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I would like to have a glimpse of understanding of grand unificaction

  1. Jul 12, 2011 #1
    I would like to have a glimpse of understanding of grand unificaction and all that.

    But before that, of course I must understand the Spin groups! I like to explicit construct them fom Clifford algebras. Below I attached three pages where it is described what to do. Sadly, I do not understand fully what goes on, especially the subscripts and superscripts do not make to much sense.

    Can some one guide me through equations (2) till (6)?

    What does (7) mean? Are the [itex]\sigma[/itex]'s the generators? What are then the [itex]\gamma[/itex]'s?

    Even more and above all, I would love to explicitly construct SO(4), SO(6) and so one. Can some help me to do that?

    thank you

    EDIT: But my ultimate goal is to understand what is a spinoral and what is vector representation and how are they related. On page three of the document, the author talks about vectors and the two spinors.

    Why are the [itex]\sigma[/itex]'s shown to be the same for the vector and both spinor representations? Are not the [itex]\sigma[/itex]'s the representations, and the vectors and spinors what the matrix representations act on?

    Attached Files:

    Last edited: Jul 12, 2011
  2. jcsd
  3. Jul 12, 2011 #2
    Re: So(10)

    Probably the problem of yours is the tensor product notation, this is not difficult once you get the point.

    Eq 2 shows how the first 2n gamma matrices of SO(2n+2) are made using the gamma matrices of SO(2n) as building blocks. The tensor notation of the expression in between of the two "=" says that you take the third Pauli matrix, tau_3, and insert for each "1" in its two non-vanishing entries the gamma matrices of SO(2n). The dimension of the new matrix is then the product of the dimension of the gamma matrices of SO(2n) (which is 2^n) and the dimension of tau_3, which is 2. Thus the gamma matrices of SO(2n+2) have dimension 2^(n+1).

    But there are two more matrices missing, since we need 2n+2 and not just 2n of such matrices. The missing ones are written in Eqs 3 and 4.

    So Eq 2-4 show how to get the gamma matrices by an iterative procedure from smaller ones. When writing the result out in terms of the smallest building blocks, namely the 2d Pauli matrices t_1,2,3, then you get what is shown in Es 5,6.
  4. Jul 12, 2011 #3
    Re: So(10)

    thanks suprised!

    I edited my post a bit and introduced additional questions, but you were too quick..

    If you had the time, could you also adress those. Very much appreciated.

    Also, the document no has three pages, not two.
  5. Jul 12, 2011 #4
    Re: So(10)

    View the generators [itex]\sigma[/itex] as abstract objects, which are defined by their commutation relations (which form the Lie algebra so(2n)).

    It turns out that there are many, in fact infinitely many possible choices of matrices (with generically different dimensions) that obey these same commutation relations. Any such given choice is called a (matrix) representation of the generators, and it acts on vectors, spinors, higher rank tensors.. by matrix multiplication.

    So the [itex]\sigma[/itex] are not the same for the vector and both spinor representations, ie, their matrix representation is different in each case, even has different dimension. But they represent the "same" abstract generators.
  6. Jul 12, 2011 #5


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    Re: So(10)

    The [itex]\gamma[/itex]'s are the grade 1 elements of the clifford algebra which are the generators of that algebra (all other elements are antisymmetric products of these).

    The [itex]\sigma[/itex]'s are grade 2 elements which form the Lie algebra generating the rotation group. They are generators of SO(2n).
    (Note that commutators of grade 2 elements are again grade 2 elements.)

    One helpful point here. We can define a simple rotation in 3-space either by defining the rotation plane (say rotation in the x-y plane where x moves toward y and y away from x) or the rotation leaving an axis invariant, (in this example the z axis).

    We use vector angles and vectors or rotation generators in 3 space but we should more properly use bi-vector angles and bi-vector rotation generators (bi-vector = rank 2 antisymmetric tensor.) In 3 dimensions these two spaces are isomorphic but that is not the case in higher dimensions. This can be confusing when you look at say the vector basis given by the (hermitian) Pauli matrices [itex]\sigma_i \sim \gamma_i[/itex] and the bi-vector basis which will be the anti-hermitian Pauli matrices:
    [itex] -i\sigma_{ij}= \frac{1}{2}[\sigma_i,\sigma_j] =[/itex] (in three dimensions)[itex]
    Now the way the paper you posted does it is to express this anti-Hermitian generator in terms of a Hermitian spin observable [itex]\sigma_{ij}[/itex] since it is aimed at a QM application.

