≥ 3.
I won’t explain this, becuse we’reout of time
You get the wrong group ifchar(F ) = 2. Again, no one cares.

Week 7, 8 March 2004, Free Quantum Field Theories
Today, I will talk about free quantum field theory, which is generally regarded as
the trivial case, although it is not really trivial, and it is the only case people can doin a really satisfactory way.
Quick summary of important examples:
We need to choose:17

1. Spacetime R1,d−1
2. A vector space Φ of “fields”. I don’t quite know how to say what these are, so
let’s just say they’re elements of Φ. Φ is a representation of Spin1,d−1 (R).
3. A mass m, for which there are two cases: m = 0, and m > 0 (negative mass
yields causality violations - generalized fields theories allow sums/integrals overdifferent masses, but they don’t seem to be used except as counterexamples).
I will now list the main examples. I have experimental evidence that this list iscomprehensive, where experiment consists of reading physics texts to see what needs
to be accounted for.
1. Hermitian scalar field: dim Φ = 1, i.e. Φ ∼= R, m = 0 or m > 0. Note: m = 0
goes wrong in dimension d = 2.
2. Vector field of mass m > 0, Φ = R1,3 . For each possible momentum, the spaceof possible states has dimension 3, which is less than dim Φ (e.g. in electroweak
theory, there are three vector bosons). For a vector field of mass m = 0 withd = 4, there is only a 2 dimensional space of states of a given momentum (e.g.
polarization of photons).
3. Spinor fields: Φ is a sum of spin or half spin representations. For example,
Dirac spinors in dimension d = 4. Here, Φ is a sum of two copies of the spinor
representation of Spin1,3 (R), so dim Φ = 8. We might expect an 8-dimensional
space of particles of a given momentum. In fact, the space has only 4 dimensions:electrons with spin up or down, and positrons with spin up or down.
As everyone knows, the remaining dimensions are killed off by the Dirac equation.
[Joel: not sure what is meant by dimension - fields given by vector fields on spacetime? Borcherds: Naı̈ve quantization gives you dim = Φ, but some other requirement
has to be satisfied. Marco: Are these the constraints given by the classical equationsof motion? Borcherds: Right now, I’m just trying to say that you get fewer fields
than you would with naı̈ve quantization. (failed to catch some of this discussion)]
More complications: If d ≡ 2, 3, 4 mod 8, then we can construct “Majorana
spinors” [AJ: pronounced Mah-yor-ah-nah]. In this case, the number of states of agiven momentum is half that of Dirac spinors. If the dimension is even and mass is0, we can construct Weyl spinors. If d ≡ 2 mod 8 and m = 0, then we get WeylMajorana spinors. It appears that free quantum field theory is a complete mess.
Problem Explain all these cases and exceptions in a uniform way.

18

Physics texts generally cover these in a case-by-case manner, and a new frameworkhas to be introduced for each new field theory (e.g. introduce the Dirac equation for
no reason at all).
We want to state:
1. What data do we need to give to define a QFT?
2. What conditions do these data have to satisfy?
I bet you’re all impressed that I know “data” is a plural word. Here is the obviousdata:
1. Spacetime R1,d−1
2. Choose m ≥ 0.
3. Choose a representation Φ of Spin1,d−1 (R).
What else do we need to choose to define a free quantum field theory? Well, I
haven’t really said what a free quantum field theory is (except a few weeks ago). The
distinguishing property is that we can split each field as a sum: φ = φ+ + φ− of a“creation” field φ+ , and an “annihilation” field φ− , that have “easy” commutation
relations.
More precisely:• [φ+ (f ), ψ + (g)] = 0 for all φ, ψ, f , and g, where [X, Y ] = XY −(−1)degXdegY Y X.
• [φ− (f ), ψ − (g)] = 0• [φ− (f ), ψ + (g)] is a scalar (depending on φ, ψ, f , and g of course).
(In general field theories, the commutator of two fields is some horrendous mess thatno one can understand.)Not really needed now: φ+ (f ) is adjoint to φ− (f ) for f real.
Remark: note that φ+ (f ) and φ− (f ) generate a Heisenberg Lie (super)algebra,
which is a Lie algebra L containing a 1-dimensional subalgebra Z with [L, L] = Z and[Z, L] = 0. The space of states is a representation of this Lie algebra, which is convenient, since the representation theory of Heisenberg algebras is basically completely
understood.
The key point is that free field theories give a Heisenberg Lie algebra, and othertheories give this huge Lie algebra that no one can understand.

19

We need to spacify a scalar. This determines the entire QFT. We will define:Z−+[φ (f ), φ (g)] = fˆ(−p)ĝ(p)(φ, ψ)p d4 pp

The ±p in f and g corresponds to some kind of translation invariance. (, ) isˆaR bilinear form on Φ depending on p. fˆ is the Fourier transform of f : f(p)
=
2πixp
e
f (x)dx. Whenever you do a Fourier transform, you have to put a 2π inxsomewhere, and this seems to be the least worst place to put it.
[Joel: p is in the dual of spacetime? Borcherds: p lies in energy momentum space.
As long as spacetime has a nondegenerate bilinear form, you can identify this withspacetime.]
What conditions must (, ) satisfy?
1. Invariance under Spin1,d−1 (R): (σφ, σψ)σp = (φ, ψ)p for σ ∈ Spin1,d−1 (R). Thisis just Lorentz invariance - not very difficult or surprising.
2. Spectral condition: (φ, ψ)p has support in {p ∈ positive cone}. Again this is
not terribly difficult.
3. Mass condition: (φ, ψ)p has support on {p2 = m2 }.
These are the obvious conditions it must satisfy. There are some subtle conditions,which I want to talk about.
Pick p, p2 = m2 , p > 0. Put Spinp1,d−1 (R) = elements fixing p. Put (, ) = (, )p , so(, ) is invariant under Spinp1,d−1 (R). Conversely, any such inner product on Φ can beextended to a set of inner products (, )p .
Actually, this isn’t quite true. It only works if m > 0 or d 6= 2. I want to explain
why we get this funny exception.
Technical point: It is not always possible to extend an invariant measure on{p2 = m2 , p > 0} to all of R1,d−1 . It is always possible if m 6= 0, as that set is closed.
It is also always possible if m = 0, d > 2 (easy to check). If m = 0, d = 2, it is notpossible essentially because the measure becomes too infinite near the origin. It is1
similar to the following problem: Extend the measure |x|dx on R \ 0 to R. This is not1
possible, as |x| is too large near 0.
Actually, you can get massless scalar fields if you don’t mind certain complications
such as non-positive-definite Hilbert spaces, etc.
What condition do we need for causality? This is the hard part of free QFT. It
seems the correct condition is not written down anywhere in the physics literature.
Correct condition: Suppose θ ∈ Spin1,d−1 (C) with θp = −p. Then (φ, ψ) =(−1)degφdegψ (θψ, θφ).
20

