This Week's Finds in Mathematical Physics (Week 253)

[John Baez]

Also available as http://math.ucr.edu/home/baez/week253.html

June 27, 2007
This Week's Finds in Mathematical Physics (Week 253)
John Baez

Yay! Classes are over! Soon I'm going to Paris for three weeks, to
talk to Paul-Andre Mellies about logic, games and category theory.
But right now I'm in a vacation mood. So, I want to take a break from
the Tale of Groupoidification, and mention some thoughts prompted by
the work of Garrett Lisi:

1) Garrett Lisi, Deferential Geometry, http://deferentialgeometry.org/

Garrett is a cool dude who likes to ponder physics while living a
low-budget, high-fun lifestyle: hanging out in Hawaii, surfing, and
stuff like that. He recently won a Foundational Questions Institute
award to think about ways to unify particle physics and gravity. This
is an institute devoted precisely to risky endeavors like this.

Lately he's been visiting California. So, before giving a talk at
Loops '07 - a loop quantum gravity conference taking place in Mexico
this week - he stopped by Riverside to explain what he's been up to.

Briefly, he's been trying to explain the 3 generations of elementary
particles using some math called "triality", which is related to the
octonions and the exceptional Lie groups. In fact, he's trying to use
the exceptional Lie group E8 to describe all the particles in the
Standard Model, together with gravity.

I'd like to know if these ideas hold water. So, I should try to
explain them! But as usual, in this Week's Finds I'll wind up
explaining not what Garrett actually did, but what it made me think
about.

For a long time, people have been seeking connections between the
messy pack of particles that populate the Standard Model and
structures that seem beautiful and "inevitable".

A fascinating step in this direction was the SU(5) grand unified
theory proposed in 1975 by Georgi and Glashow. So, I'll start by
summarizing that... and then explain how exceptional Lie groups might
get involved in this game.

What people usually call the gauge group of the Standard Model:

SU(3) x SU(2) x U(1)

actually has a bit of flab in it: there's a normal subgroup that acts
trivially on all known particles. This subgroup is isomorphic to Z/6.
If we mod out by this, we get the "true" gauge group of the Standard
Model:

G = (SU(3) x SU(2) x U(1))/(Z/6)

And, this turns out to have a neat description. It's isomorphic to the
subgroup of SU(5) consisting of matrices like this:

(g 0)
(0 h)

where g is a 3x3 block and h is a 2x2 block. For obvious reasons, I call
this group

S(U(3) x U(2))

If you want some intuition for this, think of the 3x3 block as related
to the strong force, and the 2x2 block as related to the electroweak
force. A 3x3 matrix can mix up the 3 "colors" that quarks come in -
red, green, and blue - and that's what the strong force is all about.
Similarly, a 2x2 matrix can mix up the 2 "isospins" that quarks and
leptons come in - up and down - and that's part of what the electroweak
force is about.

If this isn't enough to make you happy, go back to "week119", where
I reviewed the Standard Model and its relation to the SU(5) grand
unified theory. If even that isn't enough to make you happy, try this:

2) John Baez, Elementary particles,
http://math.ucr.edu/home/baez/qg-spring2003/elementary/

Okay - I'll assume that one way or another, you're happy with the
idea of S(U(3) x U(2)) as the true gauge group of the Standard Model!
Maybe you understand it, maybe you're just willing to nod your head
and accept it.

Now, the fermions of the Standard Model form a very nice representation
of this group. SU(5) has an obvious representation on C^5, via matrix
multiplication. So, it gets a representation on the exterior algebra
Lambda(C^5). If we restrict this from SU(5) to S(U(3) x U(2)), we get
precisely the representation of the true gauge group of the Standard Model
on one generation of fermions and their antiparticles!

This really seems like a miracle when you first see it. All sorts of
weird numbers need to work out exactly right for this trick to succeed.
For example, it's crucial that quarks have charges 2/3 and -1/3, while
leptons have charges 0 and -1. One gets the feeling, pondering this stuff,
that there really is some truth to the SU(5) grand unified theory.

To give you just a little taste of what's going on, let me show you
how the exterior algebra Lambda(C^5) corresponds to one generation of
fermions and their antiparticles. For simplicity I'll use the first
generation, since the other two work just the same:

Lambda^0(C^5) = <left-handed antineutrino>

Lambda^1(C^5) = <right-handed down quark> +
<right-handed positron, right-handed antineutrino>

Lambda^2(C^5) = <left-handed up antiquark> +
<left-handed up quark, left-handed down quark> +
<left-handed positron>

Lambda^3(C^5) = <right-handed electron> +
<right-handed up antiquark, right-handed down antiquark> +
<right-handed up quark>

Lambda^4(C^5) = <left-handed up antiquark> +
<left-handed electron, left-handed neutrino>

Lambda^5(C^5) = <right-handed neutrino>

All the quarks and antiquarks come in 3 colors, which I haven't bothered
to list here. Each space Lambda^p(C^5) is an irreducible representation of
SU(5), but most of these break up into several different irreducible
representations of S(U(3) x U(2)), which are listed as separate rows in
the chart above.

If you're curious how this "breaking up" works, let me explain - it's
sort of pretty. We just use the splitting

C^5 = C^3 + C^2

to chop the spaces Lambda^p(C^5) into pieces.

To see how this works, remember that Lambda^p(C^5) is just the vector
space analogue of the binomial coefficient "5 choose p". A basis of
C^5 consists of 5 things, and the p-element subsets give a basis for
Lambda^p(C^5).

In our application to physics, these 5 things consist of 3 "colors"
- red, green and blue - and 2 "isospins" - up and down. This gives
various possible options.

For example, suppose we want a basis of Lambda^3(C^5). Then we need to
pick 3 things out of 5. We can do this in various ways:

* We can pick 3 colors and no isospins - there's just one way to do that.

* We can pick 2 colors and 1 isospin - there are six ways to do that.

* Or, we can pick 1 color and 2 isospins - there are three ways to do that.

So, in terms of binomial coefficients, we have

(5 choose 3) = (3 choose 3)(2 choose 0) +
(3 choose 2)(2 choose 1) +
(3 choose 1)(2 choose 2)

= 1 + 6 + 3

= 10

In terms of vector spaces we have:

Lambda^3(C^5) = Lambda^3(C^3) tensor Lambda^0(C^2) +
Lambda^2(C^3) tensor Lambda^1(C^2) +
Lambda^1(C^3) tensor Lambda^2(C^2)

Taking dimensions of these vector spaces, we get 10 = 1 + 6 + 3. Finally,
in terms of the SU(5) grand unified theory, we get this:

Lambda^3(C^5) = <right-handed electron> +
<right-handed up antiquark, right-handed down antiquark> +
<right-handed up quark>

If we play this game for all the spaces Lambda^p(C^5), here's what we get:

Lambda^0(C^5) = Lambda^0(C^3) tensor Lambda^0(C^2)

Lambda^1(C^5) = Lambda^1(C^3) tensor Lambda^0(C^2) +
Lambda^0(C^3) tensor Lambda^1(C^2)

