## September 1, 2009

### Division Algebras and Supersymmetry

#### Posted by John Baez

I’ve been in love with the octonions for over a decade now. They seem downright distasteful when you first meet them: an 8-dimensional algebra where division is allowed, but multiplication is nonassociative! Only later do you realize that they’re deeply connected to a wide range of mysterious and exotic things: the icosahedron, the exceptional Lie groups, the exceptional Jordan algebra… and the way superstring theory and super-Yang–Mills theory love to live in 10 dimensions.

I wrote a paper on the octonions back in 2001, where I tried to assemble all the clues I’d found. But I didn’t get far in relating the octonions to supersymmetry. I just explained the basic facts: how octonions are naturally tied to 10-dimensional Minkowski spacetime.

Now I’m lucky to have a brave graduate student who can do calculations like a physicist and proofs like a mathematician: John Huerta. Since March — when we finished our review article on the algebra of grand unified theories — we’ve been digging into the relation between octonions and supersymmetry. As so often the case in this game, the clues are already available, lurking in existing papers. The trick is assembling them into a clear story that people can understand without going on a lengthy and painful quest.

It’s not quite done yet, but it’s close. As always, comments and corrections would be appreciated.

In a way, the story is very simple. In Minkowski spacetime of any dimension, there’s an operation

\begin{aligned} \cdot : & S \otimes S & \to& V \\ & \psi \otimes \phi & \mapsto & \psi \cdot \phi \end{aligned}

that takes two spinors and turns them into a vector. There’s also an operation

\begin{aligned} & V \otimes S & \to& S \\ & A \otimes \psi & \mapsto & A \psi \end{aligned}

whereby a vector acts on a spinor and gives another spinor. In fact these two operations are just two views of the same thing: the basic interaction between spinors and vectors. We see this interaction — very literally, we see it — whenever light hits an electron in our eyes. The photon is a vector, the electron is a spinor, and the absorption process involves the operation $A \otimes \psi \mapsto A \psi$. In Feynman diagram language, it looks like this:

Now, given a spinor $\psi$, we can define a new spinor by combining the above two maps:

$(\psi \cdot \psi) \psi$

And then comes something magic:

Theorem: $(\psi \cdot \psi) \psi = 0$ for all $\psi$ if and only if the dimension of spacetime is 3, 4, 6, or 10.

Our paper gives a proof of the ‘if’ part. The idea is this: these four magic dimensions are 2 more than the dimensions of the normed division algebras $\mathbb{R}, \mathbb{C}, \mathbb{H}$ and $\mathbb{O}$. We can use the division algebras to describe spinors and vectors in these magic dimensions. A special feature of the normed division algebras make the identity true. Namely, in these algebras the associator

$[a,b,c] = (a b)c - a(b c)$

is completely antisymmetric.

The proof is just algebra, and you don’t need to know any physics to follow it. So every mathematician in the universe should read this portion of the paper (Section 2).

We also explain how the identity $(\psi \cdot \psi) \psi = 0$ gives rise to supersymmetry in Yang–Mills theory in dimensions 3, 4, 6, and 10. This part takes a bit more physics know-how — or at least, differential geometry.

By the way, this identity $(\psi \cdot \psi) \psi = 0$ also shows up when checking supersymmetry for the Green-Schwarz Lagrangian of the classical superstring! So, it really rules the world… at least if you believe in supersymmetry.

Also by the way, I’ve been very sloppy above about left-handed versus right-handed spinors, and also Weyl, Majorana and Majorana–Weyl spinors. The paper is not sloppy in this way.

Lastly, on a more personal note: this is the last paper I need to write for a while! I did a calculation and discovered that after this one is done, I’ll have put 471 pages of papers on the arXiv since December 29th of last year. Spending all my time finishing half-written papers, some of which have been hanging over my head for years, has made me stressed-out, irritable and gloomy. But putting them off, which I’d been doing earlier, made me feel even worse. So it’s a great relief to be done, and I plan to live a happy-go-lucky, carefree, irresponsible life from now on.