Note by the way, this implies the Lorentz invariance condition, since any rotationis generated by reflections.
Question: Is (, ) symmetric or antisymmetric?
Answer: No. In general it is neither.
It seems to be hardwired into our brains that bilinear forms are either symmetricor antisymmetric, and Dirac fermions are a natural example of this not being thecase. I fell into this trap once and got a terrible headache trying to work out fermions,because when I assumed the form was symmetric I got a contradiction, and when I
assumed it was antisymmetric I got a contradiction.
Warning If (, ) is neither symmetric nor antisymmetric, the left kernel may not be
the same as the right kernel.
In general, (, ) will be degenerate, as the left kernel may be nonzero. Note:θ(left kernel) = right kernel.
Next is to prove this is necessary and sufficient for causality, but that will be done
next week, along with innumerable examples.
[Marco: Spinor field has a natural inner product - is this what you use? Borcherds:
If you’ve got a natural inner product on Φ, the first thing you do is throw it away.]
Week 8, 15 March 2004, Examples of free quantum field theories[Notes by Joel Kamnitzer - thanks, Joel!]Let p be such that p2 = m2 , p > 0. Then by invariance under Spin1,d−1 (R), a
choice of inner product (, )p is the same as a choice of (, ) invariant under Spinp1,d−1 (R),and a p, provided m > 0 and [or?] d 6= 2.
R1
If m = 0 and d = 2, then the integral ...dp won’t converge. For example, |x|dx
cannot be extended over 0.
Question: What condition do we need for causality to hold?
For all θ ∈ Spin1,d−1 (C) with θ(p) = −p, we need (φ, ψ) = (−1)degφdegψ (θψ, θφ).
In fact, this condition implies the above invariance under Spinp1,d−1 (R). Note that
there are no elements θ ∈ Spin1,d−1 (R) taking p to −p.
Proposition (φ, ψ) = (−1)degφdegψ (θψ, θφ) implies causality.
Fix θ with the above property. Then θSpin1,d−1 (R) does not necessarily map Φinto Φ. It may take Φ to iΦ, so Spin1,d−1 (R) ∪ θSpin1,d−1 (R) acts on C ⊗ Φ.
Proof We want [φ(f ), ψ(g)] = 0 if f and g [have supports that] are spacelike separated. Now,

21

=

Z

[φ+ (f ), ψ − (g)] − (−1)degφdegψ [ψ + (g), φ− (f )]=Z

f (x)g(y)e2πi(x−y)p ((φ, ψ)p − (−1)degφdegψ (ψ, φ)−p )dxdydp

so we need to show that(φ, ψ)p − (−1)degφdegψ (ψ, φ)−phas Fourier transform vanishing on spacelike vectors.
Now, a distribution h has Fourier transform vanishing on spacelike vectors if
h is invariant under Spin1,d−1 (R), and changes sign under rotations changing thedirection of time. Since if h is (invariant measure on {p|(p, p) = m2 , p > 0}− invariant
measure on {p|(p, p) = m2 , p < 0}, then ĥ vanishes on spacelike vectors, then ĥ· (anypolynomial) vanishes on spacelike vectors.
So in our case, choose θ ∈ Spin1,d−1 (C) with θ(p) = −p. Then(φ, ψ)p = (θφ, θψ)θp= (θφ, θψ)−p= (−1)degφdegψ (ψ, φ)−p
Hence, we get a free quantum field theory if we have:
1. An action of Spin1,d−1 (R) on Φ.
2. p > 0, p2 = m2
3. A complex-valued bilinear form (, ) on Φ ⊗ C, invariant under Spinp1,d−1 (R),satisfying (φ, ψ) = (−1)degφdegψ (ψ, φ) for all θ ∈ Spin1,d−1 (C) with θ(p) = −p.
4. If we want the inner product (, ) to be positive definite, then (φ, φ) ≥ 0 forφ ∈ Φ.
For example, let Φ be the vector representation R1,d−1 of Spin1,d−1 . We need tochoose (, ). The usual inner product is not positive definite. Now, Spinp1,d−1 (R) ∼=d−1
Spin0,d−1 (R) for m > 0, so Φ decomposes as R ⊕ R , where the first factor is theline spanned by p.
We can choose (, ) by:
1. (, ) nonzero on R, 0 on Rd−1 . This just gives derivatives of the free scalar field.
2. It can be 0 on R, and nonzero on Rd−1 . This gives a massive vetor field.
22

Week 9, 22 March 2004, Spring Break
Week 10, 29 March 2004, Examples of free fermionic quantum field theories
Today, I’m going to talk about more standard examples of free quantum field
theories. We will start by recalling the data we need:• Spacetime = R1,d−1• A vector p in the positive closed cone of (the dual of) spacetime. (p, p) = m2 .
m is called mass.
• A representation Φ of Spin1,d−1 (R). Φ is real, and usually finite-dimensional (inpractice, always a sum of finite-dimensional representations).
• A complex bilinear form (, ) on C ⊗ Φ. This is not necessarily symmetric,
antisymmetric, or real on φ. It must satisfy the causality condition: (φ, ψ) =(−1)degφdegψ (θφ, θψ), where θ is any element of Spin1,d−1 (C) mapping p to −pand satisfying θ∗ = τ θ, where θ∗ is the complex conjugate, and τ is the elementof order 2 in the center of Spin1,d−1 (R). The elements θ in fact lie in a cosetof Spinp1,d−1 (R). In particular, this condition implies invariance of (, ) under
Spinp1,d−1 (R).
• Extra conditions if m = 0, d = 2 (omitted).
We might want (, ) to be positive semidefinite (implying the space of states ispositive definite). This condition just means (φ, φ) ≥ 0 for all φ ∈ Φ.

In order to write an example of a free quantum field theory, it suffices to writea representation and a form, so this problem is basically reduced to one of linear
algebra. We have already mentioned the examples of scalar fields (where Φ = R) andmassive vector fields (where Φ = R1,d−1 ). Now I will mention a massless vector field.
General warning: Massless fields are usually not some sort of limit of massive
fields as m approaches zero. For example, if Φ = R1,d−1 , a vector representation of
Spin1,d−1 (R), and p is a vector of norm 0, then Spinp1,d−1 is quite different from thecase where p2 > 0, where it is Spin0,d−1 (R), which is compact, semi-simple, and all
the rest of it. If p2 = 0, you get a semidirect product Rd−2 · Spin0,d−2 (R), which inparticular is not compact.
This is an important difference. If a compact group G acts on a real vector space
Φ, we can always find a positive definite symmetric form on Φ fixed by G, by taking the
average (over G) of any positive definite form. If G is noncompact, in general, no such
positive definite form exists. In our particular case, G = Spinp1,d−1 (R), Φ = R1,d−1 ,the only G-invariant positive semi-definite form is given by projecting R1,d−1 to R
using (p, −), and taking taking the usual inner product on R. This does not give23

the quantum field theory of a massless vector field. It does give the quantum field
theory of derivatives of a massless scalar field, which is not what we want e.g. whenquantizing the electromagnetic field.
The correct way to quantize massless vectors: take Φ = R1,d−1 with the usual
inner product, which is not positive definite.
Problem 1: The resulting space of states is not positive definite. This problem isresolved using BRST operators in a rather labor intensive process about which I won’t
say any more today. This is one of the reasons why gauge theories don’t fit into theWightman axioms. [Marco: How did people resolve this problem e.g. in QED, whichcame before the discovery of BRST? Borcherds: They used ad hoc methods. AJ:
They were choosing representatives of BRST cohomology classes, but didn’t know it.]
What I wanted to do today is talk about Dirac fermions and other fermions.
We start with the following construction: Let M be any complex representation of
Spin1,d−1 (R). Suppose it has a Hermitian form h, i, invariant under Spinp1,d−1 (R), withp2 > 0. Form the space M ⊕ M † , where M † is a second copy of M - a sort of complex
conjugate. † is the antilinear operator defined by switching: (φ, ψ)† = (ψ, φ). I
haven’t quite worked out the scalar multiplication action of C on M † . It might
involve complex conjugation, but it might not. Let Φ = fixed points of †. Then Φ isa real vector space of the same real dimension as M. Define a bilinear form (, ) inM ⊕ M † by:(φ, ψ †) = hφ, ψi(φ, ψ) = (φ† , ψ † ) = 0(ψ † , φ) = (−1)degφdegψ (θφ, θ(ψ † ))Note that (, ) is neither symmetric nor antisymmetric, but satisfies the causality
condition essentially by definition. Thus, Φ together with (, ) satisfies our conditions
for constructing a free quantum field theory.
Dirac fermions come from this construction by taking M to be the complex irreducible representation of the Clifford algebra of spacetime. We need to define h, i on
M: notice that p acts on M, since p is an element in the Clifford algebra. p2 = m2 ,so p has eigenvalues ±m, and we get an eigenspace decomposition M = M+ ⊕ M− ,which is invariant under Spinp1,d−1 (R).
Dirac’s bilinear form h, i is defined to be 0 on M− , and nonzero (i.e. positive
definite) on M+ . This is basically the Dirac equation in a disguised form. Note that
this form has some interesting properties that may be disconcerting if you are used
to symmetric and antisymmetric forms: M− is the left kernel, and M+ is the right
kernel.