Lambda^2(C^5) = Lambda^2(C^3) tensor Lambda^0(C^2) +
Lambda^1(C^3) tensor Lambda^1(C^2) +
Lambda^0(C^3) tensor Lambda^2(C^2)

Lambda^3(C^5) = Lambda^3(C^3) tensor Lambda^0(C^2) +
Lambda^2(C^3) tensor Lambda^1(C^2) +
Lambda^1(C^3) tensor Lambda^2(C^2)

Lambda^4(C^5) = Lambda^3(C^3) tensor Lambda^1(C^2) +
Lambda^2(C^2) tensor Lambda^2(C^2)

Lambda^5(C^5) = Lambda^3(C^3) tensor Lambda^2(C^2)

If we interpret this in terms of physics, we get back our previous chart:

Lambda^0(C^5) = <left-handed antineutrino>

Lambda^1(C^5) = <right-handed down quark> +
<right-handed positron, right-handed antineutrino>

Lambda^2(C^5) = <left-handed up antiquark> +
<left-handed up quark, left-handed down quark> +
<left-handed positron>

Lambda^3(C^5) = <right-handed electron> +
<right-handed up antiquark, right-handed down antiquark> +
<right-handed up quark>

Lambda^4(C^5) = <left-handed up antiquark> +
<left-handed electron, left-handed neutrino>

Lambda^5(C^5) = <right-handed neutrino>

Now, all this is really cool - but in fact, even before inventing the
SU(5) theory, Georgi went a bit further, and unified all the left-handed
fermions above into one irreducible representation of a somewhat bigger
group: Spin(10). This is the double cover of the group SO(10), which
describes rotations in 10 dimensions.

If you look at the chart above, you'll see the left-handed fermions
live in the even grades of the exterior algebra of C^5:

Lambda^{even}(C^5) = Lambda^0(C^5) + Lambda^2(C^5) + Lambda^4(C^5)

This big space forms something called the left-handed Weyl spinor
representation of Spin(10). It's an irreducible representation.

Similarly, the right-handed fermions live in the odd grades:

Lambda^{odd}(C^5) = Lambda^1(C^5) + Lambda^3(C^5) + Lambda^5(C^5)

and this big space forms the right-handed Weyl spinor representation
of Spin(10). It's also irreducible.

I can't resist mentioning that there's also a gadget called the Hodge
star operator that maps Lambda^{even}(C^5) to Lambda^{odd}(C^5), and
vice versa. In terms of physics, this sends left-handed particles
into their right-handed antiparticles, and vice versa!

But if I get into digressions like these, it'll take forever to tackle the
main question: how does this "Weyl spinor" stuff work?

It takes advantage of some very nice general facts. First, C^n is
just another name for R^{2n} equipped with the structure of a complex
vector space. This makes SU(n) into a subgroup of SO(2n). So, it
makes the Lie algebra su(n) into a Lie subalgebra of so(2n).

The group SU(n) acts on the exterior algebra Lambda(C^n). So, its Lie
algebra su(n) also acts on this space. The really cool part is that
this action extends to all of so(2n). This is something you learn
about when you study Clifford algebras, spinors and the like. I don't
know how to explain it without writing down some formulas. So, for
now, please take my word for it!

This business doesn't give a representation of SO(2n) on Lambda(C^n),
but it gives a representation of the double cover, Spin(2n). This is
called the "Dirac spinor" representation. It breaks up into two
irreducible parts:

Lambda(C^n) = Lambda^{even}(C^n) + Lambda^{odd}(C^n)

and these are called the left- and right-handed "Weyl spinor"
representations.

Perhaps it's time for an executive summary of what I've said so
far:

The Dirac spinor representation of Spin(10) neatly encodes everything
about how one generation of fermions interacts with the gauge bosons
in the Standard Model, as long as we remember how S(U(2) x U(3)) sits
inside SO(10), which is double covered by Spin(10).

Of course, there's more to the Standard Model than this. There's also
the Higgs boson, which spontaneously breaks electroweak symmetry and
gives the fermions their masses. And, if we want to use this same
trick to break the symmetry from Spin(10) down to S(U(3) x U(2)), we'd
need to introduce *more* Higgs bosons. This is the ugly part of the
story, it seems. Since I'm on vacation, I'll avoid it for now.

Next: how might exceptional Lie groups get involved in this game?

When Cartan classified compact simple Lie groups, he found 3 infinite
families related to rotations in real, complex and quaternionic vector
spaces: the SO(n)'s, SU(n)'s and Sp(n)'s. He also found 5 exceptions,
which have the charming names G2, F4, E6, E7, and E8. These are all
related to the octonions. G2 is just the automorphism group of the
octonions. The other 4 are closely related to each other - thanks to
the "magic square" of Rosenfeld, Freudenthal and Tits.

I talked about the magic square a bit in "week106" and "week145", and
much more here:

3) John Baez, The magic square,
http://math.ucr.edu/home/baez/octonions/node16.html

Instead of repeating all that, let me just summarize. The magic
square gives vector space isomorphisms as follows:

F4 = so(R + O) + (R tensor O)^2

E6 = so(C + O) + (C tensor O)^2 + Im(C)

E7 = so(H + O) + (H tensor O)^2 + Im(H)

E8 = so(O + O) + (O tensor O)^2

Here F4, E6, E7 and E8 stand for the compact real forms of these
Lie algebras. R, C, H, and O are the usual suspects - the real numbers,
complex numbers, quaternions and octonions. For any real inner product
space V, so(V) stands for the Lie algebra of the rotation group of V.
And, for each of the isomorphisms above, we must equip the vector space
on the right side with a cleverly (but not perversely!) defined Lie
bracket to get the Lie algebra on the left side.

Here's another way to say the same thing, which may ring more bells:

F4 = so(9) + S_9

E6 = so(10) + S_{10}^+ + u(1)

E7 = so(12) + S_{12}^+ + su(2)

E8 = so(16) + S_{16}^+

Here S_9 means the unique irreducible real spinor representation of
so(9). In the other 3 cases, the little plus signs mean that there are
two choices of irreducible real spinor representation, and we take the
left-handed choice.

All this must seem like black magic of the foulest sort if you haven't
wasted months thinking about the octonions and exceptional groups! Be
grateful: I did it so you wouldn't have to.

Anyway: the case of E6 should remind you of something! After all, we've
just been talking about so(10) and its left-handed spinor representation.
These describe the gauge bosons and one generation of left-handed fermions
in the Spin(10) grand unified theory. But now we're seeing this stuff
neatly packed into the Lie algebra of E6!

More precisely, the Lie algebra of E6 can be chopped into 3 pieces
in a noncanonical way:

A) so(10),

B) the left-handed real spinor representation of so(10), which by now
we've given three different names:

S_{10}^+ = Lambda^{even}(C^5) = (C tensor O)^2

and

C) a copy of u(1).

The first part contains all the gauge bosons in the SO(10) grand unified
theory. The second contains one generation of left-handed fermions.
But what about the third?