    In a purely mathematical treatment the Clifford algebra is a Real algebra (there is no i) and then, depending on the dimension one gets an algebra isomorphic to an algebra of real, complex, or quaternion matrices. Then the column vectors these matrices act upon are the spinors.

    The object acting sometimes like the imaginary unit is the top element (in the cases where it is both central as with SO(2n) rather than SO(2n+1) and squares to -1, which I think occurs in SO(2n) with n odd, but may also depend on the choice of metric signature, i.e. whether you define
    [itex]\{\gamma_i,\gamma_j\}=+\delta_{ij}[/itex] or [itex] = -\delta_{ij}[/itex].)

    In higher dimensions the rotation generators ([itex]\sigma[/itex]'s) will be independent elements of the algebra from the generators of the algebra itself (the [itex]\gamma[/itex]'s). This is why it is important when extrapolating from 3 dimensional spin to realize them as grade 2 elements (antisymmetric tensors) and not grade 1 elements (vectors).

    For a detailed exposition of Clifford algebras and spin I suggest Porteus' https://www.amazon.com/Clifford-Algebras-Classical-Cambridge-Mathematics/dp/0521551773"
    Last edited by a moderator: May 5, 2017
  7. Jul 12, 2011 #6
    Re: So(10)

    Thanks so much, suprised and jambaugh!

    While I certainly have to first absorb what you all just wrote here, I can not resist, as long as you are around, to ask one more.

    I read again and again that some magic happens in SO(8), where the vector representation and the two spinor representations can be transformed into each other via an so-called outer automorphism. I first read it years ago on John Baez site http://math.ucr.edu/home/baez/twf_ascii/week61" [Broken], where he describes it quite beautifully. Unfortunately but quite naturally when towards the end he goes into the details, he is harder to understand. But still the best I saw so far on this topic.

    Often, I feel, people just show the D_4 Dynkinn diagram and say it is symmetric and voila there must exist some transformation that maps the three representation into each other.

    Is there any way to make this 'outer automorphism' more concrete and explicit? I remember back then I tried Fulton/ Harris and Porteous in the libary, as Baez refered to them in the end, but they were too hard to follow for me, I believe. Too abstract for my feeble brain...which demands concrete, explicit and worked out examples, I'm afraid...
    Last edited by a moderator: May 5, 2017
  8. Jul 12, 2011 #7
    Re: So(10)

    This is true and there are other Dynkin diagrams where one can apply such transformations.

    An easy way to visualize this is to consider SU(n+1) ~ An instead. Then the Dynkin diagram is a line which has an obvious reflection symmtry. This outer automorphism simply acts by mapping any representation into its complex conjugate. That's all to it!

    Probably you are familiar with SU(3). Then you know how the triplet and anti-triplet representations look in weight space. And then you also see what kind of rotation of weight space maps both into each other.
  9. Jul 12, 2011 #8


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    Re: So(10)

    Yes, looking at the transformations of the weights is a great way to understand the triality automorphism of so(8), and the corresponding isomorphisms between positive and negative chiral spinors and vectors.

    If you want to play with this automorphism explicitly on the Lie algebra level, as a map between generators, there isn't much literature on that. You mostly have to figure it out on your own. (I worked this out recently, for so(8) and so(4,4), and will include it in a future paper if all goes well.)
  10. Jul 13, 2011 #9


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    Re: So(10)

    Let's see. There's more detail to the matrix representations of the clifford algebras than I included in my already lengthy post.

    First recall that I mentioned that the dimension of the Clifford algebra for SO(n) has dimension [itex]2^n[/itex]. If we are then going to express it in a matrix algebra the matrices will be say, [itex]p\times p[/itex] with [itex]p[/itex] (real) dimensional pinors.
    (Pinors are extensions of spinors which allow us to include inversions as well as rotations. One is really representing O(n) not just SO(n).)

    For this to be the case [itex]p \le \sqrt{2^n} = 2^{n/2}[/itex].

    Along with this the invariants of the group must be invariants of the matrices and these will manifest differently according to dimension. You get different types of adjoints and thus different types of matrix groups into which spin group maps. It may be a form defining orthogonal (real or complex), unitary, symplectic (real or complex) group. But the pattern is periodic with cycle 8 (for real orthogonal groups).