Posted at September 1, 2009 10:09 PM UTC

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### Re: Division Algebras and Supersymmetry

I think that there are sign errors in the proof of Proposition 2 on Page 3. It should read:

“Since (ab)∗ = b∗a∗ , we have [a, b, c]∗ = -[c∗ , b∗ , a∗ ]. By alternativity this equals
[a∗, b∗ , c∗ ],
which in turn equals
−[a, b, c] by the above proposition. So, [a, b, c] is purely imaginary.”

Yours has the opposite signs written for [c∗ , b∗ , a∗ ] and [a∗, b∗ , c∗ ].

Posted by: Theo on September 2, 2009 2:55 AM | Permalink | Reply to this

### Re: Division Algebras and Supersymmetry

I wouldn’t be surprised if there were sign errors here, since I wrote up this part just yesterday and John Huerta (who is at a particle physics school in Corfu) hasn’t checked it yet. In fact I’m trying to finish off a few remaining bits in time for him to show it to the folks in Corfu.

Oh, you’re right!

Since $(a b)^* = b^* a*$ we have

\begin{aligned} [a,b,c]^* & = ((a b)c) - a(b c))^* \\ & = c^* (a b)^* - (b c)^* a^* \\ & = c^* (b^* a^*) - (c^* b^*) a^* \\ & = - [c^*, b^*, a^*] \end{aligned}

and the associator is completely antisymmetric so

$- [c^*, b^*, a^*] = [a^* , b^*, c^*]$

and the ‘above proposition’ says the associator changes sign each time we conjugate an entry, so

$[a^* , b^*, c^*] = - [a,b,c]$

so we get

$[a,b,c]^* = -[a,b,c]$

as desired.

Thanks!!! I’m sure glad you took a look at it.

Posted by: John Baez on September 2, 2009 4:15 AM | Permalink | Reply to this

### Re: Division Algebras and Supersymmetry

On page 10, you write:

“Also any subspace U ⊆ V of a super vector space becomes a super vector space, with U0 = U ∩ V0, U1 = U ∩ V1.”

This is false. Consider R^{1|1}, the supervector space with one even dimension and one odd dimension; as a real vector space, it is R^2. Then the real subspace generated by (1,1) intersects trivially with the pure-even part and with the pure-odd part.

I doubt that you actually need a good notion of sub-super-vector space. But it’s not too hard to get it: a sub-super-vector space must be a graded subspace; i.e. U \subseteq V in super-land only if U = U_0 \oplus U_1 and U_0 \subseteq V_0 and U_1 \subseteq V_1.

Recalling that as a monoidal category (but not as a braided category!), super-vect is precisely the category of Z/2 modules, one can write “sub-super-vector-space” as “sub-Z/2-representation”.

Posted by: Theo on September 2, 2009 3:21 AM | Permalink | Reply to this

### Re: Division Algebras and Supersymmetry

Another way to say that the statement is clearly false is this. The map from super-vect to vect that forgets the grading, so as to send R^{p|q} to R^{p+q}, is a faithful essentially surjective functor, but it is not full. It had better not be; otherwise vect and supervect would be the same.

But then since not all vect maps are supervect maps, it would be highly unlikely if all vect subobjects were supervect subobjects.

Posted by: Theo on September 2, 2009 3:30 AM | Permalink | Reply to this

### Re: Division Algebras and Supersymmetry

You’re right again. I know perfectly well the right concept of a subobject of a graded vector space but somehow ignored it here. Probably the psychological reason was that I only needed a pathetically simple special case of the full result. Namely, I only needed any subspace of a totally even super vector space to be a super vector space in its own right.

Thanks again!!!

Posted by: John Baez on September 2, 2009 4:28 AM | Permalink | Reply to this

### Re: Division Algebras and Supersymmetry

I also have a conceptual question.