Well, S_{10}^+ is defined to be a real representation of so(10). But,
it just so happens that the action of so(10) preserves a complex
structure on this space. It's just the obvious complex structure on
(C tensor O)^2. So, there's an action of the unit complex numbers,
U(1), on S_{10}^+ which commutes with the action of so(10).
Differentiating this, we get an action of the Lie algebra u(1):

u(1) x S_{10}^+ -> S_{10}^+

And this map gives part of the cleverly defined Lie bracket operation in

E6 = so(10) + S_{10}^+ + u(1)

All this stuff is mysterious, but suggestive. It could be mere
coincidence, or it could be the tip of an iceberg. It's more fun to
assume the latter. So, let me say some more about it...

The copy of u(1) in here:

E6 = so(10) + S_{10}^+ + u(1)

is pretty amusing from a physics viewpoint. It's if besides the gauge
bosons in so(10), there were one extra gauge boson whose sole role is
to describe the fact that the fermions form a *complex* representation
of so(10). This is funny, since one of the naive ideas you sometimes
hear is that you can take the obvious U(1) symmetry every complex Hilbert
space has and "gauge" it to get electromagnetism.

That's not really the right way to understand electromagnetism! There
are lots of different irreducible representations of U(1), corresponding
to different charges, and in physics we should think about *all* of
these, not just the obvious one that we automatically get from any
complex Hilbert space. If we only used the obvious one, all particles
would have charge 1.

But in the Spin(10) grand unified theory, the electromagnetic u(1)
Lie algebra is sitting inside so(10); it's not the u(1) you see above.
The u(1) you see above is the "obvious" one that the spinor
representation S_{10}^+ gets merely from being a complex Hilbert space.

The splitting

E6 = so(10) + S_{10}^+ + u(1)

also hints at a weird unification of bosons and fermions, something
different from supersymmetry. We're seeing E6 as a Z/2-graded Lie
algebra with so(10) + u(1) as its "bosonic" part and S_{10}^+ as its
"fermionic" part. But, this is not a Lie superalgebra, just an ordinary
Lie algebra with a Z/2 grading!

Furthermore, an ordinary Lie algebra with a Z/2 grading is precisely
what we need to build a "symmetric space". This is really cool, since
it explains what I meant by saying that the split of E6 into bosonic
and fermionic parts is "noncanonical". We'll get a space, and each
point in this space will give a different way of splitting E6 as

E6 = so(10) + S_{10}^+ + u(1)

It's also cool because it gives me an excuse to talk about symmetric
spaces... a topic that deserves a whole week of its own!

This gives me an excuse to say a word or two about symmetric spaces...
a topic that deserves a whole week of its own! Symmetric spaces are
the epitome of symmetry. A "homogeneous space" is a manifold with
enough symmetry that any point looks any other. A "symmetric space"
is a homogeneous space with an extra property: the view from any point
in any direction is the same as the view in the opposite direction!

Euclidean spaces and spheres are the most famous examples of symmetric
spaces. If an ant decides to set up residence on a sphere, any point
is just as good any other. And, if sits anywhere and looks in any
direction, the view is the same as the view in the opposite direction.

The symmetric space we get from the above Z/2-graded Lie algebra is
sort of similar, but more exotic: it's the complexified version of the
octonionic projective plane!

But let's start with the basics:

Suppose someone hands you a Lie algebra g with a Lie subalgebra h.
Then you can form the simply-connected Lie group G whose Lie algebra
is g. Sitting inside G, there's a connected Lie group H whose Lie
algebra is h. The space

G/H

is called a "homogeneous space". Such things are studied in Klein
geometry, and I've been talking about them a lot lately.

But now, suppose g is a Z/2-graded Lie algebra. Its even part will be
a Lie subalgebra; call this h. This gives a specially nice sort of
homogeneous space G/H, called a "symmetric space". This is better
than your average homogeneous space.

Why? First of all, for each point p in G/H there's a map from G/H to
itself called "reflection through p", which fixes the point p and acts
as -1 on the tangent space of p. When our point p comes from the identity
element of G, this reflection map corresponds to the Z/2 grading of the
Lie algebra, which fixes the even part and acts as -1 on the odd part.

This is what I meant by saying that in a symmetric space, "the view in
any direction is the same as the view in the opposite direction".

Second, these reflection maps satisfy some nice equations. Write p>q
for the the result of reflecting q through p. Then we have:

p>(p>q) = q

p>p = p

and

p>(q>r) = (p>q) > (p>r)

A set with an operation satisfying these equations is called an
"involutory quandle".

Let me summarize with a few theorems - I hope they're all true, because
I don't know a book containing all this stuff. We can define a "symmetric
space" to be an involutory quandle that's a manifold, where the operation
> is smooth and the reflection map

x |-> p>x

has derivative -1 at p. Any Z/2-graded Lie algebra gives a symmetric
space. Conversely, any symmetric space has a universal cover that's a
symmetric space coming from a Z/2-graded Lie algebra!

Using this correspondence, the Lie algebra E6 with the Z/2-grading I
described gives a symmetric space, roughly:

E6/(Spin(10) x U(1))

But, this guy is a lot better than your average symmetric space!

For starters, it's a "Riemannian symmetric space". This is a symmetric
space with a Riemannian metric that's preserved by all the operations
of reflection through points.

Compact Riemannian symmetric spaces were classified by Cartan, and you
can see the classification here, in a big chart:

4) Riemannian symmetric spaces, Wikipedia,
http://en.wikipedia.org/wiki/Riemannian_symmetric_space

As you'll see, there are 7 infinite families and 12 exceptional cases.
The symmetric space I'm talking about now, namely E6/(Spin(10) x U(1)),
is called EIII - it's the third exceptional case. And, as you can see
from the chart in this article, it's the complexified version of the
octonionic projective plane! For this reason, I sometimes call it

(C tensor O)P^2

In fact, this space is better than your average Riemannian symmetric
space. It's a Kaehler manifold, thanks to that copy of U(1), which
makes each tangent space complex. Moreover, the Kaehler structure is
preserved by all the operations of reflection through points. So,
it's a "hermitian symmetric space".

You're probably drowning under all this terminology unless you already
know this stuff. I guess it's time for another executive summary:

Each point in the complexified octonionic projective plane gives a
different way of splitting the Lie algebra of E6 into a bosonic part
and a fermionic part. The fermionic part is just what we need to
describe one generation of left-handed Standard Model fermions. The
bosonic part is just what we need for the gauge bosons of the Spin(10)
grand unified theory, together with a copy of u(1), which describes
the *complex structure* of the left-handed Standard Model fermions.

Another nice fact is that (C tensor O)P^2 is one of the Grassmannians
for E6. I explained this general notion of "Grassmannian" back in
"week181", and you can see this 16-dimensional one in the list near
the end of that Week.

Even better, if you geometrically quantize this Grassmannian using the
smallest possible symplectic structure, you get the 27-dimensional
representation of E6 on the exceptional Jordan algebra!

So, there's a lot of seriously cool math going on here... but since
the basic idea of relating the Standard model to E6 is only
half-baked, all the ideas I'm mentioning now are at best
quarter-baked. They're mathematically correct, but I can't tell
if they're leading somewhere interesting.

