Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Jun 29, 2007 #1
    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
     
  2. jcsd
  3. Jul 1, 2007 #2
    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)

    [...]
     
  4. Jul 1, 2007 #3
    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)

    [...]
     
  5. Jul 1, 2007 #4
    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)

    [...]
     
  6. Jul 1, 2007 #5
    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)

    [...]
     
  7. Jul 1, 2007 #6
    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
     
  8. Jul 1, 2007 #7
    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
     
  9. Jul 1, 2007 #8
    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
     
  10. Jul 1, 2007 #9
    > 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
     
  11. Jul 1, 2007 #10
    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
     
  12. Jul 1, 2007 #11
    > 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
     
  13. Jul 1, 2007 #12
    > 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
     
  14. Jul 1, 2007 #13
    > 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
     
  15. Jul 2, 2007 #14
    Re: This Week's Finds in Mathematical Physics (Week 253) -- E6, generations and triality

    > 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.
     
  16. Jul 4, 2007 #15
    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
     
  17. Jul 5, 2007 #16
    Re: This Week's Finds in Mathematical Physics (Week 253) -- E6,generations and triality

    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)
     
  18. Jul 6, 2007 #17
    Re: This Week's Finds in Mathematical Physics (Week 253) -- E6,generations and triality

    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.
     
  19. Jul 6, 2007 #18
    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
     
  20. Jul 6, 2007 #19
    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
     
  21. Jul 6, 2007 #20
    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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?