March 5, 2009

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!

Posted at March 5, 2009 8:33 PM UTC

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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.

Posted by: Robert on March 5, 2009 11:09 PM | Permalink | Reply to this

Re: Super-Yang-Mills Theory

Hey! That was a quick reply. Long time no see, Robert! You began my education on this subject almost 12 years ago.

I think you confused symmetric and anti-symmetric…

Thanks!

Posted by: John Baez on March 5, 2009 11:37 PM | Permalink | Reply to this

Re: Super-Yang-Mills Theory

Via email, someone asked me to expand on my terse reply to Robert’s comment above. Here’s what I wrote:

We need to see why

$\overline{\epsilon} F^{\mu \nu} \Gamma_\mu \partial_\nu \psi$

is proportional to

$\overline{\epsilon} F^{\mu \nu} \Gamma_\mu \Gamma_\nu \Gamma_\lambda \partial^\lambda \psi$

at least up to a total derivative.

We can do integration parts on the former term and get

$\overline{\epsilon} \partial_{\nu} F^{\mu \nu} \Gamma_\mu \psi = \overline{\epsilon} \partial_{\lambda} F_{\mu \nu} \eta^{\lambda \nu} \Gamma^{\mu} \psi$

plus a total derivative.

We can do integration by parts on the latter term and get

$\overline{\epsilon} \partial^\lambda F^{\mu \nu} \Gamma_\mu \Gamma_\nu \Gamma_\lambda \psi = \overline{\epsilon} \partial_\lambda F_{\mu \nu} \Gamma^\mu \Gamma^\nu \Gamma^\lambda \psi$

plus a total derivative.

So, it suffices to show that

$\overline{\epsilon} \partial_{\lambda} F_{\mu \nu} \eta^{\lambda \nu} \Gamma^{\mu} \psi$

is proportional to

$\overline{\epsilon} \partial_\lambda F_{\mu \nu} \Gamma^\mu \Gamma^\nu \Gamma^\lambda \psi$

In short: it suffices to show that

$\eta^{\lambda \nu} \Gamma^{\mu}$

and

$\Gamma^\mu \Gamma^\nu \Gamma^\lambda$

give proportional results when contracted with $\partial_{\lambda} F_{\mu \nu}$.

For this we must ask: what kind of symmetry does $\partial_\lambda F_{\mu \nu}$ have?

If we completely antisymmetrize $\partial_\lambda F_{\mu \nu}$, we get zero. Why? A completely antisymmetric 3-index tensor gives a 3-form. This 3-form we get from $\partial_\lambda F_{\mu \nu}$ must be the exterior covariant derivative $d_A F$. The Bianchi identity says this is zero.

On the other hand, if we completely symmetrize $\partial_\lambda F_{\mu \nu}$, we also get zero. Why? Because it’s clearly antisymmetric in the indices $\mu$ and $\nu$

So, if we completely symmetrize $\partial_\lambda F_{\mu \nu}$ we get zero, and if we completely antisymmetrize it we get zero. So, it must be the third kind of 3-index tensor!

Remember, irreps of the permutation group $S_3$ are classified by 3-box Young diagrams. There’s the tall skinny one, which means ‘completely antisymmetric’, and the short fat one, which means ‘completely symmetric’. Then there’s the third one, which Robert is calling the “2+1 hook”:

XX
X


This is the representation of $S_3$ on a 2d space that you get by putting an equilateral triangle in that space, centered at the origin, and letting $S_3$ act as symmetries of that triangle.

The theory of Young diagrams says that any tensor with this third kind of symmetry, say $S_{\mu \nu \lambda}$, can be obtained by taking a tensor $T_{\mu \nu \lambda}$, symmetrizing with respect to two of the indices, and then antisymmetrizing with respect to two other indices.

Therefore, we’ll know that the tensors

$\eta^{\lambda \nu} \Gamma^{\mu}$

and

$\Gamma^\mu \Gamma^\nu \Gamma^\lambda$

give proportional results when contracted with $\partial^{\lambda} F^{\mu \nu}$ as soon as we can show they give proportional results when we first symmetrize them with respect to two indices, and then antisymmetrize with respect to two others.