In fact, I would have kept them in the oven longer had not Garrett
Lisi brought E6's big brother E8 into the game in a tantalizing way.
I'll conclude by summarizing this... and you can look at his website for
more details. But first, let me emphasize that this E8 business is the
most recent and most speculative thing Garrett has done. So, if you
think the following idea is nuts, please don't jump to conclusions and
decide *everything* he's doing is nuts!

Briefly, his idea involves taking the description of E8 I already mentioned:

E8 = so(O + O) + (O tensor O)^2

and writing the linear transformations in so(O + O) as two 8x8 blocks
living in so(O), together with an off-diagonal block living in O tensor O.
This gives

E8 = so(O) + so(O) + (O tensor O)^3

Then, he wants to use each of the three copies of O tensor O to
describe one of the three generations of fermions, while using the
so(O) + so(O) stuff to describe bosons (including gravity).

For this, he builds on some earlier work where he sought to combine
gravity, the Standard Model gauge bosons, the Higgs and *one*
generation of Standard Model fermions in an so(8) version of
MacDowell-Mansouri gravity.

If I were really being responsible, I would describe and assess this
earlier work. But, it's summer and I just want to have fun...

In fact, the above alternate description of E8 is the one Bertram
Kostant told me about back in 1996. He said it a different way, which
is equivalent:

E8 = so(8) + so(8) + End(V_8) + End(S_8^+) + End(S_8^-)

Here V_8, S_8^+ and S_8^- are the vector, left-handed spinor, and
right-handed spinor representations of Spin(8). All three are
8-dimensional, and all are related by outer automorphisms of Spin(8).
That's what "triality" is all about. You can see more details in
"week90".

The idea of relating the three generations to triality is cute. Of
course, even if it worked, you'd need something to give the fermions
in different generations different masses - which is what happens
already in the Standard Model, thanks to the Higgs boson. It's the
bane of all post-Standard Model physics: symmetry looks nice, but the
more symmetry your model has, the more symmetries you need to explain
away! As the White Knight said to Alice:

But I was thinking of a plan
To dye one's whiskers green,
And always use so large a fan
That they could not be seen.

Someday we may think of a way around this problem. But for now, I've got
a more pressing worry. This splitting of E6:

E6 = so(10) + S_{10}^+ + u(1)

corresponds to a Z/2-grading where so(10) + u(1) is the "bosonic" or "even"
part and S_{10}^+ is the "fermionic" or "odd" part. This nicely matches the
way so(10) describes gauge bosons and S_{10}^+ describes fermions in Georgi's
grand unified theory. But, this splitting of E8:

E8 = so(8) + so(8) + End(V_8) + End(S_8^+) + End(S_8^-)

does not correspond to any Z/2-grading where so(8) + so(8) is the bosonic
part and End(V) + End(S^+) + End(S^-) is the fermionic part. There is a
closely related Z/2-grading of E8, but it's this:

E8 = so(16) + S_{16}^+

So, right now I don't feel it's mathematically natural to use this method
to combine bosons and fermions.

But, only time will tell.

Here are some more references. The SU(5) grand unified theory was published
here:

5) Howard Georgi and Sheldon Glashow, Unity of all elementary-particle
forces, Phys. Rev. Lett. 32 (1974), 438.

For a great introduction to the Spin(10) grand unified theory - which
is usually called the SO(10) GUT - try this:

6) Anthony Zee, Quantum Field Theory in a Nutshell, Chapter VII: SO(10)
unification, Princeton U. Press, Princeton, 2003.

Then, try these more advanced review articles:

7) Jogesh C. Pati, Proton decay: a must for theory, a challenge for
experiment, available as hep-ph/0005095.

8) Jogesh C. Pati, Probing grand unification through neutrino oscillations,
leptogenesis, and proton decay, available as hep-ph/0305221.

The last two also consider the gauge group "G(224)", meaning SU(2) x SU(2)
x SU(4).

By the way, there's also a cute relation between the SO(10) grand
unified theory and 10-dimensional Calabi-Yau manifolds, discussed here:

9) John Baez, Calabi-Yau manifolds and the Standard Model, available as
hep-th/0511086

This is an easy consequence of the stuff I've explained this week.

To see what string theorists are doing to understand the Standard Model
these days, see the following papers. Amusingly, they *also* use E8 -
but in a quite different way:

10) Volker Braun, Yang-Hui He, Burt A. Ovrut and Tony Pantev,
A heterotic Standard Model, available as hep-th/0501070.

A Standard Model from the E8 x E8 heterotic superstring,
hep-th/0502155.

Vector bundle extensions, sheaf cohomology, and the heterotic
Standard Model, available as hep-th/0505041.

Heterotic Standard Model moduli, available as hep-th/0509051.

The exact MSSM spectrum from string theory, available as
hep-th/0512177.

All this stuff is really cool - but alas, they get the "minimal
supersymmetric Standard Model", or MSSM, which has a lot more
particles than the Standard Model, and a lot more undetermined
parameters. Of course, these flaws could become advantages if the
next big particle accelerator, the Large Hadron Collider, sees
signs of supersymmetry.

For more on symmetric spaces, try these:

11) Sigurdur Helgason, Differential Geometry, Lie Groups, and
Symmetric Spaces, AMS, Providence, Rhode Island, 2001.

12) Audrey Terras, Harmonic Analysis on Symmetric Spaces and Applications
I, Springer, Berlin, 1985. Harmonic Analysis on Symmetric Spaces
and Applications II, Springer, Berlin, 1988.

13) Arthur Besse, Einstein Manifolds, Springer, Berlin, 1986.

They're all classics. Helgason's book will teach you differential
geometry and Lie groups before doing Cartan's classification of symmetric
spaces. Terras' books are full of fun connections to other branches of
math. Besse's book has lots of nice charts, and goes much deeper into
the Riemannian geometry of symmetric spaces.

These dig deeper into the algebraic aspects of symmetric spaces:

14) W. Bertram, The Geometry of Jordan and Lie structures,
Lecture Notes in Mathematics 1754, Springer, Berlin, 2001.

15) Ottmar Loos, Jordan triple systems, R-spaces and bounded
symmetric domains, Bull. AMS 77 (1971), 558-561.

16) Ottmar Loos, Symmetric Spaces I: General Theory, W. A. Benjamin,
New York, 1969. Symmetric Spaces II: Compact Spaces and Classification,
W. A. Benjamin, New York, 1969.

-----------------------------------------------------------------------
Previous issues of "This Week's Finds" and other expository articles on
mathematics and physics, as well as some of my research papers, can be
obtained at

http://math.ucr.edu/home/baez/

For a table of contents of all the issues of This Week's Finds, try

http://math.ucr.edu/home/baez/twfcontents.html

A simple jumping-off point to the old issues is available at

http://math.ucr.edu/home/baez/twfshort.html

If you just want the latest issue, go to

http://math.ucr.edu/home/baez/this.week.html

 

[David Bernier]

John Baez wrote:
[...]
> For a long time, people have been seeking connections between the
> messy pack of particles that populate the Standard Model and
> structures that seem beautiful and "inevitable".
>
> A fascinating step in this direction was the SU(5) grand unified
> theory proposed in 1975 by Georgi and Glashow. So, I'll start by
> summarizing that... and then explain how exceptional Lie groups might
> get involved in this game.
>
> What people usually call the gauge group of the Standard Model:
>
> SU(3) x SU(2) x U(1)
>
> actually has a bit of flab in it: there's a normal subgroup that acts
> trivially on all known particles. This subgroup is isomorphic to Z/6.