24

C ⊗ Φ = M+ ⊕ M− ⊕ M+† ⊕ M−†
If d = 4, these are all 2 dimensional complex vector spaces. In the pairing (, ), we
have M− ⊕ M−† making up the left kernel, and M+ ⊕ M+† making up the right kernel,
and there are pairings M+ ↔ M−† and M+† ↔ M− , although I might have made somemistakes in working this out.
Note that this gives a charged free field, meaning there is an action of U1 (R) on
everything. M is complex, so the action is through the unit complex numbers.
It might have occurred to you that this construction is rather complicated andarbitrary, and you might wonder if there are other ways to construct fermion fields.
The answer is yes. You can get:
• Majorana fermions when d ≡ 2, 3, or 4 mod 8.
• Weyl fermions when d is even and m = 0.
• Majorana-Weyl fermions when d ≡ 2 mod 8 and m = 0.
Why do we use Dirac fermion fields rather than others? The main reason is justexperimental evidence. From a mathematical standpoint, all four constructions areequally valid. However, each formulation gives some experimentally testable predictions, and the fermions we see behave like Dirac fermions. For a long time, it wasthought that neutrinos were Weyl fermions, but a few years ago, neutrinos acquiredmass, eliminating that possibility. It is still unknown what kind of fermions neutrinos
are, but they are suspected to be Dirac.
In the remaining five minutes I will try to describe the other constructions. This
will be necessarily rather sketchy.
Majorana: take Φ to be the real representation of the Clifford algebra Cd−1,1 (R).
The tricky part here is that 1 and d − 1 are reversed, so p2 = −m2 . The eigenvaluesof the action of p are then ±im, giving an eigenspace decomposition M ⊕ M † ⊂C ⊗ Φ, where the spaces are complex conjugates of each other. Choose h, i on M
to be Spinp1,d−1 (R)-invariant (note this group is isomorphic to Spind−1,0 (R) which iscompact). Then define (, ) on C ⊗ Φ by:(φ, ψ † ) = hφ, ψi(φ, ψ) = (φ† , ψ † ) = 0(φ† , ψ) = 0
It’s not hard to check that the causality condition holds.

25

I haven’t covered Weyl and Majorana-Weyl spinors, so you’ll have to do them as
an exercise if you are interested.
Next week, I’ll stop doing free quantum field theories, and instead I’ll explain how
Bernstein’s work on D-modules can be used to analytically continue the integrals thatarise when evaluating Feynman diagrams.
Week 11, 5 April 2004, Holonomic Functions
Today’s lecture is really quite independent of quantum field theory, so I’ll justsay what the motivation is now.
Motivation: In quantum field theory, many Feynman diagrams have values which
are given by integrals. The problem is that these integrals often don’t converge,
and this is the cause of all the problems that crop up involving renormalization
and regularization. These integrals depend on parameters (such as the dimensionof spacetime) and “converge” if the parameters are sufficiently large or small. I put“converge” in commas, because in fact they often don’t converge for any value of theparameters, but we’ll sort of ignore that. Often you can split up the integral intoa part that converges if the parameters are large, and a part that converges if theparameters are small.
Idea: Analytically continue in the parameters to define the integral for the values
you want.
The problem of analytically continuing integrals which depend on parameters is
a perfectly valid problem outside physics, so we can treat it mathematically.
The basic example which everyone knows is that of the Gamma function:Z ∞e−t ts−1 dtΓ(s) =
0

The convergence of this integral depends on the parameter s: it converges for
s > 0, as the exponential term kills everything at ∞, and we just need ts−1 to behave
well at 0. To analytically continue, we notice thatZ ∞
Γ(s − 1) =e−t ts−2 dt
0

and integrate by parts
=

Z ∞

e−t

0

=

ts−1
dt
s−1

1
Γ(s)
s−1
26

Idea: Relate Γ(s − 1) to Γ(s). This gives an analytic continuation for all s.

If you want to see why this works, the key point is that you can get a relation bydifferentiating:d s+1t= (s + 1)ts
dt
Here we’ve got a polynomial in t which happens to be t. Notice this has the followingform:dD(t, )p(t)s+1 = B(s)p(t)sdtHere, p(t) is a polynomial in t which is not terribly exciting in this case. B(s) is apolynomial in s, called the Bernstein polynomial. Here, it is s + 1. D is a polynomial
in t, dtd , and s, i.e. a differential operator.
Bernstein showed that for any polynomial p in several variables, we can always
find D and B as above, with B not identically zero. [Joel: so we’re trying to analytically continue the integral of e−t ts−1 ? Borcherds: we’re not actually going toanalytically continue it. Joel: What are we going to continue? Borcherds: I’m going
to set out tools for continuing whatever we want to continue. AJ: Does Bernstein’s
theorem apply to rational functions? Borcherds: Can’t answer that off the top of my1
in it. Borcherds: Not
head. Ogg: Probably - the continuation ends up having a B(s)really sure.]
Second basic example: If <(s) > −1, then |x|s is locally integrable, so it defines a
distrbution in x.
Problem: Can we extend the definition to all complex s?
Answer: It extends to a meromorphic (distribution-valued) function of s. It haspoles at s = −1, −2, −3, . . .

dPut |x|s = (x2 )s/2 . Then dx(x2 )s/2 = (s/2)2x(x2 )s/2−1 = sx(x2 )s/2−1 . We should
refrain from combining the last terms, since the case x < 0 is important. We differd2
2 s/2
entiate again: dx= s(s − 1)(x2 )s/2−1 . We can use this to define (x2 )s/2−1 in
2 (x )
2 s/2
terms of (x ) , as the derivative of any distribution is a distribution. Then, (x2 )s/2−1
becomes a meromorphic distribution-valued function, which picks up poles from the1
factor. In fact, we get poles of order 1 at s = −1, −2, −3, . . . . The concepts(s−1)of distribution-valued meromorphic function may be a bit strange for people who
haven’t seen it before, so we’ll calculate a residue.

Example: What is the residue of the pole at s = −1? This will be some distribution. Put s = 1 in the formula(x2 )s/2−1 =d2 2 s 21(x )s(s − 1) dx2
27

(s)
1
The pole is coming from the term s−1. In general, the residue of fs−1
at s = 1 is just
2 s/2−1
f (1), provided f is holomorphic at s = 1, so the residue of (x )at s = 1 is just:d2 2 1/2
1 d2 2 s/2(x ) = 2 (x )
s dx2
dx
d2
= 2 |x|
dx
d
g(x)
=
dx
= 2δ(s)where g(x) is the function that is 1 for positive x and −1 for negative x. Note that
this means in practice that all attempts to define a distribution |x|s at s = −1 havean ambiguity given by a multiple of δ(x). I’m not sure if that 2 is supposed to be
there, but I won’t waste time on that.
Example: We can try to take the constant term of |x|s at s = −1. The problem
is, we can also define it as the constant term of |x|s g(s), where g is holomorphic at
s = −1, g(−1) = 1. These differ by a multiple of δ(x).
I’m not just doing this to be irritating, because it turns out there is no canonicalway to define |x|−1 as a distribution. In particular, there is no way to do it that gives a
distribution that is homogeneous of degree −1. Rescaling the domain yields spurious
terms.