Indeed, they already become proportional when we symmetrize them with respect to $\nu$ and $\lambda$, thanks to the Clifford algebra relations

$\Gamma^\nu \Gamma^\lambda + \Gamma^\lambda \Gamma^\nu= -2 \eta^{\nu \lambda}$

So we’re done! Of course a more detailed analysis must work out the constant of proportionality, but that’s what grad students are for.

Posted by: John Baez on March 11, 2009 2:22 AM | Permalink | Reply to this

Re: Super-Yang-Mills Theory

can you say that without indices?
maybe even describe what it means?

Posted by: jim stasheff on March 11, 2009 2:27 PM | Permalink | Reply to this

Re: Super-Yang-Mills Theory

No; someday I might understand this stuff well enough to explain it in lucid prose, but right now it’s a computation.

Posted by: John Baez on March 11, 2009 9:49 PM | Permalink | Reply to this

Re: Super-Yang-Mills Theory

I’m mildly confused about why you say this only works in D=3,4,6 and 10, since there are theories I would call supersymmetric Yang-Mills theories in other dimensions (below 10). But I think it’s that those are the dimensions where the smallest supersymmetric multiplet containing a gauge boson turns out to contain just gauge bosons and gauginos; in other cases, there would also be adjoint scalars. Is that the condition that’s singling out those dimensions?

Posted by: Matt Reece on March 6, 2009 12:50 AM | Permalink | Reply to this

Re: Super-Yang-Mills Theory

Matt writes:

I’m mildly confused about why you say this only works in D=3,4,6 and 10, since there are theories I would call supersymmetric Yang-Mills theories in other dimensions (below 10).

Sorry, you’re right.

But I think it’s that those are the dimensions where the smallest supersymmetric multiplet containing a gauge boson turns out to contain just gauge bosons and gauginos; in other cases, there would also be adjoint scalars. Is that the condition that’s singling out those dimensions?

I guess so. Actually, all I know for sure is something much simpler: namely, the Lagrangian I wrote down is supersymmetric in the way I described only in dimensions 3, 4, 6 and 10. We can get other theories by dimensional reduction, but then I guess the gauge bosons give spin-0 particles transforming in the adjoint rep, as you mention.

Here’s the reason for the slack wording of my post. Right now I don’t even care much about the ‘only if’ theorems saying that theories of certain types only exist in certain dimensions. I’m more interested in understanding how the theories that do exist take advantage of properties of $\mathbb{R}, \mathbb{C}, \mathbb{H}$ and $\mathbb{O}$. And, I want to start with the simplest ones.

Posted by: John Baez on March 6, 2009 2:08 AM | Permalink | Reply to this

Re: Super-Yang-Mills Theory

You’re aware of the paper by Kugo and Townsend, “Supersymmetry and the Division Algebras”?

Posted by: Aaron Bergman on March 6, 2009 2:57 AM | Permalink | Reply to this

Re: Super-Yang-Mills Theory

Aaron writes:

You’re aware of the paper by Kugo and Townsend, “Supersymmetry and the Division Algebras”?

Yes, but I admit I haven’t found it very enlightening so far. I like this paper better:

• J. M. Evans, Supersymmetric Yang-Mills theories and division algebras, Nuclear Physics B, 298 (1987), 92–108.

Abstract: An explicit correspondence between simple super-Yang-Mills and classical superstrings in dimensions 3, 4, 6, 10 and the division algebras R, C, H, O is established. A gamma matrix identity necessary and sufficient for their existence is shown to yield trialities, objects which are equivalent to division algebras. The physical states lie in representations of the group of triples of the division algebra, exhibiting outer automorphisms exchanging the bosons and fermions. The identities are also interpreted in terms of projective lines over division algebras.

And I find Deligne’s treatment in Quantum Fields and Strings: A Course for Mathematicians even clearer.