If g is an element of the "true" gauge group G below, and x is a
particle, and g.x is the result of letting g act on x, then
I suppose g.x is also a particle. What does it mean, in
the "real world", to have x, and then g acting on x?

Why is it good to get rid of the flab by modding
out SU(3) x SU(2) x U(1) by the 6-element normal
subgroup to get G, the "true" gauge group?

Thanks,

David Bernier

> If we mod out by this, we get the "true" gauge group of the Standard
> Model:
>
> G = (SU(3) x SU(2) x U(1))/(Z/6)
[...]

 

[David Bernier]

John Baez wrote:
[...]
> For a long time, people have been seeking connections between the
> messy pack of particles that populate the Standard Model and
> structures that seem beautiful and "inevitable".
>
> A fascinating step in this direction was the SU(5) grand unified
> theory proposed in 1975 by Georgi and Glashow. So, I'll start by
> summarizing that... and then explain how exceptional Lie groups might
> get involved in this game.
>
> What people usually call the gauge group of the Standard Model:
>
> SU(3) x SU(2) x U(1)
>
> actually has a bit of flab in it: there's a normal subgroup that acts
> trivially on all known particles. This subgroup is isomorphic to Z/6.

If g is an element of the "true" gauge group G below, and x is a
particle, and g.x is the result of letting g act on x, then
I suppose g.x is also a particle. What does it mean, in
the "real world", to have x, and then g acting on x?

Why is it good to get rid of the flab by modding
out SU(3) x SU(2) x U(1) by the 6-element normal
subgroup to get G, the "true" gauge group?

Thanks,

David Bernier

> If we mod out by this, we get the "true" gauge group of the Standard
> Model:
>
> G = (SU(3) x SU(2) x U(1))/(Z/6)
[...]

 

[David Bernier]

John Baez wrote:
[...]
> For a long time, people have been seeking connections between the
> messy pack of particles that populate the Standard Model and
> structures that seem beautiful and "inevitable".
>
> A fascinating step in this direction was the SU(5) grand unified
> theory proposed in 1975 by Georgi and Glashow. So, I'll start by
> summarizing that... and then explain how exceptional Lie groups might
> get involved in this game.
>
> What people usually call the gauge group of the Standard Model:
>
> SU(3) x SU(2) x U(1)
>
> actually has a bit of flab in it: there's a normal subgroup that acts
> trivially on all known particles. This subgroup is isomorphic to Z/6.

If g is an element of the "true" gauge group G below, and x is a
particle, and g.x is the result of letting g act on x, then
I suppose g.x is also a particle. What does it mean, in
the "real world", to have x, and then g acting on x?

Why is it good to get rid of the flab by modding
out SU(3) x SU(2) x U(1) by the 6-element normal
subgroup to get G, the "true" gauge group?

Thanks,

David Bernier

> If we mod out by this, we get the "true" gauge group of the Standard
> Model:
>
> G = (SU(3) x SU(2) x U(1))/(Z/6)
[...]

 

[David Bernier]

John Baez wrote:
[...]
> For a long time, people have been seeking connections between the
> messy pack of particles that populate the Standard Model and
> structures that seem beautiful and "inevitable".
>
> A fascinating step in this direction was the SU(5) grand unified
> theory proposed in 1975 by Georgi and Glashow. So, I'll start by
> summarizing that... and then explain how exceptional Lie groups might
> get involved in this game.
>
> What people usually call the gauge group of the Standard Model:
>
> SU(3) x SU(2) x U(1)
>
> actually has a bit of flab in it: there's a normal subgroup that acts
> trivially on all known particles. This subgroup is isomorphic to Z/6.

If g is an element of the "true" gauge group G below, and x is a
particle, and g.x is the result of letting g act on x, then
I suppose g.x is also a particle. What does it mean, in
the "real world", to have x, and then g acting on x?

Why is it good to get rid of the flab by modding
out SU(3) x SU(2) x U(1) by the 6-element normal
subgroup to get G, the "true" gauge group?

Thanks,

David Bernier

> If we mod out by this, we get the "true" gauge group of the Standard
> Model:
>
> G = (SU(3) x SU(2) x U(1))/(Z/6)
[...]

 

[Arnold Neumaier]

John Baez schrieb:

> What people usually call the gauge group of the Standard Model:
>
> SU(3) x SU(2) x U(1)
>
> actually has a bit of flab in it: there's a normal subgroup that acts
> trivially on all known particles. This subgroup is isomorphic to Z/6.
> If we mod out by this, we get the "true" gauge group of the Standard
> Model:
>
> G = (SU(3) x SU(2) x U(1))/(Z/6)
>
> And, this turns out to have a neat description. It's isomorphic to the
> subgroup of SU(5) consisting of matrices like this:
>
> (g 0)
> (0 h)
>
> where g is a 3x3 block and h is a 2x2 block.

But this gives SU(3) x SU(2) x U(1) and not G...

Arnold Neumaier

 

[Arnold Neumaier]

John Baez schrieb:

> What people usually call the gauge group of the Standard Model:
>
> SU(3) x SU(2) x U(1)
>
> actually has a bit of flab in it: there's a normal subgroup that acts
> trivially on all known particles. This subgroup is isomorphic to Z/6.
> If we mod out by this, we get the "true" gauge group of the Standard
> Model:
>
> G = (SU(3) x SU(2) x U(1))/(Z/6)
>
> And, this turns out to have a neat description. It's isomorphic to the
> subgroup of SU(5) consisting of matrices like this:
>
> (g 0)
> (0 h)
>
> where g is a 3x3 block and h is a 2x2 block.

But this gives SU(3) x SU(2) x U(1) and not G...

Arnold Neumaier

 

[Arnold Neumaier]

John Baez schrieb:

> What people usually call the gauge group of the Standard Model:
>
> SU(3) x SU(2) x U(1)
>
> actually has a bit of flab in it: there's a normal subgroup that acts
> trivially on all known particles. This subgroup is isomorphic to Z/6.
> If we mod out by this, we get the "true" gauge group of the Standard
> Model:
>
> G = (SU(3) x SU(2) x U(1))/(Z/6)
>
> And, this turns out to have a neat description. It's isomorphic to the
> subgroup of SU(5) consisting of matrices like this:
>
> (g 0)
> (0 h)
>
> where g is a 3x3 block and h is a 2x2 block.

But this gives SU(3) x SU(2) x U(1) and not G...