Remark: (reason this comes up): Values of Feynman diagrams are meromorphic
functions of parameters whose values are distributions. This turns upon a muchlarger and more complicated scale: they cannot be defined in a “canonical” fashion.
The ambiguities you get are more or less what physicists call “anomalies”. This failureof rescaling symmetry in some massless field theories can be traced back to exactly
this phenomenon. [Ogg: Differentials have residues, but functions don’t. You need ads on the end of your expression. Borcherds: Yes, I’m being very sloppy here.]
I will discuss the proof of Bernstein’s theorem. The theorem itself is not powerfulenough to be useful in quantum field theory, but the techniques he develops are. We
start by defining D-modules. People tend to view these as sheaves on a manifold or aringed topos, but most of the fancy machinery is just for bookkeeping, and the basic
ideas are visible in the most basic example.
We consider a module over a ring K[x1 , . . . , xn , ∂1 , . . . , ∂n ]. This is a ring ofdifferential operators, but we don’t bother to include any functions to operate on. It
is defined by the following relations:• xi ‘s commute• ∂i ’s commute28

• [xi , ∂j ] = 0 for i 6= j
• ∂i xi = 1 + xi ∂i for all i
Note the analogy between ∂i and ∂x∂ i .
Example: Suppose a function f satisfies [is annihilated by] partial differential operators D1 , D2 , . . . , Dm . We can form the D-module M = K[x1 , . . . , ∂n ]/hD1 , . . . , Dm i,where this quotient is by the left or right module, and I’ve forgotten which one itis, but I’m not going to worry about it too much. There is a correspondence between functions f satisfying D1 f = D2 f = · · · = Dm f = 0 and monomorphisms of
D-modules from M to the space of smooth functions of x1 , . . . , xn , so D-modules
correspond to systems of partial differential equations. I think the whole of D-module
theory was invented by algebraists who wanted to study partial differential equations
but didn’t like analysis, so they just dispensed with the tiresome nonsense about
existence of solutions.
A holonomic D-module is one that is “small”, meaning there are “lots” of differential operators. Roughly speaking, we should have n independent differential operators in the system of differential operators corresponding to the module. If you’ve
done differential equations, you may have run across the term “overdetermined sytemof PDE’s”. I’ve never figured out what an overdetermined system is, but I’ve got astrong suspicion it has something to do with holonomic D-modules.
[Joel: could you clarify the correspondence?]

K[x1 , . . . , xn , ∂1 , . . . , ∂n ]→ {smooth functions}hD1 , . . . , Dn i
1 7→ some solution of “Di f = 0 ∀i”

This is not anything deep; it’s just a complete triviality.

I haven’t given you a terribly precise definition of D-module, since I haven’t said
what “small” means, or what “lots” of differential operators are. [Soroosh: Which Dmodules can you get by taking quotients like that? Borcherds: In fact all holonomic
D-modules can be obtained this way.]

How do we measure the size of a D-module? We do this by using Hilbert polynomials. If you don’t know what these are, I’ll be teaching a commutative algebra
course in about a year.

Quick summary: Suppose M = ⊕Mi is a finitely generated graded module overa graded ring K[x1 , . . . , xn ] (a polynomial ring, graded by deg(xi ) = 1). We canmeasure size by describing how dim(Γi ) depends on i, where Γi = ⊕j≤iMj . Thisis a polynomial in i for i sufficiently large (note that it is identically zero for i
sufficiently negative). This is called the Hilbert polynomial of the module M. Its
29

integerdegree is called the dimension of m. The leading coefficient is of the form (degree)!
,and the integer is called the multiplicity of M. Since it is already 3:00, I’ll postpone
the proof that this is a polynomial. Next week’s lecture will be essentially a primer
on D-modules and Hilbert functions and so on.

Week 12, 12 April 2004, Holonomic functions (continued)
We’ll start by recalling what we were doing last week. Our main aim was to study
the analytic continuation of Feynman integrals, and we seem to be going about thisin a roundabout way.
General fact: integrals of holonomic functions can often be analytically continued.
At the end of last lecture I was trying to tell you essentially what holonomic meant.
It means it is a solution to a holonomic system of partial differential equations, i.e.
an element of k[x1 , . . . , xn , ∂1 , . . . , ∂n ]/(PDE’s), where this is a holonomic module,
whih means it is “small”. In order to measure the size of a module, we use the Hilbert
polynomial.
Given a finitely generated module M over a (commutative) polynomial ringk[x1 , . . . xn ], M := ⊕i∈Z Mi , define Γi := ⊕m≤i Mm , so we have Γ0 ⊂ Γ1 ⊂ . . . ,and look at the dimension of Γi . Clearly, the speed at which this increases is somemeasure of the size of M. For large i, dim(Γi ) is a polynomial χ(M, i) in i. Thisis the Hilbert polynomial. The degree of χ(M, i) is called the dimension of M.
dim(M)!×(leading coefficient) is called the multiplicity of M. The factor of dim(M)!
makes this an integer.
You might be wondering why I’m spending so much time talking about stuff
that is not so clearly related to quantum field theory. The reason is that this is
not really a “quantum field theory” seminar. Rather, it is an “interesting ideas inmathematics” seminar. That is why I’m spending so much time on D-modules andHilbert polynomials.
Proof that dim(Γi ) is a polynomial for large i.
We shall use induction on n: For n = 0, M is a finitely generated module over k,i.e. a finite-dimensional vector space, so for i >> 0, dim(Γi ) = dim(M).
Now assume this has been proved for #variables < n, and look at the exact
sequence:xn
0 → ker(xn ) → Mi−1 → Mi → Mi /(image of xn ) → 0
Note that the first and last terms are homogeneous components of finitely generated
graded modules over k[x1 , . . . , xn−1 ]. We havedim(ker(xn )) − dim(Mi−1 ) + dim(Mi ) − dim(M − i/(image of xn )) = 0
30

This has something to do with Euler-Poincaré characteristics. Thendim(Mi ) − dim(Mi−1 ) = dim(M − i/(image of xn )) − dim(ker(xn ))By our inductive hypothesis, the right hand side is a polynomial in i of degree lessthan n for large i. If you’ve got a sequence of numbers whose differences comes froma polynomial of degree less than n, then the sequence comes from a polynomial of
degree n. Thus, dim(Mi ) is a polynomial of degree n for large i.
End of proof.
As you can see from the proof, this polynomial always takes integer values at integers. We recall a few facts about polynomial maps Z → Z. They do not necessarily
have integer coefficients. This is a standard mistake everyone makes once in his life
takes integer values,
out of absent-mindedness. For example, the polynomial x(x−1)
2
but does not have integer coefficients. There is a basis over Z for these polynomials
given by:
x
0
x
1
x
2
x
3

=
=
=
=
..
.

x
n

= x(x − 1) · · · (x − (n − 1))/n!



..
.

1
x/1!
x(x − 1)/2!
x(x − 1)(x − 2)/3!
..
.

