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I Bootstraping a space from its tensor square

  1. Feb 4, 2017 #1


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    By space, I mean a vector space which could be a representation of a group or even have some expanded algebraic structure. So I am not sure if this question goes here or in the Algebra subforum.

    Consider the tensor square [itex]r\otimes r[/itex] of an irreducible group representation [itex]r [/itex] with itself, and decompose it as irreducible representations. What can we said about the circumstance of finding the initial representation in the list? Or perhaps about finding its conjugate, as for instance in E6:

    [tex]27 \otimes 27 = 351 \oplus (\bar{27} \oplus \bar {351})[/tex]

    What groups have representations having this "bootstraping"? Can the irrep appear in both parts, symmetric and alternating, of the tensor square? Does it appear in an unique way, or can it be extracted from different combinations of the roots?

    Similarly, consider the tensor square [itex]A^{\otimes 2}[/itex] of an algebra. Are there situations where the new algebra does contain the initial one as a subalgebra in a non trivial way? This seems to generalise the question of generating an algebra from a finite number of elements and its n-times product, call it [itex]A^{\times n}[/itex], but perhaps it is not more general... still I wonder what can be said generically about such action. What I am expecting is that some ideal J can be chosen in [itex]A^{\otimes 2}[/itex] such that the quotient recovers the initial algebra. Or some similar mechanism, anyway.

    The motivation of the post to be in BSM is, of course, my old observation that by choosing five quarks, out of all the set of three particle generations of the standard model, and pairing them we seem to be able to recover the full three generations, and I wondering if this phenomena could be tracked to some peculiar property in mathematical representation. Thinking it also in algebraic terms is interesting because the attempts to get generations out of the exceptional jordan algebra [itex]h_3(O)[/itex] or its twin [itex]h_3(C \otimes O)[/itex] have some extra matter in the diagonal, an issue that also happens in the naive pairing.
    Last edited: Feb 4, 2017
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  3. Feb 5, 2017 #2


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    Two comments here:

    - There is a lot of casuistic. By inspecting the tables of kronecker products, we have cases where the irrep is recovered either in the symmetric part of the square
    [tex] SU(6): 15 \otimes 15 = \bar {105}^s + \bar{105}^a + \bar{15}^s [/tex]
    or in the alternating,
    [tex] SU(3): 3 \otimes 3 = 6^s + \bar 3^a [/tex]
    or not recovered at all. The tensor square of the adjoint tends to appear in the antisymmetric part, but for the lower SU(n) groups it appears in both parts, duplicated. The groups with triality, as SO(8), could even have a more complex way via the product of same dimension but different irreps.

    - the finding in the standard model has not a obvious place in the list because it is formulated in the flavour subgroup of the selected quarks, namely in SU(5), and not with an irreducible but as:
    [tex] (5 \oplus \bar 5) \otimes (5 \oplus \bar 5) \supset 24 \oplus 15 \oplus \bar{15} [/tex]
    via [itex]5 \otimes \bar 5 = 24 \oplus 1[/itex] and [itex]5 \otimes 5= \bar{15} \oplus 10[/itex] . There is the obvious hope of fitting it inside E6, perhaps adding the other [itex]\bar {24}[/itex] of the product to sum 78, or considering [itex]27 \oplus \bar {27}[/itex].
    Last edited: Feb 5, 2017
  4. Feb 7, 2017 #3


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    A related result that is not fully intuitive to me is that if the representation is faithful, then by iteratively tensoring and decomposing we will eventually arrive to any representation. Some proofs can be found in the question sites: http://math.stackexchange.com/quest...on-of-a-compact-group-occur-in-tensor-product http://mathoverflow.net/questions/5...on-of-a-compact-group-occur-in-tensor-product
    Of course, once you admit any combination of tensor product, you can just tensor the fundamental representations as many times you need to produce the weights. I think the square tensor is a funnier think because in some way it minimizes the usage of constituents.
    An intermediate step before to play with any tensor product is to consider the tensor cube. The existence of the conjugate representation in the tensor square surely helps to the existence of the singlet in the tensor cube; and the existence of the singlet in the tensor cube is a precondition for the colour group (Cvitanovic says it? See http://journals.aps.org/prd/abstract/10.1103/PhysRevD.27.616 too).
    Last edited: Feb 7, 2017
  5. Feb 9, 2017 #4


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    The appendix of the above referred article (Three-preon models of quarks and leptons and the generation problem) addresses in the appendix a related question: when does a tensor cube generate a singlet. Of course, composites need singlets, and also the question of when a tensor square generates a singlet has some interest in the faq-web: http://mathoverflow.net/questions/64925/occurence-of-trivial-representation-in-a-tensor-square . Pretty obviously if the square contains both the trivial and the original representation, also the cube will. Thus the discussion hints about necessary conditions to get [itex]R \otimes R \supset R [/itex]

    More precisely it is claimed, I guess by inspection, that [itex]R \otimes R \supset R \oplus 1 \oplus ... [/itex] happens for real or pseudoreal representations of SU(n), E6 and SO(4n+2), as well as some representations of the other groups SO(2n+1) SO(4n) Sp(2n) E7 E8 F4 and G2, and so to have a "bootstrap of a representation" is not a very selective criteria for a Lie group :frown:.

    To close the topic, there was also some nice introduction to tensor squares here: https://arxiv.org/pdf/1504.07732.pdf
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