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Recently I am reading a paper about monopoles. In several cases, it seems that writing fields in adjoint representation of the gauge group makes a difference.

Once it leads to different group after symmetry breaking when using other representation. And I also noticed statement like this, "An important open question is whether an analogous Bogomolny monopole's mass bound can be obtained if the Higgs field is not in the adjoint representation."

Can anyone kindly shed light on this. Thanks!

Update: I reckon any field (either EM field in real space or Higgs field in internal isotopic space) be in a certain type of representation space of the symmetry group associated with the Lagrangian or action. This space also dictates some constraints on the fields, e.g., specific tensor or spinor structures (anything more???). And what representation space you use contains physics as well, that is to say, we have to check it by experiments. Perhaps this question addresses on a particular case. Either does the explicit and concrete 2nd answer.

Is this understanding correct?

xiaohuamao
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    In your quantum field theory, you may choose any representation you want... The problem is to make some connection with experiments. For instance, in the Standard model of particles, fermions are in the fundamental representation (of internal symmetries), while gauge bosons are in the the adjoint representation of these internal symmetries, and the theory match perfectly with experiments. Now, in supersymmetry theories, the superpartner of a gauge boson, is a fermion (photino, wino,zino,etc...), and this fermion is also in the adjoint representation. Future experiments have to confirm this. – Trimok Nov 20 '13 at 11:34

2 Answers2

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I am not sure if I know the correct answer (as I am a student my self), but I will try (and if I am wrong, someone please correct me).

The first thing that took me some time to figure out is what they mean by adjoint representation. In Georgi's book he defines the adjoint representation of a generator as: \begin{equation} [T_i]_{jk} \equiv -if_{ijk} \end{equation} which is equivalent to the adjoint representation of a Lie algebra. However, when discussing monopole, they actually mean the adjoint representation of a Lie group . This means that $\phi$ takes values in the Lie algebra (the vector space formed by the generators) and can be expressed in terms of the generators in an arbitrary representation: \begin{equation} \phi = \phi^a t^a \end{equation} where $t^a$ denote the generators in an arbitrary representation (and there is an implicit sum over repeated indices).

Now, let us look at the simplest example, which is the bosonic part of the $\mathrm{SU(2)}$ gauge invariant Georgi-Glashow model: \begin{equation} \mathcal{L}=\frac{1}{8} \mathrm{Tr} (F_{\mu \nu} F^{\mu \nu}) - \frac{1}{4} \mathrm{Tr}(D_\mu \phi D^\mu \phi) - \frac{\lambda}{4}(1-\phi^a \phi^a )^2 \end{equation} We can write the kinetic and potential energy, $T$ and $V$, as: \begin{equation} T=\int \left( - \frac{1}{4} \mathrm{Tr} (F_{0i}F_{0i}) - \frac{1}{4} \mathrm{Tr}(D_0 \phi D_0 \phi) \right) \mathrm{d^3}x \end{equation} and: \begin{equation} V=\int \left( - \frac{1}{8} \mathrm{Tr} (F_{ij}F_{ij}) - \frac{1}{4} \mathrm{Tr}(D_i \phi D_i \phi) + \frac{\lambda}{4}(1-\phi^a \phi^a )^2 \right) \mathrm{d^3}x \end{equation} where we used $L= \int \mathcal{L} \; \mathrm{d^3}x = T-V$. In order to get finite energy solutions we have to impose boundary conditions such that the total energy of the model vanished at spatial infinity. It should be clear that one of the requirements to ensure that the energy vanishes is: \begin{equation} \phi^a \phi^a =1 \end{equation} This implies that the Higgs vacuum corresponds to an infinite amount of degenerate vacuum values lying on the surface of a unit two-sphere in field space, which we will denote by $S^2_1$. Furthermore, by imposing the aforementioned finite energy boundary condition, this gives rise to the following map: \begin{equation} \phi : S^2_\infty \to S^2_1 \end{equation} where $S^2_\infty$ denotes the two-sphere associated with spatial infinity (in 3 dimensions). This is in fact the definition of the winding number (or degree) between two two-dimensional spheres and is therefore classified by $\pi_2(S^2)=\mathbb{Z}$ (and it is in theory possible to construct topological solitons). Now, if $\phi$ was in the fundamental representation, then I don't think it is possible to construct these topological solitons.