Notice that the last polynomial vanishes as x = 0, . . . , n − 1 and takes the value
1 at n. Now, any polynomial of degree n taking integer values at integers is a linear
combination of these. It is easy to see by induction
 any integers p(0), . . . , p(n), that for
we can find an integer linear combination of x0 , . . . , nx taking these values, so if p
is a polynomial of degree n, then p − (linear combination) is 0 on 0, 1, . . . , n, and if apolynomial is degree n with n + 1 zeroes, then it must be 0.
In particular, if you notice that if an xn + · · · + a0 takes integer values, thenn!an is an integer. This explains the factor of dim(M)! in the formula mult(M) =dim(M)! × (leading coefficient). Thus, mult(M) is an integer.
One thing I haven’t bothered to check that I leave as an exercise because I won’t
do eveything is ... no, it’s an exercise. [Me: What is the exercise? Borcherds: Um.
Ogg: The exercise is to figure out what the exercise is.]Now, what we want to do is apply this to modules over noncommutative rings.
Basically, we can turn noncommutative rings into commutative rings if we’ve got a
filtration.
Suppose R is a ring k[x1 , . . . , xn , ∂1 , . . . , ∂n ], which is not commutative. We filter
R: k = R0 ⊂ R1 ⊂ . . . where Ri is spanned by all monomials in x1 , . . . , xn , ∂1 , . . . , ∂nof degree at most i, where the variables xj and ∂j are all given degree 1. We look31

at the associated graded ring R0 ⊕ R1 /R0 ⊕ R2 /R1 ⊕ · · · This turns out to be acommutative polynomial ring on 2n generators. Of course, this doesn’t work for all
noncommutative rings: we need the commutators to live in a lower filtered piece. If
M is a module over R generated by m1 , . . . , mk , we construct the Bernstein filtrationof M: We let M0 be the linear span of m1 , . . . , mk , and let Mi := Ri M0 . ThenRi Mj ⊂ Mi+j , and M0 ⊕ M1 /M0 ⊕ M2 /M1 ⊕ · · · is a module over our graded ring.
For a module over this nongraded ring, we’ve constructed a graded module over agraded ring, so we can now take the Hilbert polynomial of this to get the dimension
and multiplicity of M.
Now I get to the exercise I mentioned before: check that dim(M) and mult(M) do
not depend on our choice of generators. This basically involves showing that changing
the generators changes the Hilbert polynomial by a polynomial of degree less than n.
(Warning: other coefficients of the Hilbert polynomial do depend on the choice ofgenerators, so the other terms are more of a pain to use.)
We can find modules over the commutative ring k[x1 , . . . , xn ] of any dimension from 0 to n. This is really easy to do: for example we can take the modulek[x1 , . . . , xn ]/(x1 , . . . , xi ) which has dimension n − i.

Bernstein showed that if M is a nonzero finitely generated module over the polynomial ring k[x1 , . . . , xn , ∂i , . . . , ∂n ], then n ≤ dim(M) ≤ 2n, where the inequality on
the right is just the trivial bound.
[Joel: Does a set of generators of M become a set of generators of the graded
module? Borcherds: I think so, but I’m not sure. This is probably one of thoseamazingly plausible facts that turn out to wrong if you check the details.]
M is called holonomic if M = 0 or if dim(M) = M.
Proof of Bernstein’s theorem. Let Bi be the subspace of k[x1 , . . . , xn , ∂1 , . . . , ∂n ]spanned by monomials of degree at most i. I think this was previously called Ri but
I’ve lost track. Our aim is to show that the natural map Bi → Hom(Mi , M2i ) given
by multiplication is injective if M0 6= 0 (where Mi := Bi M0 , and M0 is spanned bya set of generators of M). Note that dim(Bi ) is a polynomial of degree 2n in i, anddim(Hom(Mi , M2i )) is a polynomial of degree at most 2 × dim(M) in i.
For i = 0, we have B0 = k, and the statement follows, as M0 6= 0. This is where
we pick up the exceptional case where M is 0.

Now, assume the statement is true for i − 1. Choose a ∈ Bi . We notice thataMi ⊂ M2i implies a ∈/ k, so a = cxα ∂ β + . . . , which is short for cxα1 1 . . . xαnn ∂1β1 . . . ∂nβn
for c ∈ k nonzero. Since a ∈/ k, some xi or some ∂i must appear in the leading term fora [under some term ordering]. We assume x1 appears (other cases are similar). Then[a, ∂1 ] 6= 0, as ∂1 reduces the number of x1 s by 1, and hence doesn’t kill everything
off. [a, ∂1 ] ∈ Bi−1 , as [Bi , Bj ] ⊂ Bi+j−2 . If aMi = 0, then[a, ∂1 ]Mi−1 = a∂1 Mi−1 − ∂1 aMi−1 = 0
32

as both the ∂1 Mi−1 and the Mi−1 on the right side next to the a’s are in Mi . By our
inductive hypothesis, [a, ∂1 ] = 0, contradicting our previous conclusion that [a, ∂1 ] 6=
0. If a contains ∂1 , we use [a, x1 ] instead.
End of proof.
As you can see, it is one of those short but tricky proofs. The actual proof is justtwo lines of algebra, but you have to get the right two lines.
I’ll just say what holonomic means. It means “M is as small as possible”. If
M = k[x1 , . . . , xn , ∂1 , . . . , ∂n ]/(some PDE’s), then M holonomic means (roughly) that
we have n “independent” PDE’s, so a holonomic function in n variables is roughly
one that satisfies n “independent” PDE’s.
If you’ve studied differential equations in the past, no doubt you’ve noticed that
there is a fundamental difference between solutions of ordinary differential equations
and solutions of partial differential equations. Solutions of ODE’s are rather rigid:
they can be determined near a point by finitely many derivatives. On the other
hand, solutions of PDE’s are incredibly floppy. You get infinite dimensional spacesof solutions, and they are not fixed by a finite amount of data. However, holonomic
systems of PDE’s behave much like ODE’s, in that their solutions have rigid behavior.
In some sense, the correct generalization of ODE’s is not PDE’s, but holonomic
systems of PDE’s. [AJ: Is there are connection between holonomy and holonomic
systems of PDE’s? Borcherds: Yes, but I haven’t quite figured out what it is. You
can take a solution to an ODE and analytically continue it around a singularity to
get holonomy. Presumably this can be done with a holonomic system of PDE’s also.]
The fundamental property of holonomic modules is that they have finite length(analogous to vector spaces having finite dimension). This means there is a sequence
0 ⊂ M 1 ⊂ M 2 · · · ⊂ M where M i /M i−1 is irreducible. Note: If R = k[x], then R
does not have finite length, as we have R ⊃ xR ⊃ x2 R ⊃ . . . .
If 0 → A → B → C → 0 is an exact sequence of holonomic modules, then:

1. mult(A)mult(C) = mult(B) (this is easy to prove, since the Hilbert polynomialof B is more or less the product of the Hilbert polynomials of A and C)
2. mult is an integer.
3. mult = 0 ⇒ dim < n ⇒ module is 0 (Bernstein).
These three properties basically imply that holonomic modules have finite length.
In fact, the length is at most the multiplicity.
We’ve pretty much finished the commutative algebra material I meant to cover.
Next week, we’ll show that the fact that holonomic modules have finite length implies
that certain integrals can be analytically continued.
33

Week 13, 19 April 2004, Holonomic functions (continued)
We’ll start by trying to remember what we did in the last seminar.
We have a Hilbert polynomial of a finitely generated graded module over a graded(commutative) ring. This gives rise to the dimension (the degree of the polynomial)and the multiplicity (leading coefficient × (degree)!, where the last bit makes themultiplicity integral).
For modules over k[x1 , . . . , xn , ∂1 , . . . , ∂n ], filter everything, look at graded rings,
which become commutative if you’ve set everything up right. Again, you get dimensions and multiplicities. [Marty: Isn’t there another natural grading on that ring?
Borcherds: Yes, you get a graded ring structure if you set the degree of ∂i to be −1
instead of 1, but then you just get a noncommutative graded ring, and there isn’t anice way to get something commutative out of it.]
A module over k[x1 , . . . , xn , ∂1 , . . . , ∂n ] is called holonomic if it is finitely generated, and of dimension ≤ n. By a theorem of Bernstein, the dimension of a nonzero
finitely generated module is at least n. This is the result that makes the theory of
D-modules possible.

