Super-Yang-Mills Theory
Posted by John Baez
I’m trying to learn the basics of supersymmetric Yang–Mills theory, and I’m stuck on what should be an elementary calculation. I was never all that good at index-juggling, and years of category theory have destroyed whatever limited abilities I had.
So, this post is basically a plea for help from experts. But I like to pay off my karmic debts in advance. So, I’ll start by explaining some basic stuff to the nonexperts out there.
(Warning: by ‘nonexpert’ I mean ‘a poor schlep like me, who knows some quantum field theory but always avoided supersymmetry’.)
The first piece of good news about supersymmetric Yang–Mills theory, for those struggling to learn it, is that it’s just Yang–Mills theory coupled to massless spin-1/2 particles. So, it’s very much like what you see in the Standard Model… except for one big difference. Namely: these fermions transform in the adjoint representation of the gauge group. Under good conditions, this permits an extra symmetry between the gauge bosons and the fermions, called supersymmetry.
What do I mean by ‘good conditions’? This is the cool part. I mean that the dimension of spacetime must be 3, 4, 6 or 10! These funny numbers 3, 4, 6 and 10 happen to be two more than my favorite powers of two: 1, 2, 4, and 8. And this is no coincidence. It’s actually special properties of the real numbers, complex numbers, quaternions and octonions that make supersymmetry tick!
Or at least that’s what everyone says. I’ve finally decided it’s time to understand this in detail, since I have a grad student — John Huerta — who is working on octonions and their applications to physics. So, we’ve been trying to work through the calculations sketched here:
- Green, Schwarz and Witten, Superstring Theory, vol. I, appendix 4.A: Super Yang–Mills Theories, pp. 244–247.
and described in more detail here:
- Pierre Deligne, Daniel Freed, et al, Quantum Fields and Strings: A Course for Mathematicians, vol. I, chapter 6: Supersymmetric Yang–Mills Theories, pp. 299–311.
I think I understand what works well in dimensions 3, 4, 6 and 10. It’s some other detail of the calculation that’s bugging me.
I will avoid writing down index-ridden expressions until the very end. Instead, I’ll try to use some slicker notation.
The basic fields in super-Yang–Mills theory are:
- a connection on the trivial principal $G$-bundle over Minkowski spacetime — let’s call this connection $A$, and
- a spinor field taking values in the Lie algebra of $G$ — let’s cal this $\psi$.
The Lagrangian is the usual thing for Yang–Mills theory coupled to spinors:
$L = -\frac{1}{4} \langle F , F \rangle + \frac{i}{2} \langle \psi, D_A \psi \rangle$
Here $F$ is the curvature of $A$, while $D_A$ is the covariant Dirac operator. I’m using the inner product of Lie-algebra-valued 2-forms to define $\langle F, F \rangle$, while $\langle \psi, D_A \psi \rangle$ is the sort of ‘inner product’ of Lie-algebra valued spinor fields that physicists typically write as $\overline{\psi} D \psi$.
The would-be generator of supersymmetries acts on $A$ and $\psi$ as follows:
$\delta A = \frac{i}{2} \epsilon \cdot \psi$
$\delta \psi = -\frac{1}{4} F \epsilon$
Here $\epsilon$ is a constant spinor field. I’m using $\cdot$ to mean the operation that eats two spinor fields and spits out a 1-form. So, physicists would write $\delta A = \epsilon \cdot \psi$ as $\delta A_{\mu} = \overline{\epsilon} \Gamma_\mu \psi$ where $\Gamma_\mu$ is a gamma matrix. Of course, $\epsilon$ is a spinor field while $\psi$ is a Lie-algebra-valued spinor field. So, $\epsilon \cdot \psi$ is really a Lie-algebra-valued 1-form. And that’s just what the connection $A$ is, too — since it’s a connection on a trivial bundle.