Throughout your exposition, you pick an identification of Minkowski spacetime with Hermitian matrices. It’s cool that this works, and you go and show that natural matrix-y operations preserve the determinant. But it seems that to make this identification is, in Weyl’s sense, “an act of violence”. In particular, the identification picks out at the minimum a basis vector in the t direction (half the trace), a basis vector in the x direction (the other diagonal coordinate), and a basis vector in another space direction (the real axis in the division algebra).

So: how does the hermitian-matrix representation change under general Lorentz action? You just say “determinant-preserving”, which is certainly true. But it seems that your whole division-algebra-inspired story requires breaking many symmetries.

And, a related question, in that above I’m asking you to be less basis-full, and now I’m asking you to be more so:

You define the representations of Spin(n+1,1) on K^2 = K\oplus K by recognizing the matrix-multiplication action of V = Hermitiian Matrices on K^2\oplus K^2 as extending to an action of Cliff(V), and then recognize Spin(n+1,1) as living inside the even part of Cliff(V). Is there a more explicit basis-dependent way to describe this? Maybe what I’m asking is: how does Spin live inside Cliff?

Posted by: Theo on September 2, 2009 3:53 AM | Permalink | Reply to this

### Re: Division Algebras and Supersymmetry

Theo wrote:

… it seems that your whole division-algebra-inspired story requires breaking many symmetries.

That’s true in a sense, but it’s pretty well understood and actually not so bad.

The set of $2 \times 2$ hermitian matrices with entries in a normed division algebra $\mathbb{K}$, say $h_2(\mathbb{K})$, is a Jordan algebra. The same is true for $h_3(\mathbb{K})$ and also for $h_n(\mathbb{K})$ with $n \gt 3$ when $\mathbb{K}$ is associative — that is, not the octonions.

By calling these guys ‘Jordan algebras’ I mean they have a unit $1$ and a product

$x \circ y = \frac{1}{2} x y + y x$

satisfying couple of identities.

What you’re saying amounts to this: the Jordan algebra automorphisms of $h_2(\mathbb{K})$ aren’t all the Lorentz transformations, but only those fixing a certain ‘time axis’.

However, people noticed this a long time ago, and then they realized that for many purposes, what ‘really matters’ about Jordan algebras is a certain lesser amount of structure, which is preserved by a larger group called the ‘structure group’. In the case of $h_n(\mathbb{R})$ this structure group is $PSL_n(\mathbb{R})$. In the case of $h_n(\mathbb{C})$ this structure group is $PSL_n(\mathbb{C})$. And in the case of $h_2(\mathbb{K})$ this structure group is the Lorentz group!

A pretty fun place to learn about this is McCrimmon’s entertaining review article Jordan algebras and their applications, especially Section III, Isotopes and structure groups.

Posted by: John Baez on September 2, 2009 4:50 AM | Permalink | Reply to this

### Re: Division Algebras and Supersymmetry

Theo wrote approximately:

… and then recognize $Spin(V)$ as living inside the even part of $Cliff(V)$. Is there a more explicit basis-dependent way to describe this? Maybe what I’m asking is: how does $Spin(V)$ live inside $Cliff(V)$?

This probably isn’t the answer you want to hear, but in all signatures and dimensions, the group $Spin(V)$ is precisely the group generated by products of pairs of unit vectors inside $Cliff(V)$. I don’t know a general way to make this more explicit using a basis of $V$. However, we have

$Spin(2,1) \cong SL(2, \mathbb{R})$ $Spin(3,1) \cong SL(2, \mathbb{C})$ $Spin(5,1) \cong SL(2, \mathbb{H})$ $Spin(9,1) \cong SL(2, \mathbb{O})$

The groups at right get a bit tricky to define in the noncommutative ($\mathbb{H}$) and nonassociative ($\mathbb{O}$) cases. However, it can be done. You can see one way here.

