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I've read the statement in countless papers, for example, here Eq. 4.2 or here Eq. 2.1 without any further explanation or reference, that the "most general renormalizable Higgs potential" for an adjoint (=24 dimensional) Higgs $\phi$ is

$$ V(\phi) = -\frac{1}{2} \mu^2 Tr(\phi^2) + \frac{1}{4} a (Tr(\phi^2))^2 + \frac{1}{2} b Tr(\phi^4) $$

where I neglected the cubic term for brevity.

In group theoretical terms this potential can be written as

$$ V(\phi) = -\frac{1}{2} \mu^2 (24\otimes 24)_{1_s} + \frac{1}{4} a (24\otimes 24)_{1_s}(24\otimes 24)_{1_s} + \frac{1}{2} b ((24\otimes 24)_{24}(24\otimes 24)_{24})_{1_s} $$

Nevertheless, pure group theory tells us there are several other quartic invariants possible. We have

$$ 24\otimes 24 = 1_s \oplus 24_s \oplus 24_a \oplus 75_s \oplus 126_a \oplus \overline{126_a} \oplus 200_s ,$$

where each representation is denoted by its dimension and the subscripts $s$ and $a$ denote symmetric and antisymmetric respectively. Naively, I would say we have 7 quartic invariants:

$$ ((24\otimes 24)_{1_s} (24\otimes 24)_{1_s} )_1 + ((24\otimes 24)_{24_s} (24\otimes 24)_{24_s} )_1 +( (24\otimes 24)_{24_a} (24\otimes 24)_{24_a})_1 + ((24\otimes 24)_{75_s} (24\otimes 24)_{75_s} )_1 +( (24\otimes 24)_{126_a} (24\otimes 24)_{126_a} )_1 +( (24\otimes 24)_{\overline{126_a}} (24\otimes 24)_{\overline{126_a}} )_1 +((24\otimes 24)_{200_s} (24\otimes 24)_{200_s})_1 ,$$

because

$$ 1_s \otimes 1_s = 1 \quad 24_s \otimes 24_s =1 \quad 75_s \otimes 75_s =1 \quad etc. $$

An thus my question: Why are all these other products missing in the "most general renormalizable potential"? Maybe only two of these seven terms are linearly independent, but, at least for me, this is far from obvious. And in addition, then why are exactly these two a suitable linearly independent choice?

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  • $\begingroup$ You representation theory reasoning fails (besides some possible errors in the calculations) because you need to consider symmetric tensor products -- you only have 1 $\phi$ and not 4 different $\phi$'s. $\endgroup$ – Peter Kravchuk Mar 20 '16 at 1:06
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The easiest answer to your question can be gleaned from a nice book by F. Iachello, Lie Algebras and Applications, Lect. Notes Phys. 708 (Springer, Berlin Heidelberg 2006), DOI 10.1007/b11785361 , ISBN-10 3-540-36236-3

SU(5) (~ A4) has rank 4, and thus 4 independent Casimir invariants (your φ transforms like the adjoint generators of the Lie algebra.) The invariants are quadratic, cubic, quartic, and quintic. The last one would provide a non-renormalizable Higgs interaction. The cubic one was rejected by fiat (the BEGN discrete iso-parity symmetry imposed)basically to simplify the model and the analysis. You thus have to take it as a mater of trust in the book that the quartic invariant is, in fact, independent of the quadratic. Unfortunately, however, you have completely garbled it.

What is multiplied by b in your second formula should have been, in your idiosyncratic language, $$ \frac{1}{2} b Tr(\phi^4) \mapsto \frac{1}{2} b(24\otimes 24 \otimes 24\otimes 24)_{1_s} .$$ That is, your four 24s need not compose pairwise to 24s, which then compose to a singlet!

Your first and second terms are the quadratic invariant and its square, but the last one is the quartic invariant, which you could convince yourself is independent of it, but, for slick detail you must hit the book(s). You might contrast this to the adjoint of SU(2), rank 1, where there is only one invariant, so, necessarily, $Tr(\phi^4)\propto (Tr\phi^2)^2$, easy to check by diagonalizing φ, as in your Ruegg reference.

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