# Gauge invariant scalar potentials

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• If $\Phi$ is a multi-component scalar field which is transforming in some representation of a gauge group say $G$ then how general a proof can one give to argue that the potential can only be a function of the G-invariant function, $\Phi^\dagger \Phi$?

This issue gets especially more confusing when one looks that the situations where $\Phi_{[ij]}$ is being thought of an anti-symmetric rank-2 tensor. Then I think the claim is that the only possible form of the potential is,

$V = \frac{m^2}{2}\Phi^{*ij}\Phi_{ij} + \frac{\lambda}{32}(\Phi^{*ij}\Phi_{ij})^2 +\frac{\lambda'}{8}\Phi^{*ij}\Phi_{jk}\Phi^{*kl}\Phi_{li}$

• Is the statement that the above is the only potential that is $G-$invariant for any $G$ and such a $\Phi$?

{..the closest thing that I could think of is that the space of all anti-symmetric rank-2 tensors, $\Phi_{[i,j]}, i,j = 1,2,..,N$, supports a natural representation of $SU(N)$ group..but so what?..}

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asked Mar 22, 2012
retagged Mar 7, 2014
The question is not very clear: in the second example, does the gauge group act on the $i,j$ indices or are they just some flavor indices? Anyway, the answer to your question is that if you contract the indices right then you've got a gauge invariant operator, and you can have more general cases than polynomials in $\Phi^\dagger \Phi$. For example, for a $SO(N)$ gauge group and a field $\phi$ transforming in the vector representation, you can have $\epsilon_{i_1, \dotsc, i_N} \phi^{i_1} \cdots \phi^{i_N}$, which is gauge invariant. The same works for $SU(N)$.

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For a general $G$ you need to know its invariants in the particular representation.

For a unitary group $SU(n)$ acting on an $n$-vector $\Phi$, the only local invariants are functions of $\Phi^*\Phi$.

In general, if you just look for local quadratic interactions, you need to split the tensor product of the representation of the field with itself into irreducible representations, and pick one of the 1-dimesnional representations or a combination of these.

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answered Mar 22, 2012 by (13,637 points)
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You want to write down the most general $G$-invariant polynomial constructed out of $\Phi$ and/or $\Phi^*$. All invariants can be identified by treating the product of several fields as a tensor product of representations, say $\Phi_{ij}\Phi^{*kl}\Phi_{mn}\Phi^{*op}$, and projecting out the singlet component. This is done by contracting all indices with $G$-invariant tensors. For example, the only (algebraically independent) invariant tensors of $SU(N)$ are $\delta^i_j$, $\epsilon_{ijk}$ and $\epsilon^{ijk}$. This tells you immediately that an invariant constructed out of a single rank-2 tensor (whether symmetric or antisymmetric) field $\Phi$ is necessarily a function of $t_n\equiv{\rm tr}(\Phi^\dagger\Phi)^n$, $\det\Phi$ and $\det\Phi^\dagger$. This already solves the problem how to write down the most general $G$-invariant function. However, the number of independent parameters in this function can be reduced by noting that the above invariants are all not independent. Since $\Phi^\dagger\Phi$ is a Hermitian positive-semidefinite matrix, all $t_n$ depend just on its $N$ real non-negative eigenvalues and thus only $t_n$ with $n=1,...,N$ are independent. They can be obtained from the generating function $$f(z)\equiv{\rm tr}\log(1+z\Phi^\dagger\Phi)=\sum_{n=1}^\infty\frac{(-1)^{n+1}}nt_nz^n.$$ In terms of the eigenvalues $\lambda_k$, this reads $f(z)=\log\prod_{k=1}^N(1+z\lambda_k)$ which can in turn be expressed in terms of $t_n$ with $n=1,...,N$ using a variant of Viete's formulas. The same strategy can be used for a field $\Phi$ in an arbitrary tensor representation as well as for other groups, taking into account the appropriate invariant tensors. Additional relations among the different invariants can occur if the tensor has some symmetry or satisfies some constraint. For example, for a traceless tensor in the adjoint representation of $SU(N)$ one has ${\rm tr}\Phi^4=\frac12({\rm tr}\Phi^2)^2$ for $N=2,3$. Of course, if you are only interested in renormalizable interactions, you will need just a small part of this machinery.

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answered Mar 23, 2012 by (115 points)

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