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October 28, 2008

Lie Theory Through Examples 4

Posted by John Baez

This week in our seminar we’ll do some examples illustrating how a representation of a simply-connected complex simple Lie group GG gives rise to a function d:L * d : L^* \to \mathbb{N} where L *L^* is the ‘weight lattice’ of GG. Wonderfully, this function completely determines the representation (up to equivalence).

In physics, the most famous example is the meson octet, corresponding to the obvious representation of SU(3)SU(3) on sl(3,)sl(3,\mathbb{C}). It looks like this…

Here we see 8 mesons classified according to their charge qq and strangeness ss — which are eigenvalues of two elements of su(3)su(3) generating a maximal torus in SU(3)SL(3,)SU(3) \subset SL(3,\mathbb{C}).

It takes a bit of practice to turn this chart into a function from the hexagonal A 3A_3 lattice to the natural numbers! Each particle counts for one. So, the 6 corners of the hexagon have d=1d = 1, while the point in the middle, drawn as two points in this chart, has d=2d = 2, since there are two mesons here with vanishing charge and strangeness: the neutral pion π 0\pi^0 and the eta η\eta. The total dimension of this representation is thus 6+2=86 + 2 = 8 — the dimension of sl(3,)sl(3,\mathbb{C}).

Here are the lecture notes:

  • Lecture 4 (Oct. 28) - Classifying representations using weights. The example of A1, which corresponds to the group SU(2). The relation between SU(2) and SO(3), and their representations. The example of A2, which corresponds to the group SU(3).

By the way, I rewrote last week’s notes.

Posted at October 28, 2008 2:54 AM UTC

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21 Comments & 0 Trackbacks

Re: Lie Theory Through Examples 4

Nice lecture, but there’s a typo in the U(1) matrix on page 1.

Posted by: Kea on October 28, 2008 6:06 AM | Permalink | Reply to this

Re: Lie Theory Through Examples 4

Wow, your reply showed up before I even checked to see if my post had appeared!

I’ll fix the typo. Thanks. By the way, the actual lecture notes have a lot more stuff than the preliminary version you just saw!

Posted by: John Baez on October 28, 2008 6:26 AM | Permalink | Reply to this

Re: Lie Theory Through Examples 4

I’ve only looked at the teaser
but no mention of the 8-fold way?

Posted by: jim stasheff on October 28, 2008 3:09 PM | Permalink | Reply to this

Re: Lie Theory Through Examples 4

Anyone who clicks my link for an explanation of ‘meson octet’ gets an introduction to the Eightfold Way.

‘Octet’ and ‘eightfold’ are both secretly references to the dimension of SU(3)SU(3). In the original Yang–Mills idea, mesons were treated as gauge bosons, and thus live in the Lie algebra of the gauge group. So, when the gauge group is the 8-dimensional group SU(3)SU(3), you get the ‘meson octet’ shown above.

Posted by: John Baez on October 28, 2008 5:09 PM | Permalink | Reply to this

Re: Lie Theory Through Examples 4

Do they have a complete description of the SU(3) weight multiplicity diagrams? It’s so easy and beautiful (to state) that I think they really should.

I think in terms of these weight multiplicity diagrams all the time. I find it weird that they are not more standardly taught.

The statement: the diagram of the rep with high weight (a,b) is a hexagon with edge lengths (a,b,a,b,a,b). The multiplicity is 1 around the outer hexagon, 2 around the next hexagon inside, 3 etc., until the hexagon degenerates to a triangle, at which point the multiplicity is constant, min(a,b)+1.

This makes it sound like there are two cases, and there are; the triangle can be a delta or a nabla.

Posted by: Allen Knutson on October 28, 2008 7:18 AM | Permalink | Reply to this

Re: Lie Theory Through Examples 4

Allen wrote:

Do they have a complete description of the SU(3) weight multiplicity diagrams?

My notes? No, they don’t yet… but someday they should.

I think in terms of these weight multiplicity diagrams all the time. I find it weird that they are not more standardly taught.

Good point — yet another thing my book should do.

Is there a standard name for the function assigning each weight its multiplicity? I suppose I could call it the ‘weight multiplicity diagram’ or ‘weight multiplicity function’. Right now I’m just calling it d:L *d : L^* \to \mathbb{N}, which gets tiresome at times.

This makes it sound like there are two cases, and there are; the triangle can be a Δ or a ∇.