Posted by: John Baez on September 2, 2009 7:19 AM | Permalink | Reply to this

### Re: Division Algebras and Supersymmetry

According to Chapter 20 of Fulton and Harris’s “Representation Theory”, Pin(V) is the set of x in Cliff(V) such that x x^* = 1 and x V x^* \subseteq V. Spin(V), of course, is the component of Pin(V) which lies in the even part of the Clifford algebra. Here * is the algebra anti-automorphism of the Clifford algebra, which acts as negation on V.

I’m not excited about the prospect of turning the second condition into coordinates, but it clearly could be done.

Posted by: David Speyer on September 2, 2009 6:37 PM | Permalink | Reply to this

### Re: Division Algebras and Supersymmetry

I bet it’s not coordinates that Theo is secretly yearning for, so much as a quick way to tell which elements of $Cliff(V)$ are in $Spin(V)$ and which ones aren’t. If so, what you just said should do the trick! That’s very nice, thanks — I hadn’t known that condition.

By the way, if you guys choose the text filter “iTeX to MathML with parbreaks” when posting comments, you can use TeX. Or at least I can fix up your comments so they use TeX. I can’t easily do this otherwise. As it stands, I’d have to delete your comment and repost it with a different text filter while pretending to be you.

If I were Jacques Distler I would have chosen a text filter that does TeX as the default. But I’m not.

Posted by: John Baez on September 2, 2009 7:32 PM | Permalink | Reply to this

### Spin

It was a little tricky for me to see what was going on in the proof at the top of page 7 as well. Maybe it would be good to include a more explicit description of what $Spin(n+1,1)$ elements are, represented as $Cliff(V)$ elements. The definition provided by David Speyer looks good, and you could even-skip the intermediary use of $Pin$. If $R$ is taken to be an element of $Spin(n+1,1)$ represented as an element of $Cliff(V)$, I think the equation in the proof should just be

$(R \Gamma(A) R^*) (R \Psi) = R (\Gamma(A) \Psi)$

It might also be useful at some point to address how $Spin(n+1,1)$ acts on $V$ in $SL(2,\mathbb{K})$ as well as on $S_+$ and $S_-$. Talking a little more about how $Spin(n+1,1)$ elements come from exponentiating Clifford bivectors may or may not hurt as well.

Posted by: Garrett on September 2, 2009 10:49 PM | Permalink | Reply to this

### Re: Division Algebras and Supersymmetry

Nice read. I didn’t catch any actual errors, but here are things that confused me.

(1) On page 5, in the displayed equation about (2/3) of the way down, you write $L_{AB}$. That actually doesn’t make sense, right? The product of two Hermitian matrices probably is not Hermitian. (Although I agree that you are making an important point here, and maybe it would confuse matters to bring up this issue.)

(2) On page 6, you say that $S_+$ and $S_-$ are modules for the even part of the Clifford algebra. Why not just say that $S_+ \oplus S_-$ is a module for the Clifford algebra?

Posted by: David Speyer on September 2, 2009 10:15 PM | Permalink | Reply to this

### Re: Division Algebras and Supersymmetry

David wrote:

On page 5, in the displayed equation about (2/3) of the way down, you write $L_{A B}$. That actually doesn’t make sense, right? The product of two Hermitian matrices probably is not Hermitian.

You’re right, thanks! — but it probably is a matrix, and that’s all John Huerta really meant when he wrote that $L_{A B} \ne L_A L_B$. I’ll make sure we say it in a way that really makes sense.

On page 6, you say that $S_+$ and $S_-$ are modules for the even part of the Clifford algebra. Why not just say that $S_+ \oplus S_-$ is a module for the Clifford algebra?

Because the Spin group sits inside the even part of the Clifford algebra, so knowing that $S_+$ and $S_-$ are modules for this even part, we know they’re representations of the Spin group. They are not modules for the whole Clifford algebra.

On the other hand, $S_+ \oplus S_-$ is a module for the whole Clifford algebra, so it becomes a representation of the Pin group, which twice as big as the Spin group. The Pin group is the double cover of the group that includes Lorentz transformations and also reflections. If you reflect a guy in $S_+$ you get a guy in $S_-$, and vice versa. That’s part of why they’re called ‘left-handed’ and ‘right-handed spinors’… though I never remember which is which.