Alternatively: Define the multiplicity of a holonomic module as above. Bernstein’s
theorem says that if the coefficient of xn in the Hilbert polynomial is 0, then themodule is 0.
Consequence: Holonomic modules have finite length, i.e. we have a chain 0 ⊂M1 ⊂ M2 ⊂ · · · ⊂ Mn = M, where each Mi /Mi−1 is irreducible. As it happens,
the length is independent of the choice of chain, and we have finite length if andonly if the module is both noetherian and artinian, for rather boring reasons. Finite
length means the modules satisfy almost any finiteness condition you could expect
to hold. One can draw an analogy, where finite length modules are to modules overk[x1 , . . . , xn , ∂1 , . . . , ∂n ] as finite dimensional vector spaces are to vector spaces.
I’m going to give some easy examples of holonomic modules, and once we’redone with those, I’ll give a construction of Bernstein which yields somewhat less easy
examples.
1. M = k[x1 , . . . , xn ] as a module over k[x1 , . . . , xn , ∂1 , . . . , ∂n ], where ∂i acts asd. What is the Hilbert polynomial? We choose 1 as a set of generators anddxitake a filtration:
basis
dim

Γ0 ⊂
Γ1
⊂
Γ2
⊂ ... ⊂
1
1, x1 , . . . , xn
1, xi , xi xj...
(n+1)(n+2)
1
n+1
...
2
n

Γk
deg ≤ k(n+1)···(n+k)
= n+k
k!
n

We have dim(Γk ) = kn! + lower terms in k, so the dimension is n and themultiplicity is 1. This module is holonomic and irreducible since we cannot
write 1 as a nontrivial sum of positive integers.
34

2. M = k[x1 , . . . , xn , ∂1 , . . . , ∂n ]. The Hilbert polynomial has degree 2n, so thismodule is not holonomic. Similarly, we can find modules of dimension betweenn and 2n by killling off some of the ∂i ’s: k[x1 , . . . , xn , ∂1 , . . . , ∂n ]/(∂i+1 , . . . , ∂n )has dimension n + i.
3. Suppose p1 , . . . , pn are nonzero polynomials, and we just look at the module
given by k[x1 , . . . , xn , ∂1 , . . . , ∂n ]/(p1 (∂1 ), p2 (∂2 ), . . . , pn (∂n )). It is easy to checkthat the dimension is n, and the multiplicity is deg(p1 ) ×deg(p2 ) ×· · ·×deg(pn ).
This is actually a module over k[x1 , . . . , xn ] generated by ∂1i1 , ∂2i2 , . . . , ∂nin , whereik ≤ deg(pk ). This is basically a system of n “independent” differential equations, and this is what holonomic systems ought to look like.
I had earlier said something about these systems being called overdetermined
systems of PDE’s, and I had slightly misspoken. These are actually “maximally
overdetermined systems of PDE’s,” and if anyone can tell me what an overdetermined
system is, I’d like to hear about it.
We’ve finished the trivial examples, and we’ll introduce a theorem of Bernstein.
Lemma Suppose M a module over k[x1 , . . . , xn , ∂1 , . . . , ∂n ] is filtered by Γ0 ⊂ Γ1 ⊂
Γ2 ⊂ . . . . Suppose dim(Γi ) ≤ pn (i) for pn some polynomial of degree at most n. Then
M is holonomic (and in particular, finitely generated).
(Note: we did not assume M is finitely generated, and in fact it is quite importantthat we did not, since the main use of this lemma is to show that modules are finitely
generated.)
Proof (surprisingly easy)
1. Every finitely-generated submodule of M is holonomic. (obvious)
2. Suppose 0 ⊂ N1 ⊂ N2 ⊂ . . . is an increasing sequence of finitely generated
submodules. Each Ni is holonomic, so they have a well-defined multiplicity,which is at most n!× degree of the coefficient of xn in pn . Thus, the sequenceof Ni ’s must eventually be constant. This implies M is finitely generated. (Ifnot, we could find an infinite strictly increasing sequence of finitely generated
submodules.)We can use this to give examples of holonomic modules. Here is an important
example:Suppose p is any nonzero polynomial. Then k[x1 , . . . , xn , p−1 ] is holonomic. If
you’re into algebraic geometry, you’d say you’ve localized at p. This is part of a very
general statement involving derived categories and complexes of sheaves, saying that
direct images of holonomic D-modules are holonomic. However, the fancy language
35doesn’t actually have any new content, except if you wanted to know what a derived
category is. Our framework contains all the meat of the ... well, we’re in Berkeley,
and there are vegetarians ... it contains all the tofu of the ...
Let m be the degree of p, and grade M as follows: Let Γk be the span of all thingsof the form f /pk . We’ll call this “stuff of degree at most k”. I’m going to write thesedown, because they are a bit tricky to work out without making a mistake. They are
just an easy calculation, or wrong:deg(f ) ≤ (m + 1)kBi Γk ⊂ Γk+i
(m + 1)k + n
dim Γk ≤n(Bi is the span of elements of degree at most i in k[x1 , . . . , xn , ∂1 , . . . , ∂n ].) By
Bernstein’s Lemma, this implies M is finitely generated and holonomic. (Note: It isfar from obvious that M is finitely generated over k[x1 , . . . , xn , ∂1 , . . . , ∂n ]. You might
expect that adding higher powers of p−1 would continue to increase the size of themodule.)We now have more or less what we need to construct the Bernstein polynomial.
Bernstein polynomial: Suppose p is any polynomial in x1 , . . . , xn . Then there is anonzero polynomial b(s) and a polynomial D(xi , ∂i , s) satisfying:b(s)ps = D(xi , ∂i , s)ps+1
with s a real variable. [Marty: Does D have to depend on s? Borcherds: Not in
trivial examples. Actually, I don’t know of any examples where s appears in D, butsince I’ve never seen a proof that D does not depend on s, I strongly suspect that s
is necessary in general.]
Proof: We work over the field K = k(s) of rational functions in s. We look at themodule M = K[x1 , . . . , xn , p−1 ]ps , with xi and ∂i acting in the obvious way: ∂i (ps ) =sps−1 + ∂i p = ps (s ∂pi p ). (This makes it clear why we need to invert p.) M is holonomic
over K[x1 , . . . , xn , ∂1 , . . . , ∂n ] (the proof of this is basically the same as for the proof
that the ring k[xa , . . . , xn , p−1] is holonomic, as the modules are basically the same
size, but this one is shifted by ps .) We look at the increasing sequence of submodulesof M generated by ps , ps−1 , ps−2, . . . . Since M is holonomic, this sequence is eventuallyconstant, so ps−k−1 = f (xi , ∂i )ps−k , where f is some polynomial with coefficients in
K = k(s). So, ps = g(xi , ∂i )ps+1 . The coefficients of g are rational functions of s, so
we can multiply by a common denominator b(s) to get b(s)ps = poly(xi , ∂i , s)ps+1 .
b(s) is the notorious Bernstein polynomial.
At this point, it seems pretty clear that you need D to depend on s, as you havevery little control over the denominators in g. In general, calculating the Bernstein36polynomial is very difficult. There is some kind of small cottage industry of people
calculating these using Newton polygons and such.
The example that everyone does is: p = x21 + x22 + · · · + x2n . Instead of going
through the trouble of calculating Hilbert polynomials, we’ll just differentiate:∂ s
p = 2xi sps−1∂xi∂2 s
p = 2sps−1 + 4x2i s(s − 1)ps−2
∂x2i
X ∂2
(
)ps = 2snps−1 + 4s(s − 1)ps−12∂xii(2sn + 4s(s − 1))ps−1 = Dps
P ∂2Hence, we have Bernstein polynomial b(s − 1) = 2sn + 4s(s − 1), and D = i ∂x2.
iThis doesn’t even contain xi terms, but that is because this is a particularly simple
example, I think.
Since I said this could be used to analytically continue
I ought to giveR1
R 1 integrals,s
you one example of analytic continuation: The integral 0 · · · 0 p(x) dx1 · · · dxn converges for <(s) ≥ 0. By Bernstein, this is equal toZ
Z
1· · · Dp(x)s+1 dx1 · · · dxnb(s)The differential operator D gives integrals over the n − 1-dimensional boundary
pieces. Inducting on n gives an analytic continuation.
I’ll have to leave this as an exercise, since it is already 3:00.
I have no idea what I’m going to do next week.
Week 14, 26 April 2004, Analytic continuation
Today, I will talk about how to analytically continue almost anything. In particular, I will show you how to analytically continue the integrals that come out of a
Feynman diagram.
We return to the example from last time, for which my explanation was too
sketchy for anyone to understand. We have some polynomial p in n variables xi , weassume p(x) to be non-negative for 0 ≤ xi ≤ 1, and we wish to continue
Z 1
0