Hunter
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  • Thanks very much for your detailed recapitulation of the theory. Could you please have a look at my Edit in the main question? Look forward to your helpful clarification. – xiaohuamao Nov 20 '13 at 03:43
  • I'm not sure if I understand your edit (especially (i)). However, note that the aforementioned Georgi-Glashow model is flawed because after symmetry breaking $\mathrm{SU(2)} \rightarrow \mathrm{U(1)}$ does not produce the neutral $Z$ boson. Furthermore, I believe (but I do not know why as I have never studied it), that some/many GUT theories start of (i.e. before symmetry breaking) with the Higgs field in the adjoint representation; for more information on this in combination with monopoles I recommend Weinberg's book. – Hunter Nov 20 '13 at 10:39
  • Finally, the $\mathrm{SU(2)}$ invariance of the Lagrangian has nothing to do with the spin of particles; it describes the weak force. – Hunter Nov 20 '13 at 10:40
  • Er...I know it has nothing to do with spin. I just wanna give a an example of common internal space, because I am confused with why we use Lie algebra sapce as the Higgs field internal space. – xiaohuamao Nov 20 '13 at 10:51
  • In fact, (1) is the most annoying one to me. I mean group G is the symmetry we want on the Lagrangian. It must have a representation space to make its elements' action doable. So I suppose the fields are just in this representation space. And you said $\phi$ expressed in terms of the generators is in an arbitrary representation. So I suppose we still doesn't fully choose the representation space. We just express the vectorial structure of the field in the Lie algebra g space. – xiaohuamao Nov 20 '13 at 10:57
  • I think the logic is clear. If the above comment is correct, then the only question is why we express the vectorial structure of the field in the Lie algebra g space, not any other space. This is just (2), which is partly answered in your comments. – xiaohuamao Nov 20 '13 at 11:01
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    @huotuichang: The fields take value in some particular representation of the Lie group. So the statement in the answer that they transform under an "arbitrary representation" is either confusing or wrong. – Olof Nov 20 '13 at 12:12
  • Ahh I understand what the confusion is now. I'm sorry. Yeah, Olof is right that field take value in a particular representation (in the case above the generators are the Pauli matrices). In the first paragraph I tried to explain that the adjoint representation that "people" sometimes refer to is not necessarily the same adjoint representation that is being discussed in most physics books (such as Georgi) on Lie groups. – Hunter Nov 20 '13 at 12:24
  • When considering other groups such as $SU(N)$, $N\geq 3$ it is possible to obtain monopoles solutions from a Higgs field in representations other than the adjoint. These are called $\mathbb Z_n$-monopoles. It is also possible to obtain Bogomolnyi solutions for these solitons. – Diracology Jul 11 '16 at 21:44
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Theories with fundamental quarks which experience spontaneous chiral symmetry breaking:

$$ SU_L(N_f) \times SU_R(N_f) \rightarrow SU_A(N_f)$$

($N_f$ is the number of flavors)

(This is the observed approximate symmetry breaking in nature where the pions are the approximate Goldstone bosons).

In contrast, theories with adjoint quarks experience the chiral symmetry breaking pattern:

$$ SU(N_f) \rightarrow SO(N_f)$$

(modulo discrete groups). Please see for example the following article Auzzi, Bolognesi, and Shifman.

The reasoning is that since the adjoint representation is real, it has only one copy of $SU(N_f)$ flavor symmetry and $SO(N_f) $ is am maximal subgroup of $SU(N_f)$, thus any symmetry breaking will start in this pattern.

The Goldstone Boson manifold will be:

$$\mathcal{M} = SU(N_f)/SO(N_f)$$

The topology of the Goldstone Boson manifold determines the existence of t’ Hooft-Polyakov monopoles, since an non trivial homotopy group $\pi_2(\mathcal{M} )$ is required for a stable monopole solution to exist. This happens in our case when $N_f =2$, in this:

$$\mathcal{M} = SU(2)/SO(2) = S^2$$

Thus $\pi_2(\mathcal{M} ) = \mathbb{Z}$ and monopoles exist.

In addition for any number of flavors there will be Skyrmions in $\mathcal{M}$ as elaborated in , Bolognesi, and Shifman's article.