    At O(8) you get a sixteen dimensional real pinor repersentation which splits into two 8 dimensional spinor representations of SO(8).

    [Side note: We say representations of the group but the spinors really give a projective representation obeying the definition of a representation only up to scalar multiples of the defining identities. They are however true representations of the underlying Lie algebras which is really what one is working with.]

    Back to Spin(8)... it gets mapped in the matrix group SO(8)xSO(8) acting on the spinors via a double covering so that 360deg rotations of the original SO(8) vectors get mapped to 180deg rotations in the SO(8)xSO(8) matrix group acting on the two conjugate spinor representations.

    Disclaimer: I'm trying to reconstruct the following from memory. I'm sure I have made some errors such as sign conventions but the outline is --I believe-- correct.

    What this means is that the (16x16) dimensional matrices corresponding to the Clifford elements [itex]\sigma_{ij}[/itex] generating SO(8) rotations will have block diagonal form:
    [itex]\left(\begin{array}{c|c} \omega_L & 0 \\ \hline \\ 0 & \omega_R\end{array}\right)[/itex]
    where the [itex]\omega,\omega^{-1}[/itex] are again generators of SO(8) (each half of the SO(8)xSO(8) group acting on the pinor:
    [itex] \Psi = \left(\begin{array}{c}\psi^{even} \\ \psi^{odd}\end{array}\right)[/itex].

    Now recall that these generators are bi-vectors under the adjoint representation in the Cl Alg. We also have the grade 1 elements which transform adjointly as vectors. Recall that I mentioned the clifford algebra represents the full orthogonal group O(8) not just SO(8). That means it includes inversions of sub-spaces. The vector generators acting adjointly (and normalized) represent inversion of a 1 dimensional subspace spanned by that vector or, depending on conventions used, inversion of the n-1 dimensional subspace orthogonal to that vector. This means the matrix representation of these vector, grade-1 elements are odd transformations and must swap the odd and even spinor subspaces of the 16-dim pinor space. They thus must have a matrix representation of the form:
    [itex]\left(\begin{array}{c|c} 0 & M \\ \hline \\ -M^{-1} & 0 \end{array}\right)[/itex]. (Note that this squares to -1. This is a choice of metric convention in the clifford algebra but is a necessary choice so we end up with the octonians below.)

    (Patience, we're almost there.)

    So now we have three actions of the SO(8) generators. Their adjoint action on the grade one elements [itex]\gamma_1,\ldots \gamma_8[/itex], the action on the even spinors [itex]\psi^{even}[/itex], and the action on the odd spinors [itex]\psi^{odd}[/itex].

    We also have a mapping defined by the action of the vector [itex]\gamma_i[/itex] mapping even spinors to odd spinors (and mapping odd spinors to even spinors). From this action we can construct the octonians. Given arbitrary bases for the vectors, even spinors, and odd spinors we define the vector's action on the spinor space by its clifford matrix in terms of the basis elements:

    [itex] \gamma_i: \psi^{even}_j \mapsto \sum_k C_{ijk}\psi^{odd}_k[/itex]
    Note that the three indexed object [itex]C_{ijk}[/itex] has each index running from 1 to 8, but each corresponds to a basis in a completely different space. We can manipulate the choice of basis so that an arbitrarily chosen one of the vectors, say [itex]\gamma_1[/itex] gives coefficients of the form:
    [itex] C_{1jk} = \delta_{jk}[/itex]
    One can further restrict the choice of basis, consistant with the above so that the coefficients are cyclically symmetric:
    [itex]C_{ijk} = C_{jki} = C_{kij}[/itex].
    In this choice we can "identify" the vectors, odd spinors, and even spinors, calling each of them [itex]e_0,\cdots,e_7[/itex]
    and the product defined by:
    [itex]e_i e_j = \sum_k C_{ijk} e_k[/itex]
    is the octonian product with [itex]e_0 = 1[/itex].

    This cyclic symmetry is the triality automorphism between vector, even spinor, and odd spinor.