I’m also assuming you know that differential forms act on spinors by Clifford multiplication, since we can use an inner product on a vector space to identify its exterior algebra with its Clifford algebra, at least as vector spaces. This is what I’m using to define $F \epsilon$. Shades of Hestenes! Of course, $F$ is really a Lie-algebra-valued 2-form, so $F \epsilon$ is really a Lie-algebra-valued spinor. And that’s just what $\psi$ is, too.
Starting from this we can use the rules of calculus to compute
$\delta L$
In dimensions 3, 4, 6, and 10 the answer should be a ‘total derivative’ — roughly speaking, a function that gives zero when we integrate it over all of spacetime. (More precisely, it’s a function that gives an exact $n$-form when we multiply it by the volume form.)
Here’s what I get for starters:
$\delta L = -\frac{1}{2} \langle \delta F, F \rangle + i \langle \delta \psi, D_A \psi \rangle + \langle \psi, (\delta A) \psi \rangle$
up to total derivatives. Then let’s use the handy rule
$\delta F = d_A \delta A$
where $d_A$ is the exterior covariant derivative of a Lie-algebra-valued differential form. We get
$\delta L = -\frac{1}{2} \langle d_A \delta A, F \rangle + i \langle \delta \psi, D_A \psi \rangle + \langle \psi, (\delta A) \psi \rangle$
Then let’s plug in the formulas for $\delta A$ and $\delta \psi$:
$\delta L = -\frac{i}{4} \langle d_A (\epsilon \cdot \psi), F \rangle - \frac{i}{4} \langle F \epsilon, D_A \psi \rangle + \langle \psi, (\epsilon \cdot \psi) \psi \rangle$
Now, the third term, the one that’s trilinear in our spinor field, is the exciting one:
$\langle \psi, (\epsilon \cdot \psi) \psi \rangle$
This only vanishes in dimensions 3, 4, 6, and 10 — thanks to the magic of the normed division algebras $\mathbb{R}, \mathbb{C}, \mathbb{H}$ and $\mathbb{O}$.
But I’m stuck on something more mundane. Why do the other two terms cancel? Why is
$\langle d_A (\epsilon \cdot \psi), F \rangle + \langle F \epsilon, D_A \psi \rangle = 0$
up to total derivatives?
Of course the two terms above look darn similar, which is promising. But if you grit your teeth and peek into their guts, you see the first term has a single gamma matrix in it, while the second has three. In index-ridden notation, the question is something like this. Why is
$\overline{\epsilon} F^{\mu \nu} \Gamma_\mu \partial_\nu \psi + \overline{\epsilon} F^{\mu \nu} \Gamma_\mu \Gamma_\nu \Gamma_\lambda \partial^\lambda \psi = 0$
up to total derivatives?
I could be off by a factor of two or something here, but that’s not what I’m worried about — as a trained mathematician, I can make factors of two come and go at will. I’m worried about the more basic problem of why these terms should have any right to cancel! Or: have I made a serious mistake?
Deligne and Freed say these terms cancel if we use the Bianchi identity together with the Clifford algebra relations. That sounds vaguely plausible. The Bianchi identity and integration by parts imply the second term is antisymmetric in $\mu, \nu,$ and $\lambda$, at least up to exact terms. And, the Clifford algebra relations say $\Gamma_\mu \Gamma_\nu + \Gamma_\nu \Gamma_\mu = -2 \eta_{\mu \nu}$ where $\eta$ is the Minkowski metric. So, it’s potentially good for turning two gamma matrices into none. However, I don’t see how it’s supposed to work.
I know physicists regard calculations like these as trivial, which is why you don’t find them in the published literature. But, I have no pride… well actually I did, but it’s going fast. I just want to know what’s going on!
Re: Super-Yang-Mills Theory
John, I think you confused symmetric and anti-symmetric: As you say, you have to use integration by parts to have the derivative act on $F$ instead of $\psi$. Then Bianchi tells you the anti-symmetric part has to vanish. What’s left over is that $\mu$, $\nu$, $\lambda$ have the symmetry of the 2+1 hook Young tableau and in fact $\lambda$ is symmetric with one of $\mu$ and $\nu$. And now you can use the Clifford rule to turn three $\Gamma$’s into one.