···

Z 1
0

p(x)s dx1 · · · dxn

which converges for <(s) ≥ 0, to a meromorphic function for all complex s. We use
37the Bernstein polynomial: b(s)p(x)s = Dp(x)s+1 , where D is some polynomial in xi ,d, and s, so
dxi
Z 1

Z 1

1
p(x) dx1 · · · dxn =
···
b(s)
0
0
s

Z

···

Z

Dp(x)s+1 dx1 · · · dxn

Note that the b(s) in the denominator gives poles at the zeroes of b(s). Now,
suppose D has a term of the form ∂x∂ 1 ×(some rubbish). Then
Z 1
0

···

Z 1
0

∂
(∗)p(x)s+1 dx1 · · · dxn =
∂x1

Z 1
0

···

Z 1

(∗)p(x)s+1

0dx2 · · · dxn

x1 =1x1 =0where (∗) is the rubbish. We hope the right side is meromorphic in s by induction onn. However, we end up having to integrate things of the form
ZZ
· · · p(x)s (some polynomial in xi )dx1 · · · dxn
We started out trying to prove a Rresult Rthat was too weak, and the inductive step
1
1
failed. We should really prove that 0 · · · 0 p(x)s poly(x)dx1 · · · dxn can be analytically continued so that our inductive
is strong
we needR ∞ hypothesisR∞
R ∞ enough.
R 1 Actually,R∞
to extend this a bit: What about 0 or −∞ ? Well, 0 dx = 0 dx + 1 dx, and for
R1
the second integral, we can change x to 1/x, so we get 0 q(x)x(∗) dx, i.e. we integratea polynomial times some (negative) power of x. This leads to a more general problem:
Can we analytically continue the following integral?
Z

···

Z

p1 (x)s1 p2 (x)s2 · · · pk (x)sk p(x)dx1 · · · dxn

The answer is yes. The proof is basically the same as for the case k = 1. Thesk−1 s1main part is the fact that k(s1 , . . . , sk )[x1 , . . . , xn , p−1
1 , . . . , pk ]p1 · · · pk is a holonomic module.
By the same argument as before, we can find Bernstein polynomials b1 , . . . , bk inseveral variables such thatb1 (s1 , . . . , sk )ps11 · · · pskk = Dps11 +1 ps22 · · · pskk
and so on. This is just like Bernstein polynomials for one variable, except there are
more variables. More generally still: p(x) can be any holonomic function. We canalso let pi (x) be non-polynomial, but then we need to be more careful dealing withsingularities.
38

Here is aR silly proof that Γ can be analytically continued: Γ(s) is defined for∞<(s) > 0 by 0 ts−1 e−t dt, and the function e−t is holonomic.
There has been a request, that people want to see a Feynman diagram some time
this semester, so here is one: [draws two vertices, with an edge between them]

We want to consider Feynman diagrams, like this mess. [slightly more complicated graph] This gives rise to some multidimensional integral in the following way:
Designate some vertices of the graph as “internal” [circles some vertices] - these are
the ones you integrate over. You integrate the following function of (x1 , . . . , xv ),where v is the number of internal vertices, and the variables xi take values in Rd or
R1,d−1 (Since integrals over Lorentz space have even more problems than those overEuclidean space, we’ll work with Euclidean space for now):
Z

···

Z Y

internal

vertices i

edges∆(xi − xj )i to jY

dxi

Where ∆ is some function of x ∈ Rd called a propagator. It measures the “amplitude” for a particle to go from the point xi to the point xj . This integral usually
does not converge.
If you want to know what an amplitude is, I would suggest you not bother,because I’ve been trying to understand them for years, and I still have no idea what
they are. This seems to be one of the ways that quantum mechanics is beyond humancomprehension. [AJ: Feynman’s book on QED has a nice explanation of amplitudes.
Borcherds: Feynman also said that no one understands quantum mechanics, and thatyou just get used to it.]
What is ∆?
pThis is often given by some Bessel function of (xi − xj )2 multiplied by some elementary function. If you’ve seen these integrals before and none of them had Bessel
functions, this is because computations with Feynman diagrams are often done in
momentum space rather than position space, and Fourier transforms of Bessel functions are often somewhat simpler. [Marty: Are these K-Bessel functions? Borcherds:
Mostly. Marty: Why do the integrals blow up? Is the problem at infinity? Borcherds:
There are problems everywhere. They come in two basic types: stuff blowing up ata point, and stuff blowing up at infinity. Ogg: Are the analytic continuations always single-valued? Borcherds: They will be single-valued in the parameters we will
introduce.]
It turns out integrals of Bessel functions often can be simplified to something
reasonable:

39

Z ∞

α

v−1

0

m2
exp(−x α − m /α)dα = 2
x2
2



2

v/2

√
Kv (2 m2 x2 )The propagator is usually more or less the left side of the equation, and v hereis the dimension of spacetime, although it is often varied for the purposes of analytic
continuation. So the Feynman integral is:
Z

···
x’s

Z Z

···
α’s

Z Y

α(∗) exp(−x2 α − m2 /α)dαdx

where (∗) is some number. By introducing these α’s, we seem to have turned acomplicated integral into an even worse one (note that the argument of the exponential
is just schematic, and some factors have been left out).
But, we can now integrate over the x’s explicitly.
Reason: The integral is of the form exp(quadratic in x’s).
R
What happens when we actually do this integral? Recall how to do e−quadratic in x dx


A B
Lemma: Let M =where A is a |I| × |I| matrix, I is some set
Bt Cparametrizing the first few rows. Then:Z

exp(−πxt Mx)

Yi∈Idxi = p1det(A)×X

det(MI∪i,I∪j )i,j ∈I/

where MI∪i,I∪j is the matrix given by adding the ith row and jth column to A.
R∞
2
Proof: −∞ e−πx dx = 1, plus lots of linear algebra. This basically involves diagonalizing a matrix, and I’ll skip that, because the details aren’t too terribly interesting.
Technical Lemma: (for evaluating various determinants) Suppose we have a graphΓ (you can think of this as having a Feynman diagram). Suppose we are given a
number te for each edge e of Γ. We are going to form the following matrix M from Γ.
mii =

X

mij = −

teone endof e is i

X

te

e has

ends i,j

Example: [draws triangle Γ with vertices 1,2,3 and edges labeled t12 , t23 , and t31 ]
t12 + t31
−t12
−t31
t12 + t23
−t23 
M =  −t12
−t31
−t23
t31 + t23
40

