Symmetry breaking and higgs representation I was wondering if there is a criterion for the representation the Higgs should change under or if it's a case by case scenario. 
For instance, electroweak symmetry breaking is done with a higgs in the fundamental representation of SU(2). Using an adjoint representation does not break the symmetries generated by the third pauli matrix (and therefore not all the gauge bosons get masses and ecetera). 
To break SU(5), we need to use a higgs in the adjoint representation. According to this paper (http://www-f1.ijs.si/~ziherl/Greljo12.pdf), the reason is: 
"Since SU(5) has 24 gauge bosons, and SM has 12, the rest of the gauge bosons should get mass after SSB. So, we need to get at least 12 Goldstone bosons. Minimal representation of the Higgs which can do the job is 24,adjoint Higgs."
I do not understand his reasoning, what is this criterion for the higgs to be able to 'do the job'? 
 A: Table III of the legendary 1974 paper by Ling-Fong Li, required canonical reading for theory students, details which low-lying Higgs representations break SU(n) groups to what subgroup and why.
The "job" is to SSBreak 12 of the 24 symmetry directions of SU(5) so the remaining 12, so far unbroken at this stage, comprise the 8+3+1=12 of the SU(3)×SU(2)×U(1) of the SM. 
Table III tells you the adjoint Higgs rep of SU(5), the 24 , breaks it to just SU(3)×SU(2)×U(1), virtually magically! (This was the "could this be a coincidence?" moment of its inceptors.) 
The smaller reps all have problems: 
The fundamental, the 5, breaks SU(5) to only SU(4), so only 9 Goldstone bosons. Taking two of those, breaks it to SU(3), so 16 Goldstone bosons—far too many—would have driven model builders mad by its dysfunctional subtlety.
The symmetric two-tensor, the 15, breaks it to SU(4), only 8 Goldstone bosons, or O(5), with 10 goldstons, not enough, in both cases.
The antisymmetric two-tensor, the 10, to SU(3), so 16 goldstons, as above, so too many. 
So the adjoint does the job indeed.
A: Actually, in the SU(2)$\times$U(1) model the Higgs field is able to provide mass to the W and Z bosons (while leaving the photon massless) irrespectively of the representation under which it transforms (as long as the representation is not the trivial one).
The most general Higgs field $\Phi$ transforming under SU(2)$\times$U(1) is characterized by two quantum numbers: its isospin $j$ and its weak hypercharge $y$. If $\Phi$ has isospin $j$ then $\Phi$ is a $(2j+1)$-dimensional complex vector. If we assume that $\Phi$ is minimally coupled to the gauge bosons and that the electric charge operator can be expressed as $Q=T_{3}+Y$ - where $T_{3}$ and $Y$ are respectively an isospin projection operator and the weak hypercharge operator - then, in order to leave the photon massless, the VEV of the Higgs field $\varphi_{0}$ must satisfy the relation
$$Q\varphi_{0}=(T_{3}+Y)\varphi_{0}=0\qquad (\star).$$
It follows that the Higgs' hypercharge $y$ must lie in the spectrum of $T_{3}$, i.e. $y$ must take one of the values $-j,-j+1,\dots,j-1,j$. Provided that this condition is fulfilled, one can easily show that the quartic Higgs-gauge boson interaction always generates a mass for the W and Z bosons, where the latter are defined in the ordinary way (e.g. using the weak mixing angle $\theta_{w}$ for the $Z$ boson). In order to see this, observe that the mass operator that appears in the Lagrangian is given by
$$
\Delta\mathcal{L}=\varphi_{0}^{\dagger}G^2\varphi_{0}\qquad(\star\star)
$$
where
$$
G_{\mu}=gT_{i}W^{i}_{\mu}+yg'B_{\mu}
$$
