The n-Category Café

Ali H. Chamseddine and Alain Connes, A dress for SM the beggar.

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Ali H. Chamseddine and Alain Connes, Why the Standard Model

Thanks for these links. I wasn’t aware of these papers.

Now I did look at them. Unfortunately, I feel less enlightened by them then I was hoping to. But that’s my fault, not theirs.

In need to think about all this a little more.

One thing that I imagine would help me is if I could see clearer on how Connes’ setup would “decompactify” to an ordinary geometry.

Let’s see, just some simple observations.

Assuming the algebra to be the tensor product $$A=C^{\mathrm{\infty }}(X)\times A_F$$ is of course simply assuming (in the case of ordinary geometry) a cartesian product space $$X\times Y.$$

Next, assuming the representation to be $$H=L^2(X,S)\otimes H_f$$ is assuming (correct me if I am wrong), in terms of ordinary geometry, that we are looking at a vector bundle over $X\times Y$ $$V\to X\times Y$$ which is the tensor product of the spinor bundle $$S\to X$$ over $X$ ($X$ is assumed to be spin and Riemannian) with some other vector bundle $$E\to Y$$ over $Y$: $$V={\pi }_X^{*}S\otimes {\pi }_Y^{*}E.$$

Sorry for wasting bandwidth with these trivialities.

But now let’s see. How should we think of $E$, conceptually?

I have the following general suspicion, which I don’t know of if it’s any good or not:

maybe a good way to think of the noncommutative algebras appearing in spectral triples is as category algebras. Because, if so, we’d reobtain a nice “geometrical” picture again from these algebras, which would nicely connect with our general idea, expounded on in a couple of discussions here, that it should be helpful to think of space(time) as a category.

So, for instance, the matrix algebra $$M_3(ℂ)$$ is just the category algebra of the groupoid $$\mathrm{Codisc}(\{a,b,c\})$$ which has three objects and precisely one morphism from every object to every other.

So, I feel like thinking of the algebra $M_3(ℂ)$ as the algebra of functions on the triangle $$\begin{array}{ccc}& & b\\ & \swarrow \nearrow & & \searrow \nwarrow \\ a& & \overleftarrow{\to }& & c\end{array}.$$

Maybe that’s not justified here in this context. But maybe it is.

From this point of view, we may think of a module of $M_3(ℂ)$ as a vector bundle with connection. In fact, it looks like a flat vector bundle with connection over a latticized disk $D^2$, namely on the standard 2-simplex.

(In general, I think, reps of $M_n(ℂ)$ would, in this logic, correspond to flat vector bundles with connection on the standard $(n-1)$-simplex.)

Of course $ℂ$ itself is the category algebra of the trivial one object one morphism category.

Is there a category such that its category algebra is the algebra of quaternions, $ℍ$? Hm, there cannot be, right?

Oh, I have to run.

Urs wrote:

In general, I think, reps of $M_n(ℂ)$ would, in this logic, correspond to flat vector bundles with connection on the standard $(n-1)$-simplex.

Yes, that sounds right. These three algebras are isomorphic:

• The algebra of $n\times n$ complex matrices, $M_n(ℂ)$.
• The category algebra of the codiscrete category on $n$ objects.
• The category algebra of the fundamental groupoid of the $(n-1)$-simplex.

So, these three sentences mean the same thing:

• $M_n(ℂ)$ is Morita equivalent to $ℂ$.
• The codiscrete category on $n$ objects is equivalent to the terminal category.
• The $n$-simplex is contractible.

To get weirder algebras, like the quaternions, into the game I think we need to do something sneaky… like maybe think about Auslander–Reiten quivers. The category algebra trick you’re discussing gives a way to get algebras from quivers. The ‘Auslander–Reiten quiver’ idea is a way to get quivers from algebras. It’s all part of some big story…

If I understood more about this story, I’d definitely explain it to you.

Oh darn. I was going to give you a smart-aleck answer, by pointing out that the Poincaré group doesn’t commute with itself… but you just ruled that out.

Anyway, I don’t know an example of the weird thing you’re noting my $E_6$ idea would seem to require. “Seem to”, because it’s really not clear yet what physical ideas the mathematics is suggesting that we explore.

But, note that this $E_6$ idea also seems to require something even weirder: a symmetry that makes us change our viewpoint about what counts as ‘multiplication by $i$’!

$E_6$ acts as isometries of the symmetric space $(ℂ\otimes 𝕆){\mathrm{P}}^2$. Each point of this space gives a different way of splitting $e_6$ as $\mathrm{so}(10)\oplus S_{10}^{+}\oplus u(1)$. The generator of $u(1)$ acts as multiplication by $i$ on the spinors in $S_{10}^{+}\cong {ℂ}^{16}$. These a

So, conjugating a guy in $e_6$ by elements of the group $E_6$, we get different ways of deciding what part of this guy lives in the Lie algebra of the gauge group $\mathrm{so}(10)$, what part lives in the spinor representation… and what part counts as ‘$i$’.

That’s pretty weird.

But, note that the spacetime symmetries are not part of this story at all, yet! The fact that $\mathrm{so}(10)$ guys are eventually supposed to be spin-1 particles while the $S_{10}^{+}$ guys are supposed to be spin-1/2 particles is completely unaccounted for, so far.

So, the whole idea is only half-baked, so far — I’m just taking a strange mathematical fact and noticing that it has some relation to particle physics, and hoping maybe it could lead somewhere… but I really don’t know where, yet.

1. isotropic is used in e.g. material science to mean the same in all directions

2. for all D ≥ 0 seems to imply it’s irrelevant

jim

Jim wrote:

isotropic is used in e.g. material science to mean the same in all directions.

That’s precisely what it means here. A Riemannian symmetric space $M$ is isotropic if when you sit at any point $p$, the view in every direction is the same. If I understand the above quote by Joseph Wolf correctly, he formalizes this in two ways and shows they’re equivalent:

• $M$ is isotropic if for point $p$, the group of isometries of $M$ fixing $p$ acts transitively on the unit sphere in the tangent space at $p$. (So, $M$ has enough symmetry to make every direction look just like every other.)
• $M$ is isotropic if for all $D\ge 0$, the group of isometries of $M$ acts transitively on the set of all pairs of points that have distance $D$ from each other.

Jim writes:

for all $D\ge 0$ seems to imply it’s irrelevant

No, the distance isn’t irrelevant. Item 2. says that for any fixed distance $D\ge 0$, there’s an isometry mapping any pair of points with distance $D$ to any other pair of points with distance $D$. But, there’s never an isometry mapping a pair of points that have distance 1 to a pair of points that have distance 2. Isometries preserve distance!