Arnold Neumaier

 

[barry.david.adams@googlemail.com]

> Anyway: the case of E6 should remind you of something! After all, we've
> just been talking about so(10) and its left-handed spinor representation.
> These describe the gauge bosons and one generation of left-handed fermions
> in the Spin(10) grand unified theory. But now we're seeing this stuff
> neatly packed into the Lie algebra of E6!
>
> More precisely, the Lie algebra of E6 can be chopped into 3 pieces
> in a noncanonical way:
>
> A) so(10),
>
> B) the left-handed real spinor representation of so(10), which by now
> we've given three different names:
>
> S_{10}^+ = Lambda^{even}(C^5) = (C tensor O)^2
>
> and
>
> C) a copy of u(1).
>
> The first part contains all the gauge bosons in the SO(10) grand unified
> theory. The second contains one generation of left-handed fermions.

Everytime I've seen the Standard model in E6, its seem to end up
with lots of extra particle states, usually an extra quark with three
colors
and charge +1/3, a heavy electron like state, and lots of singlets.
Which
look very ugly to me, e.g. arXiv:0705.0074v1

I wonder if its possible, and prettier to fit the SM into E6, just by
doubling the number of quark states. Based upon naive counting of
states,
(does the group theory work?).

E6 (27) = 15 (SM) + 12 (SM quarks only), so that you'd have

[ d(-1/3) or u(+2/3)] *(left or right) *( red, green of blue) * [A or
B] = 24 states

e(-) * (left or right)
nu (left only)

The two new quarks need to be arranged so there electric charges,
cancel out,
so packed into E6, with anti-left handed states, and normal right
handed states,
(or vice versa).

Is there any evidence for these extra quark states? Well QCD has
an ugly secret, there seem to be a set of meson states that don't
fit into the standard quark-antiquark pairs, the lightest being the
sigma(550)
which is a scalar like the pion but has the wrong parity +1, rather
than
the pions -1. This Lead to M. and S. Ishida to fitting the meson and
baryon
resonance states by doubling the number of quarks, arXiv:hep-ph/
0310062v1

If we do this then the extra states needed to make up to
E6, don't have to be super heavy (leading to the problem of
explaining why), merely very short lived.

Dr B. Adams

 

[Arnold Neumaier]

John Baez schrieb:

> What people usually call the gauge group of the Standard Model:
>
> SU(3) x SU(2) x U(1)
>
> actually has a bit of flab in it: there's a normal subgroup that acts
> trivially on all known particles. This subgroup is isomorphic to Z/6.
> If we mod out by this, we get the "true" gauge group of the Standard
> Model:
>
> G = (SU(3) x SU(2) x U(1))/(Z/6)
>
> And, this turns out to have a neat description. It's isomorphic to the
> subgroup of SU(5) consisting of matrices like this:
>
> (g 0)
> (0 h)
>
> where g is a 3x3 block and h is a 2x2 block.

But this gives SU(3) x SU(2) x U(1) and not G...

Arnold Neumaier

 

[barry.david.adams@googlemail.com]

> Anyway: the case of E6 should remind you of something! After all, we've
> just been talking about so(10) and its left-handed spinor representation.
> These describe the gauge bosons and one generation of left-handed fermions
> in the Spin(10) grand unified theory. But now we're seeing this stuff
> neatly packed into the Lie algebra of E6!
>
> More precisely, the Lie algebra of E6 can be chopped into 3 pieces
> in a noncanonical way:
>
> A) so(10),
>
> B) the left-handed real spinor representation of so(10), which by now
> we've given three different names:
>
> S_{10}^+ = Lambda^{even}(C^5) = (C tensor O)^2
>
> and
>
> C) a copy of u(1).
>
> The first part contains all the gauge bosons in the SO(10) grand unified
> theory. The second contains one generation of left-handed fermions.

Everytime I've seen the Standard model in E6, its seem to end up
with lots of extra particle states, usually an extra quark with three
colors
and charge +1/3, a heavy electron like state, and lots of singlets.
Which
look very ugly to me, e.g. arXiv:0705.0074v1

I wonder if its possible, and prettier to fit the SM into E6, just by
doubling the number of quark states. Based upon naive counting of
states,
(does the group theory work?).

E6 (27) = 15 (SM) + 12 (SM quarks only), so that you'd have

[ d(-1/3) or u(+2/3)] *(left or right) *( red, green of blue) * [A or
B] = 24 states

e(-) * (left or right)
nu (left only)

The two new quarks need to be arranged so there electric charges,
cancel out,
so packed into E6, with anti-left handed states, and normal right
handed states,
(or vice versa).

Is there any evidence for these extra quark states? Well QCD has
an ugly secret, there seem to be a set of meson states that don't
fit into the standard quark-antiquark pairs, the lightest being the
sigma(550)
which is a scalar like the pion but has the wrong parity +1, rather
than
the pions -1. This Lead to M. and S. Ishida to fitting the meson and
baryon
resonance states by doubling the number of quarks, arXiv:hep-ph/
0310062v1

If we do this then the extra states needed to make up to
E6, don't have to be super heavy (leading to the problem of
explaining why), merely very short lived.

Dr B. Adams

 

[barry.david.adams@googlemail.com]

> Anyway: the case of E6 should remind you of something! After all, we've
> just been talking about so(10) and its left-handed spinor representation.
> These describe the gauge bosons and one generation of left-handed fermions
> in the Spin(10) grand unified theory. But now we're seeing this stuff
> neatly packed into the Lie algebra of E6!
>
> More precisely, the Lie algebra of E6 can be chopped into 3 pieces
> in a noncanonical way:
>
> A) so(10),
>
> B) the left-handed real spinor representation of so(10), which by now
> we've given three different names:
>
> S_{10}^+ = Lambda^{even}(C^5) = (C tensor O)^2
>
> and
>
> C) a copy of u(1).
>
> The first part contains all the gauge bosons in the SO(10) grand unified
> theory. The second contains one generation of left-handed fermions.

Everytime I've seen the Standard model in E6, its seem to end up
with lots of extra particle states, usually an extra quark with three
colors
and charge +1/3, a heavy electron like state, and lots of singlets.
Which
look very ugly to me, e.g. arXiv:0705.0074v1

I wonder if its possible, and prettier to fit the SM into E6, just by
doubling the number of quark states. Based upon naive counting of
states,
(does the group theory work?).

E6 (27) = 15 (SM) + 12 (SM quarks only), so that you'd have

[ d(-1/3) or u(+2/3)] *(left or right) *( red, green of blue) * [A or
B] = 24 states

e(-) * (left or right)
nu (left only)

The two new quarks need to be arranged so there electric charges,
cancel out,
so packed into E6, with anti-left handed states, and normal right
handed states,
(or vice versa).

Is there any evidence for these extra quark states? Well QCD has
an ugly secret, there seem to be a set of meson states that don't
fit into the standard quark-antiquark pairs, the lightest being the
sigma(550)
which is a scalar like the pion but has the wrong parity +1, rather
than
the pions -1. This Lead to M. and S. Ishida to fitting the meson and
baryon
resonance states by doubling the number of quarks, arXiv:hep-ph/
0310062v1

If we do this then the extra states needed to make up to
E6, don't have to be super heavy (leading to the problem of
explaining why), merely very short lived.