I’ll talk about these today. (In the 8-fold way, they’re the quarks and antiquarks!)

Posted by: John Baez on October 28, 2008 5:17 PM | Permalink | Reply to this

Re: Lie Theory Through Examples 4

Is there a standard name for the function assigning each weight its multiplicity?

I’m not sure, but you could look in Guillemin, Lerman, and Sternberg’s “Symplectic fibrations and multiplicity diagrams”.

I think “T-dimension” has a nice ring to it, to give the idea that it’s the equivariant enrichment of the notion of dimension, and even to mention what group is relevant. Not that I’ve ever heard anyone use that. I often use (V λ) μ(V_\lambda)^\mu to denote the μ\mu-weight space in the irrep latexV λlatex V_\lambda, or maybe (V λ) T=μ(V_\lambda)^{T=\mu} to convey the idea that T is acting with weight μ\mu. (Note that this evokes the usual notation V TV^T for the T-fixed points.) Then you can stick dim in front of that.

Posted by: Allen Knutson on October 30, 2008 1:20 PM | Permalink | Reply to this

Re: Lie Theory Through Examples 4

Note that from the MUB point of view, one should think of 8 as (3-1)(3+1) = 2 x 4, where opposite sides of the hexagon are paired (eg. x,-x or y,-y in Carl’s notation). The 4 is the number of MUBs in 3d, and the 2 is the number of cycle types in the permutation group S3.

Posted by: Kea on October 29, 2008 9:24 AM | Permalink | Reply to this

Re: Lie Theory Through Examples 4

What the heck does ‘MUB’ mean? Massively Ugly Baloney? Mind Urown Business?

(I try to avoid acronyms: they serve to save a few keystrokes, but at the expense of limiting the discourse to people who are already familiar with what’s going on.)

Posted by: John Baez on October 30, 2008 6:15 AM | Permalink | Reply to this

Re: Lie Theory Through Examples 4

MUB = mutually unbiased bases

Posted by: Kea on October 30, 2008 6:21 AM | Permalink | Reply to this

Re: Lie Theory Through Examples 4

That sounds interesting, I’m a fan of mutually unbiased bases! So what exactly is the connection between MUBs, the hexagon and S 3S_3?

Posted by: Jamie Vicary on October 30, 2008 5:56 PM | Permalink | Reply to this

Re: Lie Theory Through Examples 4

Erm, sorry to give the impression that there’s an echo in this café, but what the heck does ‘mutually unbiased bases’ mean?

Posted by: Simon Willerton on October 30, 2008 9:48 PM | Permalink | Reply to this

Re: Lie Theory Through Examples 4

It means ‘MUBs’.

Posted by: John Baez on October 31, 2008 6:27 AM | Permalink | Reply to this

Re: Lie Theory Through Examples 4

LOL. Anyway, just google it. There are plenty of accessible papers on the subject.

Posted by: Kea on October 31, 2008 7:44 PM | Permalink | Reply to this

Re: Lie Theory Through Examples 4

Hi Jamie. S3 can be written as 6 3x3 matrices with zeroes and ones, three of which are 1-circulant and three of which are 2-circulant. The 4 MUBs in 3d can be described as eigenvectors of 4 braided S3 matrices (ie. with phases added) on which the discrete Fourier transform for 3 states acts as a 3x3 conjugation.

This is all completely analogous to the 3 Pauli operators (2x2 matrices) in 2d, which as you know give the 3 MUBs. They are necessarily all 1-circulant (a single ‘cycle type’) because there is no room for longer cycles in S2. There are many ways to look at the hexagon in this framework.

Posted by: Kea on October 30, 2008 8:53 PM | Permalink | Reply to this

Re: Lie Theory Through Examples 4

maybe a simple question about “strangeness” :
looking at the “meson octet” we see the strangeness has values 1,0,-1
but for the “baryon octet” the values are
0,-1,-2
now there’s only one 8 dimensional irrep of su(3), wouldn’t the two commuting elements have exactly the same eigenvalues in both cases? (note there’s no problem with electric charge)

Posted by: rntsai on November 4, 2008 9:24 PM | Permalink | Reply to this

Re: Lie Theory Through Examples 4

I picked up Lie5.pdf even though you didn’t blog it (interesting for me, however).

All in all, nice adjunct to Stillwell’s Naive Lie Theory (for beginners like me).

