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Invariants under group actions

  1. Sep 26, 2013 #1
    Hi,

    I am looking to find the invariants of products of fields under SU(5) and other possible gauge groups (but lets take SU(5) as an example). Take, for example, two matter fields in the 5* and 10 and two Higgses in 5 and 5* (called [itex]H_{5}[/itex] and [itex]\bar{H}_{5*}[/itex]).

    Then the term

    5* 10 [itex]\bar{H}_{5*}[/itex]

    is invariant under SU(5), as expected.

    So long as we simply multiply the fields, I suppose it is easy to find the possible combinations simply by counting the number of [itex]U[/itex]'s and [itex] U^{\dagger}[/itex]'s (for [itex]U\in[/itex]SU(5)) that will enter the term upon transforming it and checking that the number balances (in the abovementioned example, 5* will be multiplied by one [itex] U^{\dagger}[/itex], as will the Higgs. The 10 will be multiplied by two [itex]U[/itex]'s and since [itex]U U U^{\dagger}U^{\dagger}[/itex]=1, the whole term is invariant).

    So far, so good.

    But checking the literature, one notices that there seems to be another invariant term (to be found in the superpotential of SUSY SU(5)) which is not of the aforementioned form. It looks like this:

    [itex]\epsilon_{XYZWK}[/itex] 10[itex]^{XY}[/itex]10[itex]^{ZW}[/itex][itex]H_{5}^{K}[/itex]

    Firstly, I have tried checking that this term is indeed invariant under SU(5) but so far havent managed to show it... how can one see that its invariant?
    And secondly, how can one find terms like this? If I have some group G and want to construct all invariant terms under G, how would I find out that a term like above even exists?

    Thank you!
     
  2. jcsd
  3. Sep 26, 2013 #2

    fzero

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    To reason that the ##\mathbf{10}-\mathbf{10}-\mathbf{5}## term is an invariant is that there are no free ##su(5)## indices left on it. It is also important that it involves the ##\epsilon_5## invariant, since we can only use that object and the Hermitian metric ##\delta_{I\bar{I}}## to contract tensor indices.

    For a more general term, we can just take tensor objects and multiply and contract with ##\delta_{I\bar{I}}## and ##\epsilon_5## until we've saturated all indices. This involves some educated guessing as to what terms to start with. Another way to look for invariants of a Lie algebra is to construct the branching rules for products of irreducible representations. These are expressions of the form

    $$ \mathbf{r}_i \otimes \mathbf{r}_j = \oplus_k n_k \mathbf{r}_k,$$

    i.e., we construct the tensor product representation and then decompose it into a direct sum of irreducible representations.

    You are probably already familiar with some simple branching rules for ##su(2)##. If we take two spin-1/2 irrreps (this is the ##\mathbf{2}## in the notation where we use the dimension of the irrep), then in the tensor product we find a singlet and vector irrep:

    $$ \mathbf{2} \otimes \mathbf{2} = \mathbf{1} \oplus \mathbf{3},$$

    which is the well-known split into the singlet and triplet states. Furthermore, we can note that the singlet is the antisymmetric combination of spinors and the triplet is the symmetric combination.

    The utility of this approach is that the Lie algebra invariants correspond to the singlet representations that are found in the branching rules. For instance, for ##su(3)##, the product of two fundamental irreps is

    $$ \mathbf{3} \otimes \mathbf{\bar{3}} = \mathbf{1} \oplus \mathbf{8},$$

    where we recognize the singlet at corresponding to the invariants involved in constructing the kinetic or mass terms for the quarks. In fact, for ##su(N)##, the product ##\mathbf{N} \otimes \mathbf{\bar{N}}## is always the sum of the singlet and adjoint irreps.

    A standard reference for branching rules and the techniques to calculate them is the review by Slansky. However, the main technique to compute them for ##su## and ##so## groups involves Young tableau and should be well-covered already in the books by Georgi or Cahn, as well as many other texts.
     
  4. Sep 26, 2013 #3
    Hey,

    thanks a lot for that answer! I'll try to read that review by Slansky :)

    Just one short question concerning this part:

    Does that mean that [itex]\epsilon_{XYZWK}[/itex] transforms as

    [itex]\epsilon_{XYZWK}\rightarrow \epsilon_{ABCDE} (U^{\dagger})^A_X (U^{\dagger})^B_Y (U^{\dagger})^C_Z (U^{\dagger})^D_W (U^{\dagger})^E_K[/itex]

    or for short

    [itex] \epsilon \rightarrow \epsilon U^{\dagger}^5[/itex]

    I always rather thought [itex]\epsilon[/itex] was an invariant...
     
  5. Sep 26, 2013 #4

    fzero

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    Yes, [itex]\epsilon_{XYZWK}[/itex] is invariant since

    [tex]\epsilon_{XYZWK}\rightarrow \epsilon_{ABCDE} (U^{\dagger})^A_X (U^{\dagger})^B_Y (U^{\dagger})^C_Z (U^{\dagger})^D_W (U^{\dagger})^E_K = \det U^\dagger \epsilon_{XYZWK}= \epsilon_{XYZWK}.[/tex]

    Similarly, when we contract this into an expression like ##\mathbf{10}^{XY}\mathbf{10}^{ZW}\mathbf{5}^K##, because of the antisymmetry, we'll also find that the product of ##U##s that appears is precisely the determinant.

    When we use the metric ##\delta_{I\bar{J}}## we will end up with factors ##[UU^\dagger]_{K\bar{K}} = \delta_{K\bar{K}}## that lead to invariance.
     
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