I think we actually did explain this, though in a quick ‘review for those who already know, tough for those who don’t’ style:

The action of a vector swaps $S_+$ and $S_-$, so acting by vectors twice sends $S_+$ to itself and $S_-$ to itself. This means that while $S_+$ and $S_-$ are not modules for the Clifford algebra $\Cliff(V)$, they are both modules for the even part of the Clifford algebra, generated by products of pairs of vectors. The group $\Spin(n+1,1)$ lives in this even part: as is well-known, it is generated by products of pairs of unit vectors in $V$: that is, vectors $A$ with $g(A,A) = \pm 1$. As a result, $S_+$ and $S_-$ are both representations of $\Spin(n+1, 1)$.

In terms of physics, the reason we care about $S_+$ and $S_-$ individually instead of just as part of $S_+ \oplus S_-$ is that the super-Yang–Mills theory we consider only involves guys in $S_+$!

In physics jargon, guys in $S_+$ and $S_-$ are called ‘Weyl spinors’ (if they’re complex vector spaces) or ‘Majorana–Weyl spinors’ (if they’re just real). Guys in $S_+ \oplus S_-$ are called ‘Dirac spinors’ (if it’s a complex vector space) or ‘Majorana spinors’ (if it’s just real).

I have just told you something most mathematicians will never know.

Posted by: John Baez on September 2, 2009 10:52 PM | Permalink | Reply to this

### A minor suggestion

At the bottom of page 5, it would be nice to write

$\frac{1}{2} \big( \Gamma(A) \Gamma(B) + \Gamma(B) \Gamma(A) \big) = g(A,B) 1$

which may be more recognizable as the defining Clifford algebra relation.

Posted by: Garrett on September 2, 2009 10:53 PM | Permalink | Reply to this

### Re: A minor suggestion

Is that really more recognizable? I see this just about as often:

$\Gamma(A)^2 = g(A,A) 1$

It’s equivalent.

You’ve got to admit it’s cuter.

And it says what’s really going on when we create a Clifford algebra: we’re making up an associative algebra that includes vectors, where when you square a vector, you get a number that’s the square of its length!

All that $\frac{1}{2}$ business and $\Gamma(A) \Gamma(B) + \Gamma(B) \Gamma(A)$ business just cloaks the basic idea. People who prefer that formulation deserve to be punished.

Posted by: John Baez on September 2, 2009 11:03 PM | Permalink | Reply to this

### Re: Division Algebras and Supersymmetry

A general word of appreciation: Thanks, this paper describes supersymmetry in a very coherent way.

And a question: This work depends on a Minkowski spacetime background, but there’s a lot of machinery in place that generalizes to curved spacetime. How difficult would it be to generalize this work to supersymmetric Yang-Mills in curved spacetime?

Posted by: Garrett on September 2, 2009 10:56 PM | Permalink | Reply to this

### Re: Division Algebras and Supersymmetry

To generalize to curved backgrounds all you need is a nonzero covariantly constant spinor field to take the place of the constant spinor field we’re calling ε.

Look for examples and very soon you’ll be talking about Calabi–Yau manifolds.

Posted by: John Baez on September 2, 2009 11:10 PM | Permalink | Reply to this

### Minor typo

on p.3 “Im(x) = …” should be “Im(a) = …”

Posted by: RodMcGuire on September 6, 2009 9:13 PM | Permalink | Reply to this

### Re: Minor typo

Thanks!

Here’s a further list of errata from John Huerta, which I’m about to fix — just so people who already nabbed the paper know about these problems.

I have scoured our paper for mistakes. Most of them are just typos, but I have a rather annoying discussion of the signs in the Hodge star at the end.

First, the typos:

At the end of the second paragraph, need a period.

In the second paragraph on the second page, need spaces in the list of dimensions.