Thank you for your efforts.

Posted by: Maya Incaand on April 5, 2009 3:11 PM | Permalink | Reply to this

Re: Lie Theory Through Examples 4

This week in our seminar we’ll do some examples illustrating how a representation of a simply-connected complex simple Lie group G gives rise to a function

(1)d:L * d: L^* \rightarrow \mathbb{N}

where L *L^* is the ‘weight lattice’ of G. Wonderfully, this function completely determines the representation (up to equivalence).

That was great, I always knew I would come back to these lectures and recently I reread them. Did it end after lecture 5? I feel cheated, I was hoping you’d go on! What is the resolution of the mystery? What kind of weighting functions

(2)d:L * d: L^* \rightarrow \mathbb{N}

correspond to representations of GG? I once heard a talk by Ocneanu where he seemed to be talking in these terms. Maybe this is standard stuff but I am struggling to find a presentation like yours in the books. What is the condition on dd? Is there some kind of elegant “differential operator”-like thing \nabla on the lattice LL such that:

A function d:L *d : L^* \rightarrow \mathbb{N} corresponds to a representation of GG if and only if d=0\nabla d = 0??

That would be great.

P.S. I didn’t understand the paragraph about the hexagons. How can a hexagon have side length “a,b,a,b,a,b”… doesn’t a hexagon have equal sides? And you said: “draw a big hexagon”, like there was only one, and then you spoke about different hexagons…?

Posted by: Bruce Bartlett on October 19, 2009 12:42 PM | Permalink | Reply to this

Re: Lie Theory Through Examples 4

Bruce wrote:

Did it end after lecture 5?

The course kept on going, but I got too busy to write notes in LaTeX. Maybe I can get John Huerta to cough up his handwritten notes. He may even have already given them to me!

2008 was the year when I was trying to do way too many things at once…

What kind of weighting functions

d:L * d : L^* \to \mathbb{N}

correspond to representations of GG?

This is the moment when Allen Knutson should materialize and explain it, or at least point you to an online reference. For now I will just tell you that it’s a very fun long story that leads through things like the Weyl character formula, the Kostant multiplicity formula, and a bunch of symplectic geometry.

That sounds a bit intimidating. But I wish you could just browse through Fulton and Harris’ book on group representation theory and look at some pictures of these weighting function! That would be the right way to start.

The function dd is not the kernel of some differential operator: its graph is piecewise linear and forms the top of a beautiful polytope.

How can a hexagon have side lengths “a,b,a,b,a,ba,b,a,b,a,b“… doesn’t a hexagon have equal sides?

A hexagon is any shape with 6 sides. A regular hexagon has all 6 sides of equal length, and all angles equal. But the support of the weighting function is not usually a regular hexagon.

In general, for any simple group GG, here’s what the support of dd looks like for an irreducible representation. Take a point in the weight lattice and act on it by the Weyl group. Take the convex hull of the resulting set of points. This will be a convex polytope. The points in this polytope that lie in L *L^* are the places where dd is nonzero.

What does this amount to for SU(3)SU(3)? The Weyl group is the 6-element dihedral group: the symmetry group of an equilateral triangle. The weight lattice is a lattice having this group as symmetries. So, if you take a point in this lattice and act on it by this group, typically you get 6 points. And these are the corners of a hexagon with side lengths a,b,a,b,a,ba,b,a,b,a,b as you march around, with all interior angles equal to 120 degrees.

Try it!

(Sometimes a=0a = 0 or b=0b = 0 and then your hexagon degenerates to an equilateral triangle. And if you’re really unlucky, a=b=0a = b = 0, so your hexagon degenerates to a single point.)

Posted by: John Baez on October 19, 2009 9:29 PM | Permalink | Reply to this

Through Examples 4

Ok, thanks. Yes, those notes will be extremely handy, if ever you or John Huerta are able to scan them in!

I always knew I had to take a decent look at Fulton and Harris someday. I can’t keep running forever. I have looked at the pictures now, and indeed they look very informative.

Posted by: BB Bartlett on October 19, 2009 10:34 PM | Permalink | Reply to this

Lie Theory Through Examples notes

I posted the notes on my webpage, under Notes. JB may eventually choose to post them on his own site.

Posted by: John Huerta on October 27, 2009 11:57 PM | Permalink | Reply to this

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