    OK, now we'll start with the octonians and reconstruct the vector and two spinor representations. First recall that octonian multiplication is NOT associative and we need an associative algebra to represent the associative group. Here's what we do:
    For each octonian [itex] e_k[/itex] define the left adjoint action:
    [itex]L_k: e_j \mapsto e_k e_j[/itex]
    and the (negative?) right adjoint action: [itex] R_k: e_j \mapsto - e_j (e_k)^{-1}[/itex] (Not sure if this is exactly right!!!)

    Now the left action corresponds to the action of vector clifford element on even spinors (yielding odd spinors) and the right action corresponds to the action of a vector clifford element on odd spinors (yielding even spinors).

    Now also note that an algebra's product is associative if and only if the left and right adjoint actions commute.
    [itex]L_aR_b c = a(cb) = (ac)b= R_b L_a c[/itex]
    Since it is non-associative they will not commute, and since the vector clifford generators swap odd and even spinors we have to represent products of these actions by alternating left and right actions.

    So to get the spinor representations we use:
    [itex]2\omega^{even}_{jk} = R_j L_k - R_k L_j[/itex]
    [itex]2\omega^{odd}_{jk} = L_j R_k - L_k R_j[/itex]
    And finally the vector representation either by:
    [itex][\omega^{odd}_{jk}, R_i][/itex]

    which, if I haven't made one of many errors should be equivalent.

    Further disclaimer. I need to double check this exposition (done from memory) and am sure I've made an error or two. But errors aside it is "I hope" a way to better understand triality in spin(8).
    Last edited by a moderator: May 5, 2017
  11. Jul 14, 2011 #10
    Re: So(10)

    Suprised, Garrett, thank you!

    Jambaugh, thank you very very much for the effort. Much appreciated. I plan studying it all the coming weekend.

    I always felt that this topic or this web of interrelated topics (Lie theory, clifford algebras, spin groups, octonions, Dynkin diagrams, lattices, triality, etc.) is soo beautiful.

    That on top even some parts show up and other parts might show up in the explanation of Nature, makes it just irresistible...

    Hopefully, one day I will understand inside out.

    But till then, thank you again!
    Last edited: Jul 14, 2011
  12. Jul 24, 2011 #11
    Re: So(10)

    Talking a little bit more about so(8).

    I read that the simple roots of so(8) can be written as

    a_1 = (1,-1, 0, 0)
    a_2 = (0, 1,-1, 0)
    a_3 = (0, 0, 1,-1)
    a_4 = (0, 0, 1, 1)

    whereas the fundamental weights can be written as

    b_1 = (1, 0, 0, 0)
    b_2 = (1, 1, 0, 0)
    b_3 = 1/2(1, 1, 1,-1)
    b_4 = 1/2(1, 1, 1, 1)

    Also, the Cartan matrix of so(8) is

    A_ij = (2 -1 0 0; -1 2 -1 -1; 0, -1, 0, 0; 0, -1, 0, 2)

    But then I also read that a_i nand b_j are related by the Cartan matrix simply via a_i = A_ij b_j. Unfortunately, that does not work for my given a_i and b_j.

    What is wrong?

    thank you
  13. Jul 25, 2011 #12
    Re: So(10)

    I got the simple roots

    a_1 = (1,-1, 0, 0)
    a_2 = (0, 1,-1, 0)
    a_3 = (0, 0, 1,-1)
    a_4 = (0, 0, 1, 1)

    and fundamental weights

    b_1 = (1, 0, 0, 0)
    b_2 = (1, 1, 0, 0)
    b_3 = 1/2(1, 1, 1,-1)
    b_4 = 1/2(1, 1, 1, 1)

    from the attached text

    How are they realted? By the Cartan matrix?

    Attached Files:

  14. Jul 25, 2011 #13
    Re: So(10)

    Yes they are related by the Cartan matrix. And this is a) very simple, b) written in any textbook.. so why don't you try to figure this out yourself as an easy homework exercise?
  15. Jul 25, 2011 #14


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    Re: So(10)

    Your roots and fundamental weights are correct ....
    Note that for [itex]C[/itex] the Cartan matrix, it is not that [itex]C \mathbf{b}_j = \mathbf{a}_j[/itex].

    Form a matrix B using the fundamental weight vectors as rows. Look then at CB and it will be the matrix of root vectors as rows. The rows of the Cartan matrix are the Dynkin coeffients of the simple roots.