Section 1, last paragraph: $A \cdot \psi$ should be $A \psi$.

Theorem 10 works for $S_-$ also, with the same proof.

Ditto Theorem 11.

Section 6, 1st paragraph: “merely function” should be “functions”

Middle of page 12: Need period on $\cdot : S_+ \otimes S_+ \to V^*.$

Ditto the next displayed equation.

We stopped saying we would assume $\psi in S_+$ for convenience.

In proof of Prop 12: “pair of Hodge star” to “pair of Hodge stars”

In proof of Prop 13: $\star d\star$ should be $\star d_A \star$ in the product rule.

Reference to Gauge Fields in proof of Prop 12: It’s actually exercise 68. Whoops.

Now, on to this Hodge star business:

Is this formula in Gauge Fields quite right? The formula you have is:

$\star^2 = (-1)^{p(n-p) + s}$

Here, $s$ seems to be the number of +’s, rather than the number of -’s in the metric. I think the formula should read:

$\star^2 = (-1)^{p(n-p) + n - s}$

(Or, even better, $s$ should be the number of minus signs in the metric instead of the number of plus signs. Sigh — yet another error in the 2nd printing of Gauge Fields, Knots and Gravity.)

$p(n-p)$ signs for straightening out indices, $n - s$ minus signs from the $n - s$ minus signs in the metric. For example, consider $\star^2$ on 0 forms:

$\star^2 1 = \star vol = (-1)^{n-s}$

since there are no permutations necessary, and we eat $n - s$ vectors with norm $-1$ in the second step.

Anyway, if this is true, and I’m quite sure it is, the formula in Prop 12 should actually read:

$\delta \langle F, F\rangle = (-1)^n 2 \langle \psi, (d_A \star F) \epsilon \rangle + divergence$

where $n$ is the dimension of spacetime, or equivalently the dimension of $\mathbb{K}$, since they are of the same parity.

Change the $\star^2$ line in the proof to read “$\star^2 = (-1)^n$”. And add a factor of $(-1)^n$ to subsequent equations, or remark “up to a sign”.

It still has to cancel because of Snygg’s formula. There just needs to be a $(-1)^n$ factor in front of $\star d \star F$ in this formula:

$\partial (F \epsilon) = (d F) \epsilon + (-1)^n (\star d \star F) \epsilon$

at least for 2-forms and under the signatures under consideration. I’ve been trying to get the sign down for all $p$-forms, dimensions, and signatures, and I have an expression, specifically

$\partial (F \epsilon) = (d F) \epsilon + (-1)^{n+q+1} (\star d \star F) \epsilon$

that I arrived at by churning through Snygg’s calculation, all of which looks right. Yet I’m slightly mistrustful of this formula, because the sign is independent of $p$. These signs are really annoying!

Anyhow, back to physics.

John

(John is in Corfu now, taking a massive dose of physics classes from 9 am to 8:30 pm each day, which explains his final remark here.)

Posted by: John Baez on September 7, 2009 12:11 AM | Permalink | Reply to this

### Re: Minor typo

Sigh — yet another error in the 2nd printing of Gauge Fields, Knots and Gravity.)

In Exercise 68, the signature should be (n-s,s) instead of (s,n-2). In other words, there should be s minus signs in the metric, not s plus signs.

But surely the original was

(s,n-s)

??? Otherwise it was an even bigger error!

Posted by: Toby Bartels on September 7, 2009 4:55 AM | Permalink | Reply to this

### Re: Minor typo

Thanks, now I’ve fixed that typo. Even my errata have errata!

Posted by: John Baez on September 7, 2009 6:34 AM | Permalink | Reply to this
Read the post Division Algebras and Supersymmetry II
Weblog: The n-Category Café
Excerpt: The real numbers, complex numbers, quaternions and octonions give Lie 2-superalgebras that describe the parallel transport of superstrings, and Lie 3-superalgebras that describe the parallel transport of 2-branes!
Tracked: March 14, 2010 7:44 PM

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