Posted by: John Baez on March 6, 2009 6:38 AM | Permalink | Reply to this

Re: Super-Yang-Mills Theory

We can get other theories by dimensional reduction, but then I guess the gauge bosons give spin-0 particles transforming in the adjoint rep, as you mention.

Right. If you think through the dimensional reduction I think it becomes apparent why 3, 4, 6, and 10 are the dimensions where the multiplets are simplest. They’re the largest dimension in which the smallest spinor rep has a given number of components: 16 components in 6 < D <= 10, 8 in 4 < D <= 6, etc.

So e.g. when reducing from 6D to 5D, a 6D Weyl spinor has the same number of components as a 5D Dirac spinor, so the fermion doesn’t get reduced in a nontrivial way; A_6 becomes a scalar, and has to live in the same multiplet. On the other hand when reducing from 5D to 4D, the number of spinor components drops, the 5D gaugino splits into two 4D fermions, and it becomes possible for the A_mu and A_5 component to fit into different 4D multiplets.

I’m not sure if I phrased that clearly.

Posted by: Matt Reece on March 6, 2009 3:26 AM | Permalink | Reply to this

Re: Super-Yang-Mills Theory

Matt writes:

They’re the largest dimension in which the smallest spinor rep has a given number of components: 16 components in 6 < D <= 10, 8 in 4 < D <= 6, etc.

Right. And that’s why they’re related to division algebras: one can show that there’s a division algebra of dimension $n$ iff the double cover of the rotation group $SO(n)$ has a spinor rep whose dimension equals $n$. The dimensions of the spinor reps eventually outstrip $n$, but they match at $n = 1,2,4,8$.

More precisely, the real dimension $d$ of the smallest spinor rep of the double cover of $SO(n)$ goes like this: $n = 1: d = 1$ $n = 2: d = 2$ $n = 3: d = 4$ $n = 4: d = 4$ $n = 5: d = 8$ $n = 6: d = 8$ $n = 7: d = 8$ $n = 8: d = 8$ Note the matchup at $n = 1,2,4,8$. These dimensions are the largest ones for which $d$ takes a fixed value.

You’re talking about the double cover of the Lorentz group in $D$ dimensions, so your story is a bit different — but related. The smallest spinor rep of this group is twice the dimension of the smallest spinor rep of the double cover of $SO(n)$ when $D = n+2$, so in physics the magic numbers are not $1,2,4,8$ but $3,4,6,10$.

Posted by: John Baez on March 6, 2009 6:49 AM | Permalink | Reply to this

indexless Super-Yang-Mills Theory

John writes:

I will avoid writing down index-ridden expressions until the very end.

This is essentially an unintentional paraphrase of Yuri Rainich’s statement in his book: Mathematics fo Relativity:
“The last thing you want to do is write things down in indices”.

Knowing Yuri’s penchant for language as well (he illustrated similarity geometry in class by comparing Alice (in Wonderland )and Gulliver as to the similarity transformations implied in the words and also as to their gender related psychology.

Posted by: jim stasheff on March 6, 2009 1:20 AM | Permalink | Reply to this

Re: Super-Yang-Mills Theory

Verifying that the Lagrangian of $N=1$ SUSY Yang-Mills in 10D boils down to showing the Fierz identity 4.A.7 in Green-Schwarz-Witten (GSW). As an added bonus we can dimensionally reduce to 4D and check that the theory we get has $N=4$ SUSY in 4D.

Fierz identities have a much nicer interpretation in representation theory than physicists often emphasize.

Following GSW, we can drop the $m$ and $n$ indices of 4.A.7 if we introduce two new (Weyl or Majorana-Weyl?) spinors $\bar{\epsilon}_1$ and $\epsilon_2.$ Then the Fierz identity we need to show is

$(\bar{\epsilon}_1 \Gamma^{\mu} \epsilon_2)(\bar{\psi}_1 \Gamma_{\mu} \psi_2) + (\bar{\epsilon}_1 \Gamma^{\mu} \psi_1)(\bar{\psi}_2 \Gamma_{\mu} \epsilon_2) - (\bar{\epsilon}_1 \Gamma^{\mu} \psi_2)(\bar{\psi_1} \Gamma_{\mu} \epsilon_2) = 0$

where $\psi_1$ and $\psi_2$ are Majorana-Weyl spinors.