    For any element of any representation, its weight will be its Dynkin coefficients times the fundamental weights:
    [itex] \mathbf{x} = \sum_j \xi^j \mathbf{b}_j[/itex]
  16. Jul 25, 2011 #15
    Re: So(10)

    Hint: check your "Cartan matrix"
  17. Jul 25, 2011 #16
    Re: So(10)


    Any textbook? Which one?

    In which book do I find the crisp and clear information like that:

  18. Jul 26, 2011 #17
    Re: So(10)

    Since this is an SO(10) thread, I wonder if I might make some remarks about Frank Wilczek's recent talk at http://www-conference.slu.se/strings2011/programme_NEW.html" [Broken]. Much of the talk was about the SO(10) GUT.

    I never had an informed opinion about which GUT, if any, to prefer, and then more recently I liked arivero's unorthodox ideas about a hidden supersymmetry in the non-unified standard model, which made it seem less likely that all the work on GUTs and SUSY-GUTs led anywhere real. But maybe they can coexist after all, and in any case, the gauge couplings unify exactly with SUSY-GUT, which is an indication that something is right; and Wilczek in his talk says that there are other, quantitative indications that SO(10) is right. So it's worth trying to identify them in this thread.

    I refer to the http://www-conference.slu.se/strings2011/presentations/4%20Thursday/1140_Wilczek.pdf" [Broken].

    Slide 12: SO(10) brings order to the fermion representations. As he puts it, there is one big multiplet for the fermions ("one material") and one big multiplet for the bosons ("one force").

    Slides 16 through 25 discuss the gauge coupling unification that occurs if you add supersymmetry.

    Slide 29 is the interesting one. He mentions bottom-tau unification and bottom-top-tau unification. This refers to the masses of those particles becoming the same at high energies. That is, he thinks that, given the low energy masses of these particles, bottom-tau unification at high energies is very likely, and bottom-top-tau unification is quite likely; and both scenarios have implications for the way that supersymmetry is broken in the supersymmetric SO(10) GUT.

    What I'd like to know is (1) whether I should also be agreeing with him on the significance and interpretation of the low-energy mass relations (2) what the specific consequences for supersymmetry breaking are, if we assume these mass unifications. In a http://arxiv.org/abs/hep-ph/0101187" [Broken] I find Wilczek saying

    "The observed ratio mb/mtau plausibly derives from equality at the unified scale, as required in all unification schemes extending SU(5). The large value of the “Dirac” neutrino mass, fixed to mt through an SO(10) relation, played a role in our earlier discussion of neutrino masses."

    ... which at least offers a lead on the details of both relations, but no more than that. I guess I'll post more details if I run across them in the literature.
    Last edited by a moderator: May 5, 2017
  19. Jul 27, 2011 #18
    Re: So(10)

    Since suprised said in an earlier post that an easy way to see the outer automorphism of the three representations of so(8) is to look at the weight diagrams, I would now like to dothat. But again I need a little help...

    Since I got for so(8) the fundamental weights

    b_1 = (1, 0, 0, 0)
    b_2 = (1, 1, 0, 0)
    b_3 = 1/2(1, 1, 1,-1)
    b_4 = 1/2(1, 1, 1, 1)

    how to get the weight diagrams? Any book to recommend that explains well what do to?

    Or can I just look it up somewhere?

  20. Jul 27, 2011 #19


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    Re: So(10)

    The fundamental weight b_1 corresponds to the vector representation. All the other weights will be permutations of the 4 coordinates and their negatives.
    [itex](\pm 1,0,0,0);\, (0,\pm 1,0,0); \, (0,0,\pm 1,0);\, (0,0,0,\pm 1)[/itex]

    The fundamental weights b_3 and b_4 correspond to the spinor representations and the other weights are obtained by changing signs in pairs of the coordinates.

    Here is a rough picture, in w,x,y,z coordinates.
    Note that the adjoint representation (b_2) will correspond to the bi-vectors are all the sums of distinct vector weights. (8 choose 2 = 28). These will occur on the edges of the middle cube (w=0) and faces of the left and right cubes (w=+/-1) in the upper row, (plus the 4 zero weights occuring at the center of the middle cube w=0)

    Attached Files:

  21. Jul 27, 2011 #20


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    Re: So(10)

    Could anyone say some words about the breaking of SO(10) and specifically how many different breaking scales there are and how many different masses can be assigned to the gauge bosons?
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