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Can we get the SM from 11D Kaluza Klein?

  1. Aug 24, 2014 #1

    arivero

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    I would like to interest the PF community in a collective activity based on some topics we have discussed from time to time.

    We know that there are seven dimensional manifolds whose isometry group is the same that the standard model gauge group. But they are not enough to get the standard model. Two questions can be raised:

    - Why such manifolds are choosen instead the highest symmetric one, the seven-sphere. This is to ask, why the Standard Model gauge group is not SO(8)?

    - How to get different representations for left and right spinors, given the non-go theorems of compacification over manifolds?

    The guiding idea is that both questions could have a common answer, by finding some orbifold (orientifold?) from a discrete quotient of the seven-sphere. The clue could be the concept of "branched covering", discussed in some papers of Atiyah, that relates CP2 and S4.
     
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  3. Aug 24, 2014 #2

    ChrisVer

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    http://cds.cern.ch/record/331309/files/9708206.pdf
    quoting from here:
    Also for the general discussing, I think this can be helpful? Although I currently don't have the time to read it myself. :frown:
    http://arxiv.org/pdf/1309.0266v1.pdf

    Also for the 2nd line... I know this might sound primitive (since I haven't ever checked what happens to 11D, I only did that on 5D), but I think this holds in general for the odd higher dimensional theories, which have a [itex]\Gamma_{5} \propto 1 [/itex] have the problem of distinguishing between left and right movers when the compactification happens on a sphere (in 5D that was the 5th dimension was compactified on a circle S1), because that's what you get by the solution of the fermionic modes. The solution ocurred for that case using Orbifolds (S1/Z2).
     
    Last edited: Aug 24, 2014
  4. Aug 24, 2014 #3

    arivero

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    Indeed this is a reason to start this work with aim towards publication. Some people seems to think that given that S7 is all the history. Point is, we do not need 28 generators, as SO(8) has. We can we happy with SU(3)xSU(2)xSU(2), with 14 generators, and of course with SU(3)xSU(2)xU(1), with only 12. The question is how the manifolds (or orbifolds) with such symmetries relate to S7, if they do.

    It seems that at least a first topic should be to clarify what spaces are we interested on.

    S7 can be seen as the Hopf fibration of S3 over S4. A topic that aficionados of Quaternions and Octonions can enjoy.

    I am not sure if S7 can be seen also as a U(1) fibration over S4 x S2, or if there is a multiparametric family of such fibrations including -or not- S7.

    What I have heard is the spaces we are interested on, those with the symmetry group of the Standard Model, can be seen as a family of U(1) fibrations over CP2 x CP1. Of course CP1=S2, the point of the branched covering being the relationship between CP2 and S4.
     
  5. Aug 24, 2014 #4

    arivero

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    Yep, thas is the thing I propose to aim to!

    Point is, here in 4D we have a discrete quotient dumbing down S4 to CP2. The project would be to use this technique up in 7D to downgrade S7 to a U(1) fiber bundle over CP2 x CP1 or to some bundle of S3 over CP2, instead of the standard (Hopf) S3 over S4. So the orbifolding of S7 would produce the SM.
     
  6. Aug 26, 2014 #5
    Sounds close to what Tony Smith does. From Smith:

     
  7. Aug 26, 2014 #6

    arivero

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    (EDIT: I see that some discussion was generated in a moderated forum nine years ago, when someone did a related question. That should be 2005. http://sci.physics.strings.narkive.com/uM0jlRnd/standard-model-from-string-theory and at that time perhaps I was not fully aware of Witten's article)



    Yep, except that last statement, taken at facial value, is wrong, isn't it?


    Tony seems to be describing what this "correspond" should really mean. Also, note that he is including space-time in the 8 dimensionald manifold RP1 x S7.
     
    Last edited: Aug 26, 2014
  8. Aug 27, 2014 #7
    Tony is talking from the point of view of his model which like Garrett Lisi's uses two D4s from a split E8 for bosons.
     
  9. Aug 27, 2014 #8
    I read somewhere that S7 is a U(1) fibration over CP3, and that CP3 is (I think) an S2 fibration over S4.

     
  10. Aug 27, 2014 #9

    arivero

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    In any case, it is true that the thing we need to understand is how to relate the 12 generators of SU(3)xSU(2)xU(1), or the 14 generators of SU(3)xSU(2)xSU(2), with the 28 generators of SO(8).

    Or, thinking with manifolds, about how to enhance S7 to become a Witten Space. Which, given the Hopf fibration, I guess it will be very similar to enhancing S4 to become CP2. It is a bit puzzling that "enhancing" the space, i.e. removing a discrete quotient, halves the number of generators of the isometry group in the S7<->Witten case. In the second case, from SO(5) to SU(3), it is less extreme, ten against eight.

    (Note: Witten Space is, as LM pointed time ago, not a recognized terminology. I took it from Cristine M. Escher, who also defines "Generalised Witten Manifolds". In the heroic times there was some discussion about how many integer parameters are needed to define uniquely a manifold of this kind.)
     
  11. Aug 27, 2014 #10
    After a quick look at that Witten Space stuff about the closest thing Tony does to that is to calculate a Weinberg angle while calculating weak boson masses. Tony starts with a Hopf fibration of S3 for this.

    For Tony, the required thing for SU(3)xSU(2)xU(1) is that there are 4 quantum numbers/Casimir Operators just like for D4. The extra D4 generators for Tony create duals for creation/annihilation operators for a bosonic Fock space. The Gravity duals are in the Standard Model D4 and the Standard Model duals are in the Gravity D4.
     
  12. Aug 27, 2014 #11

    arivero

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    Seems the right thing to do. Lens spaces.
     
  13. Sep 2, 2014 #12
    By definition, CPn (complex projective space) is the set of non-zero complex vectors (C^(n+1)\{0}) up to the phase. But C^(n+1)\{0} is (biholomophic to) S^(2n+1) or
    S^(2n+1) is a S^1 fibration over CPn. Now take n=3, then
    S^7 is a S^1 fibration over CP3.
    Michell is right.
     
  14. Sep 2, 2014 #13

    ohwilleke

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    Call me a pessimist, but a lot of really, really smart people have been pouring over every possible Standard Model gauge group unification for the last forty years and have nothing to show for it but a lot of null results. This looked promising when I was middle school in early 1980s. At this point, my intuition is that the whole enterprise must be barking up the wrong tree.
     
  15. Sep 3, 2014 #14

    arivero

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    I deny the major premise. Most schools do realistic, short to middle term, assigments for their research, particulartly if you want the student to have a decent PhD. So at the end only a few specialized groups have enough brain power to address the task, and most times they have some other research projects to run in paralell with this.
     
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