The n-Category Café

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.

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.

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!!!

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.

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.

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.

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.

Thanks, glad you liked it!

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.

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.)