Dr B. Adams

 

[barry.david.adams@googlemail.com]

> Anyway: the case of E6 should remind you of something! After all, we've
> just been talking about so(10) and its left-handed spinor representation.
> These describe the gauge bosons and one generation of left-handed fermions
> in the Spin(10) grand unified theory. But now we're seeing this stuff
> neatly packed into the Lie algebra of E6!
>
> More precisely, the Lie algebra of E6 can be chopped into 3 pieces
> in a noncanonical way:
>
> A) so(10),
>
> B) the left-handed real spinor representation of so(10), which by now
> we've given three different names:
>
> S_{10}^+ = Lambda^{even}(C^5) = (C tensor O)^2
>
> and
>
> C) a copy of u(1).
>
> The first part contains all the gauge bosons in the SO(10) grand unified
> theory. The second contains one generation of left-handed fermions.

Everytime I've seen the Standard model in E6, its seem to end up
with lots of extra particle states, usually an extra quark with three
colors
and charge +1/3, a heavy electron like state, and lots of singlets.
Which
look very ugly to me, e.g. arXiv:0705.0074v1

I wonder if its possible, and prettier to fit the SM into E6, just by
doubling the number of quark states. Based upon naive counting of
states,
(does the group theory work?).

E6 (27) = 15 (SM) + 12 (SM quarks only), so that you'd have

[ d(-1/3) or u(+2/3)] *(left or right) *( red, green of blue) * [A or
B] = 24 states

e(-) * (left or right)
nu (left only)

The two new quarks need to be arranged so there electric charges,
cancel out,
so packed into E6, with anti-left handed states, and normal right
handed states,
(or vice versa).

Is there any evidence for these extra quark states? Well QCD has
an ugly secret, there seem to be a set of meson states that don't
fit into the standard quark-antiquark pairs, the lightest being the
sigma(550)
which is a scalar like the pion but has the wrong parity +1, rather
than
the pions -1. This Lead to M. and S. Ishida to fitting the meson and
baryon
resonance states by doubling the number of quarks, arXiv:hep-ph/
0310062v1

If we do this then the extra states needed to make up to
E6, don't have to be super heavy (leading to the problem of
explaining why), merely very short lived.

Dr B. Adams

 

[Zurab Silagadze]

> The idea of relating the three generations to triality is cute.
The idea that three generations are related to triality was examined
with somewhat different perspective (but with SO(10), E6, octonions
and exceptional Jordan algebra) in my paper
"SO(8) color as possible origin of generations" published in Phys.
Atom. Nucl. 58 (1995) 1430-1434.
An electronic version:
http://uk.arxiv.org/abs/hep-ph/9411381 (but it does not include
figures)
Here is the KEK Library scanned version of more detailed preprint:
http://ccdb4fs.kek.jp/cgi-bin/img_index?9405634

With best regards, Zurab Silagadze.

 

[Marc Nardmann]

Arnold Neumaier wrote:
> John Baez schrieb:
>
>
>> What people usually call the gauge group of the Standard Model:
>>
>> SU(3) x SU(2) x U(1)
>>
>> actually has a bit of flab in it: there's a normal subgroup that acts
>> trivially on all known particles. This subgroup is isomorphic to Z/6.
>> If we mod out by this, we get the "true" gauge group of the Standard
>> Model:
>>
>> G = (SU(3) x SU(2) x U(1))/(Z/6)
>>
>> And, this turns out to have a neat description. It's isomorphic to the
>> subgroup of SU(5) consisting of matrices like this:
>>
>> (g 0)
>> (0 h)
>>
>> where g is a 3x3 block and h is a 2x2 block.
>>
>
> But this gives SU(3) x SU(2) x U(1) and not G...
>

I do not understand what you are saying here. Consider the map

f: SU(3) x SU(2) x U(1) --> S(U(2) x U(3))

given by

f(A,B,z) = diag(z^2.A, z^{-3}.B).

One can check easily that f is well-defined, a group homomorphism, and
surjective. Its kernel is

K = {(A,B,z) | z^6=1, A=z^{-2}.id, B=z^3.id},

which is isomorphic to Z/6Z. Thus G = (SU(3) x SU(2) x U(1)) / K is
isomorphic to S(U(2) x U(3)).

-- Marc Nardmann

 

[arivero@unizar.es]

On 2 jul, 10:11, Zurab Silagadze <silaga...@inp.nsk.su> wrote:

> "SO(8) color as possible origin of generations" published in Phys.
> Atom. Nucl. 58 (1995) 1430-1434.
> An electronic version:http://uk.arxiv.org/abs/hep-ph/9411381(but it does not include
> figures)
> Here is the KEK Library scanned version of more detailed preprint:http://ccdb4fs.kek.jp/cgi-bin/img_index?9405634
> With best regards, Zurab Silagadze.

It could be worthwhile to mention some other flirts with triality in
the Arxiv. Not to speak of octonions.

H. Mkrtchyan, R. Mkrtchyan 10d N=1 Massless BPS supermultiplets
http://arxiv.org/abs/hep-th/0312281
Lee Smolin The exceptional Jordan algebra and the matrix string
http://arxiv.org/abs/hep-th/0104050
Stephen L. Adler Frustrated SU(4) as the Preonic Precursor of the
Standard Model http://arxiv.org/abs/hep-th/9610190
F. D. T. Smith SU(3)xSU(2)xU(1), Higgs, and Gravity from Spin(0,8)
Clifford Algebra Cl(0,8)

 

[markwh04@yahoo.com]

On Jul 2, 3:11 am, Zurab Silagadze <silaga...@inp.nsk.su> wrote:
> > The idea of relating the three generations to triality is cute.
> The idea that three generations are related to triality was examined
> with somewhat different perspective (but with SO(10), E6, octonions
> and exceptional Jordan algebra) in my paper
> "SO(8) color as possible origin of generations" published in Phys.
> Atom. Nucl. 58 (1995) 1430-1434.

I put up the Lagrangian and development of the underlying Yang-Mills-
Higgs-(Einstein) theory in

The Standard Model Lagrangian
http://federation.g3z.com/Physics/index.htm#StandardModel

I made brief reference there to what is probably the best account of
the origin of generations: Tsou's duality-based theory. It reproduces
the details of the mass matrices to a high degree of accuracy. This is
also briefly described in Penrose's 2006 Road to Reality.

The article presents some key points not seen in the general context
of Yang-Mills-Higgs or (more generally) gauge theory.

(a) the metrics normally hidden in the notation to the fermionic,
scalar and vector parts of the Lagrangian are fully brought out in the
open
(b) the Higgs coupling is developed from the assumption of
trilinearity of the fermion-scalar interaction
(c) the gauge invariance of the extra metrics is pointed out and
promoted to a principle. However, the fermionic metric is NOT gauge
invariant and cannot be made so, unless parity is restored as a broken
symmetry
(d) the 2 Wong invariants in the fermion spectrum
(d) the fermion spectrum, itself, along with the 2 Wong invariants
both point to an underlying parity-symmetric origin
(e) the 5 "qubit" charges are developed
(f) the gravity sector is worked from a general metric-affine
assumption. In addition to the curvature scalar and cosmological
constant, a third term can arise in the Lagrangian that has non-
trivial coupling with spin
(g) the invariants and spectrum point to (SU(2)_L x SU(2)_R) x
(SO(4)), and one of the two "fermion space" bases (the "Casimir
basis") is essentially based on this correspondence.