and $\varphi_{0}$ solves equation $(\star)$. When acting on $\varphi_{0}$, thanks to equation $(\star)$,
$$
G_{\mu}=\frac{g}{\sqrt{2}}(T_{+}W^{+}_{\mu}+T_{-}W^{-}_{\mu})+y(g'B_{\mu}-gW_{\mu}^{3})=\frac{g}{\sqrt{2}}(T_{+}W^{+}_{\mu}+T_{-}W^{-}_{\mu})-y\sqrt{g^2+g^{\prime\,2}}Z_{\mu}
$$
where $T_{\pm}$ are the SU(2) raising/lowering operators. In particular, the photon field disappears from the mass operator $(\star\star)$ so that the photons remain massless. The W and Z masses, on the other hand, can be computed explicitly and are given by
$$
m_{W}^{2}=\frac{e^{2}v^{2}}{2\sin^{2}\theta_{w}}\ [j(j+1)-y^{2}]\qquad\qquad m_{Z}^{2}=\frac{e^{2}v^{2}y^{2}}{\sin^{2}\theta_{w}\cos^{2}\theta_{w}}
$$
where $e$ is the electromagnetic coupling and $v$ is the Higgs VEV (the calculation is quite straightforward from ($\star$) and ($\star\star$) using $T_{+}T_{-}+T_{-}T_{+}=2(T^{2}-T_{3}^{2})\to 2(j(j+1)-y^2)$). Observe that, as opposed to the standard $j=y=1/2$ result,
$$
m_{W}^{2}=m_{Z}^{2}\ \cos^{2}\theta_{w}
$$
in a general $(j,y)$-representation one has
$$
m_{W}^{2}=\frac{j(j+1)-y^{2}}{2y^2}\ m_{Z}^{2}\ \cos^{2}\theta_{w}
$$
so that the experimental ratio between the bosons' masses is not reproduced for a general representation. Nonetheless, the choice $j=y=1/2$ is not unique in this respect (for example $j=3$, $y=2$ also provides the experimental ratio).
A much stricter criterion for the choice of the representation comes from the requirement that the Higgs field can be coupled to the fermions in such a way as to generate their mass. If one insists that only one Higgs field exists, then the gauge-invariance of couplings such as $\overline{E}_{L}\Phi e_{R}$ forces $\Phi$ to live in the $j=y=1/2$ representation. Actually, I cheated a little bit in obtaining the bosons' masses: the assumption $Q=T_{3}+Y$ is only justified in light of the relations $Q\nu=0$, $Q e=-e$, etc. that specify the fermions' electric charges, with the left-handed fermions living in the $(j,y)=(1/2,1/2)$ representation, the charged leptons living in the $(j,y)=(0,0)$ representation, etc. To sum up: if we assume that the fermions behave as they do in the SM then the SU(2)$\times$U(1) representation of the Higgs field is fixed; more generally, any representation of SU(2)$\times$U(1) is able to provide the necessary pattern of mass generation for the bosonic sector (albeit not necessarily with the experimental $m_{W}/m_{Z}$ ratio).
For larger gauge groups $G$ containing SU(2)$\times$U(1), one must choose the representation in such a way that SSB also involves the $\text{dim}(G)-4$ generators which do not generate SU(2)$\times$U(1), so as to generate a large mass for the respective gauge bosons. Therefore the simple analysis given above in general does not hold. Without going into detail, since the longitudinal degrees of freedom for the massive gauge bosons are acquired by "eating" real components (Goldstone bosons) from the Higgs field, the SSB of a group with dimension $n$ that leaves one gauge boson massless requires a Higgs field which has real dimension at least $n-1$ (one real field to be eaten for every longitudinal component), and this requirement constrains the representation in which it can live. For instance, if $G=$SU(2)$\times$U(1) ($n=4$) then $\text{dim}_{R}(\Phi)\geq 3$ and indeed in the SM $\text{dim}_{R}(\Phi)=4$ (with the spare degree of freedom manifesting itself as the Higgs boson); if $G=$SU(5) ($n=24$) then $\text{dim}_{R}(\Phi)\geq 23$; if $G=$SO(10) ($n=45$) then $\text{dim}_{R}(\Phi)\geq 44$, etc. Observe, however, that this is only a necessary condition.