If we denote the fundamental 10 dimensional representation of $SO(1,9)$ by V and the 16 dimensional (reducible) Weyl representation by $S,$ then we note that the projection map from $S \otimes S \rightarrow Sym^2 S$ factors through the exterior algebra (of dimension 1024) $\bigoplus_{j=0}^{10} \wedge^j V$ and it actually factors through $\bigoplus_{j=1,5,9} \wedge^j V$

Check that $dim Sym^2 S = 16*17/2 = 136 = dim V + dim \wedge^5 V = 10 + 126$

Checking the Fierz identity 4.A.7 is really equivalent to showing that the intertwining map from $S \otimes S \rightarrow S \otimes S$ is the zero map. This has a nice pictorial representation that looks like a 4-point function or a conformal block. However I don’t know of a nicer way to show the vanishing of the intertwining map than to keep translating from GSW. Intentionally or not, it seems very appropriate for intertwiners to make an appearance at the n-Category Cafe.

Posted by: Richard on March 6, 2009 7:25 AM | Permalink | Reply to this

Re: Super-Yang-Mills Theory

Thanks for explaining the Fierz identity approach! But I’m afraid I still find Fierz identities rather scary — to me they still deserve the name ‘fierce identities’.

In particular, how do you ‘note’ that the projection $S \otimes S \to Sym^2 S$ factors through $\wedge^1 V \oplus \wedge^5 V \oplus \wedge^9 V$? Where do the numbers 1,5,9 come from, and how would the story work in different dimensions?

However I don’t know of a nicer way to show the vanishing of the intertwining map than to keep translating from GSW.

There’s a nice way that uses the division algebras $\mathbb{R}, \mathbb{C}, \mathbb{H}$ and $\mathbb{O}$. This has already been explained in the octonionic (10-dimensional) case by Jörg Schray — see equation 30 of his paper The general classical solution of the superparticle. John Huerta and I are working on beautifying this argument and generalizing it to all four division algebras.

Intentionally or not, it seems very appropriate for intertwiners to make an appearance at the $n$-Category Cafe.

Intentionally! Another of our projects may be to develop a Feynman diagram approach to spinors and vectors in the magic dimensions: 1,2,4,8 in the Euclidean case, 3,4,6,10 in the Lorentzian. Cvitanovic already has a nice diagrammatic approach to Fierz identities in arbitrary dimensions — but I don’t understand it well enough yet.

Posted by: John Baez on March 6, 2009 4:34 PM | Permalink | Reply to this

Re: Super-Yang-Mills Theory

Middle dimensional forms are always in the symmetric square(s) of the irreducible spinor(s), even in odd dimensions (where the two middle dimensional form representations are isomorphic). The forms appearing in any particular component of spinors tensor spinors have period 4 in the degree of the form because the Clifford algebra anti-involutions have period 4. Further, the Hodge star identifies isomorphic representations.

In odd dimensions, this is enough to determine which degree forms appear in which components of spinors tensor spinors. In even dimensions, it suffices also to know that the forms adjacent to the middle dimension appear in the tensor product of irreducible spinors with opposite chirality.

This structure drives and determines period 8 observations about spinors, independent of signature or even the underlying field.

In particular, the above is enough information to “note” that in ten dimensions the symmetric square of the two irreducible spin reps yield self-dual and anti-self-dual forms in 1-forms+5-forms+9-forms. But it is incorrect to say that this is the symmetric square of the non-irreducible direct sum of these two spin reps.

- davidmjc

Posted by: davidmjc on March 21, 2009 12:46 AM | Permalink | Reply to this
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Re: Super-Yang-Mills Theory

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