There's a lot of other novel features that I think make this worth
taking a look at.

In a way, the SU(2)_L x SU(2)_R x SO(4) part might be viewed as a
least common denominator to any unification. However, it entails
promoting Baryon number to a gauge symmetry. I haven't yet included
the theta term (which is easy to do, in fact) nor the Majorana term
(for neutrino see-saw). The Majorana term can be fit in the Casimir
basis, but I just haven't look at this in detail let.

One interesting feature of the see-saw Majorana term is that if the
corresponding coupling coefficient is promoted to a field, it would
provide the extra generators for the SU(2)_R part of the parity-
symmetric SU(2)_L x SU(2)_R, complementing the isospin SU(2)_L sector.

 

[Arnold Neumaier]

Marc Nardmann schrieb:
> Arnold Neumaier wrote:
>> John Baez schrieb:
>>
>>
>>> What people usually call the gauge group of the Standard Model:
>>>
>>> SU(3) x SU(2) x U(1)
>>>
>>> actually has a bit of flab in it: there's a normal subgroup that acts
>>> trivially on all known particles. This subgroup is isomorphic to Z/6.
>>> If we mod out by this, we get the "true" gauge group of the Standard
>>> Model:
>>>
>>> G = (SU(3) x SU(2) x U(1))/(Z/6)
>>>
>>> And, this turns out to have a neat description. It's isomorphic to the
>>> subgroup of SU(5) consisting of matrices like this:
>>>
>>> (g 0)
>>> (0 h)
>>>
>>> where g is a 3x3 block and h is a 2x2 block.
>>>
>> But this gives SU(3) x SU(2) x U(1) and not G...
>>
> I do not understand what you are saying here. Consider the map
> f: SU(3) x SU(2) x U(1) --> S(U(2) x U(3))
> given by
> f(A,B,z) = diag(z^2.A, z^{-3}.B).
> One can check easily that f is well-defined, a group homomorphism, and
> surjective. Its kernel is
> K = {(A,B,z) | z^6=1, A=z^{-2}.id, B=z^3.id},
> which is isomorphic to Z/6Z. Thus G = (SU(3) x SU(2) x U(1)) / K is
> isomorphic to S(U(2) x U(3)).

Thanks for the details; I was mistaken; now I understand.

Arnold Neumaier

 

[Arnold Neumaier]

Marc Nardmann schrieb:
> Arnold Neumaier wrote:
>> John Baez schrieb:
>>
>>
>>> What people usually call the gauge group of the Standard Model:
>>>
>>> SU(3) x SU(2) x U(1)
>>>
>>> actually has a bit of flab in it: there's a normal subgroup that acts
>>> trivially on all known particles. This subgroup is isomorphic to Z/6.
>>> If we mod out by this, we get the "true" gauge group of the Standard
>>> Model:
>>>
>>> G = (SU(3) x SU(2) x U(1))/(Z/6)
>>>
>>> And, this turns out to have a neat description. It's isomorphic to the
>>> subgroup of SU(5) consisting of matrices like this:
>>>
>>> (g 0)
>>> (0 h)
>>>
>>> where g is a 3x3 block and h is a 2x2 block.
>>>
>> But this gives SU(3) x SU(2) x U(1) and not G...
>>
> I do not understand what you are saying here. Consider the map
> f: SU(3) x SU(2) x U(1) --> S(U(2) x U(3))
> given by
> f(A,B,z) = diag(z^2.A, z^{-3}.B).
> One can check easily that f is well-defined, a group homomorphism, and
> surjective. Its kernel is
> K = {(A,B,z) | z^6=1, A=z^{-2}.id, B=z^3.id},
> which is isomorphic to Z/6Z. Thus G = (SU(3) x SU(2) x U(1)) / K is
> isomorphic to S(U(2) x U(3)).

Thanks for the details; I was mistaken; now I understand.

Arnold Neumaier

 

[Arnold Neumaier]

Marc Nardmann schrieb:
> Arnold Neumaier wrote:
>> John Baez schrieb:
>>
>>
>>> What people usually call the gauge group of the Standard Model:
>>>
>>> SU(3) x SU(2) x U(1)
>>>
>>> actually has a bit of flab in it: there's a normal subgroup that acts
>>> trivially on all known particles. This subgroup is isomorphic to Z/6.
>>> If we mod out by this, we get the "true" gauge group of the Standard
>>> Model:
>>>
>>> G = (SU(3) x SU(2) x U(1))/(Z/6)
>>>
>>> And, this turns out to have a neat description. It's isomorphic to the
>>> subgroup of SU(5) consisting of matrices like this:
>>>
>>> (g 0)
>>> (0 h)
>>>
>>> where g is a 3x3 block and h is a 2x2 block.
>>>
>> But this gives SU(3) x SU(2) x U(1) and not G...
>>
> I do not understand what you are saying here. Consider the map
> f: SU(3) x SU(2) x U(1) --> S(U(2) x U(3))
> given by
> f(A,B,z) = diag(z^2.A, z^{-3}.B).
> One can check easily that f is well-defined, a group homomorphism, and
> surjective. Its kernel is
> K = {(A,B,z) | z^6=1, A=z^{-2}.id, B=z^3.id},
> which is isomorphic to Z/6Z. Thus G = (SU(3) x SU(2) x U(1)) / K is
> isomorphic to S(U(2) x U(3)).

Thanks for the details; I was mistaken; now I understand.

Arnold Neumaier

 

[Arnold Neumaier]

Marc Nardmann schrieb:
> Arnold Neumaier wrote:
>> John Baez schrieb:
>>
>>
>>> What people usually call the gauge group of the Standard Model:
>>>
>>> SU(3) x SU(2) x U(1)
>>>
>>> actually has a bit of flab in it: there's a normal subgroup that acts
>>> trivially on all known particles. This subgroup is isomorphic to Z/6.
>>> If we mod out by this, we get the "true" gauge group of the Standard
>>> Model:
>>>
>>> G = (SU(3) x SU(2) x U(1))/(Z/6)
>>>
>>> And, this turns out to have a neat description. It's isomorphic to the
>>> subgroup of SU(5) consisting of matrices like this:
>>>
>>> (g 0)
>>> (0 h)
>>>
>>> where g is a 3x3 block and h is a 2x2 block.
>>>
>> But this gives SU(3) x SU(2) x U(1) and not G...
>>
> I do not understand what you are saying here. Consider the map
> f: SU(3) x SU(2) x U(1) --> S(U(2) x U(3))
> given by
> f(A,B,z) = diag(z^2.A, z^{-3}.B).
> One can check easily that f is well-defined, a group homomorphism, and
> surjective. Its kernel is
> K = {(A,B,z) | z^6=1, A=z^{-2}.id, B=z^3.id},
> which is isomorphic to Z/6Z. Thus G = (SU(3) x SU(2) x U(1)) / K is
> isomorphic to S(U(2) x U(3)).

Thanks for the details; I was mistaken; now I understand.

Arnold Neumaier