This is the matrix tha turns up when evaluating the Feynman diagram arisingfrom the graph Γ.
Suppose I and J are subsets of the vertices of Γ, with |I| = |J| = n for some n.
By MIJ we mean the n × n matrix of elements mij , i ∈ I, j ∈ J.
The problem we want to solve is: Find a nice formula for det(MIJ ). The answer,
somewhat surprisingly, is that it actually has a reasonably nice formula:det(MIJ ) =

X
T

±|T |

T is a set of n edges of Γ sith the following property:
1. Every point of I is connected by T to a point not in I.
2. Every point of J is connected by T to a point not in J.
For example, if I is the whole graph, the determinant is 0. For the above formula,
we define:|T | =Y

tee∈T

Now, I’ve got to tell you what the sign is. T is a union of trees. Each tree either
joins a point of I − J to a unique point of J − I, or it has no points in I − J or
J − I. We get an isomorphism from I − J to J − I by connectivity in T . Choose afixed isomorphism from I − J to J − I. The sign ± is the sign of the permutation of
I − J given by composition of one isomorphism with the inverse of the other. Sincechanging your isomorphism by an odd permutation changes the sign of both sides, it
doesn’t really matter which isomorphism you choose [not sure I heard this correctly].
Proof Induction on |I∩J|. We actually only need the cases where I = J or |I−J| = 1,
but the induction involves starting with I and J disjoint, and seeing what happens
when vertices are moved into the intersection.
Week 15, 3 May 2004, Analytic continuation (continued)
Last lecture we were in the middle of trying to compute the determinant of arather hairy looking matrix, so we’ll just recall what the matrix is.
Suppose we’re given a graph Γ with n vertices, and define a n × n matrix Mij asfollows: choose a variable te for each edge e, and either take a sum of singular 2 × 2
matrices:X i j ite −teM=edges j−te te41

where the sum is over all vertices i and j, and all edges e from i to j, or we can takeone big matrix:
Mij =

X

edges e:i→j

Mii =

X

−te if i 6= jteone end of e is i

Given two subsets I and J of the same size m, we can form the minor MIJ withrows in I and columns in J.
Problem: What is det(MIJ )?
P
A: It is given by T ±|T |, where
1. T has m edges.
2. Everything in I is connected by T to something not in I.
3. Everything in J is connected by Y to something not in J.
Note that any graph T with properties 2 and 3 has at least m edges, so these
conditions are quite tight. In particular, they force T to be a forest (i.e. a union oftrees, a graph with no loops). They also force each point of I − J to be connected toa unique point of J − I, so we get an isomorphism of I − J with J − I.
Q
|T | = e∈T te with sign ± being the sign of the isomorphism from I − J to J − I
in some fixed total order of I − J and J − I, although the actual order chosen doesn’tchange the sign of the determinant.
It’s actually rather surprising that the determinant turns out so nicely. One’s
first reaction is that this is going to be a horrible mess.
Proof by induction on I ∩ J:

Note: for applications, we only need the cases I ∩ J = ∅ or I ∩ J = point.

Suppose I ∩ J = ∅. I and J each have m points, and for each i ∈ I, there is asingle edge of T connecting i to something in J. The formula for det(MIJ ) turns out
to be (essentially) the usual formula for the determinant of n × n matrices, althoughΓ may be missing some edges that would contribute some terms.
Now, suppose there is some i ∈ I ∩ J. Pick an edge e joining i to some j 6= i,and form a new matrix N from M by adding the ith row to the jth row, then theith column to the jth column. Then det(NIJ ) = det(MIJ ) as i ∈ I ∩ J, and te only42

occurs in position (i, i) of N:
M=i
j



i
j

te + ∗ −te + ∗−te + ∗ te + ∗N=ij

 i j te + ∗ ∗∗∗Here, ∗ = some junk. So, det(NIJ ) = det(MIJ ) is linear in te .

What is the coefficient of te ? It is given by ± the minor of N with column androw i removed. What we are left with is the matrix corresponding to the graph Γ/e,meaning we identify the vertices i and j and discard the edge e. The fact that we’ve
added the ith row and column to the jth row and column means all the adjacency
information in the ith vertex has been transferred to the jth vertex.
By induction, the coefficient of te is the sum over forests of Γ/e of size m − 1
connecting all points of I − i to something not in I − i and so on, which correspondto forests of Γ containing e and satisfying the usual conditions.
This proves the result, as it is true for all edges e.
I should note a couple of slightly remarkable things: For the case I = J, we have
I J 
M=I
A B
J
C D
P
We notice that if I = J, then det(A) =|T |, where the sum is over all forests with
|I| edges such that everything in I is connected to something not in I. Things in I
should be thought of as “internal vertices”, and vertices not in I should be thoughtof as “external vertices”. [Noah: Should there be a ± in the sum? Borcherds: No.
This is the second interesting point.]
You will notice here that there is no sign. This is because the sign of any
permutation of the empty set is +1. This is because the group of permutations of theempty set is the trivial group. So det(A) is a polynomial in te with non-negative
coefficients. This is far from obvious if you just look at the terms of the determinant
itself. So, det(A) ≥ 0 (if we set te ≥ 0).
The same is true if |I − J| = |J − I| = 1, since all permutations of a one-element
set have sign +1. det(MIJ ) has all terms of the same sign.

When the difference is at least two, you no longer have this property. It is rather
convenient that the only cases that come up when evaluating Feynman integrals arethose for which all coefficients are positive.
Now, recall that all Feynman integrals could be reduced to:Z Z
α∗ exp(−x2 α − α/m2 )dαdx
α

x

43

where the α’s and x’s vary over d-dimensional spacetime. We do the integral over thex’s using the fact thatPZ
Y
(xi , xj ) det(MI∪i,J∪j )1texp
exp(−πx Mx)
dxi = pdet(A)
det(A)i∈I
I horribly garbled this last time, as I forgot the exponential on the right and the det(A)in the denominator of the exponential, but it doesn’t really matter, andthe importantp
A B.
term is the det(A) in the denominator, where A is given by M =Bt D
We use this theorem to integrate over the x’s, so the Feynman integral becomesof the form:
XZ
1
det(MI∪i,J∪j ) Y∗expα(xi , xj )dαidet(A)d/2
det(A)αThe i’s and j’s in this integral correspond to external vertices, i.e. those not in I. In
the physical description, the amplitude should depend on the positions of the particles
before and after the interaction.
This integral is now of a form that can be analytically continued using Bernstein’s
ideas:ZZZ ∞(non-negative polys in α)∗ × (reasonable function)dαα=0

The polynomials in α are either powers of α or det(A), and we know that det(A)is positive for αi ≥ 0, as it is given by a sum over trees, so there is no problem taking
positive powers of it.
Note: This initially converges for d negative with large absolute value, and we analytically continue to the dimension of spacetime that we want. This is closely related towhat physicists call “dimensional regularization”, which involves treating the dimension of spacetime as some arbitrary real number. Unfortunately, the integral usually
has poles at d = 1, 2, 3, 4, . . . and is otherwise regular. It can be rather frustrating tohave done all this work and get a pole for your answer. [Ogg: Why don’t you takeresidues? Borcherds: That is the question that mathematicians always like to ask why not take the constant term? The short reply is that you get the wrong answer.]In general, renormalization is used to cancel out the terms you get from poles.
This is a rather bizarre process by which terms are inserted that depend on the
coupling constant, so that the resulting poles exactly cancel out the poles you getfrom the analytic continuation. The main justification for why it works is that theanswers you get agree with experiment to some absurd number of significant figures.
Well, that concludes the seminar for today, and for the term. There is anotherday of classes next week, but I’m rather exhausted.
44